ECE342 Test 2 Solutions, Nov 4, :008:00pm, Closed Book (one page of notes allowed)


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1 ECE342 Test 2 Solutions, Nov 4, :008:00pm, Closed Book (one page of notes allowed) Please use the following physical constants in your calculations: Boltzmann s Constant: Electron Charge: Free space Permittivity: Thermal Voltage: k = ev/k = J/K q = coulomb ɛ 0 = F/cm V T = kt/q = 25.8 mv at 300K Silicon Properties at 300K: Bandgap: E G = 1.12 ev Relative Permittivity: ɛ r = 11.7 Intrinsic carrier density: n i = p i = /cm 3 Intrinsic Si Electron Mobility: µ n = 1350 cm 2 /V s Intrinsic Si Hole Mobility: µ p = 480 cm 2 /V s Carrier Mobility Trends for Silicon Mobility ( cm 2 /V sec ) Electron Mobility Hole mobility in n type material with N D =10 17 (Problem 1h) 320 cm 2 /(Vsec) 400 Hole Mobility Total Impurity Concentration (cm 3 ) Intrinsic Carrier Density (per cm 3 ) Intrinsic Silicon Carrier Density Absolute Temperature (K)
2 1. A silicon PN junction diode is formed using an acceptor concentration of /cm 3 and a donor population of /cm 3. The junction area is 400 µm 2. (a) Complete the diagram above. Show the depletion region, and indicate the polarity of any bound charges which remain in the depletion region. Clearly indicate which side of the depletion region is wider. Show the direction of the electric field in the depletion region. (b) Find the builtin potential for the device. Solution: ( ) NA N D V 0 = V T ln n 2 i ( ) = (25.8 mv) ln ( ) 2 = V (c) Find the depletion region width for 5 V reverse bias applied. Solution: w D5 = ( ) 2(11.7)( F/cm) (5 V V) C cm cm 3 = cm (= µm) (d) Find the smallsignal capacitance associated with a 5 V reverse bias. Solution: C J5 = ɛa = (11.7)( F/cm)( cm 2 ) w D = F (= pf) cm (e) Show on the diagram where the electric field is a maximum. Find the magnitude of the electric field for a 5 V reverse bias at this location. Solution: The electric field as a function of position is as shown on the graph below. The maximum occurs at the metallurgic junction. The area under the curve gives the (known) junction voltage: Area = 1 2 E maxw D5 = 5 V + V 0 E max = 2( V) cm = V/cm. For silicon, this result is close to the critical electric field for causing an avalanche breakdown. At 5 volts, this diode is either broken down, or is about to break down. Notice that the Efield magnitude is primarily set by N D (the lower of the two doping levels), since the lower doping level largely determines the depletion region width. One reason the collector of a BJT is lightly doped is to achieve high breakdown voltages for the basecollector junction.
3 (f) Which type of carrier (conductionband electrons or valanceband holes) are the dominant charge carrier in the junction of this diode under forward bias. (Justify) Solution: Holes dominate, since N A N D. We derived the diode equation as [ i D = n 2 Dn i qa + D ] p (e v D/V T 1) L n N A L p N D In this case, D n and D p have the same order of magnitude, L n and L p have the same order of magnitude, and N A is 50 times N D. So the second term in the brackets above (due to the movement of holes) dominates. (g) Which type of carrier (conductionband electrons or valanceband holes) are the dominant charge carrier in the junction of this diode under reverse bias. (Justify) Solution: Again, the holes dominate (for exactly the same reason as in 1f). When the diode is reverse biased, the minority carriers in each material carry the leakage current. In this case the electron density in the pmaterial (n 2 i /N A) is much smaller than the hole density in the nmaterial (n 2 i /N D). (h) Diode measurements have indicated that n = 1 and I s = A at T = 300 K. Assuming that all of the current is being carried by the dominant carrier from part 1f, estimate the diffusion length (L n or L p... whichever is appropriate). Solution: Using the dominant term for I s gives I s = n 2 i qa [ Dp L p N D The diffusivity of the holes (D p ) should be obtained for the ntype material (the holes are the minority carriers, diffusing into the ntype region). From the graph on the first page of the test, Solving for L p gives D p = µ p V T = (320 cm 2 /V sec)(25.8 mv). L p = ( cm 3 ) 2 ( C)( cm 2 )(320 cm 2 /V sec)( V) (10 16 A)(10 17 cm 3 ) = cm (= µm) ]
4 2. For each of the circuits shown in this problem, assume that a 0.7 V diode drop model is valid, and that the Early effect can be neglected (V A = ). (a) Determine the mode of operation of the BJT, and provide the indicated bias voltages and currents. Show on the circuit diagram the direction of the various bias currents. V E : 9 V V C : 6.13 V V B : 8.3 V BJT Mode I E : 6.13 ma I C : 6.01 ma I B : 0.12 ma Forward Active Solution: Assuming the BJT is forward active gives V B = = 8.3 V. This gives I B, since I C = βi B = 50I B. 8.3 = I B (18 kω) + (I B + 50I B )(1 kω) I B = (8.3 V)/(18 kω + 51 kω) = 0.12 ma Now use I C = βi B and I E = I B + I C : I C = 50I B = 6.01 ma I E = 51I B = 6.13 ma Knowing I C and I B gives the collector voltage V C = (6.13 ma)(1 kω) = 6.13 V. Since V EC = > 0.2 V, the ForwardActive assumption has been justified. (b) Determine the value of R C which will result in V CE = 3 V. Provide the indicated bias voltages and currents. Show on the circuit diagram the direction of the various bias currents. V E : 2.72 V V C : 5.72 V V B : 3.42 V R C : 52.9 kω I E : µa I C : µa I B : 1.61 µa Solution: First, replace the base bias network with its Thevenin equivalent, as shown on the circuit diagram below. We ll eventually select R C so that transistor is forward active. Using KVL on the baseemitter loop, with I E = (β + 1)I B = 51I B, gives 3.46 V = (23.53 kω)i B (33 kω)(51i B ) Solving gives the bias currents: I B = 1.61 µa I C = 50I B = µa I E = 51I B = µa The resulting emitter and base voltages are V E = (33 kω)(82.48 µa) = 2.72 V V B = V E = 3.42 V To get V CE = 3 V, select R C to give V C = = 5.72 V. R C = µa = 52.9 kω
5 3. One disadvantage of the current source/sink circuits that we re examining in Lab 4 is that the current depends upon the supply voltage V CC. This problem investigates a technique for creating currents which are largely independent of the supply. The circuit idea is shown below. The pnp transistors Q 1 and Q 2 match each other, and the npn transistors Q 3 and Q 4 are identical except that the emitter area of Q 3 is 10 times that of Q 4 (so that I s3 = 10I s4 ). You can assume that all transistors operate at the same temperature. In implementation, a separate startup circuit (not shown) is needed to ensure that the circuit turns on, so that I 1, I 2 0. For this problem, you can assume that V CC is large enough that all transistors are in the forwardactive mode of operation, and that β and V A are large for all transistors. (a) Give approximate values of V 1 and V 2 Solution: From Q 1 : V 1 V CC 0.7 V. From Q 4 : V V. (b) First consider only Q 1 and Q 2, and develop a relationship between I 1 and I 2. (Show how the currents are related to each other, and explain why the relationship must hold.) Solution: Since the transistors are matched, and they share the same value of v EB, the currents will also be matched. I 1 = I 2 (1)
6 (c) Now consider Q 3 and Q 4, and develop a second relationship which relates I 1 and I 2. (Once again, be sure to explain where the relationship comes from!) Solution: From Q 4, v BE4 = V T ln(i 2 /I s4 ). So the baseemitter voltage of Q 3 is v BE3 = v BE4 I 1 R = V T ln(i 2 /I s4 ) I 1 R. This base emitter voltage determines I 1 : I 1 = I s3 e v BE3/V T = 10I s4 e ln(i2/is4) I1R/V T = (10I s4 )(I 2 /I s4 )e I1R/V T Cancelling I s4 gives I 2 = 1 ( ) 10 I I1 R 1 exp V T (2) (d) Using your solutions from parts 3b and 3c, solve for the current I 1 in terms of the resistor value. Find the value of R which will result in I 1 = 1 ma. Solution: Substituting (2) into (1) gives I 1 = 1 ( ) 10 I I1 R 1 exp We know I 1 0 due to the startup circuit. Cancelling I 1 and solving for the current gives V T I 1 = V T R ln(10) For I 1 = 1 ma, set R = 25.8 mv 1 ma ln(10) = 59.4 Ω. (e) Assuming that I 1 is set to 1 ma, and that Q 5 matches Q 1 and Q 6 matches Q 3. Determine the load currents I C5 and I C6. Solution: Q 1 and Q 5 are matched, and share the same emitterbase voltage. So I C5 = I C1 = 1 ma. Q 6 and Q 4 share the same baseemitter voltage, but the emitter area of Q 6 is ten times that of Q 4. So I C6 = 10I C4 = 10 ma. (f) What is the minimum supply voltage for which your design of part 3d will operate correctly? (Justify) Solution: All transistors must remain forwardactive. Looking at the left side of the circuit (Q 1 and Q 3 ) gives V E3 = (59.4 Ω)(1 ma) = 59.4 mv To stay forward active, require V CE3 > 0.2 V. So V 1 V CC 0.7 > V. V CC > 0.96 V (Looking at the right side, Q 2 and Q 4, requires V CC > 0.9 V.)
7 4. For the following, either circle the underlined word to indicate that you think it is correct as written, or cross it out and replace it with the correct word or phrase. (a) In designing a BJT, one way that β might be increased is to increase the doping level in the emitter collector. (Justify) Solution: Increasing the emitter doping level improves the injection efficiency. That is, a larger percentage of the current in the baseemitter junction is due to carriers injected by the emitter... which are largely swept out through the collector. Less base current is required for the same collector current, so β is increased. (b) For a npn transistor in the forwardactive mode of operation, most of the current through the baseemitter junction is due to diffusion of holes electrons across the junction. (c) A significant portion of the base current in a forwardactive pnp BJT is due to diffusion of electrons across the baseemitter junction. (d) For an npn transistor in the forwardactive mode of operation, most of the current through the basecollector junction is due to drift of electrons holes across the junction. (e) The baseemitter junction of a BJT has a lower higher breakdown voltage than the basecollector junction. (Justify) Solution: A more lightly doped collector implies a wider depletion region in the basecollector junction, resulting in a smaller electric field for a given applied voltage. (Larger voltages can be applied to the basecollector junction before the Efield is large enough to cause breakdown.) The basecollector junction has a higher breakdown voltage.
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