Chapter Introduction to Electronics Section - Atomic Structure. An atom with an atomic number of 6 has 6 electrons and 6 protons.. The third shell of an atom can have n = (3) = 8 electrons. Section - Materials Used in Electronics 3. The materials represented in Figure in the textbook are (a) insulator (b) semiconductor (c) conductor 4. An atom with four valence electrons is a semiconductor. 5. In a silicon crystal, each atom forms four covalent bonds. Section -3 Current in Semiconductors 6. When heat is added to silicon, more free electrons and holes are produced. 7. Current is produced in silicon at the conduction band and the valence band. Section -4 N-Type and P-Type Semiconductors 8. Doping is the carefully controlled addition of trivalent or pentavalent atoms to pure (intrinsic) semiconductor material for the purpose of increasing the number of majority carriers (free electrons or holes). 9. Antimony is a pentavalent (donor) material used for doping to increase free electrons. Boron is a trivalent (acceptor) material used for doping to increase the holes. Section -5 The PN Junction 0. The electric field across the pn junction of a diode is created by donor atoms in the n region losing free electrons to acceptor atoms in the p region. This creates positive ions in the n region near the junction and negative ions in the p region near the junction. A field is then established between the ions.. The barrier potential of a diode represents an energy gradient that must be overcome by conduction electrons and produces a voltage drop, not a source of energy.
Chapter Diode Applications Section - Diode Operation. To forward-bias a diode, the positive terminal of a voltage source must be connected to the p region.. A series resistor is needed to limit the current through a forward-biased diode to a value that will not damage the diode because the diode itself has very little resistance. Section - oltage-current Characteristic of a Diode 3. To generate the forward bias portion of the characteristic curve, connect a voltage source across the diode for forward bias and place an ammeter in series with the diode and a voltmeter across the diode. Slowly increase the voltage from zero and plot the forward voltage versus the current. 4. A temperature increase would cause the barrier potential of a silicon diode to decrease from 0.7 to 0.6. Section -3 Diode Models 5. (a) The diode is reverse-biased. (b) The diode is forward-biased. (c) The diode is forward-biased. (d) The diode is forward-biased. 6. (a) R = 5 8 = 3 (b) F = 0.7 (c) F = 0.7 (d) F = 0.7 7. (a) R = 5 8 = 3 (b) F = 0 (c) F = 0 (d) F = 0 8. Ignoring r R : (a) R 5 8 = 3 (b) I F = 00 0.7 = 74 ma 560 0 F = I F r d + B = (74 ma)(0 ) + 0.7 =.44
Chapter (c) I tot = 30 30 = 6.9 ma 4.85 k R tot 6.9 ma I F = = 3. ma F = I F r d + 0.7 = (3. ma)(0 ) + 0.7 = 0.73 (d) Approximately all of the current from the 0 source is through the diode. No current from the 0 source is through the diode. I F = 0 0.7 =.9 ma 0 k0 F = (.9 ma)(0 ) + 0.7 = 0.79 Section -4 Half-Wave Rectifiers 9. See Figure -. Figure - 0. (a) PI = p = 5 (b) PI = p = 50. AG = p 00 = 63.7. (a) (b) I I F F ( ( p) in p) in 0.7 R 0.7 R 5 0.7 4.3 = 9.5 ma 47 47 50 0.7 49.3 = 4.9 ma 3.3 k 3.3 k 3. n (0.)0 = 4 rms sec pri 4. n (0.5)0 = 60 rms sec pri p(sec) =.44(60 ) = 84.8 p( sec) 84.8 avg ( sec) = 7.0 P L( p) P Lavg ( ) p( sec) 0.7 (84. ) R L avg ( sec) (7.0 ) R L 0 0 = 3. W = 3.3 W 3
Chapter Section -5 Full-Wave Rectifiers p 5. 5 (a) AG = =.59 p (00 ) (b) AG = = 63.7 (0 ) (c) AG = p 0 + 0 = 6.4 (40 ) (d) AG = p 5 5 = 0.5 6. (a) Center-tapped full-wave rectifier (b) p(sec) = (0.5)(.44)0 = 4.4 p sec ( ) 4.4 (c) =. (d) See Figure -. RL =. 0.7 = 0.5 Figure - p( sec) (e) I F = 0.7 0.5 = 0.5 ma R.0 k L (f) PI =. + 0.5 = 4.7 7. AG = 0 p AG = = 60 for each half p = AG = (60 ) = 86 8. See Figure -3. Figure -3 4
Chapter 9. PI = p = AG( out ) (50 ) = 78.5 0. PI = p(out) =.44(0 ) = 8.3. See Figure -4. Figure -4 Section -6 Power Supply Filters and Regulators. r(pp) = 0.5 r ( pp ) 0.5 r = = 0.00667 75 DC 3. r(pp) = p( in) fr L C 30 (0 Hz)(600 )(50 F) = 8.33 pp DC = 30 ( ) (40 Hz)(600 )(50 F) p in frlc = 5.8 pp ( ) 8.33 4. %r = r 00 00 = 3.3% DC 5.8 5. r(pp) = (0.0)(8 ) = 80 m r(pp) = p( in) frlc C = ( ) 8 (0 Hz)(.5 k )(80 m) p in frlr = 556 F 6. p( in) 80 r ( pp) = 6.67 fr C (0 Hz)(0 k)(0 F) L DC = 80 ( ) (40 Hz)(0 k )(0 F) p in frlc r ( pp ) 6.67 r = = 0.087 76.7 DC = 76.7 5
Chapter 7. p(sec) = (.44)(36 ) = 50.9 r(rect) = p(sec).4 = 50.9.4 = 49.5 Neglecting R surge, r(pp) = p( rect) 49.5 frlc (0 Hz)(3.3 k )(00 F) r ( pp) DC = p( rect) p( rect) frlc = 49.5 0.65 = 48.9 =.5 8. p(sec) =.44(36 ) = 50.9 See Figure -5. Figure -5 NL FL 5.5 4.9 9. Load regulation = 00% 00% FL 4.9 = 4% 30. FL = NL (0.005) NL = (0.005) =.94 Section -7 Diode Limiters and Clampers 3. See Figure -6. Figure -6 6
Chapter 3. Apply Kirchhoff s law at the peak of the positive half cycle: (b) 5 = R + R + 0.7 R = 4.3 4.3 R = =.5 out = R + 0.7 =.5 + 0.7 =.85 See Figure -7(a)..3 (c) R = = 5.65 out = R + 0.7 = 5.65 + 0.7 = 6.35 See Figure -7(b). 4.3 (d) R = =.5 out = R + 0.7 =.5 + 0.7 =.85 See Figure -7(c). Figure -7 7
Chapter 33. See Figure -8. Figure -8 34. See Figure -9. 35. See Figure -0. Figure -9 Figure -0 30 0.7 36. (a) I p. k (b) Same as (a). = 3.3 ma 8
Chapter 37. (a) (b) (c) (d) I p I p I p I p 30 ( 0.7 ). k = 7.86 ma 30 ( 0.7 ). k = 8.5 ma 30 (.3 ). k = 8.8 ma 30 (.7 ) = 9.4 ma. k 38. See Figure -. Figure - 39. (a) A sine wave with a positive peak at 0.7, a negative peak at 7.3, and a dc value of 3.3. (b) A sine wave with a positive peak at 9.3, a negative peak at 0.7, and a dc value of +4.3. (c) A square wave varying from +0.7 to 5.3 with a dc value of 7.3. (d) A square wave varying from +.3 to 0.7 with a dc value of +0.3. 40. (a) A sine wave varying from 0.7 to +7.3 with a dc value of +3.3. (b) A sine wave varying from 9.3 to +7.3 with a dc value of +4.3. (c) A square wave varying from 0.7 to +5.3 with a dc value of +7.3. (d) A square wave varying from.3 to +0.7 with a dc value of 0.3. Section -8 oltage Multipliers 4. OUT = p(in) = (.44)(0 ) = 56.6 See Figure -. Figure - 9
Chapter 4. OUT(trip) = 3 p(in) = 3(.44)(0 ) = 84.8 OUT(quad) = 4 p(in) = 4(.44)(0 ) = 3 See Figure -3. Figure -3 Section -9 The Diode Datasheet 43. The PI is specified as the peak repetitive reverse voltage = 00. 44. The PI is specified as the peak repetitive reverse voltage = 000. 45. I F(AG) =.0 A R L(min) = 50.0 A = 50 Section -0 Troubleshooting 46. (a) Since D = 5 = 0.5 S, the diode is open. (b) The diode is forward-biased but since D = 5 = S, the diode is open. (c) The diode is reverse-biased but since R =.5 = 0.5 S, the diode is shorted. (d) The diode is reverse-biased and R = 0. The diode is operating properly. 47. A = S = +5 B = S 0.7 = 5 0.7 = +4.3 C = S + 0.7 = 8 + 0.7 = +8.7 D = S = +8.0 48. If a bridge rectifier diode opens, the output becomes a half-wave voltage resulting in an increased ripple at 60 Hz. 0
Chapter p (5 )(.44) 49. avg = 04 The output of the bridge is correct. However, the 0 output from the filter indicates that the surge resistor is open or that the capacitor is shorted. 50. (a) Correct (b) Incorrect. Open diode. (c) Correct (d) Incorrect. Open diode. 5. sec = 0 = 4 rms 5 p(sec) =.44(4 ) = 33.9 The peak voltage for each half of the secondary is p ( sec ) 33.9 = 7 The peak inverse voltage for each diode is PI = (7 ) + 0.7 = 34.7 The peak current through each diode is p( sec) 0.7 7.0 0.7 I p = 49.4 ma RL 330 The diode ratings exceed the actual PI and peak current. The circuit should not fail. Application Activity Problems 5. (a) Not plugged into ac outlet or no ac available at outlet. Check plug and/or breaker. (b) Open transformer winding or open fuse. Check transformer and/or fuse. (c) Incorrect transformer installed. Replace. (d) Leaky filter capacitor. Replace. (e) Rectifier faulty. Replace. (f) Rectifier faulty. Replace. 53. The rectifier must be connected backwards. 54. 6 with 60 Hz ripple Advanced Problems 55. r = p( in) frlc C = ( ) 35 (0 Hz)(3.3k )(0.5 ) p in = 77 F frlr
Chapter 56. DC = p( in) frlc DC p( in) frlc DC frlc p ( in) = C DC fr L p( in) C = = 6. F (40 Hz)(.0 k)( 0.933) (40 Hz)(.0 k)(0.067) Then r = pin ( ) 5 = frlc (0 Hz)(.0 k )(6. F) 57. The capacitor input voltage is p(in) = (.44)(4 ).4 = 3.5 p( in) 3.5 R surge = = 65 m Isurge 50 A The nearest standard value is 680 m. 58. See Figure -4. The voltage at point A with respect to ground is A =.44(9 ) =.7 Therefore, B =.7 0.7 = r = 0.05 B = 0.05( ) = 0.6 peak to peak C = B (0 Hz)(680 )(0.6 ) frlr The nearest standard value is 70 F. Let R surge =.0. I surge(max) = A.0 I F(A) = = 7.6 ma 680 PI = p(out) + 0.7 = 4.7 = 45 F Figure -4
Chapter 59. See Figure -5. I L(max) = 00 ma 9 R L = = 90 00 ma r =.44(0.5 ) = 0.354 r = (0.35 ) = 0.7 peak to peak r = 9 (0 Hz)(90 ) C 9 C = = 74 F (0 Hz)(90 )(0.7) Use C = 00 F. Each half of the supply uses identical components. N400 diodes are feasible since the average current is (0.38)(00 ma) = 3.8 ma. R surge =.0 will limit the surge current to an acceptable value. Figure -5 60. See Figure -6. 6. C = (.44)(0 ) 0.7 = 70 C = (.44)(0 ) (0.7 ) = 338 Figure -6 3
Chapter MultiSim Troubleshooting Problems The solutions showing instrument connections for Problems 6 through 79 are available from the Instructor Resource Center. The faults in the circuit files may be accessed using the password book (all lowercase). To access supplementary materials online, instructors need to request an instructor access code. Go to www.pearsonhighered.com/irc to register for an instructor access code. Within 48 hours of registering, you will receive a confirming e-mail including an instructor access code. Once you have received your code, locate your text in the online catalog and click on the Instructor Resources button on the left side of the catalog product page. Select a supplement, and a login page will appear. Once you have logged in, you can access instructor material for all Prentice Hall textbooks. If you have any difficulties accessing the site or downloading a supplement, please contact Customer Service at http://47.prenhall.com. 6. Diode shorted 63. Diode open 64. Diode open 65. Diode shorted 66. No fault 67. Diode shorted 68. Diode leaky 69. Diode open 70. Diode shorted 7. Diode shorted 7. Diode leaky 73. Diode open 74. Bottom diode open 75. Reduced transformer turns ratio 76. Open filter capacitor 77. Diode leaky 78. D open 79. Load resistor open 4
Chapter 3 Special-Purpose Diodes Section 3- The Zener Diode. See Figure 3-.. I ZK 3 ma Z 9 Figure 3-3. Z Z = I Z Z 5.65 5.6 30 ma 0 ma 0.05 0 ma = 5 4. I Z = 50 ma 5 ma = 5 ma Z = I Z Z Z = (+5 ma)(5 ) = +0.375 Z = Z + Z = 4.7 + 0.375 = 5.08 5. T = 70C 5C = 45C (6.8 )(0.0004/ C) Z = 6.8 + 45C = 6.8 C + 0. = 6.9 Section 3- Zener Diode Applications 6. IN(min) = Z + I ZK R = 4 + (.5 ma)(560 ) = 4.8 7. Z = (I Z I ZK )Z Z = (8.5 ma)(0 ) = 0.57 OUT = Z Z = 4 0.57 = 3.43 IN(min) = I ZK R + OUT = (.5 ma)(560 ) + 3.43 = 4.3 8. Z = I Z Z Z = (40 ma 30 ma)(30 ) = 0.3 Z = + Z = + 0.3 =.3 IN Z 8.3 R = = 43 40 ma 40 ma 5
Chapter 3 9. Z + 0.3 =.3 See Figure 3-. Figure 3-0. Z(min) = Z I Z Z Z = 5. (49 ma ma)(7 ) = 5. (48 ma)(7 ) = 5. 0.336 = 4.76 R = 8 4.76 = 3.4 I T = R 3.4 = 47 ma R I L(max) = 47 ma ma = 46 ma Z(max) = 5. + (70 ma 49 ma)(7 ) = 5. + 47 m = 5.5 R = 8 5.5 =.75 I T =.75 = 5 ma I L(min) = 5 ma 70 ma = 55 ma. % Load regulation = Z(max) Z(min) Z(min) 00% = 5.5 4.76 4.76 00% = 0.3%. With no load and IN = 6 : IN Z 6 5. I Z = 3 ma R Z 9 Z OUT = Z I Z Z Z = 5. (49 ma 3 ma)(7 ) = 5. 0.6 = 4.97 With no load and IN = : IN Z 5. I Z = 38 ma R Z 9 Z OUT = Z + I Z Z Z = 5. + (38 ma 49 ma)(7 ) = 5. +.3 = 6.4 OUT % Line regulation = 00% = 6.4 4.97 00% = 4.% 6 IN 3. % Load regulation = NL FL FL 00% = 8.3 7.98 7.98 00% = 3.3% 4. % Line regulation = OUT IN 00% = 0. 0 5 00% = 4% 5. % Load regulation = NL FL FL 00% = 3.6 3.4 3.4 00% = 5.88% 6
Chapter 3 Section 3-3 The aractor Diode 6. At 5, C = 0 pf At 0, C = 0 pf C = 0 pf 0 pf = 0 pf (decrease) 7. From the graph, R = 3 @ 5 pf 8. f r = LC T C T = =.7 pf 4 Lf r 4 ( mh)( MHz) Since they are in series, each varactor must have a capacitance of C T = 5.4 pf 9. Each varactor has a capacitance of 5.4 pf. Therefore, from the graph, R must be slightly less than 3. Section 3-4 Optical Diodes 4 0. I F = = 35.3 ma 680 From the graph, the radiant power is approximately 80 mw.. See Figure 3-3. R = 5. = 97 30 ma The nearest standard % value is 97.6 or the nearest standard 5% value is 9. Figure 3-3 7
Chapter 3. F. for I F = 0 ma 9 Maximum LEDs/branch =. 4 Select 3 LEDs/branch: Number of branches = 48 3 = 6 9 3(. ) R LIMIT = = 0 0 ma Use sixteen 0 resistors. 3. F.5 for I F = 30 ma Maximum LEDs/branch = 4.5 9.6 Select 5 LEDs/branch: Number of branches = 00 5 = 0 4 5(.5 ) R LIMIT = = 383 30 ma See Figure 3-4. Figure 3-4 4. I R = 0 00 k = 50 A 8
Chapter 3 5. (a) R = I S 3 00 A = 30 k (b) R = I S 3 350 A = 8.57 k (c) R = I S 3 50 A = 5.88 k 6. The microammeter reading will increase. Section 3-5 Other Types of Diodes 7. R = I 5 m 00 m 75 m 0.5 ma 0.5 ma 0.0 ma = 750 8. Tunnel diodes are used in oscillators. 9. The reflective ends cause the light to bounce back and forth, thus increasing the intensity of the light. The partially reflective end allows a portion of the reflected light to be emitted. Section 3-6 Troubleshooting 30. (a) All voltages are correct. (b) 3 should be. Zener is open. (c) should be 0. Fuse is open. (d) Capacitor C is open. (e) R is open or D 5 is shorted. 3. (a) With D 5 open, OUT 30. (b) With R open, OUT = 0. (c) With C leaky, OUT has excessive 0 Hz ripple limited to. (d) With C open, OUT is full wave rectified voltage limited to. (e) With D 3 open, OUT has 60 Hz ripple limited to. (f) With D open, OUT has 60 Hz ripple limited to. (g) With T open, OUT = 0. (h) With F open, OUT = 0. Application Activity Problems 3. (a) Faulty regulator 33. Incorrect transformer secondary voltage 34. LED open, limiting resistor open, faulty regulator, faulty bridge rectifier 35. I L = k = ma; reg = 6 = 4 P reg = (4 )( ma) = 48 mw 9
Chapter 3 Datasheet Problems 36. From the datasheet of textbook Figure 3-7: (a) @ 5C: P D(max) =.0 W for a N4738A (b) For a N475A: @ 70C; P D(max) =.0 W (6.67 mw/c)(0c) =.0 W 33 mw = 867 mw @ 00C; P D(max) =.0 W (6.67 mw/c)(50c) =.0 W 333 mw = 667 mw (c) I ZK = 0.5 ma for a N4738A (d) @ 5C: I ZM = W/7 = 37.0 ma for a N4750A (e) Z Z = 700 7.0 = 693 for a N4740A 37. From the datasheet of textbook Figure 3-4: (a) I F(max) = 00 ma (b) C max = pf C 00 pf (c) C 0 = = 5.4 pf; range is 00 pf 5.4 pf for an 836A. CR 6.5 38. From the datasheet of textbook 3-34: (a) 9 cannot be applied in reverse across a TSMF000 because R(max) = 5. (b) When 5. is used to forward-bias the TSMF000 for I F = 0 ma, F.3 5..3 3.8 R = = 90 0 ma 0 ma (c) At 5C maximum power dissipation is 90 mw. If F =.5 and I F = 50 ma, P D = 75 mw. The power rating is not exceeded. (d) For I F = 40 ma, radiant intensity is approximately 0.9 mw/sr. (e) For I F = 00 ma and = 0, radiant intensity is 40% of maximum or (0.4)(5 mw/sr) = 0 mw/sr 39. From the datasheet of textbook Figure 3-47: (a) With no incident light and a 0 k series resistor, the typical voltage across the resistor is approximately R = ( na)( k) =. (b) Reverse current is greatest at about 940 nm. (c) Sensitivity is maximum for 830 nm. 0
Chapter 3 Advanced Problems 40. See Figure 3-5. Figure 3-5 4. OUT() 6.8, OUT() 4 4. For a 0 k load on each output: OUT I OUT() = R 6.8 = 0.68 ma 0 k OUT I OUT() = R 4 =.4 ma 0 k R 0 6.8 = 3. I Z = 3. 0.68 ma =.5 ma k R 0 4 = 96 I Z = 96.4 ma = 93.6 ma k I T = 0.68 ma +.4 ma +.5 ma + 93.6 ma = 09. ma The fuse rating should be 50 ma or /4 A. 43. See Figure 3-6. Use a N4738A zener. I T = 35 ma + 3 ma = 66 ma 4 8. R = = 39 66 ma Figure 3-6
Chapter 3 44. C max = = 03.4 pf 4 Lf 4 ( mh)(350 khz) min C min = = 7.5 pf 4 Lfmax 4 ( mh)(850 khz) To achieve this capacitance range, use an 86A varactor and change to 30. 45. See Figure 3-7. From datasheet, F =. for red LED. D. R = = 495 I 0 ma Use standard value of 50. Figure 3-7 46. See Figure 3-8. Figure 3-8
Chapter 3 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 47 through 50 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 47. Zener diode open 48. Capacitor open 49. Zener diode shorted 50. Resistor open 3
Chapter 4 Bipolar Junction Transistors Section 4- Bipolar Junction Transistor (BJT) Structure. Majority carriers in the base region of an npn transistor are holes.. Because of the narrow base region, the minority carriers invading the base region find a limited number of partners for recombination and, therefore, move across the junction into the collector region rather than out of the base lead. Section 4- Basic BJT Operation 3. The base is narrow and lightly doped so that a small recombination (base) current is generated compared to the collector current. 4. I B = 0.0I E = 0.0(30 ma) = 0.6 ma I C = I E I B = 30 ma 0.6 ma = 9.4 ma 5. The base must be negative with respect to the collector and positive with respect to the emitter. 6. I C = I E I B = 5.34 ma 475 A = 4.87 ma Section 4-3 BJT Characteristics and Parameters 7. DC = I I C E 8.3 ma 8.69 ma = 0.947 8. DC = I I C B 5 ma 00 A = 5 9. I B = I E I C = 0.5 ma 0.3 ma = 0. ma = 00 A IC 0.3 ma DC = I 00 A = 0.5 B 0. I E = I C + I B = 5.35 ma + 50 A = 5.40 ma I DC = C 5.35 ma = 0.99 I 5.40 ma E. I C = DC I E = 0.96(9.35 ma) = 8.98 ma 4
Chapter 4 R C 5. I C = RC.0 k IC 5 ma DC = I 50 A B = 5 ma = 00 3. DC = DC DC 00 0 = 0.99 BB BE 4. I B = R B 3 0.7 00 k = 3 A I C = DC I B = 00(3 A) = 4.6 ma I E = I C + I B = 4.6 ma + 3 A = 4.6 ma CE = CC I C R C = 0 (4.6 ma)(.0 k) = 5.4 5. I C = does not change. For CC = 0 : CE = CC = I C R C = 0 (4.6 ma)(.0 k) = 5.4 For CC = 5 : CE = 5 (4.6 ma)(.0 k) = 0.7 CE = 0.7 5.4 = 5.3 increase 6. I B = I C = BB CC R B R C BE CE 4 0.7 3.3 = 70 A 4.7 k 4.7 k 4 8 = 34 ma 470 I E = I C + I B = 34 ma + 70 A = 34.7 ma IC 34 ma DC = = 48.4 I 70 A 7. (a) BE = 0.7 BB I B = R B B BE 4.3 =. ma 3.9 k I C = DC I B = 50(. ma) = 55 ma CE = CC I C R C = 5 (55 ma)(80 ) = 5.0 CB = CE BE = 5.0 0.7 = 4.40 (b) BE = 0.7 BB I B = R B BE 3 ( 0.7 ).3 = 85. A 7 k 7 k I C = DC I B = 5(85. A) = 0.7 ma CE = CC I C R C = 8 (0.7 ma)(390 ) = 3.83 CB = CE BE = 3.83 (0.7 ) = 3.3 5
Chapter 4 8. (a) I C(sat) = I B = R BB CC C R B 5 80 BE = 83.3 ma 5 0.7 =. ma 3.9 k I C = DC I B = 50(. ma) = 55 ma I C < I C(sat) Therefore, the transistor is not saturated. (b) I C(sat) = I B = R BB CC C R B 8 390 BE = 0.5 ma 3 0.7 7 k = 85. A I C = DC I B = 5(85. A) = 0.7 ma I C < I C(sat) Therefore, the transistor is not saturated. 9. B = E = B BE = 0.7 =.3 E.3 I E = =.3 ma R.0 k E I C = DC I E = (0.98)(.3 ma) =.7 ma DC 0.98 DC = = 49 0.98 DC I B = I E I C =.3 ma.7 ma = 30 A 0. (a) B = BB = 0 C = CC = 0 E = B BE = 0 0.7 = 9.3 CE = C E = 0 9.3 = 0.7 BE = 0.7 CB = C B = 0 0 = 0 (b) B = BB = 4 C = CC = E = B BE = 4 (0.7 ) = 3.3 CE = C E = (3.3) = 8.7 BE = 0.7 CB = C B = (4 ) = 8. For DC = 00: B BE 0 0.7 I E = = 930 A RE 0 k DC 00 DC = = 0.990 0 DC I C = DC I E = (0.990)(930 A) = 9 A 6
Chapter 4 For DC = 50: I E = 930 A DC 50 DC = 5 DC = 0.993 I C = DC I E = (0.993)(930 A) = 94 A I C = 94 A 0.9 A = 3 A. P D(max) = CE I C P CE(max) = I D(max) C. W 50 ma = 4 3. P D(max) = 0.5 W (75C)( mw/c) = 0.5 W 75 mw = 45 mw Section 4-4 The BJT as an Amplifier 4. out = A v in = 50(00 m) = 5 5. A v = out in 0 300 m = 33.3 RC 560 6. A v = = 56 r e 0 c = out = A v in = 56(50 m) =.8 BB BE 7. I B = R B.5 0.7 = 8 A 00 k I C = DC I B = 50(8 A) = 4.5 ma CC CE 9 4 R C = =. k I 4.5 ma C 8. (a) DC current gain = DC = 50 (b) DC current gain = DC = 5 7
Chapter 4 Section 4-5 The BJT as a Switch CC 5 9. I C(sat) = = 500 A RC 0 k IC(sat) 500 A I B(min) = = 3.33 A 50 I B(min) = DC IN(min) 0.7 R B R B I B(min) = IN(min) 0.7 IN(min) = R B I B(min) + 0.7 = (3.33 A)(.0 M) + 0.7 = 4.03 5 30. I C(sat) = =.5 ma. k I.5 ma I B(min) = C(sat) = 50 A 50 R B(min) = DC IN(cutoff) = 0 IN 0.7 4.3 = 7. k I 50 A B(min) Section 4-6 The Phototransistor 3. I C = DC I = (00)(00 A) = 0 ma 3. I = (50 lm/m )( A/lm/m ) = 50 A I E = DC I = (00)(50 A) = 5 ma 33. I out = (0.30)(00 ma) = 30 ma 34. I OUT I IN I IN = = 0.6 I OU 0 ma T 0.6 0.6 = 6.7 ma Section 4-7 Transistor Categories and Packaging 35. See Figure 4-. 36. (a) Small-signal (b) Power (c) Power (d) Small-signal (e) RF Figure 4-8
Chapter 4 Section 4-8 Troubleshooting 37. With the positive probe on the emitter and the negative probe on the base, the ohmmeter indicates an open, since this reverse-biases the base-emitter junction. With the positive probe on the base and the negative probe on the emitter, the ohmmeter indicates a very low resistance, since this forward-biases the base-collector junction. 38. (a) Transistor s collector junction or terminal is open. (b) Collector resistor is open. (c) Operating properly. (d) Transistor s base junction or terminal open (no base or collector current). 39. 5 0.7 (a) I B = 68 k = 63. A 9 3. I C = 3.3 k =.76 ma IC.76 ma DC = I 63. A = 7.8 B 4.5 0.7 (b) I B = 7 k = 4 A 4 6.8 I C = 470 = 5.3 ma IC 5.3 ma DC = I 4 A = 09 Application Activity Problems B 40. Q OFF, Q ON I R = 0, P R = 0 mw I R = 0, P R = 0 mw 0.7 I R3 = I R4 = = 304 A. k36 k P R3 = (304 A) (. k) = 0 W P R4 = (304 A) (36 k) = 3.3 mw I R5 = 0.76 = 9 ma 60 P R5 = (9 ma) (60 ) = 4 mw 9
Chapter 4 Q ON, Q OFF I R = 0.7 = 5 A 75 k P R = (5 A) (75 k) =.7 mw (0.7 ) P R = = 0.49 W.0 M I R4 0. = 9.9 ma. k P R4 = (9.9 ma) (. k) = 8 mw I R3 0, P R3 = 0 mw I R5 = 0, P R5 = 0 mw 4. I C(max) = 00 ma CC R L(min) = = 60 I 00 ma C(max) 4. See Figure 4-. Figure 4- Datasheet Problems 43. From the datasheet of textbook Figure 4-0: (a) For a N3904, CEO(max) = 40 (b) For a N3904, I C(max) = 00 ma (c) For a N3904 @ 5C, P D(max) = 65 mw (d) For a N3904 @ T C = 50C, P D(max) = 65 mw 5 mw/c(5c) = 65 mw 5 mw = 500 mw (e) For a N3904 with I C = ma, h FE(min) = 70 30
Chapter 4 44. For an MMBT3904 with T A = 65C: P D(max) = 350 mw (65C 5C)(.8 mw/c) = 350 mw 40C(.8 mw/c) = 350 mw mw = 38 mw 45. For a PZT3904 with T C = 45C: P D(max) = W (45C 5C)(8 mw/c) = W 0C(8 mw/c) = W 60 mw = 840 mw 46. For the circuits of textbook Figure 4-66: 3 0.7.3 (a) I B = = 6.97 ma 330 330 Let h FE = 30 I C = 30(6.97 ma) = 09 ma CC CE(sat) 30 0. I C(sat) = = 0 ma RC 70 The transistor is saturated since I C cannot exceed 0 ma. P D = (0. )(0 ma) = mw At 50C, P D(max) = 350 mw (50C 5C)(.8 mw/c) = 80 mw No parameter is exceeded. (b) CEO = 45 which exceeds CEO(max). 47. For the circuits of textbook Figure 4-67: 5 0.7 4.3 (a) I B = = 4.30 A 0 k 0 k h FE(max) = 300 I C = 300(4.30 A) = 9 ma 9 I C(sat) = = 9 ma.0 k The transistor is saturated. 3 0.7.3 (b) I B = = 3 A 00 k 00 k h FE(max) = 300 I C = 300(3 A) = 6.90 ma I C(sat) = =.4 ma 560 The transistor is not saturated. 48. I B(min) = I B(max) = h I C FE(max) I h C FE(min) 0 ma 50 0 ma 50 = 66.7 A = 00 A 3
Chapter 4 49. For the circuits of textbook Figure 4-69: 8 0.7 7.3 (a) I B = = 07 A 68 k 68 k h FE = 50 I C = 50(07 A) = 6. ma C = 5 (6. ma)(680 ) = 5 0.95 = 4.05 CE = 4.05 0.7 = 3.35 P D = (3.35 )(6. ma) = 53.9 mw At 40C, P D(max) = 360 mw (40C 5C)(.06 mw/c) = 39 mw No parameters are exceeded. 5 0.7 4.3 (b) I B = = 95 A 4.7 k 4.7 k h FE = 300 I C = 300(95 A) = 74 ma 35 0.3 I C(sat) = 73.8 ma 470 The transistor is in hard saturation. Assuming CE(sat) = 0.3, P D = (0.3 )(73.8 ma) =. mw No parameters are exceeded. Advanced Problems 50. DC = DC DC DC DC DC = DC DC = DC ( + DC ) DC DC = ) ( DC 5. I C = 50(500 A) = 75 ma CE = 5 (80 )(75 ma) =.5 Since CE(sat) = 0.3 @ I C = 50 ma, the transistor comes out of saturation. 5. From the datasheet, DC(min) = 5 (for I C = 00 ma) 50 ma I B(max) = = 0 ma 5 3 0.7.3 R B(min) = = 30 0 ma 0 ma Use the standard value of 40 for R B. To avoid saturation, the load resistance cannot exceed about 9 = 53.3 50 ma See Figure 4-3. Figure 4-3 3
Chapter 4 53. Since I B = 0 ma for I C = 50 ma, 9 0.7 8.3 R B(min) = = 830 0 ma 0 ma Use 90. The load cannot exceed 53.3. See Figure 4-4. Figure 4-4 54. R C(min) = Av r e = 50(8 ) = 400 (Use 430 ) 5 I C = = 6.3 ma 430 Assuming h FE = 00, 6.3 ma I B = = 63 A 00 4 0.7 R B(max) = = 0.3 k (Use 8 k) 63 A See Figure 4-5. Figure 4-5 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 55 through 6 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 55. R B shorted 56. R C open 57. Collector-emitter shorted 58. Collector-emitter open 59. R E leaky 60. Collector-emitter shorted 6. R B open 6. R C open 33
Chapter 5 Transistor Bias Circuits Section 5- The DC Operating Point. The transistor is biased too close to saturation.. I C = DC I B = 75(50 A) =.3 ma CE = CC I C R C = 8 (.3 ma)(.0 k) = 8.3 = 6.75 Q-point: CEQ = 6.75, I CQ =.3 ma 3. I C(sat) R CC C 8.0 k = 8 ma 4. CE(cutoff) = 8 5. Horizontal intercept (cutoff): CE = CC = 0 ertical intercept (saturation): CC 0 I C(sat) = ma R 0 k 6. I B = BB C 0.7 R B BB = I B R B + 0.7 = (0 A)(.0 M) + 0.7 = 0.7 I C = DC I B = 50(0 A) = ma CE = CC I C R C = 0 ( ma)(0 k) = 0 7. See Figure 5-. CE = CC I C R C CC CE 0 4 R C = =. k I 5 ma I B = I C DC C 5 ma 00 = 0.05 ma 0 0.7 R B = = 86 k 0.05 ma P D(min) = CE I C = (4 )(5 ma) = 0 mw Figure 5-34
Chapter 5 8. I B = I C(sat) = BB R R B CC C BE.5 0.7 = 80 A 0 k 8 390 = 0.5 ma I C = DC I B = 75(80 A) = 6 ma The transistor is biased in the linear region because 0 < I C < I C(sat). 9. (a) I C(sat) = 50 ma (b) CE(cutoff) = 0 (c) I B = 50 A I C = 5 ma CE = 5 0. (a) I C 4 ma (b) Interpolating between I B = 400 A and I B = 500 A I B 450 A (c) CE.5 See Figure 5-. Section 5- oltage-divider Bias. R TH CC R R 6.7 k Figure 5- R 4.7 k 5.64 RR ( k )(4.7 k ) TH R R 6.7 k TH BE IE E TH DC 3.87 k.64 0.7 =.75 ma R R / 680 3.87 k / 50 B = I E R E + BE = (.75 ma)(680 + 0.7 =.57 IERIN(BASE) ( IE)(0 R) (.75 ma)(47 k ) DC(min).57 B B 35
Chapter 5. I C(sat) = CC R R C E 5 = 6.88 ma.8 k E(sat) = I C(sat) R E = (6.88 ma)(680 ) = 4.68 B = E(sat) + 0.7 = 4.68 + 0.7 = 5.38 R RIN(BASE) CC B R R RIN(BASE) DCB (50)(5.38 ) RIN(BASE) 7 k IE 6.88 ma ( R RIN(BASE) ) CC B ( R R RIN(BASE) ) ( R RIN(BASE) ) CC ( R RIN(BASE) ) B R B ( R R )( ) R IN(BASE) CC B B R B ( R RIN(BASE) ).3 k CC B R RIN(BASE).3 k 7.3 S R.3 k 7 k R 7.3 S = 3.7 kω R k 3. B = CC 5 =.5 R R 4 k E =.5 0.7 = 0.55 E 0.55 I E = = 809 A R 680 4. TH = E I C 809 A CE = CC I C R C E = 5 (809 A)(.5 k + 680 ) = 3. R 5 k 9 =.8 CC R R 6 k RR (47 k )(5 k ) R TH = =.4 k R R 6 k TH BE.8 0.7 =.34 ma RE RTH / DC k.4 k / 0 I C I E =.34 ma C = CC I C R C = 9 (.34 ma)(. k) = 6.05 E = I E R E = (.34 ma)(.0 k) =.34 B = E + BE =.34 + 0.7 =.04 36
Chapter 5 5. See Figure 5-3. Figure 5-3 6. (a) R 5.6 k TH = CC ( ) R R =.74 38.6 k RR (5.6 k )(33 k ) R TH = = 4.79 k R R 38.6 k TH BE.74 0.7 = 4. ma R R / 560 4.79 k / 50 E TH DC B = I E R E + BE = (4. ma)(560 k + 0.7 =.3 + 0.7 =.6 TH BE.74 0.7 (b) = 3.7 ma R R / 560 4.79 k / 50 E TH DC B = I E R E + BE = (3.7 ma)(560 k + 0.7 =.08 + 0.7 =.38 7. (a) EQ = B + 0.7 =.6 + 0.7 = 0.9 EQ 0.9 I CQ I E = =.63 ma R 560 E CQ = CC I C R C = (.63 ma)(.8 k) = 9.07 CEQ = CQ EQ = 9.07 (0.9 ) = 8.6 (b) P D(min) = I CQ CEQ = (.63 ma)(8.6 ) = 3.3 mw 8. B =.6 CC B (.6 ) I = = 35 A R 33 k I = R B.6 =.88 A 5.6 k I B = I I = 35 A 88 A = 7 A 37
Chapter 5 Section 5-3 Other Bias Methods 9. Using Equation 5-9: EE BE ( 5 ) 0.7 4.3 I E = RE RB/ DC. k0 k /00. k0. k I C I E =.86 ma IC.86 ma I B = = 8.6 A 00 B = I B R B = (8.6 A)(0 k) = 0.86 E = B 0.7 = 0.86 0.7 = 0.886 C = CC I C R C = 5 (.86 ma)(.0 k) = 3.4 =.86 ma 0. Assume CE 0 at saturation. E = 0.886 so C(sat) = 0.886 CC C(sat) 5 ( 0.886 ) I C(sat) = RC.0 k RE 4. R E(min) = = 698 I 5.89 ma C(sat) = 5.89 ma. At 00C: BE = 0.7 (.5 m/c)(75c) = 0.53 EE BE ( 5 ) 0.53 4.49 I E = R R /. k0 k /00.3 k E B DC At 5C: I =.86 ma (from problem 9) E I E =.95 ma.86 ma = 0.09 ma =.95 ma. A change in DC does not affect the circuit when R E >>R B / DC. Since EE BE I E = R R / E B DC In the equation, if R B / DC is much smaller than R E, the effect of DC is negligible. 3. Assume DC = 00. EE E 0 0.7 I C I E = = 6.3 ma R R / 470 0 k /00 E B CE = EE CC I C (R C + R E ) = 0 3. = 6.95 4. B = 0.7 CC BE I C = R R C B / DC 3 0.7 =.06 ma.8 k 33 k /90 C = CC I C R C = 3 (.06 ma)(.8 k) =.09 38
Chapter 5 5. I C =.06 ma from Problem 4. I C =.06 ma (0.5)(.06 ma) = 0.795 ma CC BE I C = R R / C R C = CC B BE DC ICR I C B / DC 3 0.7 (0.795 ma)(33 k) /90 0.795 ma =.53 k 6. I C = 0.795 ma from Problem 5. CE = CC I C R C = 3 (0.795 ma)(.53 k) = 0.989 P D(min) = CE I C = (0.989 )(0.795 ma) = 786 W 7. See Figure 5-4. CC BE I C = R R C B / DC 0.7 = 7.87 ma. k 47 k / 00 C = CC I C R C = (7.87 ma)(. k) =.56 Figure 5-4 8. BB = CC ; E = 0 CC 0.7 0.7.3 I B = = 54 A R k k B I C = DC I B = 90(54 A) = 46.3 ma CE = CC I C R C = (46.3 ma)(00 ) = 7.37 9. I CQ = 80(54 A) = 9.5 ma CEQ = (9.5 ma)(00 ) =.75 30. I C changes in the circuit with a common CC and BB supply because a change in CC causes I B to change which, in turn, changes I C. 3. I B = BB R B BE 9 0.7 = 553 A 5 k CC 9 I C(sat) = = 90 ma RC 00 For DC = 50: I C = DC I B = 50(553 A) = 7.7 ma CE = CC I C R C = 9 (7.7 ma)(00 ) = 6.3 For DC = 5: I C = DC I B = 5(553 A) = 69. ma CE = CC I C R C = 9 (69. ma)(00 ) =.08 Since I C < I C(sat) for the range of DC, the circuit remains biased in the linear region. 39
Chapter 5 CC 9 3. I C(sat) = = 90 ma RC 00 At 0C: DC = 0 0(0.5) = 55 CC BE 9 0.7 I B = = 553 A R 5 k B I C = DC I B = 55(553 A) = 30.4 ma CE = CC I C R C = 9 (30.4 ma)(00 ) = 5.96 At 70C: DC = 0 + 0(0.75) = 93 I B = 553 A I C = DC I B = 93(553 A) = 07 ma I C > I C(sat), therefore the transistor is in saturation at 70C. I C = I C(sat) I C(0) = 90 ma 30.4 ma = 59.6 ma CE CE(0) CE(sat) = 5.96 0 = 5.96 Section 5-4 Troubleshooting 33. The transistor is off; therefore, = 0, = 0, 3 = 8. 34. = 0.7, = 0 8 0.7 0.7 I B = = A 70 A = 5 A 33 k 0 k I C = 00(5 A) = 30. ma 8 I C(sat) = = 3.64 ma, so C E = 0. k If the problem is corrected, 0 k = 8 =.86 0 k 33 k = E =.86 0.7 =.6.6 I E = =.6 ma.0 k 3 = C = 8 (.6 ma)(. k) = 5.45 35. (a) Open collector (b) No problems (c) Transistor shorted from collector-to-emitter (d) Open emitter 36. For DC = 35: 4.5 k B = ( 0 ) = 3. 4.5 k For DC = 00: 5.7 k B = ( 0 ) = 3.4 5.7 k The measured base voltage at point 4 is within the correct range. 40
E = 3. + 0.7 =.4.4 I C I E = = 3.53 ma 680 Chapter 5 C = 0 (3.53 ma)(.0 k) = 6.47 Allowing for some variation in BE and for resistor tolerances, the measured collector and emitter voltages are correct. 37. (a) The 680 resistor is open: Meter : 0 Meter : floating 5.6 k Meter 3: B = ( 0 ) 5.6 k Meter 4: 0 = 3.59 (b) The 5.6 k resistor is open. 9.3 I B = = 75 A 0 k 35(680 ) I C = 35(75 A) = 9.6 ma 0 I C(sat) = = 5.95 ma 680 The transistor is saturated. Meter : 0 Meter : (5.95 ma)(680 ) = 4.05 Meter 3: 4.05 + 0.7 = 4.75 Meter 4: 0 (5.95 ma)(.0 k) = 4.05 (c) The 0 k resistor is open. The transistor is off. Meter : 0 Meter : 0 Meter 3: 0 Meter 4: 0 (d) The.0 k resistor is open. Collector current is zero. Meter : 0 Meter :.7 0.7 = 0.57 5.6 k 680 Meter 3: (0 ) + 0.7 = 0.57 + 0.7 =.7 0 k 5.6 k 680 Meter 4: floating (e) A short from emitter to ground. Meter : 0 Meter : 0 Meter 3: 0.7 (0 0.7 ) 9.3 I B = 0.93 ma 0 k 0 k I C(min) = 35(0.93 ma) = 3.6 ma 4
Chapter 5 0 I C(sat) = = 0 ma.0 k The transistor is saturated. Meter 4: 0 (f) An open base-emitter junction. The transistor is off. Meter : 0 Meter : 0 5.6 k Meter 3: (0 ) = 3.59 5.6 k Meter 4: 0 Application Activity Problems 38. With R open: B = 0, E = 0, C = CC = 9. 39. Faults that will cause the transistor of textbook Figure 5-9(a) to go into cutoff: R open, R shorted, base lead or BE junction open. 40. At 45C: R Therm =.7 k R Therm.7 k B 9 9 = 3.8 R R Therm 7.4 k E = B 0.7 =.58 E.58 I E = I C = R = 5.49 ma 470 3 C = OUT = 9 (5.49 ma)( k) = 3.5 At 48C: R Therm =.78 k.78 k B = 9 =.47 6.48 k E =.47 0.7 =.77 I E = I C =.77 = 3.77 ma 470 C = OUT = 9 (3.77 ma)( k) = 5.3 At 53C: R Therm =.8 k.8 k B = 9 =.93 5.98 k E =.93 0.7 =.3 I E = I C =.3 =.6 ma 470 C = OUT = 9 (.6 ma)( k) = 6.38 4
Chapter 5 4. The following measurements would indicate an open CB junction: C = CC = +9. B normal E 0 Datasheet Problems 4. For T = 45C and R =.7 k R IN(base) =.7 k (30)(470 ).7 k 4.k =.7 k min R IN(base) =.7 k (300)(470 ).7 k 4k =.65 k max.7 k.7 k B(min) = 9. 9. =.6.7 k 5.6 k 7.87 E(min) =.6 0.7 =.9.9 So, I C I E = = 4.09 ma 470 C(max) = 9. (4.09 ma)(.0 k) = 5.0.65 k.65 k B(max) = 9. 9. =.9.65 k 5.6 k 8.5 k E(max) =.9 0.7 =.. So, I C I E = = 4.73 ma 470 C(min) = 9. (4.73 ma)(.0 k) = 4.37 For T = 55C and R =.4 k: R IN(base) =.4 k (30)(470 ).4 k 4.k =.4 k min R IN(base) =.4 k (300)(470 ).4 k 4k =.3 k max.4 k.4 k B(min) = 9. 9. =.54.4 k 5.6 k 6.74 k E(min) =.54 0.7 = 0.839 0.839 So, I C I E = =.78 ma 470 C(max) = 9. (.78 ma)(.0 k) = 7.3.3 k.3 k B(max) = 9. 9. =.64.3 k 5.6 k 6.83 k E(max) =.64 0.7 = 0.938 0.938 So, I C I E = =.0 ma 470 C(min) = 9. (.0 ma)(.0 k) = 7.0 43
Chapter 5 43. At T = 45C for minimum DC : P D(max) = (5.0.9 )(4.09 ma) = (3.09 )(4.09 ma) =.6 mw At T = 55C for minimum DC : P D(max) = (7.3 0.839 )(.78 ma) = (6.48 )(.78 ma) =.5 mw For maximum beta values, the results are comparable and nowhere near the maximum. P D(max) = 65 mw (5.0 m/c)(30c) = 475 mw No ratings are exceeded. 44. For the datasheet of Figure 5-49 in the textbook: (a) For a NA, I C(max) = A continuous (b) For a N8A, EB(max) = 6.0 45. For a NA @ T = 00C: P D(max) = 0.8 W (4.57 mw/c)(00c 5C) = 0.8 W 343 mw = 457 mw 46. If I C changes from ma to 500 ma in a N9A, the percentage change in DC is 30 50 DC = 00% = 40% 50 Advanced Problems 47. See Figure 5-5. CC CEQ 5 5 R C = = k ICQ 5 ma Assume DC = 00. I 5 ma I BQ = CQ = 50 A 00 R B = DC CC I BQ BE 5 0.7 = 86 k Figure 5-5 50 A 48. See Figure 5-6. Assume DC = 00. I 0 ma I BQ = CQ = 50 A DC 00 Let R B =.0 k (50 A)(.0 k) 0.7.3 R E = =.3 k 0 ma 0 ma (.3 4 ) 8.7 R C = = 870 0 ma 0 ma 870 and.3 k are not standard values. R C = 80 and R E =. k give I CQ 9.38 ma, CEQ 5.05. Figure 5-6 44
Chapter 5 49. See Figure 5-7. DC(min) 70. Let R E =.0 k. E = I E R E =.5 ma(.0 k) =.5 B =.5 + 0.7 =. CC CEQ E 9.5 3 R C = = 3 k ICQ.5 ma CC 9 R + R = =.57 k min ICC(max) ICQ 5 ma.5 ma Asssume DC R E >>R. The ratio of bias resistors equals the ratio of the voltages as follows. R 6.8 = 3.09 R. R = 3.09R R + R = R + 3.09R =.57 k 4.09R =.57 k.57 k R = = 68 4.09 So, R 60 and R =.9 k k. Figure 5-7 From this, DCB (70)(. ) R IN(base) = = 03 kw >> R I.5 ma E 60 so, B = 9 =.3.6 k E =.3 0.7 =.43.43 I CQ I E = =.43 ma.0 k CEQ = 9 (.43 ma)(.0 k + 3 k) = 3.8 50. See Figure 5-8. DC 75. 0 ma I BQ = = 33 A 75 CC CE 5.5 R C = = 350 (use 360 ) I 0 ma R B = CE CQ 0.7.55 0.7 = 6 k (use 6. k) I 33 A BQ 5 0.7 I CQ = = 9.7 ma 360 6. k / 75 CEQ = C = 5 (9.7 ma)(360 ) =.50 Figure 5-8 5. The N3904 in textbook Figure 5-47 can be replaced with a NA and maintain the same voltage range from 45C to 55C because the voltage-divider circuit is essentially independent and the DC parameters of the two transistors are comparable. 45
Chapter 5 5. For the NA using the datasheet graph in textbook Figure 5-50 at I C = 50 ma and CE =.0 : At T = 55C, h FE(min) = (0.45)(50) =.5 At T = 5C, h FE(min) = (0.63)(50) = 3.5 At T = 75C, h FE(min) = (0.53)(50) = 6.6 53. If the valve interface circuit loading of the temperature conversion circuit changes from 00 k to 0 k, the Q-point will have a reduced CEQ because the current through R C will consist of the same I C and a larger I L. I CQ is unaffected in the sense that the transistor collector current is the same, although the collector resistance current is larger. The transistor saturates sooner so that lower temperatures do not register as well, if at all. 54. It is not feasible to operate the circuit from a 5. dc supply and maintain the same range of output voltages because the output voltage at 60C must be 6.478. Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 55 through 60 are available from the Instructor s Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 55. R C open 56. R B open 57. R open 58. Collector-emitter shorted 59. R C shorted 60. Base-emitter open 46
Chapter 6 BJT Amplifiers Section 6- Amplifier Operation. Slightly greater than ma minimum. From the graph of Figure 6-4, the highest value of dc collector current is about 6 ma. Section 6- Transistor AC Models 3. 5 m 5 m r e = 8.33 I 3 ma E 4. ac = h fe = 00 5. I C = DC I B = 30(0 A) =.3 ma I.3 ma I E = C =.3 ma 0.99 6. DC = DC 5 m 5 m r e = 9 I.3mA ac = I I E C B I I C B ma = 33 5 A 0.35 ma = 7 3 A Section 6-3 The Common-Emitter Amplifier 7. See Figure 6-. Figure 6-47
Chapter 6 R 4.7 k 8. (a) B = CC 5 =.64 R R 6.7 k (b) E = B 0.7 =.64 0.7 =.94 E.94 (c) I E = R =.94 ma E.0 k (d) I C I E =.94 ma (e) C = CC I C R C = 5 (.94 ma)(. k) =.6 9. I CC = I BIAS + I C B.64 I BIAS = R = 56 A 4.7 k I CC = 56 A +.94 ma =.50 ma P = I CC CC = (.5 ma)(5 ) = 37.5 mw 4.7 k 0. (a) B = 5 =.64 4.7 k k E =.64 0.7 =.94.94 I E = =.94 ma.0 k 5 m 5 m r e =.9 IE.94 ma = 00(0.9 ) 0 k R in(base) = (b) R in = R in(base) E r e ac r e R E R 0k k 4.7 k = 3.73 k R RC. k (c) A v = =.7 R. 0. (a) R in(base) = ac r e = 00(.9 ) =.9 k (b) R in =.9 k k 4.7 k = 968 r e RC. k (c) A v = = 7. 9. (a) R in(base) = ac r e = 00(.9 ) =.9 k (b) R in =.9 k k 4.7 k = 968 (c) A v = R RC R. k 0 k c L = 40 r r. 9 e e 48
Chapter 6 3. (a) TH = R k 8 = 3.66 CC R R 59 k RR (47 k )( k ) R TH = = 9.56 k R R 59 k TH BE 3.66 0.7 =.63 ma R R / k9.56 k / 75 E TH DC (b) E = I E R E = (.63 ma)( k) =.63 (c) B = E + BE =.63 + 0.7 = 3.76 (d) I C I E =.63 ma (e) C = CC I C R C = 8 (.63 ma)(3.3 kw) = 9.3 (f) CE = C E = 9.3.63 = 6.69 4. From Problem 3, I E =.63 ma 5 m 5 m (a) R in(base) = acre ac 70 I = 665 E.63 ma (b) R in = R R R ( ) 47 k k 665 = 6 in base RC RL 3.3 k 0 k (c) A v = = 6 re 9.5 (d) A i = ac = 70 (e) A p = A v A i = (6)(70) = 8,70 R 640 5. b = in in 640 600 Rin Rs Attenuation of the input network is R in 640 = 0.56 Rin Rs 640 600 Av 0. 56A v = 0.56(53) = 3 = 80 6. TH = R 3.3 k 8 =.73 CC R R 5.3 k RR ( k )(3.3 k ) R TH = =.59 k R R 5.3 k TH BE.73 0.7 = 8.78 ma R R / 00.59 k /50 E TH DC 49
Chapter 6 5 m 5 m r' e = =.85 IE 8.78 ma Maximum gain is at R e = 0 RC 330 v(max) = = 6 re.85 Minimum gain is at R e = 00. RC 330 v(min) = = 3. R r.85 7. TH = E e R 3.3 k 8 =.73 CC R R 5.3 k RR ( k )(3.3 k ) R TH = R R 5.3 k =.59 k TH BE.73 0.7 RE RTH / DC 00.59 k /50 = 8.78 ma 5 m 5 m r e = =.85 IE 8.78 ma Maximum gain is at R e = 0 RC RL 330 600 v(max) = = 74.7 re.85 Minimum gain is at R e = 00. RC RL 3 v(min) = R r 0.85 =.07 E e 8. R in = R R ac r e 3.3 k k 50(3.5 ) = 40 Attenuation of the input network is Rin 40 = 0.578 R R 40 300 in s R 330.0 k c A v = = 76.3 re 3. 5 A 0. 5777 = 0.578(76.3) = 44. v A v 9. See Figure 6-. 5 m r e = 9.8.55 ma R e 0 r e Set R e = 00. The gain is reduced to RC 3.3 k A v = R e r e 09. 8 = 30. Figure 6-50
Chapter 6 Section 6-4 The Common-Collector Amplifier R 4.7 k 0. B = CC 5.5 =.76 R R 4.7 k B 0.7.76 0.7 I E = =.06 ma R.0 k E E 5 m r e = 3.6.06 ma RE.0 k A v = = 0.977 R.0 k 3.6 r e = 3. k. R in = R R r R R R 0 k 4.7 k 00 k ac e E acre R OUT = B 0.7 = R R CC 4.7 k 0.7 5.5 0.7 =.06 4.7 k. The voltage gain is reduced because A v = Re. R r e e R 4.7 k 3. B = CC 5.5 =.76 R R 4.7 k B BE.76 0.7 I E = =.06 ma R.0 k E 5 m 5 m r e = 3.6 I.06 ma A v = A e E R E r R E R L R re RE RL RE RL E RL Av RE RL Av re v R RE RL Av Av re L Av re 0.9(3.6 ) R E RL =.4 ( Av ) 0.9 R L R E =.4R L +.4R E R L R E.4R L =.4R E.4RE (.4 )(000 ) R L = = 70 R.4 000. 4 E 5
Chapter 6 4. (a) C = 0 R k B = CC 0 = 4 R R 55 k E = B 0.7 = 4 0.7 = 3.3 C = 0 B = E = 3.3 E = B 0.7 = 3.3 0.7 =.6 (b) DC = DC DC = (50)(00) = 5,000 E 0.7.6 (c) I E = = 7.3 A DCRE 00(.5 k ) 5 m 5 m r e =.45 k I 7.3 A I E = R E E E.6 =.73 ma.5 k 5 m 5 m r e = 4.5 I.73 m E (d) R in = R R R in ( base) R in(base) = ac ac R E = (50)(00)(.5 k) =.5 M R in = 33 k k.5 M = 3. k 5. R in(base) = ac ac R E = (50)(00)(.5 k) =.5 M R in = R R R in ( base) = k 33 k.5 M = 3. k I in = R in in I in(base) = = 75.8 A 3. k R in in(base) = 44.4 na.5 M I e ac ac I in(base) = (50)(00)(44.4 na) = 667 A I e 667 A A i = 8.8 I 75.8 A in Section 6-5 The Common-Base Amplifier 6. The main disadvantage of a common-base amplifier is low input impedance. Another disadvantage is unity current gain. 5
Chapter 6 R 0 k 7. E = CC BE 4 0.7 = 6.8 R R 3 k 6.8 I E = = 0.97 ma 60 R in(emitter) = r e 5 m 5 ma r e =.8 I 0.97 ma RC. k A v = = 56. 8 A i A p = A i A v 56 E 8. (a) Common-base (b) Common-emitter (c) Common-collector Section 6-6 Multistage Amplifiers 9. A v = A v A v = (0)(0) = 400 30. A v( db) = 0 db + 0 db + 0 db = 30 db 0 log A v = 30 db 30 log A v = =.5 0 A v = 3.6 R 8. k 3. (a) E CC BE 5 0.7 =.9 R R 33 k 8. k E.9 I E = =.9 ma R.0 k E 5 m 5 m r e = 0.9 I.9 ma E R in() = R6 R5 acr e = 8. k 33 k 75(0.9 ) =.48 k A v = R r e C Rin() 3.3 k.48 k = 93.6 r 0. 9 e RC 3.3 k A v = = 303 0. 9 (b) A v = A v A v = (93.6)(303) = 8,36 (c) A v(db) = 0 log(93.6) = 39.4 db A v(db) = 0 log(303) = 49.6 db A = 0 log(8,36) = 89. db v(db) 53
Chapter 6 3. (a) A v = A v = R C Rin() 3.3 k.48 k = 93.6 r 0. 9 e e RC RL 3.3 k 8 k = 56 r 0. 9 (b) R in() = R R acr e = 33 k 8. k 75(0.9 ) =.48 k Attenuation of the input network is Rin().48 k = 0.95 Rin() Rs.48 k 75 A = (0.95)A v A v = (0.95)(93.6)(56) =,764 v (c) A v(db) = 0 log(93.6) = 39.4 db A v(db) = 0 log(56) = 48. db A = 0 log(,764) = 87. db v(db) 33. R k B = CC R R k =.6 E = B 0.7 =.46 E.46 I C I E = R 4.7 k = 0.3 ma 4 C = CC I C R 3 = (0.3 ma)( k) = 5.6 B = C = 5.6 E = B 0.7 = 5.6 0.7 = 4.46 E 4.46 I C I E = = 0.446 ma R 0 k 6 C = CC I C R 5 = (0.446 ma)(0 k) = 7.54 5 m 5 m r e = 56 IE 0.446 ma R in() = ac r e = (5)(56 ) = 7 k 5 m 5 m r e = 80.4 IE 0.3mA R3 Rin() k 7 k A v = = 66 r 80.4 e R5 0 k A v = = 79 r e 56 A = A v A v = (66)(79) =,84 v 34. (a) 0 log() =.6 db (b) 0 log(50) = 34.0 db (c) 0 log(00) = 40.0 db (d) 0 log(500) = 68.0 db 54
Chapter 6 35. (a) 0 log = 3 db (b) 0 log = 6 db (c) 0 log = 0 db 3 6 log = 0.5 log 0 = 0.3 log 0 0 = 0.5 0 =.4 = = 3.6 (d) 0 log = 0 db (e) 0 log = 40 db 0 log 40 = log 0 = 0 = 0 = 00 Section 6-7 The Differential Amplifier 36. Determine I E for each transistor: R 4.3 E I R = 6.5 ma E RE. k I RE IE( Q) IE( Q) = 3.5 ma Determine I C for each transistor: I I = 0.980(3.5 ma) = 3.85 ma C( Q) E( Q) I I = 0.975(3.5 ma) = 3.69 ma C( Q) E( Q) Calculate the collector voltages: C( Q) = 5 (3.85 ma)(3.3 k) = 4.49 C( Q) = 5 (3.69 ma)(3.3 k) = 4.54 The differential output voltage is: OUT = C( Q) C( Q) = 4.54 4.49 = 0.05 = 50 m 37. measures the differential output voltage. measures the non-inverting input voltage. 3 measures the single-ended output voltage. 4 measures the differential input voltage. I measures the bias current. 55
Chapter 6 38. Calculate the voltage across each collector resistor: R C = (.35 ma)(5. k) = 6.89 R C = (.9 ma)(5. k) = 6.58 The differential output voltage is: OUT = ( ) ( ) C( Q) C( Q) CC RC CC RC RC RC = 6.89 6.58 = 0.3 = 30 m 39. (a) Single-ended differential input, differential output (b) Single-ended, differential input, single-ended output (c) Double-ended differential input, single-ended output (d) Double-ended differential input, differential output Section 6-8 Troubleshooting R 0 k 40. E = 0 0.7 0 0.7 =.05 R R 57 k E.05 I E = =.05 ma R.0 k 4 C = 0 (.05 ma)(4.7 k) = 5.07 CE = 5.07.05 = 4.0 CE 4.0 r CE = 3.83 k IE.05 ma With C shorted: R IN() = R6 DCR8 0 k 5(.0 k) = 9.6 k Looking from the collector of Q : r CE R4 RIN() (3.83 k.0 k ) 9.6 k = 3.7 k 3.7 k C = 0 = 4.03 3.7 k 4.7 k 4. Q is in cutoff. I C = 0 A, so C = 0. 4. (a) Reduced gain (b) No output signal (c) Reduced gain (d) Bias levels of first stage will change. I C will increase and Q will go into saturation. (e) No signal at the Q collector (f) Signal at the Q base. No output signal. 56
Chapter 6 43. r e = 0.9 R in =.48 k A v = 93.6 A v = 30 Test Point DC olts AC olts (rms) Input 0 5 A Q base.99 0.8 Q emitter.9 0 Q collector 7.44.95 m Q base.99.95 m Q emitter.9 0 Q collector 7.44 589 m Output 0 589 m Application Activity Problems 44. For the block diagram of textbook Figure 6-46 with no output from the power amplifier or preamplifier and only one faulty block, the power amplifier must be ok because the fault must be one that affects the preamplifier's output prior to the power amplifier. Check the input to the preamplifier. 45. (a) No output signal (b) Reduced output signal (c) No output signal (d) Reduced output signal (e) No output signal (f) Increased output signal (perhaps with distortion) 46. R 7 = 0 will bias Q off. 47. (a) Q is in cutoff. (b) C = EE (c) C is unchanged and at 5.87. Datasheet Problems 48. From the datasheet in textbook Figure 6-63: (a) for a N3947, ac(min) = h fe(min) = 00 (b) For a N3947, r e( min) cannot be determined since h re(min) is not given. (c) For a N3947, r c( min) cannot be determined since h re(min) is not given. 49. From the N3947 datasheet in Figure 6-63: (a) For a N3947, ac(max) = 700 4 hre 00 (b) For a N3947, r e (max) = 40 h 50 S oe 57
Chapter 6 (c) For a N3947, 4 hre 00 r c (max) = 0 k h 50 S oe 50. For maximum current gain, a N3947 should be used. Advanced Problems 5. In the circuit of textbook Figure 6-6, a leaky coupling capacitor would affect the biasing of the transistors, attenuate the ac signal, and decrease the frequency response. 5. See Figure 6-3. Figure 6-3 58
Chapter 6 53. For the nd stage: 30 30 I R6-7 = R R = 435 A 69 k 6 7 B = CC I R6-7 R 6 = 5 (435A)(47 k) = 5 0.5 = 5.5 E 5.5 0.7 I E = =. ma R9 R0 5.3 k 5 m r e = 0.7. ma With R 0 = 0 for max gain: R8 6.8 k A v() = = 45. (unloaded) R9 r e 50.7 With a 0 k load: R8 RL 6.8 k 0 k 4.05 k A v() = = 6.9 R9 re 50.7 50.7 To keep unloaded gain: 4.05 k 45. R9 0.7 4.05 k = 45.(R 9 + 0.7 ) = 45.R 9 + 934 4.05 k934 R 9 = = 69. 45. 54. R C > (00)(330 ) = 33 k To prevent cutoff, C must be no greater than (00)(.44)(5 m) = 8.46 In addition, C must fall no lower than 8.46 3.54 = 4.93 to prevent saturation. R C = 00 RE r e 5 m r e IE I C R C = 8.46 I C R C = 3.54 I C 00( R E r e ) = 3.54 5 m I C 00 330 3.54 IC (33 k)i C +.5 = 3.54 I C = 3.4 A 5 m r e = 797 3.4 A R C = 00(330 + 797 ) = 3 k Let R C = 0 k. C = (3.4 A)(0 k) = 8.3 C(sat) = 8.3 3.54 = 4.69 59
Chapter 6 RE(tot) 4.69 RC 7.3 R = (0.64)(0 k) = 77 k. Let R E = 68 k. E(tot) E = (3.4 A)(68 k) =.4 B =.4 + 0.7 =.84 R.84 = 0.30 R 9.6 R = 0.30R. If R = 0 k, R = 6. k. The amplifier circuit is shown in Figure 6-4. From the design: 6. k B = =.84 6. k E =.4.4 I C I E = = 3.3 A 68.3 k 5 m r e = 798 3.3 A 0 k A v = = 06 or 40.5 db 795 330 C = (3.3 A)(0 k) = 8.4 The design is a close fit. Figure 6-4 55. See Figure 6-5. R in = 0 k 0 k (00)(5.k) = 53.6 k minimum Figure 6-5 Figure 6-6 56. See Figure 6-6. 60
Chapter 6 57. See Figure 6-7. 6 0.7 I C = = 0 ma 50 k/00 5 m r e =.5 0 ma 80 A v = = 7.4.5 This is reasonably close (3.3% off) and can be made closer by putting a 7.5 resistor in series with the 80 collector resistor. Figure 6-7 58. Assuming ac = 00, C = fr c (00 Hz)(330 k330 k (00 34 k )) = = 0.0 F (00 Hz)(6 k ) C = fr c (00 Hz)( k47 k k (00 5.3 k )) = = 0.043 F (00 Hz)(36.98 k ) 59. I C I E RC A v = r e RC 5 m/ I E RC 5 m/ I C RCI C 5 m R C 5 m = 40 RC Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 60 through 65 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 60. C open 6. C shorted 6. R E leaky 63. C open 64. C open 65. C 3 open 6
Chapter 7 Power Amplifiers Section 7- The Class A Power Amplifier R 330. (a) B = CC 5 = 3.7 R R.0 k 330 k E = B BE = 3.7 0.7 = 3.0 E 3.0 I CQ I E = = 68.4 ma R R 8. 36 E E CEQ = CC (I C )(R E + R E + R L ) = 5 (68.4 ma)(8. + 36 + 00 ) = 5.4 RL 00 (b) A v = =.7 R r 8. 0. 37 E e R in = ac (R E + r e ) R R = 00 (8. + 0.37 ) 330.0 k= 9 A p = R in 9 A.7 v = 63 RL 00 The computed voltage and power gains are slightly higher if r e is ignored.. (a) If R L is removed, there is no collector current; hence, the power dissipated in the transistor is zero. (b) Power is dissipated only in the bias resistors plus a small amount in R E and R E. Since the load resistor has been removed, the base voltage is altered. The base voltage can be found from the Thevenin equivalent drawn for the bias circuit in Figure 7-. Figure 7- Applying the voltage-divider rule and including the base-emitter diode drop of 0.7 result in a base voltage of.. The power supply current is then computed as I CC = CC. R 5..0 k = 3.8 ma Power from the supply is then computed as P T = I CC CC = (3.8 ma)(5 ) = 07 mw 6
Chapter 7 (c) A v =.7 (see problem (b)). in = 500 m pp = 77 m rms. out = A v in = (.7)(77 m) =.07 out.07 P out = = 4.8 mw R 00 L 3. The changes are shown in Figure 7-. The advantage of this arrangement is that the load resistor is referenced to ground. Figure 7-4. A CC amplifier has a voltage gain of approximately. Therefore, A p = R 50 5. (a) TH = = 5.4 CC R R 90 RR (680 )(50 ) R TH = R R 90 = 9 TH BE 5.4 0.7 = 54 ma RE RTH / DC 79.7 9 / 5 I C I E = 54 ma C = CC I C R C = (54 ma)(00 ) = 6.6 E = I E R E = (54 ma)(79.7 ) = 4.3 CE = C E = 6.6 4.3 =.3 (b) TH = R 4.7 k = 3.38 CC R R 6.7 k RR ( k )(4.7 k ) R TH = = 3.38 k R R 6.7 k TH BE 3.38 0.7 = 5.7 ma RE RTH / DC 4 3.38 k / 0 I C I E = 5.7 ma C = CC I C R C = (5.7 ma)(470 ) = 4.6 E = I E R E = (5.7 ma)(4 ) =.3 CE = C E = 4.6.3 =.39 R R in out. k 50 = 44 63
Chapter 7 6. The Q-point does not change because R L is capacitively coupled and does not affect the DC values. 7. For the circuit in Figure 7-43(a): From Problem 5(a), I CQ = 54 ma; CEQ =.3 R e = RC R L = 00 00 = 50 ce(cutoff) = CEQ + I CQ R c =.3 + (54 ma)(50 ) = 5 Since CEQ is closer to saturation, I c is limited to CEQ I c(p) = R =.3 = 46 ma c 50 out is limited to out(p) = CEQ =.3 For the circuit in Figure 7-43(b): From Problem 5(b), I CQ = 5.7 ma; CEQ =.39 R e = RC R L = 470 470 = 35 ce(cutoff) = CEQ + I CQ R c =.39 + (5.7 ma)(35 ) = 6.08 Since CEQ is closer to saturation, I c is limited to CEQ I c(p) = R =.39 = 0. ma c 35 out is limited to out(p) = CEQ =.39 R 8. (a) A p = in A v RL R RC R 00 00 c L 50 A v = 0.6 R R 4.7 4.7 R in = R E E R Rin( base) R R acre R in = 680 50 (5)(4.7 ) 680 50 588 = 95 A p = (0.6) 95 = 9 00 (b) R RC R 470 470 c L 35 A v = 0.7 RE RE R in = k 4.7 k (0)( ) k 4.7 k.64 k =.48 k A p = (0.7).48 k = 36 470 64
Chapter 7 9. TH = R k 4 = 4. CC R R 5.7 k RR (4.7 k )( k ) R TH = = 85 R R 5.7 k TH BE 4. 0.7 = 5 ma RE RTH / DC 30 85 / 90 I C I E = 5 ma C = CC I C R C = 4 (5 ma)(560 ) = 0 E = I E R E = (5 ma)(30 ) = 3.5 CEQ = C E = 0 3.5 = 6.75 P D(min) = P DQ = I CQ CEQ = (5 ma)(6.75 ) = 69 mw 0. From Problem 9: I CQ = 5 ma and CEQ = 6.75 ce(cutoff) = CEQ + I CQ R c = 6.75 + (5 ma)(64 ) = 3.5 P out = 0.5I CQ R c = 0.5(5 ma) (64 ) = 8.5 mw Pout Pout Pout 8.5 mw = = 0.38 P I I (4 )(5 ma) DC CC CC CC CQ Section 7- The Class B and Class AB Push-Pull Amplifiers. (a) B(Q ) = 0 + 0.7 = 0.7 B(Q) = 0 0.7 = 0.7 E = 0 CC ( CC).4 9 ( 9 ).4 I CQ = R R.0 k.0 k CEQ(Q) = 9 CEQ(Q) = 9 = 8.3 ma (b) out = in = 5.0 rms ( out ) 5.0 P out = R 50 L = 0.5 W CC 9.0. I c(sat) = R L 50 ce(cutoff) = 9 = 80 ma These points define the ac load line as shown in Figure 7-3. The Q-point is at a collector current of 8.3 ma (see problem ) and the dc load line rises vertically through this point. Figure 7-3 65
Chapter 7 3. R in = ac ( re RL ) R R From Problem, I CQ = 8.3 ma so, I E 8.3 ma 5 m r e = 3 8.3 ma R in = 00(53 ).0 k.0 k = 5300.0 k.0 k= 457 4. The DC voltage at the output becomes negative instead of 0. 5. (a) B(Q) = 7.5 + 0.7 = 8. B(Q) = 7.5 0.7 = 6.8 5 E = = 7.5 CC.4 5.4 I CQ = R R.0 k.0 k CEQ(Q ) = 5 7.5 = 7.5 CEQ(Q) = 0 7.5 = 7.5 = 6.8 ma (b) in = out = 0 pp = 3.54 rms ( L ) (3.54 ) P L = = 67 mw 75 R L 6. (a) Maximum peak voltage = 7.5 p. 7.5 p = 5.30 rms ( L ) (5.30 ) P L(max) = = 375 mw 75 R L (b) Maximum peak voltage = p. p = 8.48 rms ( L ) (8.48 ) P L(max) = = 960 mw 75 R L 7. (a) C open or Q open (b) power supply off, open R, Q base shorted to ground (c) Q has collector-to-emitter short (d) one or both diodes shorted 8. R in = ac ( re RL ) R R From Problem 5: I CQ = 6.8 ma so, I E 6.8 ma 66
Chapter 7 5 m 5 m = 3.68 I 6.8 ma r e E R in = 00(78.7 ) k k= 485 485 b = = 0.9 rms 485 50 Section 7-3 The Class C Amplifier t 9. P D(avg) = on CE(sat) I C(sat) T = (0.)(0.8 )(5 ma) = 450 W 0. f r = LC (0 mh)(0.00f) = 50.3 khz. out(pp) = CC = ( ) = 4. P out = CC R c 0.5 0.5(5 ) 50 =.5 W t P D(avg) = on CE(sat) I C(sat) = (0.)(0.8 )(5 ma) = 0.45 mw T Pout.5 W = = 0.9998 P P.5 W 0.45 mw out D(avg) Section 7-4 Troubleshooting 3. With C open, only the negative half of the input signal appears across R L. 4. One of the transistors is open between the collector and emitter or a coupling capacitor is open. 5. (a) No dc supply voltage or R open (b) Diode D open (c) Circuit is OK (d) Q shorted from collector to emitter Application Activity Problems 6. For the block diagram of textbook Figure 7-34 with no signal from the power amplifier or preamplifier, but with the microphone working, the problem is in the power amplifier or preamplifier. It must be assumed that the preamp is faulty, causing the power amp to have no signal. 7. For the circuit of Figure 7-35 with the base-emitter junction of Q open, the dc output will be approximately 5 with a signal output approximately equal to the input. 67
Chapter 7 8. For the circuit of Figure 7-35 with the collector-emitter junction of Q 5 open, the dc output will be approximately +5 with a signal output approximately equal to the input (some distortion possible). 9. On the circuit board of Figure 7-49, the vertically oriented diode has been installed backwards. Datasheet Problems 30. From the BD35 datasheet of textbook Figure 7-50: (a) DC(min) = 40 @ I C = 50 ma, CE = DC(min) = 5 @ I C = 5 ma, CE = (b) For a BD35, CE(max) = CEO = 45 (c) P D(max) =.5 W @ T C = 5C (d) I C(max) =.5 A 3. P D = 0 W @ 50C from graph in Figure 7-50. 3. P D = W @ 50C. Extrapolating from the case temperature graph in Figure 7-50, since P D =.5 W @ 5C ambient. This derating gives W. 33. As I C increases from 0 ma to approximately 5 ma, the dc current gain increases. As I C increases above approximately 5 ma, the dc current gain decreases. 34. h FE 89 @ I C = 0 ma Advanced Problems 35. T C is much closer to the actual junction temperature than T A. In a given operating environment, T A is always less than T C. 36. 4 4 I C(sat) = 330 00 430 = 55.8 ma CE(cutoff) = 4.0 k BQ = 4 = 4..0 k 4.7 k EQ = 4. 0.7 = 3.5 3.5 I EQ I CQ = 00 = 35. ma R c = 330 330 = 65 CQ = 4 (35. ma)(330 ) =.4 CEQ =.4 3.5 = 8.90 8.90 I c(sat) = 35. ma + = 89. ma 65 ce(cutoff) = 8.90 + (35. ma)(65 ) = 4.7 See Figure 7-4. Figure 7-4 68
Chapter 7 37. See Figure 7-5. 5 I R I R = = 74 ma 86 8 B 5 = 3.4 86 E = 3.4 0.7 =.44.44 I E I C = = 503 ma 4.85 C = 5 (0 )(503 ma) = 9.97 CE = 7.53 5 m r e = 0.05 503 ma The ac resistance affecting the load line is R c + R e + r e = 0 Figure 7-5 ac = DC 00 7.53 I c(sat) = 503 ma + =.4 A 0. ce(cutoff) = 7.53 + (503 ma)(0. ) =.7 The Q-point is closer to cutoff so P out = (0.5)(503 ma) (0. ) =.9 W As loading occurs, the Q-point will still be closer to cutoff. The circuit will have P out W for R L 37.7. (39 standard) 38. Preamp quiescent current: 30 I = I = = 45 A 660 k I 3 = I 4 = I 5 = 5 0.7 = 4 A 34 k I 6 = I 7 = 30 = 435 A 69 k B = 5 (435 A)(47 k) = 5.45 5 ( 5.45 0.7 ) I 8 = I 9 = I 0 = =.73 ma 5.3 k I tot = 45 A + 4 A + 435 A +.73 ma =.63 ma Power amp quiescent current: I 0 5.7 3(0.7 ) 3.6 I = = 3.6 ma.0 k.0 k 5 ( 0.7 ) 4.3 I 3 = = 65 ma 0 0 I tot = 3.6 ma + 65 ma = 78.6 ma 69
Chapter 7 Signal current to load: Scope shows 9.8 peak output. 0.707(9.8 ) I L = = 866 ma 8 I tot(sys) =.63 ma + 78.6 ma + 866 ma = 947 ma Amp. hrs = 947 ma 4 hrs = 3.79 Ah Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 39 through 43 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 39. C in open 40. R E open 4. Q collector-emitter open 4. D shorted 43. Q drain-source open 70
Chapter 8 Field-Effect Transistors (FETs) Section 8- The JFET. (a) A greater GS narrows the depletion region. (b) The channel resistance increases with increased GS.. The gate-to-source voltage of an n-channel JFET must be zero or negative in order to maintain the required reverse-bias condition. 3. See Figure 8-. 4. See Figure 8-. Figure 8- Section 8- JFET Characteristics and Parameters 5. DS = P = 5 at point where I D becomes constant. 6. GS(off) = P = 6 The device is on, because GS =. 7. By definition, I D = I DSS when GS = 0 for values of DS > P. Therefore, I D = 0 ma. 8. Since GS > GS(off), the JFET is off and I D = 0 A. Figure 8-7
Chapter 8 9. P = GS(off) = (4 ) = 4 The voltmeter reads DS. As DD is increased, DS also increases. The point at which I D reaches a constant value is DS = P = 4. 0. I D = I DSS GS GS(off) 0 I D = 5 ma = 5 ma 8 I D = 5 ma = 3.83 ma 8 I D = 5 ma =.8 ma 8 3 I D = 5 ma =.95 ma 8 4 I D = 5 ma =.5 ma 8 5 I D = 5 ma = 0.703 ma 8 6 I D = 5 ma = 0.33 ma 8 7 I D = 5 ma = 0.078 ma 8 8 I D = 5 ma = 0 ma 8 See Figure 8-3. Figure 8-3 7
Chapter 8. I D = I I I DSS GS GS(off) GS GS(off) GS GS(off) I I D DSS D DSS GS = GS(off) I D IDSS GS = 8.5 ma = 8 (0.39) =.63 5 ma. g m = GS 4 g m 0 300 S = 600 S GS(off) 8 3. g m = GS g m 0 000 S = 49 S GS(off) 7 g fs = g m = 49 S 4. R IN = I GS GSS 0 5 na = 000 M 5. GS = 0 : I D = GS I DSS = 8 ma( 0) = 8 ma GS(off) GS = : I D = 8 ma = 8 ma( 0.) = 8 ma(0.8) = 5. ma 5 GS = : I D = 8 ma = 8 ma( 0.4) = 8 ma(0.6) =.88 ma 5 3 GS = 3 : I D = 8 ma = 8 ma( 0.6) = 8 ma(0.4) =.8 ma 5 4 GS = 4 : I D = 8 ma = 8 ma( 0.8) = 8 ma(0.) = 0.30 ma 5 5 GS = 5 : I D = 8 ma = 8 ma( ) = 8 ma(0) = 0 ma 5 Section 8-3 JFET Biasing 6. GS = I D R S = ( ma)(00 ) =. 73
Chapter 8 7. R S = I GS D 4 5 ma = 800 8. R S = I GS D 3.5 ma =. k 9. (a) I D = I DSS = 0 ma (b) I D = 0 A (c) I D increases 0. (a) S = ( ma)(.0 k) = (b) S = (5 ma)(00 ) = 0.5 D = ( ma)(4.7 k) = 7.3 D = 9 (5 ma)(470 ) = 6.65 G = 0 G = 0 GS = G S = 0 = GS = G S = 0 0.5 = 0.5 DS = 7.3 = 6.3 DS = 6.65 0.5 = 6.5 (c) S = (3 ma)(470 ) =.4 D = 5 (3 ma)(. k) = 8.4 G = 0 GS = G S = 0 (.4 ) =.4 DS = 8.4 (.4 ) = 6.99. From the graph, GS at I D = 9.5 ma. GS R S = = I 9.5 ma D I 4 ma. I D = DS S = 7 ma GS(off) 0 GS = =.93 3.44 3.44 R S = GS.93 = 49 (The nearest standard value is 430.) I 7 ma D DD D 4 R D = =.7 k (The nearest standard value is.8 k.) ID 7 ma Select R G =.0 M. See Figure 8-4. Figure 8-4 74
Chapter 8 3. R IN(total) = R G RIN R IN = I GS GSS 0 = 500 M 0 na R IN(total) = 0 M 500 M = 9.8 M 4. For I D = 0, GS = I D R S = (0)(330 ) = 0 For I D = I DSS = 5 ma GS = I D R S = (5 ma)(330 ) =.65 From the graph in Figure 8-69 in the textbook, the Q-point is GS 0.95 and I D.9 ma 5. For I D = 0, GS = 0 For I D = I DSS = 0 ma, GS = I D R S = (0 ma)(390 ) = 3.9 From the graph in Figure 8-70 in the textbook, the Q-point is GS. and I D 5.3 ma 6. Since R D = 9 5 = 4, R 4 D I D = = 0.85 ma R 4.7 k D S = I D R S = (0.85 ma)(3.3 k) =.8 R. M G = DD 9 =.6 R R. M GS = G S =.6.8 =.9 Q-point: I D = 0.85 ma, GS =.9 7. For I D = 0, R. M GS = G = DD = 4.8 R R 5.5 M For GS = 0, S = 4.8 G S GS 4.8 I D = =.45 ma RS RS 3.3 k The Q-point is taken from the graph in Figure 8-7 in the textbook. I D.9 ma, GS =.5 Section 8-4 The Ohmic Region DS 0.8 8. R DS = = 4 k I 0.0 ma D 75
Chapter 8 0.4 9. R DS = =.67 k 0.5 ma 0.6 R DS = =.33 k 0.45 ma R DS =.67 k.33 k =.34 k GS 30. g m = gm0.5 ms GS(off) 3.5 =.5 ms(0.74) =.07 ms 3. r ds = g.07 ms = 935 m Section 8-5 The MOSFET 3. See Figure 8-5. Figure 8-5 33. An n-channel D-MOSFET with a positive GS is operating in the enhancement mode. 34. An E-MOSFET has no physical channel or depletion mode. A D-MOSFET has a physical channel and can be operated in either depletion or enhancement modes. 35. MOSFETs have a very high input resistance because the gate is insulated from the channel by an SiO layer. Section 8-6 MOSFET Characteristics and Parameters ID(on) 0 ma 36. K = ( ) ( 3 ) GS GS(th) = 0. ma/ I D = K( GS GS(off) ) = (0. ma/ )(6 + 3 ) =.08 ma GS 37. I D = DSS I GS(off) ID 3 ma I DSS = GS 0 GS(off) = 4.69 ma 76
Chapter 8 38. (a) n channel GS (b) I D = DSS 5 I I D = 8 ma = 0 ma GS(off) 5 4 3 I D = 8 ma = 0.3 ma I D = 8 ma =.8 ma 5 5 I D = 8 ma =.88 ma I D = 8 ma = 5. ma 5 5 0 I D = 8 ma = 8 ma I D = 8 ma =.5 ma 5 5 3 I D = 8 ma = 5.7 ma I D = 8 ma = 0.5 ma 5 5 4 5 I D = 8 ma = 5.9 ma I D = 8 ma = 3 ma 5 5 (c) See Figure 8-6. Figure 8-6 77
Chapter 8 Section 8-7 MOSFET Biasing 39. (a) Depletion (b) Enhancement (c) Zero bias (d) Depletion 40. 0 M (a) GS = 0 = 6.8 4.7 M This MOSFET is on..0 M (b) GS = (5 ) =.7 M This MOSFET is off. 4. Since GS = 0 for each circuit, I D = I DSS = 8 ma. (a) DS = DD I D R D = (8 ma)(.0 k) = 4 (b) DS = DD I D R D = 5 (8 ma)(. k) = 5.4 (c) DS = DD I D R D = 9 (8 ma)(560 ) = 4.5 4. (a) I D(on) = 3 ma @ 4, GS(th) = R 4.7 M GS = DD 0 = 3. R R 4.7 M ID(on) 3 ma 3 ma K = = 0.75 ma/ ( ) (4 ) ( ) GS GS(th) I D = K( GS GS(th) ) = (0.75 ma/ )(3. ) =.08 ma DS = DD I D R D = 0 (.08 ma)(.0 k) = 0.08 = 8.9 (b) I D(on) = ma @ 3, GS(th) =.5 R 0 M GS = DD 5 =.5 R R 0 M ID(on) ma ma K = ( ) (3.5 ) (.5 ) GS GS(th) = 0.89 ma/ I D = K( GS GS(th) ) = (0.89 ma/ )(.5.5 ) = 0.89 ma DS = DD I D R D = 5 (0.89 ma)(.5 k) = 5.34 = 3.66 43. (a) DS = GS = 5 DD DS 5 I D = R. k D (b) DS = GS = 3. DD DS 8 3. I D = R 4.7 k D = 3.8 ma =.0 ma 44. DS = DD I D R D = 5 ( ma)(8. k) = 6.8 GS = DS I G R G = 6.8 (50 pa)( M) = 6.799 78
Chapter 8 Section 8-8 The IGBT 45. The input resistance of an IGBT is very high because of the insulated gate structure. 46. With excessive collector current, the parasitic transistor turns on and the IGBT acts as a thyristor. Section 8-9 Troubleshooting 47. When I D goes to zero, the possible faults are: R D or R S open, JFET drain-to-source open, no supply voltage, or ground connection open. 48. If I D goes to 6 ma, the possible faults are: The JFET is shorted from drain-to-source or DD has increased. 49. If DD is changed to 0, I D will change very little or none because the device is operating in the constant-current region of the characteristic curve. 50. The device is off. The gate bias voltage must be less than GS(th). The gate could be shorted or partially shorted to ground. 5. The device is saturated, so there is very little voltage from drain-to-source. This indicates that GS is too high. The.0 M bias resistor is probably open. Application Activity Problems 5. (a) 500 m (b) 00 m (c) 0 m (d) 400 m 53. At GS = 6, I D 0 ma At GS =, I D 5 ma 54. GS = sensor = 400 m OUT = 9.048 DD OUT 9.048 I D = R R 0 3 4 =.64 ma GS = sensor = 300 m OUT = 7.574 I D = 7.574 = 3.95 ma 0 GS = sensor = 00 m OUT = 5.930 I D = 5.930 = 5.4 ma 0 79
Chapter 8 GS = sensor = 00 m OUT = 4.890 I D = 4.890 = 6.35 ma 0 GS = sensor = 0 m OUT = 4.97 I D = 4.97 = 6.97 ma 0 GS = sensor = 00 m OUT = 3.56 I D = 3.56 = 7.53 ma 0 GS = sensor = 00 m OUT =.960 I D =.960 = 8.07 ma 0 GS = sensor = 300 m OUT =.38 I D =.38 = 8.59 ma 0 See Figure 8-7. Figure 8-7 R 50 k 55. GS = DD = 4 R R 50 k From the graph in Figure 8-79 in the textbook for GS = 0 and GS = 4 : I D 8 ma OUT = (8 ma)(0 ) = 3.04 80
Chapter 8 Datasheet Problems 56. The N5457 is an n-channel JFET. 57. From the datasheet in textbook Figure 8-4: (a) For a N5457, GS(off) = 0.5 minimum (b) For a N5457, DS(max) = 5 (c) For a N5458 @ 5C, P D(max) = 30 mw (d) For a N5459, GS(rev) = 5 maximum 58. P D(max) = 30 mw (.8 mw/c)(65c 5C) = 30 mw 3 mw = 97 mw 59. g m0(min) = g fs = 000 S 60. Typical I D = I DSS = 9 ma 6. From the datasheet graph in textbook Figure 8-80: I D.4 ma at GS = 0 6. For a N3796 with GS = 6, I D = 5 ma 63. From the datasheet graph in textbook Figure 8-80: At GS = +3, I D 3 ma At GS =, I D 0.4 ma 64. y fs = 500 S at f = khz and at f = MHz for both the N3796 and N3797. There is no change in g fs over the frequency range. 65. For a N3796, GS(off) = 3.0 typical Advanced Problems 66. For the circuit of textbook Figure 8-8: GS I D = DSS I where GS = I D R S GS(off) From the N5457 datasheet: I DSS(min) =.0 ma and GS(off) = 0.5 minimum I D = 66.3 A GS = (66.3 A)(5.6 k) = 0.37 DS = (66.3 A)(0 k + 5.6 k) =.0 8
Chapter 8 67. For the circuit of textbook Figure 8-8: 3.3 k C = 9 = (0.48)(9 ) =.3 3.3 k From the equation, GS I D = DSS I where GS = G I D R S GS(off) I D is maximum for I DSS(max) and GS(off) max, so that I DSS = 6 ma and GS(off) = 8.0 I D = 3.58 ma GS =.3 (3.58 ma)(.8 k) =.3 6.45 = 4. 68. From the N5457 datasheet: I DSS(min) =.0 ma and GS(off) = 0.5 minimum I D(min) = 66.3 A DS(max) = (66.3 A)(5.6 k) =.0 and I DSS(max) = 5.0 ma and GS(off) = 6.0 maximum I D(max) = 677 A DS(min) = (677 A)(5.6 k) =.4 69. ph = +300 m I D = (.9 ma)( + 0.3 /5.0 ) = (.9 ma)(.06) = 3.6 ma DS = 5 (3.6 ma)(.76 k) = 5 8.99 = +6.0 70. ma = ma = I DSS (ma) RS GS(off) (ma) RS.9 ma 0.5 (ma) RS 0.345 = 0.5 (ma) RS 0.587 = 0.5 ( ma) RS 0.43 = 0.5 R S =.06 k Use R S =. k. Then I D = 963 A GS = S = (963 A)(. k) =.9 So, D =.9 + 4.5 = 6.6 R D = 9 6.6 =.47 k 963 A Use R D =.4 k. So, DS = 9 (963 A)(4.6 k) = 4.57 8
Chapter 8 7. Let I D = 0 ma. 4 R D = = 00 0 ma Let S =. R S = = 00 0 ma ID(on) 500 ma K = = 6.7 ma/ ( GS(on) GS(th) ) (0 ) Let I D = 0 ma. ( GS ) 0 = = 3.4 6.7 ma/ GS =.8 GS =.8 G = S +.8 = 4.8 Figure 8-8 For the voltage divider: R 7. =.5 R 4.8 Let R = 0 k. R = (.5)(0 k) = 5 k See Figure 8-8. Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 7 through 80 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 7. R S shorted 73. R D shorted 74. R G shorted 75. R open 76. Drain-source open 77. R D open 78. R shorted 79. Drain-source shorted 80. R shorted 83
Chapter 9 FET Amplifiers and Switching Circuits Section 9- The Common-Source Amplifier. (a) I d = g m gs = (6000 S)(0 m) = 60 A (b) I d = g m gs = (6000 S)(50 m) = 900 A (c) I d = g m gs = (6000 S)(0.6 ) = 3.6 ma (d) I d = g m gs = (6000 S)( ) = 6 ma. A v = g m R d A v 0 R d = = g 3500 S m = 5.7 k RDr (4.7 k)( k) 3. A v = ds gm 4. ms RD rds 6.7 k 4. R d = R r 4.7 k k = 3.38 k D m ds gmrd (4. ms)(3.38 k) A v = =.73 g R (4. ms)(.0 k) s = 4. 5. (a) N-channel D-MOSFET with zero-bias. GS = 0. (b) P-channel JFET with self-bias. GS = I D R S = (3 ma)(330 ) = 0.99 (c) N-channel E-MOSFET with voltage-divider bias. GS = R R R DD 4.7 k 4.7 k = 3.84 6. (a) G = 0, S = 0 D = DD I D R D = 5 (8 ma)(.0 k) = 7 (b) G = 0 S = I D R D = (3 ma)(330 ) = 0.99 D = DD + I D R D = 0 + (3 ma)(.5 k) = 5.5 R 4.7 k (c) G = DD = 3.84 R R 4.7 k S = 0 D = DD I D R D = (6 ma)(.0 k) = 6 84
Chapter 9 7. (a) n-channel D-MOSFET (b) n-channel JFET (c) p-channel E-MOSFET 8. From the curve in Figure 9-6(a) in the textbook: I d(pp) 3.9 ma.3 ma =.6 ma 9. From the curve in Figure 9-6(b) in the textbook: I d(pp) 6 ma ma = 4 ma From the curve in Figure 9-6(c) in the textbook: I d(pp) 4.5 ma.3 ma = 3. ma 0. D = DD I D R D = (.83 ma)(.5 k) = 7.76 S = I D R S = (.83 ma)(.0 k) =.83 DS = D S = 7.76.83 = 4.93 GS = G S = 0.83 =.83. A v = g m R d = R 5000 S.5 k 0 k g = 6.5 m D R L pp(out) = (.88)(50 m)(6.5) = 90 m. A v = g m R d R d =.5 k.5 k = 750 A v = (5000 S)(750 ) = 3.75 out = A v in = (3.75)(50 m) = 88 m rms 3. (a) A v = g m R d = mrd RL 3.8 ms. k k (b) A v = g m R d = R R 5.5 ms. k 0 k 4. See Figure 9-. g = 3.8 ms(38 ) = 4.3 g = 5.5 ms(.8 k) = 9.9 m D L 5. I D = I DS 5 ma S = 7.5 ma Figure 9-85
Chapter 9 6. GS = (7.5 ma)(0 ) =.65 IDSS (5 ma) g m0 = = 7.5 ms 4 GS(off) g m = (7.5 ms)(.65 /4 ) = 4.4 ms g (4.4mS)(80 3.3 k) m Rd (4.4mS)(657 ) A v = =.47 g RS (4.4mS)(0 ) 0.97 m 7. A v = g m R d = ( 4.4mS)(80 3.3 k 4.7 k) = (4.4 ms)(576 ) =.54 8. I D = I DS S 9 ma = 4.5 ma GS = I D R S = (4.5 ma)(330 ) =.49 DS = DD I D (R D + R S ) = 9 (4.5 ma)(.33 k) = 3 9. A v = g m R d = R 3700 S.0 k 0 k g = 3700 S(909 ) = 3.36 m D R L out = A v in = (3.36)(0 m) = 33.6 m rms R 6.8 k 0. GS = DD 0 = 5.48 R R 4.8 k ID(on) 8 ma K = = 0.3 ma/ ( ) (0.5 ). R IN = GS GS(th) I D = K( GS GS(th) ) = 0.3 ma/ (5.48.5 ) =.84 ma DS = DD I D R D = 0 (.84 ma)(.0 k) = 7. I GS GSS 5 = 600 M 5 na R in = 0 M 600 M = 9.84 M. A v = g m R d = ms.0 k 0 M 48 4.8 out = A v in = 4.8(0 m) = 48 m rms I D = I DSS = 5 ma D = 4 (5 ma)(.0 k) = 9 See Figure 9-. Figure 9-86
Chapter 9 R 47 k 3. GS = DD 8 = 9 R R 94 k ID(on) 8 ma K = = 0.5 ma/ ( ) ( 4 ) GS GS(th) I D(on) = K( GS GS(th) ) = 0.5 ma/ (9 4 ) = 3.3 ma DS = DD I D R D = 8 (3.5 ma)(.5 k) = 3.3 A v = g m R D = 4500 S(.5 k) = 6.75 ds = A v in = 6.75(00 m) = 675 m rms Section 9- The Common-Drain Amplifier 4. R s =. k k 545 gmrs (5500 S)(545 ) A v = g R (5500 S)(545 ) = 0.750 R IN = I GS m GSS R in = 0 M s 5 = 3 0 50 pa 30 0 M 5. R s =. k k 545 gmrs (3000 S)(545 ) A v = g R (3000 S)(545 ) = 0.60 R IN = I GS m GSS R in = 0 M s 5 = 3 0 50 pa 30 0 M 6. (a) R s = 4.7 k 47 k = 4.7 k gmrs (3000 S)(4.7 k) A v = g R (3000 S)(4.7 k) m s (b) R s =.0 k 00 = 90.9 gmrs (4300 S)(90.9 ) A v = g R (4300 S)(90.9 ) 7. (a) R s = 4.7 k 0 k = 3. k m s gmrs (3000 S)(3. k) A v = g R (3000 S)(3. k) m s = 0.98 = 0.8 = 0.906 (b) R s = 00 0 k = 99 gmrs (4300 S)(99 ) A v = = 0.99 g R (4300 S)(99 ) m s 87
Chapter 9 Section 9-3 The Common-Gate Amplifier 8. A v = g m R d = 4000S(.5 k) = 6.0 9. R in(source) = g m 4000 S = 50 30. A v = g m R d = 3500S(0 k) = 35 R in = R S. k g m 3500 S = 53 3. X L = fl = (00 MHz)(.5 mh) = 943 k A v = g m(cg) X L = (800 S)(943 k) = 640 GS 5 R in = R3 5 M IGSS na = 5 M 500 M = 4.6 M Section 9-4 The Class D Amplifier 3. A v = (9 ) 8 = 3600 5 m 5 m 33. P out = ( )(0.35 A)= 4. W P int = (0.5 )(0.35 A) + 40 mw = 87.5 mw + 40 mw = 7.5 mw Pout 4. W = 0.95 P P 4. W + 7.5 mw out int Section 9-5 MOSFET Analog Switching 34. G p(out) = GS(Th) p(out) = G GS(Th) = 8 4 = 4 pp(in) = p(out) = 4 = 8 35. f min = 5 khz = 30 khz 36. R = fc f = RC (0 k )(0 pf) = 0 MHz 37. R = fc (5 khz)(0.00 F) = 40 k 88
Chapter 9 Section 9-6 MOSFET Digital Switching 38. out = + 5 when in = 0 out = 0 when in = + 5 39. (a) out = 3.3 (b) out = 3.3 (c) out = 3.3 (d) out = 0 40. (a) out = 3.3 (b) out = 0 (c) out = 0 (d) out = 0 4. The MOSFET has lower on-state resistance and can turn off faster. Section 9-7 Troubleshooting 4. (a) D = DD ; No signal at Q drain; No output signal (b) D 0 (floating); No signal at Q drain; No output signal (c) GS = 0 ; S = 0 ; D less than normal; Clipped output signal (d) Correct signal at Q drain; No signal at Q gate; No output signal (e) D = DD ; Correct signal at Q gate; No Q drain signal or output signal 43. (a) out = 0 if C is open. (b) A v = g m R d = 5000 S(.5 k) = 7.5 gmrd 7.5 A v = =.4 gmrs (5000 S)(470 ) A v = A v A v = (7.5)(.4) = 6.8 out = A v in = (6.8)(0 m) = 68 m (c) GS for Q is 0, so I D = I DSS. The output is clipped. (d) No out because there is no signal at the Q gate. Datasheet Problems 44. The N3796 FET is an n-channel D-MOSFET. 45. (a) For a N3796, the typical GS(off) = 3.0 (b) For a N3797, DS(max) = 0 (c) At T A = 5C, P D(max) = 00 mw (d) For a N3797, GS(max) = 0 46. P D = 00 mw (.4 mw/c)(55c 5C) = 66 mw 47. For a N3796 with f = khz, g m0 = 900 S minimum 48. At GS = 3.5 and DS = 0, I D(min) = 9.0 ma, I D(typ) = 4 ma, I D(max) = 8 ma 49. For a zero-biased N3796, I D(typ) =.5 ma 89
Chapter 9 50. A v(max) = (800 S)(. k) = 3.96 Advanced Problems 5. R d(min) =.0 k 4 k = 800 A v(min) = (.5 ms)(800 ) =.0 R d(max) =.0 k 0 k = 909 A v(min) = (7.5 ms)(909 ) = 6.8 5. I DSS(typ) =.9 ma R D + R S = = 4.4 k.9 ma = 435 g m 300 S If R S = 0, then R D 4 k (3.9 k standard) A v = (300 S)(3.9 k) = 8.97 DS = 4 (.9 ma)(3.9 k) = 4.3 =.7 The circuit is a common-source zero-biased amplifier with a drain resistor of 3.9 k. 53. To maintain DS = for the range of I DSS values: For I DSS(min) = ma R D = = 6 k ma For I DSS(max) = 6 ma R D = = k 6 ma To maintain A v = 9 for the range of g m (y fs ) values: For g m(min) = 500 S 9 R D = 500 S = 6 k For g m(max) = 3000 S 9 R D = 3000 S = 3 k A drain resistance consisting of a. k fixed resistor in series with a 5 k variable resistor will provide more than sufficient range to maintain a gain of 9 over the specified range of g m values. The dc voltage at the drain will vary with adjustment and depends on I DSS. The circuit cannot be modified to maintain both DS = and A v = 9 over the full range of transistor parameter values. See Figure 9-3. Figure 9-3 90
Chapter 9 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 54 through 6 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 54. Drain-source shorted 55. C open 56. C open 57. R S shorted 58. Drain-source open 59. R open 60. R D open 6. R open 6. C open 9
Chapter 0 Amplifier Frequency Response Section 0- Basic Concepts. If C = C, the critical frequencies are equal, and they will both cause the gain to decrease at 40 db/decade below f c.. At sufficiently high frequencies, the reactances of the coupling capacitors become very small and the capacitors appear effectively as shorts; thus, negligible signal voltage is dropped across them. 3. BJT: C be, C bc, and C ce FET: C gs, C gd, and C ds 4. Low-frequency response: C, C, and C 3 High-frequency response: C bc, C be, and C ce R 4.7 k 5. E CC 0.7 0 0.7 =.79 R R 37.7 k E.79 I E = = 3. ma R 560 E 5 m r e = 7.8 3. ma R. k 5.6 k c A v = = 0 re 7. 8 C in(miller) = C bc (A v + ) = 4 pf(0 + ) = 8 pf A 6. C out(miller) = v 03 C bc 4 pf = 4 pf Av 0 7. I D = 3.36 ma using Eq. 9 and a programmable calculator. GS = (3.36 ma)(.0 k) = 3.36 (0 ma) g m0 = =.5 ms 8 3.36 g m = (.5 ms) =.45 ms 8 A v = g m R d = (.45 ms) (.0 k 0 k) =.3 C gd = C rss = 3 pf C in(miller) = C gd (A v + ) = 3 pf(.3) = 6.95 pf A C out(miller) = v.3 C gd 3 pf = 5.8 pf Av.3 9
Chapter 0 Section 0- The Decibel 8. A p = P P out in 5 W 0.5 W = 0 P A p(db) = 0 out log = 0 log 0 = 0 db Pin out. 9. in = = 4 m rms Av 50 A v(db) = 0 log(a v ) = 0 log 50 = 34.0 db 5 0. The gain reduction is 0 log = 8.3 db 65 mw. (a) 0 log = 3.0 dbm mw mw (b) 0 log = 0 dbm mw 4 mw (c) 0 log = 6.0 dbm mw 0.5 mw (d) 0 log = 6.0 dbm mw 4.7 k. B = 0 =.79 37.7 k.79 I E = = 3.0 ma 560 5 m r e = 7.8 3. ma 5.6 k. k A v = = 0 7.8 A v(db) = 0 log(0) = 46. db At the critical frequencies, A v(db) = 46. db 3 db = 43. db Section 0-3 Low-Frequency Amplifier Response 3. (a) f c = (b) f c = RC RC = 38 Hz (00 )(5 F) =.59 khz (.0 k)(0. F) 93
Chapter 0 4. R IN(BASE) = DC R E =.5 k R R IN(BASE) 4.7 k.5 k E = 9 0.7 9 0.7 =.3 R k 4.7 k.5 k R R IN(BASE) E.3 I E = = 3 ma R 00 E 5 m r e =.9 3 ma R in(base) = = (5)(.9 ) = 40 ac r e R in = 50 + R in(base) R R = 50 + 40 k 4.7 k = 74 For the input circuit: f c = = 58 Hz R in C (74 )( F) For the output circuit: f c = = 77 Hz ( RC RL ) C3 (900 )( F) For the bypass circuit: R TH = f c = R R R k 4.7 k 50 49.3 s R / R C (.3 )(0 F) r e TH DC E = 6.89 khz RC RL 0 680 A v = = 86.6 re. 9 A v(db) = 0 log(86.6) = 38.8 db The bypass circuit produces the dominant low critical frequency. See Figure 0-. Figure 0-94
Chapter 0 5. From Problem 4: A v(mid) = 86.6 A v(mid) (db) = 38.8 db For the input RC circuit: f c = 578 Hz For the output RC circuit: f c = 77 Hz For the bypass RC circuit: f c = 6.89 khz The f c of the bypass circuit is the dominant low critical frequency. At f = f c = 6.89 khz: A v = A v(mid) 3 db = 38.8 db 3 db = 35.8 db At f = 0.f c : A v = 38.8 db 0 db = 8.8 db At 0f c (neglecting any high frequency effects): A v = A v(mid) = 38.8 db 6. At f = f c = X C = R = tan X C tan () = 45 R At f = 0.f c, X C = 0R. = tan (0) = 84.3 At f = 0f c, X C = 0.R. = tan (0.) = 5.7 7. R in(gate) = I GS GSS 0 = 00 M 50 na R in = R G R in ( gate) 0 M 00 M = 9.5 M For the input circuit: f c = = 3.34 Hz R in C (9.5 M)(0.005 F) For the output circuit: f c = = 3.0 khz ( RD RL ) C (560 0 k)(0.005 F) The output circuit is dominant. See Figure 0-. (A v is determined in Problem 8.) Figure 0-95
Chapter 0 (5 ma) 8. g m = g m0 = = 5 ms 6 A v(mid) = g ( R D ) 5 ms(560 0 k) =.65 m R L A v(mid) (db) = 8.47 db At f c : A v = 8.47 db 3 db = 5.47 db At 0.f c : A v = 8.47 db 0 db =.5 db At 0f c : A v = A v(mid) = 8.47 db (if 0f c is still in midrange) Section 0-4 High-Frequency Amplifier Response 9. From Problems 4 and 5: r e =.9 and A v(mid) = 86.6 Input circuit: C in(miller) = C bc (A v + ) = 0 pf(87.6) = 876 pf C tot = Cbe Cin( miller) = 5 pf + 876 pf = 90 pf f c = = 4.3 MHz Rs R R acr ectot 50 k 4.7 k 40 90 pf Output circuit: A C out(miller) = v 87.6 C bc 0 pf = 0. pf Av 86.6 f c = R c C = 94.9 MHz out( miller ) (66 )(0.pF) Therefore, the dominant high critical frequency is determined by the input circuit: f c = 4.3 MHz. See Figure 0-3. Figure 0-3 96
Chapter 0 0. At f = 0.f c = 458 khz: A v = A v(mid) = 38.8 db At f = f c = 4.58 MHz: A v = A v(mid) 3 db = 38.8 db 3 db = 35.8 db At f = 0f c = 45.8 MHz: A v = A v(mid) 0 db = 38.8 db 0 db = 8.8 db At f = 00f c = 458 MHz: The roll-off rate changes to 40 db/decade at f = 94.6 MHz. So, for frequencies from 45.8 MHz to 94.6 MHz, the roll-off rate is 0 db/decade and above 94.6 MHz it is 40 db/decade. The change in frequency from 45.8 MHz to 94.6 MHz represents 94.6 MHz 45.8 MHz 00% =.8% 458 MHz 45.8 MHz So, for.8% of the decade from 45.8 MHz to 458 MHz, the roll-off rate is 0 db/decade and for the remaining 88.% of the decade, the roll-off rate is 40 db/ decade. A v = 8.8 db (0.8)(0 db) (0.88)(40 db) = 8.8 db.36 db 35.3 db = 8.9 db. C gd = C rss = 4 pf C gs = C iss C rss = 0 pf 4 pf = 6 pf Input circuit: C in(miller) = C gd (A v + ) = 4 pf(.65 + ) = 4.6 pf C tot = Cgs Cin( miller ) = 6 pf + 4.6 pf = 0.6 pf f c = =.9 MHz RC s tot (600 )(0.6 pf) Output circuit: A C out(miller) = v.65 C gd 4 pf = 5.5 pf Av.65 f c = R d C = 54.5 MHz out( miller ) (530 )(5.5pF) The input circuit is dominant.. From Problem : For the input circuit, f c =.9 MHz and for the output circuit, f c = 54.5 MHz. The dominant critical frequency is.9 MHz. At f = 0.f c =.9 MHz: A v = A v(mid) = 8.47 db, = 0 At f = f c =.9 MHz: A v = A v(mid) 3 db = 8.47 db 3 db = 5.47 db, = tan () = 45 At f = 0f c = 9 MHz: From.9 MHz to 54.5 MHz the roll-off is 0 db/decade. From 54.5 MHz to 9 MHz the roll-off is 40 db/decade. The change in frequency from.9 MHz to 54.5 MHz represents 54.5 MHz.9MHz 00% = 35.8% 9 MHz.9MHz So, for 35.8% of the decade, the roll-off rate is 0 db/decade and for 64.% of the decade, the rate is 40 db/decade. A v = 5.47 db (0.358)(0 db) (0.64)(40 db) = 7.4 db At f = 00f c = 90 MHz: A v = 7.4 db 40 db = 67.4 db 97
Chapter 0 Section 0-5 Total Amplifier Frequency Response 3. f cl = 36 Hz f cu = 8 khz 4. From Problems 4 and 9: f cu = 4.3 MHz and f cl = 6.89 khz BW = f cu f cl = 4.3 MHz 6.89 khz = 4.33 MHz 5. f tot = (BW)A v(mid) ftot 00 MHz BW = = 5.6 MHz Avmid ( ) 38 Therefore, f cu BW = 5.6 MHz 6. 6 db/octave roll-off: At f cu : A v = 50 db 6 db = 44 db At 4f cu : A v = 50 db db = 38 db 0 db/decade roll-off: At 0f cu : A v = 50 db 0 db = 30 db Section 0-6 Frequency Response of Multistage Amplifiers 7. Dominant fcl = 30 Hz Dominant fcu =. MHz 8. BW =. MHz 30 Hz. MHz 9. fcl = cu 400 Hz / 400 Hz 0.643 / = 6 Hz f (800 khz) = 0.643(800 khz) = 55 khz BW = 55 khz 6 Hz 54 khz 30. 50 Hz 50 Hz f cl = 98. Hz /3 0.50 3. 5 Hz 5 Hz fcl = / 0.643 = 94 Hz fcu.5 MHz BW =.5 MHz 94 Hz.5 MHz 98
Chapter 0 Section 0-7 Frequency Response Measurements 3. f cl = f cu = 0.35 t f 0.35 t r 0.35 ms 0.35 0 ns = 350 Hz = 7.5 MHz 33. Increase the frequency until the output voltage drops to 3.54 (3 db below the midrange output voltage). This is the upper critical frequency. 34. t r 3 div 5 s/div = 5 s t f 6 div 0. ms/div = 600 s 0.35 0.35 f cl = = 583 Hz 600 s t f 0.35 0.35 f cu = = 3.3 khz 5 s t r BW = 3.3 khz 583 Hz =.7 khz Application Activity Problems 35. Q stage: f cl(input) = =.55 Hz ( R R R ) C (6.3 k ) F ac 4 f cl(bypass) = = 5.9 Hz RC ( k )0 F 4 f cl(output) = = 4.30 Hz ( R R R ( R R ) C (37 k ) F 5 6 7 ac 9 0 3 Q stage: f cl(input) = = 7.9 Hz ( R5 R6 R7 ac ( R9 R0) C3 (8.9 k ) F f cl(bypass) = = 0.006 Hz R (08 )00 F 6 R 7 R9 C4 ac f cl(output) = = 4.45 Hz ( R8 RL ) C5 (35.8 k ) F The dominant critical frequency of 5.9 Hz is set by the Q bypass circuit. 36. Changing to F coupling capacitors does not significantly affect the overall bandwidth because the upper critical frequency is much greater than the dominant lower critical frequency. 99
Chapter 0 37. Increasing the load resistance on the output of the second stage has no effect on the dominant lower critical frequency because the critical frequency of the output circuit will decrease and the critical frequency of the first stage input circuit will remain dominant. 38. The Q stage bypass circuit set the dominant critical frequency. f cl(bypass) = = 5.9 Hz RC 4 ( k )0 F This frequency is not dependent on ac and is not affected. Datasheet Problems 39. C in(tot) = (5 + )4 pf + 8 pf = pf 40. BW min = f A T v( mid ) 300 MHz 50 = 6 MHz 4. C gd = C rss =.3 pf C gs = C iss C rss = 5 pf.3 pf = 3.7 pf C ds = C d C rss = 5 pf.3 pf = 3.7 pf Advanced Problems 4. From Problem : r e = 7.8 and I E = 3. ma C 0 (3. ma)(. k) = 3 dc The maximum peak output signal can be approximately 6. The maximum allowable gain for the two stages is 6 A v(max) = = 44.44(0 m) For stage : R c =. k 33 k 4.7 k (50)(7.8) = 645 645 A v = = 8.6 7.8 For stage : R c =. k 5.6 k =.58 k.58 k A v = = 0 7.8 A v(tot) = (8.6)(0) = 6,685 The amplifier will not operate linearly with a 0 m rms input signal. The gains of both stages can be reduced or the gain of the second stage only can be reduced. 00
Chapter 0 One approach is leave the gain of the first stage as is and bypass a portion of the emitter resistance in the second stage to achieve a gain of 44/8.6 = 5.3. Rc A v = = 5.3 Re r e R 5.3.58 k 40. R e = c re = 300 5.3 5.3 Modification: Replace the 560 emitter resistor in the second stage with an unbypassed 300 resistor and a bypassed 60 resistor (closest standard value is 70 ). 43. From Problems 7, 8, and : C tot = Cgs Cin( miller ) = 0.6 pf.65 C out(miller) = 4 pf = 5.5 pf.65 Stage : f cl(in) = = 3.34 Hz R in C (9.5 M)(0.005 F) f cl(out) = = 3.34 Hz since R in() >> 560 (9.5 M)(0.005 F) f cu(in) = =.9 MHz (600 )(0.6 pf) f cu(out) = = 0.9 MHz (560 )(0.6 pf 5.5pF) Stage : f cl(in) = = 3.34 Hz C (9.5 M)(0.005 F) R in f cl(out) = = 3.0 khz (0.6 k)(0.005 F) f cu(in) = = 0.9 MHz (560 )(0.6 pf 5.5pF) f cu(out) = = 54.5 MHz (560 0 k )(5.5 pf) Overall: f cl(in) = 3.34 khz and f cu(in) = 0.9 MHz BW 0.9 MHz 0
Chapter 0 44. R in() = k (00)(30 ) = 3 k 3 k B() = =.38, E() = 0.68 3 k I E() = 0.68 30 =.3 ma, r e =.7 R c() = 4.7 k 33 k k (00)(00 ) =.57 k.57 k A v() = = 3 R in() = k (00)(00 ) = 8 k 8 k B() = = 4.4, E() = 3.54 5k 3.54 I E() = = 3.5 ma, r e = 7.3.0k R c() = 3 k 0 k =.3 k.3k A v() = = 4 maximum 07.3.3k A v() = =.7 minimum 0k 7.3 A v(tot) = (3)(4) = 55 maximum A v(tot) = (3)(.7) = 5.3 minimum This is a bit high, so adjust R c() to 3 k, then 3 k k 33 k 0k A v() = =.4 Now, A v(tot) = (.3)(4) = 53 maximum A v(tot) = (.3)(.7) = 48.5 minimum Thus, A v is within 3% of the desired specifications. Frequency response for stage : R in = k 00 k 3 k =.5 k f cl(in) = =.38 Hz (.5 k)(0 F) R emitter = 0 (00.7 ( k 00 k /00) = 5 f cl(bypass) = =.7 Hz (5 )(00 F) R out = 3 k + ( 33 k k (00)(07 ) ) = 8.9 k f cl(out) = =.79 Hz (8.9k)(0 F) 0
Chapter 0 Frequency response for stage : f cl(in) =.79 Hz (same as f cl(out) for stage ) R out = 3 k + 0 k = 3 k f cl(out) = =. Hz (3 k)(0 F) This means that C E() is the frequency limiting capacitance. R emitter 90 (00 7 ( k 33 k 3 k) /00) = 5 For f cl = khz: C E() = =.38 F (5 )( khz).5 F is the closest standard value and gives f cl(bypass) = = 9 Hz (5 )(.5 F) This value can be moved closer to khz by using additional parallel bypass capacitors in stage to fine-tune the response. Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 45 through 48 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 45. R C open 46. Output capacitor open 47. R open 48. Drain-source shorted 03
Chapter Thyristors Section - The Four-Layer Diode. A = BE + CE(sat) = 0.7 + 0. = 0.9 5 0.9 4. R S I A = R S R S BIAS A 4. = 4. ma.0 k AK 5. (a) R AK = = 5 M I A A (b) From 5 to 50 for an increase of 35. Section - The Silicon-Controlled Rectifier (SCR) 3. See Section - in the textbook. 4. Neglecting the SCR voltage drop, R max = 30 0.7 =.93 k 0 ma 5. When the switch is closed, the battery causes illumination of the lamp. The light energy causes the LASCR to conduct and thus energize the relay. When the relay is energized, the contacts close and 5 ac are applied to the motor. 6. See Figure -. Figure - 04
Chapter Section -3 SCR Applications 7. Add a transistor to provide inversion of the negative half-cycle in order to obtain a positive gate trigger. 8. D and D are full-wave rectifier diodes. 9. See Figure -. Section -4 The Diac and Triac Figure - 0. in(p) =.44 in(rms) =.44(5 ) = 35.4 35.35 I p = in(p) = = 35.4 ma.0 k 0 Current at breakover = = 0 ma.0 k See Figure -3. Figure -3 05
Chapter 5. I p = = 3.9 ma 4.7 k See Figure -4. Figure -4 Section -5 The Silicon-Controlled Switch (SCS). See Section -5 in the text. 3. Anode, cathode, anode gate, and cathode gate Section -6 The Unijunction Transistor (UJT) 4. = r B rb r B =.5 k.5 k 4 k = 0.385 5. p = BB + pn = 0.385(5 ) + 0.7 = 6.48 6. BB I v v R 0.8 R 5 ma BB 747 < R < 00 k I p P 0 0 A Section -7 The Programmable UJT (PUT) R3 0 k 7. (a) A = B 0.7 0 + 0.7 = 9.79 R R3 k R3 47 k (b) A = B 0.7 9 + 0.7 = 5. R R3 94 k 06
Chapter 8. (a) From Problem 7(a), A = 9.79 at turn on. 9.79 I = 470 = 0.8 ma at turn on 0 I p = 470 =.3 ma See Figure -5. Figure -5 (b) From Problem 7(b), A = 5. at turn on. 5. I = = 5.8 ma at turn on 330 0 I p = = 30.3 ma 330 See Figure -6. Figure -6 07
Chapter R3 0 k 9. A = 6 0.7 6 R R3 0 k R A = 3.7 at turn on. See Figure -7. + 0.7 = 3.7 at turn on Figure -7 5 k 0. A = 5 k 6 + 0.7 = 4.3 at turn on R A = 4.3 See Figure -8. Figure -8 Application Activity Problems. The motor runs fastest at 0 for the motor speed control circuit.. If the rheostat resistance decreases, the SCR turns on earlier in the ac cycle. 3. As the PUT gate voltage increases in the circuit, the PUT triggers on later in the ac cycle causing the SCR to fire later in the cycle, conduct for a shorter time, and decrease the power to the motor. 08
Chapter Advanced Problems 4. D : 5 zener (N4744) R : 00, W R : 00, W Q : Any SCR with a A minimum rating (.5 A would be better) R 3 : 50, W 5. See Figure -9. Figure -9 6. p = BB + pn = (0.75)( ) + 0.7 = 9.7 I v = 0 ma and I p = 0 A 9.7 R < = 5 k 0 A R > =. k 0 ma Select R = 5 k as an intermediate value. During the charging cycle: t RC (t) = F ( e / F 0 ) 9.7 = ( ) t.3 ln R C t RC e / t = R.3 C ln =.56R C = 79.8 0 3 C 09
Chapter During the discharging cycle (assuming R >> R B ): t RC (t) = F ( e / F 0 ) t RC = 0 (0 9.3 ) e / t ln R C 9.3 t = R C ln 9.3 =.3R C Let R = 00 k, so t = 3 0 3 C. Since f =.5 khz, T = 400 s T = t + t = 79.8 0 3 C + 3 0 3 C = 303 0 3 C = 400 s 400 s C = = 0.003 F 3 3030 See Figure -0. Figure -0 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 7 through 9 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 7. Cathode-anode shorted 8. Gate-cathode open 9. R shorted 0
Chapter The Operational Amplifier Section - Introduction to Operational Amplifiers. Practical op-amp: High open-loop gain, high input impedance. low output impedance, and high CMRR. Ideal op-amp: Infinite open-loop gain, infinite input impedance, zero output impedance, and infinite CMRR.. Op amp is more desirable because it has a higher input impedance, a lower output impedance, and a higher open-loop gain. Section - Op-Amp Input Modes and Parameters 3. (a) Single-ended differential input (b) Double-ended differential input (c) Common-mode 4. CMRR (db) = 0 log(50,000) = 08 db Aol 75,000 5. CMRR (db) = 0 log 0log Acm 0.8 = 0 db 6. CMRR = A cm = A A ol cm A ol CMRR 90,000 300,000 = 0.3 7. I BIAS = 8.3 A 7.9 A = 8. A 8. Input bias current is the average of the two input currents. Input offset current is the difference between the two input currents. I OS = 8.3 A 7.9 A = 400 na 9. Slew rate = 4 5 s =.6 / s 0. out 0 t = 40 s slew rate 0.5 / s
Chapter Section -4 Op-Amps with Negative Feedback. (a) oltage-follower (b) Noninverting (c) Inverting. B = i R i R R f.0 k = 9.90 0 3 0k f = B out = (9.90 0 3 )5 = 0.0495 = 49.5 m 3. (a) A cl(ni) = = 374 B.5 k / 56.5 k (b) out = A cl(ni) in = (374)(0 m) = 3.74 rms.5 k (c) f = 3. 74 = 9.99 m rms 56.5 k 4. (a) A cl(ni) = = B 4.7 k / 5.7 k (b) A cl(ni) = = 0 B 0 k /.0M (c) A cl(ni) = = 47.8 B 4.7 k / 4.7 k (d) A cl(ni) = = 3 B.0 k / 3 k R f 5. (a) + = Acl(NI) Ri R f = R i (A cl(ni) ) =.0 k(50 ) = 49 k R f (b) = Acl(I) R i R f = R i (A cl(i) ) = 0 k(300) = 3 M (c) R f = R i (A cl(ni) ) = k(7) = 84 k (d) R f = R i (A cl(i) ) =. k(75) = 65 k 6. (a) A cl(f) = R f 00 k (b) A cl(i) = = Ri 00 k (c) A cl(ni) = R 47 k i R 47 k.0 M i R f R f 330 k (d) A cl(i) = = 0 Ri 33 k =
7. (a) out in = 0 m, in phase R f (b) out = A cl in = in R = ()(0 m) = 0 m, 80 out of phase i (c) out = in in 0 m = 3 m, in phase R 47 k R 047 k i R f Chapter (d) out = R f Ri in 330 k 0 m 33 k = 00 m, 80 out of phase 8. in (a) I in = Rin. k = 455 A (b) I f I in = 455 A (c) out = I f R f = (455 A)( k) = 0 R f k (d) A cl(i) = = 0 Ri. k Section -5 Effects of Negative Feedback on Op-Amp Impedances.7 k 9. (a) B = = 0.0048 56.5 k Z in(ni) = ( + A ol )Z in = [ + (75,000)(0.0048)]0 M = 8.4 G Z out 75 Z out(ni) = = 89. m A B (75,000)(0.0048) ol.5 k (b) B = = 0.03 48.5 k Z in(ni) = ( + A ol B)Z in = [ + (00,000)(0.03)] M = 6.0 G Z out 5 Z out(ni) = = 4.04 m A B (00,000)(0.03) ol 56 k (c) B = = 0.053.056 M Z in(ni) = ( + A ol B)Z in = [ + (50,000)(0.053)] M = 5.30 G Z out 50 Z out(ni) = = 9.0 m A B (50,000)(0.053) ol 3
Chapter 0. (a) Z in(f) = ( + A ol )Z in = ( + 0,000)6 M =.3 0 =.3 T Z out 00 Z out(f) = = 455 A 0,000 ol (b) Z in(f) = ( + A ol )Z in = ( + 00,000)5 M = 5 0 = 500 G Z out 60 Z out(f) = = 600 A 00,000 ol (c) Z in(f) = ( + A ol )Z in = ( + 50,000)800 k = 40 G Z out 75 Z out(f) = =.5 m A 500,000 ol. (a) Z in(i) R i = 0 k Ri 0 k B = = 0.065 R R 60 k i f Z out 40 Z out(i) = A B (5,000)(0.065) ol = 5. m (b) Z in(i) R i = 00 k 00 k B = = 0.09.M Zout 50 Z out(i) = AB (75,000)(0.9) ol = 7.3 m (c) Z in(i) R i = 470 470 B = = 0.045 0,470 Z out 70 Z out(i) = A B (50,000)(0.045) ol = 6. m Section -6 Bias Current and Offset oltage. (a) R comp = R in = 75 placed in the feedback path. I OS = 4 A 40 A = A (b) OUT(error) = A v I OS R in = ()( A)(75 ) = 50 3. (a) R c = R.7 k 560 k =.69 k (b) R c = (c) R c = i R f R.5 k 47 k =.45 k i R f R 56 k.0 M = 53 k i R f See Figure -. Figure - 4
Chapter 4. OUT(error) = A v IO = ()( n) = n 5. OUT(error) = ( + A ol ) IO OUT(error) 35 m IO = 00,000 A ol = 75 n Section -7 Open-Loop Frequency and Phase Responses 6. A cl = 0 db 50 db = 70 db 7. The gain is ideally 75,000 at 00 Hz. The midrange db gain is 0 log(75,000) = 05 db The actual gain at 00 Hz is A v (db) = 05 db 3 db = 0 db 0 A v = log = 5,89 0 BW ol = 00 Hz 8. f f c X C = X R C Rf c f (.0 k)(5 khz) 3 khz =.67 k 9. (a) (b) (c) (d) (e) out in out in out in out in out in f f c f f c f f c f f c f f c khz khz 5 khz khz khz khz 0 khz khz 00 khz khz = 0.997 = 0.93 = 0.707 = 0.55 = 0.9 5
Chapter 30. (a) A ol = (b) A ol = (c) A ol = (d) A ol = Aol( mid ) 80,000 00 Hz f khz ( ) f c ol Aol( mid ) 80,000 khz f khz ( ) f c ol Aol( mid ) 80,000 0 khz f khz ( ) f c ol Aol( mid ) 80,000 MHz f khz ( ) f c ol = 79,603 = 56,569 = 7960 = 80 3. (a) f c = (b) f c = (c) f c = RC RC RC f khz =.59 khz; = tan (0 k)(0.0 F) tan = 5.5 f c.59 khz f khz = 5.9 khz; = tan (.0 k)(0.0 F) tan = 7.7 f c 5.9 khz f khz = 59 Hz; = tan (00 k)(0.0 F) tan = 85.5 f c 59 Hz f 00 Hz 3. (a) = tan tan = 0.674 f c 8.5 khz f 400 Hz (b) = tan tan =.69 f c 8.5 khz f 850 Hz (c) = tan tan = 5.7 f c 8.5 khz f 8.5 khz (d) = tan tan = 45.0 f c 8.5 khz f 5 khz (e) = tan tan = 7. f c 8.5 khz 6
Chapter f (f) = tan f c See Figure -. tan 85 khz 8.5 khz = 84.3 Figure - 33. (a) A ol(mid) = 30 db + 40 db + 0 db = 90 db f 0 khz (b) = tan tan = 86.6 f c 600 Hz f 0 khz = tan tan =.3 f c 50 khz f 0 khz 3 = tan tan =.86 f c 00 khz tot = 86.6.3.86 80 = 8 34. (a) 0 db/decade (b) 0 db/decade (c) 40 db/decade (d) 60 db/decade Section -8 Closed-Loop Frequency Response R f 68 k 35. (a) A cl(i) = = 30.9; A cl(i) (db) = 0 log(30.9) = 9.8 db Ri. k (b) A cl(ni) = = 5.7; A cl(ni) (db) = 0 log(5.7) = 3.9 db B 5 k / 35 k (c) A cl(f) = ; A cl(f) (db) = 0 log() = 0 db These are all closed-loop gains. 36. BW cl = BW ol ( + BA ol(mid) ) = 500 Hz[ + (0.05)(80,000)] = 4.05 MHz 7
Chapter 37. A ol (db) = 89 db A ol = 8,84 A cl f c(cl) = A ol f c(ol) Aol f c( ol) (8,84)(750 Hz) A cl = f 5.5 khz c( cl) A cl (db) = 0 log(3843) = 7.7 db = 3843 38. A cl = A ol f f c( ol) c( cl) (8,84)(750 Hz) 5.5 khz = 3843 f T = A cl f c(cl) = (3843)(5.5 khz) =. MHz 39. (a) A cl(f) = BW = f c(cl) = f A T cl 8 MHz =.8 MHz 00 k (b) A cl(i) =. k = 45.5.8 MHz BW = 45.5 = 6.6 khz k (c) A cl(ni) = = 3.0 k.8 MHz BW = = 5 khz 3 M (d) A cl(i) = 5.6 k = 79.8 MHz BW = 79 = 5.7 khz 50 k 40. (a) A cl = = 6.8 k Aol f c( ol) (0,000)(50 Hz) f c(cl) = =.65 MHz Acl 6.8 BW = f c(cl) =.65 MHz.0 M (b) A cl = = 00 0 k Aol f c( ol) (95,000)(50 Hz) f c(cl) = = 97.5 khz Acl 00 BW = f c(cl) = 97.5 khz 8
Chapter Section -9 Troubleshooting 4. (a) Faulty op-amp or open R (b) R open, forcing open-loop operation 4. (a) Circuit becomes a voltage-follower and the output replicates the input. (b) Output will saturate. (c) No effect on the ac; may add or subtract a small dc voltage to the output. (d) The voltage gain will change from 0 to 0.. 43. The gain becomes a fixed 00 with no effect as the potentiometer is adjusted. Application Activity Problems 44. The push-pull stage will operate nonlinearly if a diode is shorted, a transistor is faulty, or the opamp stage has excessive gain. 45. If a 00 k resistor is used for R, the gain of the op amp will be reduced by a factor of 00. 46. If D opens, the positive half of the signal will appear on the output through Q 3 and Q 4. The negative half is missing due to the open diode. Datasheet Problems 47. From the datasheet of textbook Figure -77: 470 B = = 0.0099 47 k 470 A ol = 00,000 (typical) Z in =.0 M (typical) Z in(ni) = ( + 0.0099)(00,000)( M) = ( + 980) M = 3.96 G 48. From the datasheet in Figure -77: R f 00 k Z in(i) = R i = = k A 00 49. A ol = 50 /m = cl 50 m 50. Slew rate = 0.5 /s = 8 (8 ) = 6 6 t = 0.5 / s = 3 s 50,000 = 50,000 9
Chapter Advanced Problems 5. Using available standard values of R f = 50 k and R i =.0 k, 50 k A v = + = 5.0 k.0 k B = = 6.6 0 3 5k Z in(ni) = ( + (6.6 0 3 )(50,000))300 k = 99.6 M The compensating resistor is R c = R i R f 50 k.0 k = 993 See Figure -3. Figure -3 5. See Figure -4. % tolerance resistors are used to achieve a 5% gain tolerance. 53. From textbook Figure -78: f c = 0 khz at A v = 40 db = 00 In this circuit 33 k A v = + = 00. 00 333 The compensating resistor is R c = 33 k 333 = 330 See Figure -5. Figure -4 Figure -5 0
Chapter 54. From textbook Figure -79: For a 0 output swing minimum, the load must be 600 for a 0 and 60 for 0. So, the minimum load is 60. 55. For the amplifier, 00 k A v = = 50 k The compensating resistor is R c = 00 k k =.96 k k See Figure -6. Figure -6 56. From textbook Figure -78 the maximum 74 closed loop gain with BW = 5 khz is approximately 60 db (0 db)log(5 khz/ khz) = 60 db (0 db)(0.7) = 46 db A v(db) = 0 log A v A v(db) 46 A v = log log = 00 0 0 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 57 through 7 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 57. R f open 58. R i open 59. R f leaky 60. R i shorted 6. R f shorted 6. Op-amp input to output open 63. R f leaky
Chapter 64. R i leaky 65. R i shorted 66. R i open 67. R f open 68. R f leaky 69. R f open 70. R f shorted 7. R i open 7. R i leaky
Chapter 3 Basic Op-Amp Circuits Section 3- Comparators. out(p) = A ol in = (80,000)(0.5 m)(.44) = 7 Since is the peak limit, the op-amp saturates. out(pp) = 4 with distortion due to clipping.. (a) Maximum negative (b) Maximum positive (c) Maximum negative R 8 k 3. UTP = ( 0 ) 0 =.77 R R 65 k R 8 k LTP = ( 0 ) ( 0 ) =.77 R R 65 k 4. HYS = UTP LTP =.77 (.77 ) = 5.54 5. See Figure 3-. Figure 3-3
Chapter 3 R 8 k out ( max) R R 5k LTP = 3.88 HYS = UTP LTP = 3.88 (3.88 ) = 7.76 6. (a) UTP = (b) UTP = = 3.88 R 68 k out ( max) R R 8 k LTP = 3.43 HYS = UTP LTP = 3.43 (3.43 ) = 6.86 7. When the zener is forward-biased: 8 k out = out 0.7 8 k 47 k out = (0.77) out 0.7 out ( 0.77) = 0.7 0.7 out = = 0.968 0.77 When the zener is reverse-biased: 8 k out = out + 6. 8 k 47 k out = (0.77) out + 6. out ( 0.77) = +6. 6. out = = +8.57 0.77 8. 0 k out = out 0 k 47 k (4.7 + 0.7 ) out = (0.75) out 5.4 5.4 out = = 6.55 0.75 UTP = (0.75)(+6.55 ) = +.5 LTP = (0.75)(6.55 ) =.5 See Figure 3-. = 3.43 Figure 3-4
Chapter 3 Section 3- Summing Amplifiers R f 9. (a) OUT = (.5 ) = ( +.5 ) =.5 R i R f k (b) OUT = ( 0. 0.5 ) = (.6 ) = 3.5 R 0 k 0. (a) R = R =.8 i (b) I R = k = 45.5 A.8 I R = k = 8.8 A I f = I R + I R = 45.5 A + 8.8 A = 7 A (c) OUT = I f R f = (7 A)( k) =.8 R f. 5 in = in R R f = 5 R R f = 5R = 5( k) = 0 k. See Figure 3-3. Figure 3-3 R f R f R f R f 3. OUT = 3 4 R R R3 R4 0 k 0 k 0 k 0 k = 3 3 6 0 k 33 k 9k 80 k = ( + 0.9 + 0.33 + 0.33 ) = 3.57 I f = OUT R f 3.57 = 357 A 0 k 5
Chapter 3 4. R f = 00 k Input resistors: R = 00 k, R = 50 k, R 3 = 5 k, R 4 =.5 k, R 5 = 6.5 k, R 6 = 3.5 k Section 3-3 Integrators and Differentiators dout IN 5 5. dt RC (56 k )(0.0 F) = 4.06 m/s 6. See Figure 3-4. Figure 3-4 7. I = C pp (0.00 F)(5 ) = ma T / 0 s / pp 8. out = RC (5 k )(0.047 F) T / 0.5 ms See Figure 3-5. =.8 Figure 3-5 6
Chapter 3 9. For the 0 ms interval when the switch is in position : out IN 5 5 = 50 /s = 50 m/ms t RC (0 k)(0 F) 0.s out = (50 m/ms)(0 ms) = 500 m = 0.5 For the 0 ms interval when the switch is in position : out IN 5 5 = +50 /s = +50 m/ms t RC (0 k)(0 F) 0.s out = (+50 m/ms)(0 ms) = +500 m = +0.5 See Figure 3-6. Section 3-4 Troubleshooting Figure 3-6 0. R B = out ( Z R R 0.7 ) ( B = Z 0.7 ) R R R Normally, B should be (4.3 0.7 ) B = 0.5 = 0 Since the negative portion of B is only.4, zener D must be shorted: (0 0.7 ) B = 0.5 =.4. The output should be as shown in Figure 3-7. has no effect on the output. This indicates that R is open. Figure 3-7 7
Chapter 3. A v =.5 k 0 k = 0.5 The output should be as shown in Figure 3-8. An open R ( is missing) will produce the observed output, which is incorrect. Figure 3-8 3. The D input is missing (acts as a constant 0). This indicates an open 50 k resistor. Application Activity Problems 4. The first thing that you should always do is visually inspect the circuit for bad contacts or loose connections, shorts from solder splashes or wire clippings, incorrect components, and incorrectly installed components. After careful inspection, you have found nothing wrong. Measurements are now necessary to isolate a component s fault. 5. An open decoupling capacitor can make the circuit more susceptible to power line noise. 6. If a.0 k resistor is used for R, the inverting input would be increased, causing the pulse width to narrow for a given setting of the potentiometer. Advanced Problems 4 7. I R--3 = = 39. A 6 k Minimum setting of R : IN = (39. A)(56 k) = 9.8 v = p sin v 9.8 sin = 0 = 0.98 p = sin v sin (0.98) = 78.5 (on positive half cycle) p Angle from 78.5 to 90 = 90 78.5 =.5 Angle from 90 to next point at which v = 9.8 : =.5 Angle from first point at which v = 9.8 to second point at which v = 9.8 on sine wave is =.5 +.5 = 3 8
Chapter 3 3 min. duty cycle = 00 = 6.39% 360 See Figure 3-9(a). Maximum setting of R : = (39. A)(556 k) = 9.8 IN v sin = p 9.8 = 78.5 (on negative half cycle) 0 3603 max. duty cycle = 00 = 93.6% 360 See Figure 3-9(b). (a) (b) Figure 3-9 8. Let IN = 4.8 Let I = 39. A IN = I R I R = 4.8 I R = 7. 7. R = = 84 k 39. A Change R and R 3 to 84 k. 9
Chapter 3 9. 00 m/s = 5 /R i C 5 R i C = = 50 s 00 m/ s For C = 3300 pf: 50 s R i = = 5.5 k = 5 k + 50 3300 pf For a 5 peak-peak triangle waveform: 5 t ramp up = t ramp down = = 50 s 00 m/ s = (50 s) = 00 s f in = /00 s = 00 khz See Figure 3-0. Figure 3-0 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 30 through 39 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 30. R open 3. Op-amp inputs shorted together 3. Op-amp + input to output shorted 33. D shorted 34. Top 0 k resistor open 35. Middle 0 k resistor shorted 36. R f leaky 37. R f open 38. C leaky 39. C open 30
Chapter 4 Special-Purpose Op-Amp Circuits Section 4- Instrumentation Amplifiers. A v() = + A v() = + R R R R G G 00 k.0 k 00 k.0 k = 0 = 0. A cl = + R R G 00 k.0 k = 0 3. out = A ) = 0(0 m 5 m) =.005 cl ( in( ) in() 4. A v = + G R R G R = Av R R G = R (00 k) 00 k 000 999 A v = 00. 00 50.5 k 5. R G = A v 50.5 k A v =.0 k + = 5.5 6. Using the graph in textbook Figure 4-6, BW 300 khz 7. Change R G to 50.5 k 50.5 k R G = 4 A v. k 8. R G = 50.5 k 50.5 k.7 k 0 A v 3
Chapter 4 Section 4- Isolation Amplifiers 9. A v(total) = (30)(0) = 300 R f 8 k 0. (a) A v = + = 3. Ri 8. k R f 50 k A v = Ri 5 k + = A v(tot) = A v A v = (3.)() = 35. R f 330 k (b) A v = + = 33 Ri.0 k R f 47 k A v = Ri 5 k + = 4.3 A v(tot) = A v A v = (33)(4.3) =,367. A v = (from Problem 0(a)) A v A v = 00 R f 00 Av R = 9.09 i R f = (9.09 )R i = (8.09)(8. k) = 66 k Change R f (8 k) to 66 k. Use 68 k % standard value resistor.. A v = 33 (from Problem 0(b)) A v A v = 440 R f 440 Av =.33 Ri 33 Change R f (47 k) to 3.3 k. Change R i (5 k) to 0 k. 3. Connect pin 6 to pin 0 and pin 4 to pin 5. Make R f = 0. Section 4-3 Operational Transconductance Amplifiers (OTAs) 4. g m = I out in 0 A 0 m = ms 5. I out = g m in = (5000 S)(00 m) = 500 A out = I out R L = (500 A)(0 k) = 5 3
Chapter 4 6. g m = I out in I out = g m in = (4000 S)(00 m) = 400 A out 3.5 R L = = 8.75 k I 400 A out ( ) 0.7 ( ) 0.7 3.3 7. I BIAS = R BIAS 0 k 0 k From the graph in Figure 4-57: g m = KI BIAS (6 S/A)(06 A) =.70 ms out I out RL A v = = g m R L = (.70 ms)(6.8 k) =.6 in in = 06 A 8. The maximum voltage gain occurs when the 0 k potentiometer is set to 0 and was determined in Problem 7. A v(max) =.6 The minimum voltage gain occurs when the 0 k potentiometer is set to 0 k. ( ) 0.7 3.3 I BIAS = = 0 A 0 k 0 k 30 k g m (6 S/A)(0 A) =.6 ms A v(min) = g m R L = (.6 ms)(6.8 k) =.0 9. The MOD waveform is applied to the bias input. The gain and output voltage for each value of MOD is determined as follows using K = 6 S/A. The output waveform is shown in Figure 4-. For MOD = +8 : 8 ( 9 ) 0.7 6.3 I BIAS = = 48 A 39 k 39 k g m = KI BIAS (6 S/A)(48 A) = 6.69 ms out I out RL A v = = g m R L = (6.69 ms)(0 k) = 66.9 in in out = A v in = (66.9)(00 m) = 6.69 For MOD = +6 : 6 ( 9 ) 0.7 4.3 I BIAS = = 367 A 39 k 39 k g m = KI BIAS (6 S/A)(367 A) = 5.87 ms out I out RL A v = = g m R L = (5.87 ms)(0 k) = 58.7 in in out = A v in = (58.7)(00 m) = 5.87 33
Chapter 4 For MOD = +4 : 4 ( 9 ) 0.7.3 I BIAS = = 35 A 39 k 39 k g m = KI BIAS (6 S/A)(35 A) = 5.04 ms out I out RL A v = = g m R L = (5.04 ms)(0 k) = 50.4 in in out = A v in = (50.4)(00 m) = 5.04 For MOD = + : ( 9 ) 0.7 0.3 I BIAS = = 64 A 39 k 39 k g m = KI BIAS (6 S/A)(64 A) = 4. ms out I out RL A v = = g m R L = (4. ms)(0 k) = 4. in in out = A v in = (4.)(00 m) = 4. For MOD = + : ( 9 ) 0.7 9.3 I BIAS = = 38 A 39 k 39 k g m = KI BIAS (6 S/A)(38 A) = 3.8 ms out I out RL A v = = g m R L = (3.8 ms)(0 k) = 38. in in out = A v in = (38.)(00 m) = 3.8 Figure 4-9 ( 9 ) 0.7 7.3 0. I BIAS = = 444 A 39 k 39 k TRIG(+) = I BIAS R = (444 A)(0 k) = +4.44 TRIG() = I BIAS R = (444 A)(0 k) = 4.44 34
Chapter 4. See Figure 4-. Figure 4- Section 4-4 Log and Antilog Amplifiers. (a) ln(0.5) = 0.693 (b) ln() = 0.693 (c) ln(50) = 3.9 (d) ln(30) = 4.87 3. (a) log 0 (0.5) = 0.30 (b) log 0 () = 0.30 (c) log 0 (50) =.70 (d) log 0 (30) =. 4. Antilog x = 0 x or e x, depending on the base used. IN ln = e.6 = 4.95 IN log = 0.6 = 39.8 5. The output of a log amplifier is limited to 0.7 because the output voltage is limited to the barrier potential of the transistor s pn junction. in 6. out ( 0.05 )ln I s Rin 3 = ( 0.05 )ln = (0.05 )ln(365.9) = 48 m (00 na)(8 k) in 7. out ( 0.05 )ln I EBO R in.5 = ( 0.05 )ln = (0.05 )ln(53.9) = 57 m (60 na)(47 k) 35
Chapter 4 in 8. out = R f I EBO antilog = R f I 5 m 0.5 EBO e in 5 m 5 m out = (0 k)(60 na) e = (0 k)(60 na)e 9 = (0 k)(60 na)(803) = 4.86 in 9. out(max) (0.05 ) ln (0.05 ) ln I EBO Rin (60 na)(47 k) = (0.05 )ln(354.6) = 47 m in 00 m out(min) (0.05 ) ln (0.05 ) ln I EBO Rin (60 na)(47 k) = (0.05 ) ln(35.5) = 89. m The signal compression allows larger signals to be reduced without causing smaller amplitudes to be lost (in this case, the peak is reduced 85% but the 00 m peak is reduced only 0%). Section 4-5 Converters and Other Op-Amp Circuits 30. (a) IN = Z = 4.7 IN 4.7 I L =.0 k R i = 4.7 ma 0 k (b) IN = = 6 0 k R i = 0 k 0 k 00 = 5. k I L = IN R i 3. See Figure 4-3. 6 =.8 ma 5.k Figure 4-3 36
Chapter 4 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 3 through 36 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 3. R G leaky 33. R open 34. R f open 35. Zener diode open 36. Lower 0 k resistor open 37
Chapter 5 Active Filters Section 5- Basic Filter Responses. (a) Band-pass (b) High-pass (c) Low-pass (d) Band-stop. BW = f c = 800 Hz 3. f c = = 48. Hz RC (. k)(0.005 F) No, the upper response roll-off due to internal device capacitances is unknown. 4. The roll-off is 0 db/decade because this is a single-pole filter. 5. BW = f c f c = 3.9 khz 3. khz = 0.7 khz = 700 Hz f 0 = f f (3. khz)(3.9 khz) = 3.53 khz Q = f c 0 BW c 3.53 khz 700 Hz = 5.04 f 6. Q = 0 BW f 0 = Q(BW) = 5( khz) = 5 khz Section 5- Filter Response Characteristics 7. (a) nd order, stage R3. k DF = = = Not Butterworth R. k 4 (b) nd order, stage R3 560 DF = R.0 k = 0.56 =.44 Approximately Butterworth 4 (c) 3rd order, stages, st stage ( poles): R3 330 DF = =.67 R4.0 k nd stage ( pole): R 6 DF = =.67 Not Butterworth R 7 38
8. (a) From Table 5- in the textbook, the damping factor must be.44; therefore, R 3 = 0.586 R 4 R 3 = 0.586R 4 = 0.586(. k) = 703 Nearest standard value: 70 Chapter 5 (b) R 3 = 0.56 R4 This is an approximate Butterworth response (as close as you can get using standard 5% resistors). (c) From Table 5-, the damping factor of both stages must be, therefore R 3 = R 4 R 3 = R 4 = R 6 = R 7 = k (for both stages) 9. (a) Chebyshev (b) Butterworth (c) Bessel (d) Butterworth Section 5-3 Active Low-Pass Filters 0. High Pass st stage: R3.0 k DF = =.85 R4 6.8 k nd stage: R7 6.8 k DF = = 0.786 R8 5.6 k From Table 5- in the textbook: st stage DF =.848 and nd stage DF = 0.765 Therefore, this filter is approximately Butterworth. Roll-off rate = 80 db/decade. f c = RRCC R5R6C3C4 (4.7 k)(6.8 k)(0. F)(0.F) = 90 Hz. R = R = R = R 5 = R 6 and C = C = C = C 3 = C 4 Let C = 0. F (for both stages). f c = R C RC R = = 3.8 k C (90 Hz)(0. F) f c Choose R = 3.9 k (for both stages) 39
Chapter 5 3. Add another identical stage and change the ratio of the feedback resistors to 0.068 for first stage, 0.586 for second stage, and.48 for third stage. See Figure 5-. Figure 5-4. See Figure 5-. Section 5-4 Active High-Pass Filters Figure 5-5. Exchange the positions of the resistors and the capacitors. See Figure 5-3. Figure 5-3 40
Chapter 5 6. f c = RC 90 Hz f 0 = = 95 Hz R = = 765 f c C (95 Hz)(0. F) Let R = 7.5 k. Change R, R, R 5 and R 6 to 7.5 k. 7. (a) Decrease R and R or C and C. (b) Increase R 3 or decrease R 4. Section 5-5 Active Band-Pass Filters 8. (a) Cascaded high-pass/low-pass filters (b) Multiple feedback (c) State variable 9. (a) st stage: f c = = 3.39 khz RC (.0 k)(0.047 F) nd stage: f c = = 7.3 khz RC (.0 k)(0.0 F) f 0 = f (3.39 khz)(7.3 khz) = 4.95 khz c f c BW = 7.3 khz 3.39 Hz = 3.84 khz R R3 47 k.8 k (b) f 0 = C RR R (0.0 F) (47 k )(.8 k )(50 k ) 3 Q = f 0 CR = (449 Hz)(0.0 F)(50 k) = 4.66 f 449 Hz BW = 0 = 96.4 Hz Q 4.66 = 449 Hz (c) For each integrator: f c = = 5.9 khz RC (0 k)(0.00 F) f 0 = f c = 5.9 khz R5 560 k Q = (56 ) 3 = 9 R6 3 0 k 3 f BW = 0 5.9 khz = 838 Hz Q 9 4
Chapter 5 R5 0. Q = 3 R6 Select R 6 = 0 k. R5 R5 R6 Q = 3R6 3 3R6 3R 6 Q = R 5 + R 6 R 5 = 3R 6 Q R 6 = 3(0 k)(50) 0 k = 500 k 0 k = 490 k f 0 = =.33 khz ( k)(0.0f) BW = f 0.33 khz = 6.6 Hz Q 50 Section 5-6 Active Band-Stop Filters. See Figure 5-4. Figure 5-4. f 0 = f c = RC Let C remain 0.0 F. R = = 33 k f 0C (0 Hz)(0.0 F) Change R in the integrators from k to 33 k. Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 3 through 3 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 3. R 4 shorted 4. R 3 open 5. C 3 shorted 6. R 5 open 4
Chapter 5 7. R open 8. R shorted 9. R open 30. C open 3. R 7 open 43
Chapter 6 Oscillators Section 6- The Oscillator. An oscillator requires no input other than the dc supply voltage.. Amplifier and positive feedback circuit Section 6- Feedback Oscillators 3. Unity gain around the closed loop is required for sustained oscillation. A cl = A v B = B = = 0.033 75 A v 4. To ensure startup: A cl > since A v = 75, B must be greater than /75 in order to produce the condition A v B >. For example, if B = /50, A v B = 75 =.5 50 Section 6-3 Oscillators with RC Feedback Circuits 5. out in 3. out = in = 733 m 3 3 6. f r = RC (6. k)(0.0 F) =.8 khz 7. R = R R 00 k R = = 50 k 8. When dc power is first applied, both zener diodes appear as opens because there is insufficient output voltage. This places R 3 in series with R, thus increasing the closed-loop gain to a value greater than unity to assure that oscillation will begin. 9. R f = (A v )(R 3 + r ds ) = (3 )(80 + 350 ) =.34 k 44
Chapter 6 0. f r = (.0 k)(0.05 F) = 0.6 khz. B = 9 A cl = B = 9 A cl = R R f i R f = A cl R i = 9(4.7 k) = 36 k f r = = 68 Hz 6(4.7 k )(0.0 F) Section 6-4 Oscillators with LC Feedback Circuits. (a) Colpitts: C and C 3 are the feedback capacitors. f r = L C T CC3 (00 F)(000 pf) C T = = 90.9 pf C C 00 pf f r = (b) Hartley: f r = L T C 3 (5mH)(90.9 pf) = 36 khz L T = L + L =.5 mh + 0 mh =.5 mh f r = = 68.5 khz (.5 mh)(470 pf) 47 pf 3. B = 470 pf = 0. The condition for sustained oscillation is A v = B 0. = 0 Section 6-5 Relaxation Oscillators 4. Triangular waveform. R f = 4R C R3 4( k)(0.0 56 k =.6 khz F) 8 k 45
Chapter 6 5. Change f to 0 khz by changing R : R f = 4RC R3 R 56 k R = 4 fc R3 4(0 khz)(0.0 F) 8 k = 3.54 k p F 6. T = IN RC R5 47 k p = = 3.84 R4 R5 47 k PUT triggers at about +3.84 (ignoring the 0.7 drop) Amplitude = +3.84 =.84 R k IN = ( ) ( ) =.6 R R k 3.84 T = = 89 s.6 (00 k )(0.00 F) f = T 89 s = 3.46 khz See Figure 6-. Figure 6-46
Chapter 6 7. G = 5. Assume AK =. R 5 = 47 k R5 G = R4 R5 Change R 4 to get G = 5. 5 (R 4 + 47 k) = (47 k) R 4 (5 ) = (47 k) (47 k)5 ( 5 )47 k R 4 = = 65.8 k 5 p F 8. T = IN RC IN 3 p = T F 0 s RC (4.7 k)(0.00 F) pp(out) = p F = 7.38 = 6.38 + = 7.38 Section 6-6 The 555 Timer as an Oscillator 9. (0 ) 3 CC 3 = 3.33 (0 ) 3 CC 3 = 6.67 0. f = ( R.44 R ) C.44 (.0 k 6.6 k)(0.047 ext F) = 4.03 khz. f = C ext =.44 ( R R ) ( R.44 R C ext ) f.44 (.0 k 6.6 k )(5 khz) = 0.0076 F. R R Duty cycle (dc) = R R 00% dc(r + R ) = (R + R )00 75(3.3 k + R ) = (3.3 k + R )00 75(3.3 k) + 50R = 00(3.3 k) + 00R 50R 00R = 00(3.3 k) 75(3.3 k) 50R = 5(3.3 k) 5(3.3 k ) R = 50 =.65 k 47
Chapter 6 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 3 through 8 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 3. Drain-to-source shorted 4. C 3 open 5. Collector-to-emitter shorted 6. R open 7. R open 8. R leaky 48
Chapter 7 oltage Regulators Section 7- oltage Regulation OUT m. Percent line regulation = 00% 00% IN 6 = 0.0333% OUT / OUT m/8. Percent line regulation = 00% 00% IN 6 NL / FL 0 9.90 3. Percent load regulation = 00% 00% FL 9.90 = 0.0047%/ =.0% 4. From Problem 3, the percent load regulation is.0%. For a full load current of 50 ma, this can be expressed as.0% = 0.00404%/mA 50 ma Section 7- Basic Linear Series Regulators 5. See Figure 7-. Figure 7- R 33 k 6. OUT = REF.4 R3 0 k = 0.3 R 5.6 k 7. OUT = REF.4 = 8.5 R3. k 49
Chapter 7 8. For R 3 =. k: R 5.6 k OUT = REF.4 = 8.5 R3. k For R 3 = 4.7 k: R 5.6 k OUT = REF.4 = 5.6 R3 4.7 k The output voltage decreases by 3.4 when R 3 is changed from. k to 4.7 k. R 5.6 k 9. OUT = REF.7 R3. k = 9.57 0. I L(max) = 0.7 R L(max) 4 0.7 R 4 = I 0.7 ma 50 ma =.8 P = I L(max) R 4 = (50 ma).8 = 0.75 W, Use a 0.5 W.. R 4 = I L(max) =.8 =.4 0.7 R 4 0.7.4 = 500 ma Section 7-3 Basic Linear Shunt Regulators. Q conducts more when the load current increases, assuming that the output voltage attempts to increase. When the output voltage tries to increase due to a change in load current, the attempted increase is sensed by R 3 and R 4 and a proportional voltage is applied to the opamp s non-inverting input. The resulting difference voltage increases the op-amp output, driving Q more and thus increasing its collector current. 3. I C = R R 00 = 0 ma R3 0 k 4. OUT = REF 5. R4 3.9 k OUT 8. I L = = 8. ma R k I L = L OUT R L 8. = 5. ma. k I L = 5. ma 8. ma = 3.0 ma I S = I L = 3.0 ma = 8. 50
Chapter 7 5. I L(max) = R IN 5 00 = 50 ma P R = I L(max) R = (50 ma) 00 = 6.5 W Section 7-4 Basic Switching Regulators t 6. OUT = on IN T t on = T t off T = = 0.000 s = 00 µs f 0 khz 40 s OUT = 00 s = 4.8 7. f = 00 Hz, t off = 6 ms T = = 0 ms f 00 Hz t on = T t off = 0 ms 6 ms = 4 ms duty cycle = t 4 ms on = 0.4 T 0 ms percent duty cycle = 0.4 00% = 40% 8. The diode D becomes forward-biased when Q turns off. 9. The output voltage decreases. Section 7-5 Integrated Circuit oltage Regulators 0. (a) 7806: +6 (b) 7905.: 5. (c) 788: +8 (d) 794: 4 R 0 k. OUT = REF I ADJ R.5 R.0 k = 3.7 + 0.5 = 4.3 + (50 A)(0 k) 5
Chapter 7. OUT(min) = R R R (min) = 0.5 ( 0) 0 =.5 (min) REF I ADJ R( min) OUT(min) = R(max) 0 k OUT(max) = REF I ADJ R( max).5 (50 A)(0 k ) R 470.5 (.8) 0.5 = 8.4 = 3. The regulator current equals the current through R + R. OUT 4.3 I REG =.3 ma R R k 4. IN = 8, OUT = I REG(max) = ma, REF =.5 R = REF.5 = 65 I REG ma Neglecting I ADJ : R =.5 = 0.8 R = R 0.8 = 5.4 k I REG ma For R use 60 and for R use either 5600 or a 0 k potentiometer for precise adjustment to. Section 7-6 Applications of IC oltage Regulators 5. Rext(min) = 0.7 0.7 0.7 R ext = I 50 ma max =.8 6. OUT = + I L = = 00 ma =. A 0 I ext = I L I max =. A 0.5 A = 0.7 A P ext = I ext ( IN OUT ) = 0.7 A(5 ) = 0.7 A(3 ) =. W 7. Rlim(min) = 0.7 0.7 0.7 R lim(min) = I ext A See Figure 7-. = 0.35 Figure 7-5
Chapter 7.5 8. R = =.5 500 ma See Figure 7-3. Figure 7-3 9. I = 500 ma 8 R = = 6 500 ma See Figure 7-4. Figure 7-4 30. Connect pin 7 to pin 6. Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 3 through 34 are available from the Instructor Resource Center. See Chapter for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 3. R leaky 3. Zener diode open 33. Q collector-to-emitter open 34. R open 53
Chapter 8 Basic Programming Concepts for Automated Testing Section 8- Programming Basics. The five basic instruction types are simple instructions, conditional instructions, loop instructions, branching instructions, and exception instructions.. The flowchart symbols are () subroutine, () decision, (3) continuation, (4) begin/end, (5) input/output, and (6) task. 3. One possible pseudocode description is program PrintAverage begin input first number input second number input third number sum is first number plus second number plus third number average is sum divided by 3 print average end PrintAverage Section 8- Automated Testing Basics 4. The automated test system components are () test controller, () test equipment and instrumentation, (3) test fixture, (4) switching control, (5) switching circuitry, and (6) unit under test (UUT). 5. Advantages of electromechanical relays are good electrical isolation and high current and voltage handling ability. Section 8-3 The Simple Sequential Program 5. 6. The process flow for a simple sequential program is linear in which the program begins, executes a series of instructions in sequence, and then terminates. 7. The flowchart represents a simple sequential program, as none of the program operations alters the sequence of program execution from its linear process flow. 54
Chapter 8 Section 8-4 Conditional Execution 8. Because the start value is 7 and not less than 5, the top-level IF statement is skipped. The program enters the top-level ELSE, sets the output value to the 7 = 5 and tests whether the start value is greater than 7. The start value of 7 is not greater than 7, so the nested top-level IF statement is skipped. The program enters the top-level nested ELSE statement and tests whether the start value is greater than 6. The start value of 7 is greater than 6, so the output value is set to 5 / 3 =.6. The program exits the nested IF-THEN-ELSE instruction and prints.6 as the output value. 9. One possible pseucode description is NewIdentifyKeyalue begin input key value case (key value) begin case : print Key value equals break : print Key value equals break 3 : print Key value equals 3 break default: print Key value is more than 3 break end case end NewIdentifyKeyalue Section 8-5 Program Loops 0. Loop corresponds to a WHILE-DO loop, Loop corresponds to a FOR-TO-STEP loop, and Loop 3 corresponds to a REPEAT-UNTIL loop.. The index value starts at and the step value increases the index value by for each loop, so the FOR-TO-STEP loop will execute once for each odd index value from through 9 and exit on an index value of. The sum begins at 0 and each loop will add the index value plus, so sum = (+) + (3+) + (5+) + (7+) + (9+) = + 4 + 6 + 8 + 0 = 30. The outer loop will execute three times and the inner loop will execute a number of times equal to the index, the outer loop index value. The sum begins at 0 and each loop will add to the sum, so Index equals : Index does not exceed end value of 3, so program enters inner loop. Inner loop executes time, so sum = 0 + =. 55
Chapter 8 Index adjusted by step value of to + =. Index equals : Index does not exceed end value of 3, so program enters inner loop. Inner loop executes times, so sum = + + = 3. Index adjusted by step value of to + = 3. Index equals 3: Index does not exceed end value of 3, so program enters inner loop. Inner loop executes 3 times, so sum = 3 + + + = 6. Index adjusted by step value of to 3 + = 4. Index equals 4: Index exceeds end value of 3, so program skips outer loop and prints the sum value of 6. Section 8-6 Branches and Subroutines 3. One reason that programs should avoid using unconditional branches is that unconditional branches too often encourage poor programming practices by compensating for poorly designed code. A second reason is that unrestricted unconditional branching typically results in spaghetti code that is difficult to modify and maintain. A third reason is that replacing the branch instruction with the desired target instruction(s) produces the same result as the unconditional branch, so that unconditional branches are often unnecessary. 4. One possible pseudocode description is procedure Comparealues(FirstResistor, SecondResistor) begin input FirstResistor value input SecondResistor value print First resistor value is and FirstResistor value print Second resistor value is and SecondResistor value if (FirstResistor value is greater than SecondResistor value) then begin if print First resistor value is greater than second resistor value end if else begin else print Second resistor value is greater than first resistor value end else end Comparealues 56
Chapter 9 (Website) Electronic Communications Systems and Devices Section 9- Basic Receivers. See Figure 9-. Figure 9-. See Figure 9-. Figure 9-3. f LO = 680 khz + 455 khz = 35 khz 4. f LO = 97. MHz + 0.7 MHz = 07.9 MHz 5. f RF = 0.9 MHz 0.7 MHz = 9. MHz f IF = 0.7 MHz (always) 57
Chapter 9 Section 9- The Linear Multiplier 6. (a) out.5 (b) out.6 (c) out +.0 (d) out +0 7. out = K X Y = 0.5(+3.5 )(.9 ) =.7 8. Connect the two inputs together. 9. (a) out = K = (0.)(+ )(+.4 ) = +0.8 (b) out = K = K ( 0.)( 3. ) = +.04 (6. ) (c) out = = +.07 3 (d) out = 6. = +.49 Section 9-3 Amplitude Modulation 0. f diff = f f = 00 khz 30 khz = 70 khz f sum = f + f = 00 khz + 30 khz = 30 khz 9 cycles. f = = 9000 cycles/s = 9 khz ms cycle f = = 000 cycles/s = khz ms f diff = f f = 9 khz khz = 8 khz f sum = f + f = 9 khz + khz = 0 khz. f c = 000 khz f diff = 000 khz 3 khz = 997 khz f sum = 000 khz + 3 khz = 003 khz 8 cycles 3. f = =.8 MHz 0 s cycle f = = 00 khz 0 s f diff = f f =.8 MHz 00 khz =.7 MHz f sum = f + f =.8 MHz + 00 khz =.9 MHz f c =.8 MHz 4. f c =. MHz by inspection f m = f c f diff =. MHz.955 MHz = 4.5 khz 58
Chapter 9 f diff f sum 8.47 khz 853 KHz 5. f c = = 850 khz f m = f c f diff = 850 khz 847 khz = 3 khz 6. f diff(min) = 600 khz 3 khz = 597 khz f diff(max) = 600 khz 300 Hz = 599.7 khz f sum(min) = 600 khz + 300 khz = 600.3 khz f sum(max) = 600 khz + 3 khz = 603 khz See Figure 9-3. Figure 9-3 Section 9-4 The Mixer 7. (sin A)(sin B) = [cos( A B) cos( A B)] in() = 0. sin [(00 khz)t] in() = 0.5 sin [(3300 khz)t] in() in() = (0. )(0.5 ) sin [(00 khz)t] sin [(3300 khz)t] (0. )(0.5 ) out = [cos (3300 khz 00 khz)t cos (3300 khz + 00 khz)t] out = 5 m cos [(00 khz)t] 5 m cos [(5500 khz)t] 8. f IF = f LO f c = 986.4 khz 980 khz = 6.4 khz 59
Chapter 9 Section 9-5 AM Demodulation 9. See Figure 9-4. Figure 9-4 0. See Figure 9-5.. See Figure 9-6. Figure 9-5 Figure 9-6 Section 9-6 IF and Audio Amplifiers. f c f m =. MHz 8.5 khz =.95 MHz f c + f m =. MHz + 8.5 khz =.085 MHz f c =. MHz f LO f m = 455 khz 8.5 khz = 446.5 khz f LO + f m = 455 khz + 8.5 khz = 463.5 khz f LO = 455 khz 3. The IF amplifier has a 450 khz to 460 khz passband. The audio/power amplifiers have a 0 Hz to 5 khz bandpass. 60
Chapter 9 4. C 4 between pins and 8 makes the gain 00. With R set for minimum input, in = 0. out(min) = A v in(min) = 00(0 ) = 0 With R set for maximum input, in = 0 m rms. out(max) = A v in(max) = 00(0 m) = rms Section 9-7 Frequency Modulation 5. The modulating input signal is applied to the control voltage terminal of the CO. As the input signal amplitude varies, the output frequency of the CO varies proportionately. 6. An FM signal differs from an AM signal in that the information is contained in frequency variations of the carrier rather than amplitude variations. 7. aractor Section 9-8 The Phase-Locked Loop (PLL) 8. See Figure 9-7. Figure 9-7 9. (a) The CO signal is locked onto the incoming signal and therefore its frequency is equal to the incoming frequency of 0 MHz. (b) c = i o (50 m)(400 m) cos e cos(30 5) = (0.050)(0.966) = 48.3 m 30. f o = +3.6 khz, c = +0.5 f o 3.6 khz K = 0.5 = 7. khz/ c 3. K =.5 khz/, c = +0.67 f o K = c f o = K c = (.5 khz/)(+0.67 ) = 005 Hz 6