Feedback in Electronic Circuits

Transcription

1 ROCHESTER INSTITUTE OF TECHNOLOGY MICROELECTRONIC ENGINEERING Feedback in Electronic Circuits Dr. Lynn Fuller Webpage: 82 Lomb Memorial Drive Rochester, NY Tel (585) MicroE webpage: Feedback.ppt Page 1

2 OUTLINE Introduction dvantages of Feedback Generalized pproach Voltage Shunt Feedback Voltage Series Feedback Current Shunt Feedback Current Series Feedback Examples for each References Homework Questions Old Exam Questions Page 2

3 INTRODUCTION XS XE XF = XL/XE = XF/XL Consider the feedback amplifier shown. XL = XE = (XS XF) = (XS XL) = XS XL or XL(1)= XS f = XL/XS = /(1) XL In this analysis, is defined as the gain of the amplifier without feedback, known as the open loop gain, is the gain of the feedback network, T= is known as the loop gain, and f is the gain of the amplifier with feedback, known as the closed loop gain. Page 3

4 INTRODUCTION NEGTIVE FEEDCK: If the loop gain is negative for any circuit then that circuit has negative feedback. For negative feedback the closed loop gain is: f = 1 POSITIVE FEEDCK: If the loop gain is positive for any circuit then that circuit has positive feedback. f = / (1 ). Note that if = 1 the gain f = infinity and the circuit will oscillate. That is no input is needed. Which is useful if you want an oscillator. We will continue by discussing only negative feedback. Page 4

5 NEGTIVE FEEDCK DECRESES SENSITIVITY Decrease in Sensitivity Suppose changes by x%. How much would f change? To answer this question lets compute the sensitivity of f with respect to : Sensitivity symbol (not integral) Divide both sides by f Df = Df f f Incremental change in f = = Incremental change in df d D = d d 1 (1) = 2 D = f Df /f D / 1 D = 1 (1) 2 D 1 (1) 2 D 1 D = (1) (1) f = 1 (1) So we see that the sensitivity of the gain f to changes in gain is less than 1 Microelectronic which is Engineering an improvement over an amplifier without feedback. Page 5

6 NEGTIVE FEEDCK DECRESES SENSITIVITY Decrease in Sensitivity R2 = 10K R1=1K Vs R3 Vo 10 4 < < 10 5 The gain without feedback,, is large but not precise (may change from one op amp to the next). The gain with feedback, f, is R2/R1 = 10 which is precise (and lower than ) Page 6

7 NEGTIVE FEEDCK DECRESES DISTORTION Reduction in Distortion: Suppose and amplifier consists of two stages, 1 and 2 such that = 1 2 and that distortion is modeled as the addition of an unwanted signal XS as shown in the figure below. XS XS XE 1 = XI/XE XI XE 2 = XL/XE XL XF = XF/XL XL = 2 XE = 2 (XI XS ) XL = 2(1XE XS ) = 2 (1(XSXF) XS ) XL = 2(1(XSXL) XS ) Page 7

8 NEGTIVE FEEDCK DECRESES DISTORTION XL = 2 XE = 2 (XI XS ) XL = 2(1XE XS ) = 2 (1(XSXF) XS ) XL = 2(1(XSXL) XS ) XL = 21XS 21XL 2XS 12 XL = (112) XS XS 1 If Xs were some unwanted signal, say distortion, we could decrease the effect of Xs by introducing it as close as possible to the output of the amplifier. In other words make, as large as possible, use feedback to achieve the desired gain and also reduce distortion introduced after 1. In many amplifiers distortion is introduced in the output stage, (crossover, etc.) Page 8

9 NEGTIVE FEEDCK DECRESES DISTORTION V V Vx Crossover Distortion t Vin V Vo Rload Vo Without Feedback Dashed Line V t With Feedback Solid Line Page 9

10 NEGTIVE FEEDCK INCRESES NDWITDTH andwidth is the frequency range where the amplifier gain is flat. XS =(jw) XL (jw) = = 1000 = 60d f = 10 = 40d 20d 0d = 0.1 no feedback (jw) f (jw) with feedback f1 100f1 f (jw) = hz j f/f1 1 = = Let f1 = 1000 hz jf/f jf/f jf/f1 100 = f (jw) 101 1jf/(101 f1) Page 10

11 SUMMRY OF DVNTGES OF NEGTIVE FEEDCK Negative feedback reduces the gain of an amplifier compared to the gain without feedback: However, an amplifier with negative feedback has the following improvements 1. The gain with feedback is less sensitive to the amplifier gain value itself 2. Feedback can reduce unwanted distortion. 3. Feedback increases the bandwidth Page 11

12 REVIEW OF TWO PORT EQUIVLENT CIRCUITS I1 I1 = y 11 V1 y 12 V2 I2 = y 21 V1 y 22 V2 yparameter I2 I1 V1 = h 11 I1 h 12 V2 I2 = h 21 I1 h 22 V2 hparameter I2 V1 I1 y 11 y 12 V2 y 21 V1 gparameter y 22 I1 = g 11 V1 g 12 I2 V2 = g 21 V1 g 22 I2 I2 V2 V1 h 11 h 12 V2 h 21 I1 h 22 V1 = z 11 I1 z 12 I2 V2 = z 21 I1 z 22 I2 11 zparameter I2 V2 V1 g 11 g 12 I2 g 21 V1 g 22 V2 V1 z 11 z 12 I2 z 21 I1 z 22 V2 Page 12

13 GENERLIZED PPROCH Generalized pproach for the nalysis of Feedback mplifiers: The components of a feedback amplifier are: 1. n mplifier (y,g,h,z parameter twoport model) 2. Feedback Network (y,g,h,z parameter twoport model) 3. Sampling Network (series or parallel connections) 4. Mixing or Comparing Network (connections) 5. Load (a resistor) 6. Source (Thevenin or Norton equivalent) Page 13

14 MPLIFIER MODELS ie yparameter il ie hparameter il Ve VL Ve h 11 VL y 11 y 12 VL y 21 Ve y 22 h 12 VL h 21 ie h 22 ie gparameter il ie zparameter il Ve g 22 z 11 VL Ve z 22 VL g 11 g 12 il g 21 Ve z 12 il z 21 ie Page 14

15 FEEDCK NETWORK MODELS ie yparameter il ie hparameter il Ve VL Ve h 11 VL y 11 y 12 VL y 21 Ve y 22 h 12 VL h 21 ie h 22 ie gparameter il ie zparameter il Ve g 22 z 11 VL Ve z 22 VL g 11 g 12 il g 21 Ve z 12 il z 21 ie Page 15

16 SMPLING, MIXING, LOD, SOURCE 3. The sampling network consists of the wires used to connect the feedback network to the amplifier. Sampling can be done in parallel (voltage sampling) or in series (current sampling) with the load. 4. The mixing network consists of the wires used to connect the feedback network to the amplifier. Mixing can be in parallel (shunt) or series with the source. 5. The load is a resistor connected to the output of the amplifier. 6. The source is represented by its Thevinin or Norton equivalent circuit. VS Or IS RS RS Thevinin IS Or VS/RS Norton RS Page 16

17 FEEDCK MPLIFIER CONFIGURTIONS Since and are being considered as twoport networks there are four ways in which these twoport networks can be connected to provide feedback as shown below: VoltageShunt Feedback (use y parameters) S RL VoltageSeries Feedback (use h parameters) S RL CurrentShunt Feedback (use g parameters) S RL CurrentSeries Feedback (use z parameters) S RL Page 17

18 VOLTGESHUNT FEEDCK Since voltage shunt feedback implies parallel connections for the sampling and mixing networks. We will select, yparameter two port models and a Norton model for the source. The resulting equivalent circuit can be greatly simplified by combining parallel current sources and parallel conductances. S RL Page 18

19 VOLTGESHUNT FEEDCK Select yparameters and Norton equivalent circuits for voltageshunt feedback: IS Or VS/RS RS Ve VL RL y 11 y 12 VL y 21 Ve y 22 Ve VL y 11 y 12 VL y 21 Ve y 22 Page 19

20 SIMPLIFIED VOLTGESHUNT FEEDCK From the previous page we combine current sources in parallel and conductances in parallel. (the equivalent circuit is simplified) IS Or VS/RS Ve VL y 11 y 12 VL y 21 Ve y 22 y 11 = y 11 y 11 1/Rs y 22 = y 22 y 22 1/RL y 12 = y 12 y 12 y 21 = y 21 y 21 Page 20

21 NLYSIS OF VOLTGESHUNT FEEDCK IS Or VS/RS Ve VL y 11 y 12 VL y 21 Ve y 22 KCL gives IS = y 11 Ve y 12 VL 0 = y 21 Ve y 22 VL this eqn gives Ve = y 22 VL/y 21 IS = y 11 (y 22 VL/y 21 ) y 12 VL = [y 11 y 22 /y 21 y 12 ]VL 1 VL/IS = [y12 y 11 y 22 /y 21 ] VL/IS = y 21 y 11 y 22 y 12 y 1 21 y 11 y 22 f = 1 Page 21

22 NLYSIS OF VOLTGESHUNT FEEDCK Rf = VL/IS is the gain with feedback : Notice that the gain with feedback, f, for voltageshunt feedback has units of ohms (W). f is not a voltage gain, it is not a current gain, it is a transresistance. Suppose we wanted a voltage gain instead of transresistance. Recall that IS came from the Norton equivalent of the source. Thus VS = IS RS Rf = VL/IS Vf = VL/VS = (VL/IS) (1/RS) = Rf (1/RS) If = IL/IS = (VL/IS) (1/RL) = Rf (1/RL) Page 22

23 NLYSIS OF VOLTGESHUNT FEEDCK R = y 21 y 11 y 22 is the gain of the feedback amplifier with the feedback disabled. R is not the gain of the feedback amplifier with the feedback disconnected. Rf = y 21 y 11 y 22 y 12 y 1 21 y 11 y 22 y 12 = 0 y = 21 y11 y 22 Page 23

24 NLYSIS OF VOLTGESHUNT FEEDCK = y 12 : this quantity is very important for estimating the gain of the amplifier with feedback Rf ~ 1 = 1 y 12 This approximation is good as y 21 goes to infinity Rf = y 21 y 11 y 22 y 12 y 1 21 y 11 y 22 Exact gain T = = y 12 y 21 y 11 y 22 is the loop gain T Page 24

25 NLYSIS OF VOLTGESHUNT FEEDCK Input dmittance, Y If = IS/Ve IS Or VS/RS Ve y 11 y 12 VL y 21 Ve y 22 IS = y 11 Ve y 12 VL 0 = y 21 Ve y 22 VL VL = y 21 /y 22 Ve VL IS = y 11 Ve y 12 y 21 /y 22 Ve = y 11 Ve 1 y 12 y 21 y 11 y 22 Y If = IS/Ve = y 11 1 y 12 y 21 y 11 y 22 = y 11 1 T Page 25

26 NLYSIS OF VOLTGESHUNT FEEDCK Output dmittance, Y Of = Io/Vo with Is= zero Io Ve VL Vo y 11 y 12 VL y 21 Ve y 22 0 = y 11 Ve y 12 Vo Io = y 21 Ve y 22 Vo Ve = y 12 /y 11 Vo Io = y 21 y 12 /y 11 Vo y 22 Vo = y 22 Vo 1 y 12 y 21 y 11 y 22 Y Of = Io/Vo = y 22 1 y 12 y 21 y 11 y 22 = y 22 1 T Page 26

27 EXMPLE VOLTGESHUNT FEEDCK Example: vs 1K Vcc 10K 1K 1K 100 VL ssume DC analysis is good b = 100, V=infinite, rp=1k block ac eq. ckt 1. Identify block, block mixing network, sampling network, source and load. vs RS vbe rp block gmvbe or bib 10K Rc RL vo Page 27

28 EXMPLE VOLTGESHUNT FEEDCK 2. Find two port parameters for block and block. I1 I2 yparameter I1 = y 11 V1 y 12 V2 I2 = y 21 V1 y 22 V2 V1 V2 y 11 y 12 V2 y 21 V1 y 22 y 11 = 1/1K = 1E3 y 12 = 0 (almost always) y 21 = gm=/rp=0.1 y 22 = 1/1K = 1E3 y 11 = 1/10K = 1E4 y 12 = 1/10K = 1E4 y 21 = 1/10K = 1E4 y 22 = 1/10K = 1E4 Page 28

29 EXMPLE VOLTGESHUNT FEEDCK 3. Find the combined parameters y 11 = y 11 y 11 1/RS = 1e3 1e4 1e3 = 2.1e3 y 12 = y 12 y 12 = 0 1e4 = 1e4 y 21 = y 21 y 21 = 0.1 1e4 = =0.1 y 22 = y 22 y 22 1/RL =1e3 1e4 1e3 = 2.1e3 4. Compute quantities of interest 4.1 Gain with feedback (transresistance) Rf = y 21 y 11 y 22 y 12 y 1 21 y 11 y Voltage gain with feedback = (2.1e3)(2.1e3) 0.1(1e4) (2.1e3)(2.1e3) = 6940 ohms Vf = Rf (1/RS) = 6940 (1/1000) = 6.94 Page 29

30 EXMPLE VOLTGESHUNT FEEDCK 4.3 Current Gain with Feedback If = Rf (1/RL) = 6940 (1/1000) = pproximate Gain Rf ~ = 1/y 12 = 1/1e4 = ohms Vf ~ = (1/RS) = 10 If ~ = (1/RL) = Loop gain = T = y 12y 21 y 11 y 22 = (1e4)(0.1)/(2.1e3)(2.1e3) = 2.27 Page 30

31 EXMPLE VOLTGESHUNT FEEDCK y 4.6 Input admittance YIf = y 11 (1 T) = y 11 ( 1 12 y 21 ) y 11 y 22 YIf = (2.1e3)(1 2.27) = 6.86 ms Input impedance ZIf = 1/YIf = 146 ohms Note: this ZIf = 146 ohms is equal to the 1Kohm RS in parallel with Zin the amplifier input impedance. So 1000//Zin = 146 therefore we can find Zin = 171 ohms IS Or VS/RS RS 1K Zin 171 ohms Page 31

32 EXMPLE VOLTGESHUNT FEEDCK 4.7 Output Impedance Zof = 1/Yof Yof = y 22 (1T) = y 22 (1 2.27) = 6.86 ms Zof = 1 / Yof = 146 ohms Note: this 146 includes the 1000 ohm RL so Zo (without RL) is = 171 ohms Page 32

33 VOLTGE SERIES FEEDCK Since voltage series feedback implies parallel connection at the load and series connection at the source, we will select, hparameter two port models and a Thevenin model for the source. The resulting equivalent circuit can be greatly simplified by combining appropriate components S RL Page 33

34 VOLTGE SERIES FEEDCK ie VS Or IS RS RS h 11 h 22 VL h 12 VL h 21 ie h 11 h 22 h 12 VL h 21 ie Page 34

35 SIMPLIFIED VOLTGESERIES FEEDCK From the previous page we combine appropriate components to get the equivalent circuit shown ie VS Or IS RS Ve h 11 h 12 VL h 21 ie h 22 VL h 11 = h 11 h 11 Rs h 22 = h 22 h 22 1/RL h 12 = h 12 h 12 h 21 = h 21 h 21 Page 35

36 NLYSIS OF VOLTGESERIES FEEDCK ie VS Or IS RS Ve h 11 h 12 VL h 21 ie h 22 VL KVL gives VS = h 11 ie h 12 VL KCL gives 0 = h 21 ie h 22 VL this eqn gives ie = h 22 VL / h 21 VS = h 11 (h 22 VL/h 21 ) h 12 VL = [h 11 h 22 /h 21 h 12 ]VL 1 VL/VS = [h12 h 11 h 22 /h 21 ] VL/VS = h 21 h 11 h 22 h 12 h 1 21 h 11 h 22 f = 1 Page 36

37 VOLTGE SERIES FEEDCK Gain with Feedback Vf = h 21 h 11 h 22 h 12 h 1 21 h 11 h 22 This is a voltage gain (other gains can be found) Gain with Feedback Disabled V = h 21 h 11 h 22 pproximate Gain ~= 1 h 12 Loop Gain h 12 h 21 T = h 11 h 22 Input Impedance Output dmittance ZIf = h 11 (1T) Yof = h 22 (1T) Page 37

38 EXMPLE VOLTGESERIES FEEDCK Example: vs 1K 100K 1K Vcc 1K vs 1. Identify block, block mixing network, sampling network, source and load. VL ssume DC analysis is good b = 100, V=infinite, rp=1k RS source R block vbe rp block gmvbe or bib Re ac eq. ckt RL vo Page 38

39 EXMPLE VOLTGESERIES FEEDCK 2. Find two port parameters for block and block. V1 = h 11 I1 h 12 V2 I2 = h 21 I1 h 22 V2 I1 I2 hparameter V1 h 11 h 11 = rp = 1000 h 12 = 0 h 21 = b = 100 h 22 = 1/ro = 0 h 12 V2 h 21 I1 h 22 V2 h 11 = 0 h 12 = 1 h 21 = 1 h 22 = 1/Re = 1E3 Page 39

40 EXMPLE VOLTGESERIES FEEDCK 3. Find the combined parameters h 11 = h 11 h 11 RS//R = = 2000 h 12 = h 12 h 12 = 0 1 = 1 h 21 = h 21 h 21 = = 101 h 22 = h 22 h 22 1/RL = 0 1e3 1e3 = 2e3 4. Compute quantities of interest 4.1 Gain with feedback (Voltage Gain) h 21 h 11 h 22 Vf = = = h 12 h 1 21 h 11 h (2000)(2e3) (101)(1) (2000)(2e3) Page 40

41 EXMPLE VOLTGESERIES FEEDCK 4.2 pproximate Voltage Gain Vf ~ = 1/h 12 = Loop gain = T = h 12h 21 h 11 h 22 = (101)(1)/(2000)(2e3) = 25.3 Page 41

42 EXMPLE VOLTGESERIES FEEDCK h 4.4 Input impedance ZIf = h 11 (1 T) = h 11 ( 1 12 h 21 ) h 11 h 22 ZIf = (2000)(1 25.3) = 52.6Kohm Note: this ZIf = 52.6K ohms is equal to the 1Kohm RS//R in series with Zin the amplifier input impedance. So Zin = 51.6K ohms Vth = VS R/(RRS) 1K Vth = RS//R 51.6K ohms Zin Page 42

43 EXMPLE VOLTGESERIES FEEDCK 4.5 Output Impedance Zof = 1/Yof Yof = h 22 (1T) = 2E3 (1 25.3) = 50.6 ms Zof = 1 / Yof = 19.8 ohms Note: this 19.8 includes the 1000 ohm RL so Zo (without RL) is = 20.1 ohms Page 43

44 EXMPLE RIT OP MP Lets use feedback to get the exact voltage gain, approximate voltage gain, input impedance and output impedance for the circuit below. The Op mp is the one you built in lab with a differential amplifier, level shift stage and output stage. The overall voltage gain was 600 V/V and the differential input resistance was ~2K ohms. The output resistance was ~200 ohms. RS=2K RIT Vo Vs R2 = 10K RL = 1K R1=1K Identify the feedback and make sure it is negative feedback. Page 44

45 EXMPLE RIT OP MP The small signal ac equivalent circuit of the feedback amplifier on the previous page is: RS = 2K Rin 2K Vin Rout = Vin RL = 1K R1=1K R2 = 10K Voltage Series h parameters Page 45

46 EXMPLE VOLTGESERIES FEEDCK Find two port parameters for block and block. I1 hparameter I2 V1 = h 11 I1 h 12 V2 I2 = h 21 I1 h 22 V2 Rin 2K Inverting mp Vin Rout = Vin V1 h 11 h 12 V2 h 11 = 2K h 12 = 0 h 21 = 6000 h 22 = 1/200=5E3 h 21 I1 h 22 V2 h 11 = 1K//10K=909 h 12 = 1/11= h 21 = h 22 =1/11K=9.09E5 Page 46

47 EXMPLE VOLTGESERIES FEEDCK Find the combined parameters h 11 = h 11 h 11 RS = = 4909 h 12 = h 12 h 12 = = h 21 = h 21 h 21 = 6000 = 6000 h 22 = h 22 h 22 1/RL = 1/2001/11K 1/1K = 6.091E3 Compute quantities of interest 1. Exact Gain with feedback (Voltage Gain) h 21 h 11 h 22 Vf = = = 10.4 h 12 h 1 21 h 11 h (4909)(6.091e3) ( 6000)(0.0909) (4909)(6.09e3) Page 47

48 EXMPLE VOLTGESERIES FEEDCK 2. pproximate Voltage Gain Vf ~ = 1/h 12 = 1/ = 11 which agrees with ideal op amp theory 3. Loop gain = T = h 12h 21 h 11 h 22 = Input Impedance Z If = h 11 1 T = 4909 (1 18.2) = 94.5K but includes RS = 94.5K2K=92.5K without RS 5. Output Impedance Y Of = h 22 1 T = 6.09E3 (1 18.2) = 0.117S but includes RL Zout = 1/0.117 = 8.53 ohms but includes RL Zout=Z out//rl=8.53 ohms Z out = 8.61 ohm without RL Page 48

49 SPRED SHEET Page 49

50 SUMMRY OF FEEDCK MPLIFIERS Input dmittance Y If = y 11 1 T Y Of = y 22 1 T Output dmittance Input Impedance Z If = h 11 1 T Y Of = h 22 1 T Output dmittance Loop Gain y 12 y 21 T = y 11 y 22 Rf ~ = 1 y 12 ~ Gain Loop Gain h 12 h 21 T = h 11 h 22 Vf ~ = 1 h 12 ~ Gain VoltageShunt y 21 y 11 y 22 y 12 y 1 21 y 11 y 22 Transresistance Rf = VoltageSeries Exact Gain h 21 h 11 h 22 h 12 h 1 21 h 11 h 22 Voltage gain Vf = Exact Gain Voltage Gain Vf = Rf (1/RS) If = Rf (1/RL) Current Gain Voltage Gain Vf = Vf If = Vf (RS/RL) Current Gain Page 50

51 SUMMRY OF FEEDCK MPLIFIERS CurrentShunt Input dmittance Loop Gain Exact Gain Voltage Gain g 12 g 21 g 21 Y Of = g 22 1 T T = g 11 g g 11 g Vf = If (RL/RS) If = g 12 g Z 1 21 If = g 11 1 T If ~ = 1 g 12 g 11 g If = If 22 Output Impedance ~ Gain Current Gain Current Gain CurrentSeries Input Impedance Loop Gain Exact Gain Voltage Gain z 12 z z 21 Z 21 Of = z 22 1 T T = z 11 z z 11 z 22 Vf = Gf (RL) 22 Gf = z 12 z Z 1 21 If = z 11 1 T Gf = ~ 1 z 12 z 11 z If = Gf (RS) 22 Output Impedance ~ Gain Transconductance Current Gain Page 51

52 REFERENCES 1. Sedra and Smith, Device Electronics for Integrated Circuits, 2nd Edition, Kamins and Muller, John Wiley and Sons, The ipolar Junction Transistor, 2nd Edition, Gerald Neudeck, ddisonwesley, Page 52

53 V1 HOMEWORK PROLEM 1 1. Find all 16 two port parameters for each of the following circuits. 1a) 1b) I1 V1 I1 R2 R R1 V2 I2 I2 V2 1c) I1 V1 R Vcc RC I2 V2 Page 53

54 HOMEWORK SOLUTION FOR PROLEM 1 1a) y11 = 1/(R1 // R2) y12 = 1/R1 y21 = 1/R1 y22 = 1/R1 z11 = R2 z12 = R2 z21 = R2 z22 = R1 R2 h11 = R1 // R2 h12 = R2 / (R1R2) h21 = R2 / (R1R2) h22 = 1/ (R1R2) g11 = 1/R2 g12 = 1 g21 = 1 g22 = R1 1b) y11 = 1/R y12 = 1/R y21 = 1/R y22 = 1/R z11 = infinity z12 = infinity z21 = infinity z22 = infinity h11 = R h12 = 1 h21 = 1 h22 = 0 g11 = 0 g12 = 1 g21 = 1 g22 = R 1c) y11 = 1/(R // rp) y12 = 0 y21 = gm y22 = 1/RC z11 = R // rp z12 = 0 z21 = b RC z22 = RC h11 = (R // rp) h12 = 0 h21 = b h22 = 1/RC g11 = 1/(R // rp) g12 = 0 g21 = gm RC g22 = RC Page 54

55 HOMEWORK PROLEM 2 ND SOLUTION f 2 a) Find If changes by 20% how much does f change. b) Redesign the feedback amplifier in a) so XS that f is reduced by a factor of 100. Solution: = 100 = 1 XL 2 a) f = 1 1 = 1/101 =.0099 If changes by 20% then f changes by 20% x = 0.198% ~0.2% b) If we want the sensitivity to be 1/10001 instead of 1/101 then we increase the gain of the amplifier to 10,000 giving f = /(1) = 1 (same as before) and sensitivity = 1/(1) = thus a 20% change in is 0.002% change in gain with feedback Page 55

56 HOMEWORK PROLEM 3 ND SOLUTION 3. XS =(jw) = 0.1 XL (jw) = j f/f1 Let f1 = 1000 hz Draw a ode plot of the amplitude part of the gain function for (jw) and f (jw) Solution: 60d 40d 20d 0d (jw) f (jw) 1K 100K 10K hz f = 1 = = jf/ jf/ jf/ = jf/ Page 56

57 HOMEWORK PROLEM 4 4) Refer to the feedback amplifiers shown below. For each identify the type of feedback and determine if the feedback is negative or positive. Vcc 4a) 4b) vs VL Vin R1 R2 Vo Page 57

58 HOMEWORK PROLEM 4 (CONTINUED) 4c) Vcc 4d) Vcc RS vs Re Vo vs Re Vo Page 58

59 HOMEWORK PROLEM 4 (CONTINUED) 4e) Vcc Vs Vo Page 59

60 HOMEWORK PROLEM 4 SOLUTION 4. ll are negative feedback a. Voltage Shunt b. Voltage Shunt c. Voltage Series d. Voltage Shunt e. Voltage Series Page 60

61 HOMEWORK PROLEM 5 5) For each of the circuits shown in problems 7, 8, and 9, estimate the gain with feedback. xf ~= 1/ and = X12, where X12 is the appropriate combined two port parameters. Identify the type of gain, transconductance, transresistance, current or voltage. Convert xf to voltage gain with feedback, vf. Page 61

62 HOMEWORK PROLEM 5 SOLUTION 5a) for the amplifier in problem 6 we have voltage shunt feedback which uses y parameter two port equivalent circuits for the analysis. y 12 = y 12 y 12 = 0 1/10K Rf ~= 1/y 12 = 10K ohms Vf = Rf 1/RS = 10K/600 = b) for the amplifier in problem 7 we have current shunt feedback which uses g parameter two port equivalent circuits for the analysis. g 12 = g 12 g 12 = 0 200/(20010k) = If ~= 1/g 12 = 1/ = 51 Vf = If ( RL)/RS = 51 (2K/1K) = 102 5c) for the amplifier in problem 8 we have voltage series feedback which uses h parameter two port equivalent circuits for the analysis. h 12 = h 12 h 12 = 0 100/(10010k) = Vf ~= 1/h 12 = 1/ = 101 Page 62

63 HOMEWORK PROLEM 6 6) Find the exact gain with feedback for each of the circuit shown below. b = 100 rp = 1k V = infinite K 1K 300K 300K 1K 1K vs 2K 10K gm = Ic/VT=b/rp = 100/1K = 100mS Find the two port parameters for the and networks and compute Vf (exactly) Page 63

64 HOMEWORK PROLEM 6 SOLUTION IS RS or 600 VS/RS rp=1k Vin1 300K gmvin1 1K rp=1k Vin2 300K gmvin2 10K rp=1k 1K 1K Vin3 300K gmvin3 2K RL y11=1/(300k//1k)= y12=0 y22=1/1k=0.001 y21=gm3(gm2(1k//300k//1k))(gm1(1k//300k//1k) = 250 y11=1/10k= y12=1/10k y21=1/10k y22=1/10k Page 64

65 HOMEWORK PROLEM 6 SOLUTION y 21 y 11 y 22 y 12 y 1 21 y 11 y 22 Transresistance Rf = Exact Gain rf = 250/(.00277)(.0016) 1 (0.0001)(250)/(.00277)(.0016) = 9998ohms vf = rf 1/RS = 9998/600 = 16.7 Note: rf ~ 1/y12 = ohms y11= y11 y111/rs = = y12 = y12 y12 = = y21= y21 y21 = = 250 y22 = y22 y22 1/RL = = Page 65

66 HOMEWORK PROLEM 7 7) Find the exact gain with feedback for the circuit shown below. rp = 2K = 100 ro = 30K K K 2K Vo vs 10K K Page 66

67 HOMEWORK PROLEM 7 SOLUTION 7 cont.) Find the exact gain with feedback for the circuit shown below. The ac equivalent circuit below is useful in separating the block of the feedback amplifier. IS Or VS/RS RS 1K Ve 130K rp=2k ro=30k 2K VL 10K Ve 200 VL Page 67

68 HOMEWORK PROLEM 7 SOLUTION 7 cont.) g11=0.5e3 g12=0 g21=13.7e3(30k)=3020 g22=30k g11=0.098e3 g12= g21= g22=193 1/RS = 1E3 RL=2K Exact Gain (current gain) If = g 21 g 11 g 22 g 12 g 1 21 g 11 g 22 = 27.3 g11=g11g111/rs = 1.598E3 g12 = g12 g12 = g21 = g21 g21 = 3120 g22 = g22 g22 RL = 32.19K Note: if ~ 1/g12 = 51 vf = if (RL)/RS = 27.3 (2K/1K) = 54.6 Page 68

69 HOMEWORK PROLEM 8 8) Find the exact gain with feedback for the circuit shown below. 1K Rb 8K V 4K rp = 2K = 100 ro = infinite Rb >>1K Rz >>8K vs 100 Rz Vbb 100 2K Vo 10K Page 69

70 HOMEWORK PROLEM 8 SOLUTION VS RS 1K rp=1k Vin1 gmvin1 1K 100 rp=1k Vin2 8K gmvin K rp=1k Vin3 gmvin3 4K RL 2K 10K Note: the block and block are not completely separate, with proper mixing and sampling networks. The following equivalent circuit can be used Page 70

71 HOMEWORK PROLEM 8 SOLUTION First consider the10k and 100 ohm as the block 10K// K K Vo X Vo h12 x Vo h21 x I1 We can not just take the 100 ohm out of the emitter of the first transistor because the transistor will not work correctly in the block, Note: h21 >> h21 so neglect h21 Page 71

72 HOMEWORK PROLEM 8 SOLUTION In the emitter of the first transistor put the left hand part of the block 2 port equivalent circuit rp=1k gmvin1 gmvin1 RS RS rp=1k 1K Vin1 1K VS VS 10K//100 h12 x Vo h12 x Vo h12 x Vo 10K//100 Note: the right h12vo is in series with a current source and can be eliminated. Note: left h12vo is in a series loop to left of the two X s (next page) and can be moved anywhere in that loop. Page 72

73 HOMEWORK PROLEM 8 SOLUTION VS RS 1K rp=1k Vin1 X X Ib1 or gmvin1 1K 10K//100 rp=1k Vin2 8K Ib2 or gmvin2 1K gmvin2 100 rp=1k Vin3 4K Ib3 or gmvin3 gmvin3 RL 2K 10.1K Vo h12 x Vo Page 73

74 HOMEWORK PROLEM 8 SOLUTION Finally we have an ac equivalent circuit where the block and block are separate and proper mixing and sampling networks exist. h11=rp (1)10K//100=12K h12=0 h22=1/(2k//(rp4k)/(1))= h21 = h21=i2/i1withv2=0 I2 = (1)Ib3 so I2/Ib3 = (1) = 101 Ib3 = Ib2 (4K/(4Krp)) so Ib3/Ib2 = 66.7 Ib2 = Ib1 (8K/(8K(rp(1)100)) so Ib2/Ib1=39.8 h12 = I2/I1 = I2/Ib3 x Ib3/Ib2 x Ib2/Ib1 = Page 74

75 HOMEWORK PROLEM 8 SOLUTION h11 = 0,already included h12 = 100/(10010K) = h21 = neglected compared to h21 of block h22 =1/(10K100) =1/10.1K RS = 1K h11 = h11 h11 = 13K h12 = h12 h12 = h21 = h21 h21 = h22 = h22 h22 = RL = infinite, included in block Vf = Exact Gain h 21 h 11 h 22 h 12 h 1 21 h 11 h 22 Voltage gain (13K)(0.0099) = = (0.0099) 1 (13K)(0.0099) Page 75

76 OLD EXM QUESTION Pro 2. 12V eta = 150 vs u 4K 1uf 2K vo 12K Find Exact Gain With Feedback, pproximate Gain with Feedback, Voltage Gain, Current Gain, Input Resistance Page 76

77 OLD EXM QUESTION Pro 2. vs 700K 2K C1 5K 56K T1 C2 10K Vcc =20 3K C3 T2 4K vo 1K C4 ssume eta = 150 V= 100 Calculate dc value of IC for T1 and T2 Pro 3. Calculate gm, rp and ro for T1 and T2 Calculate the voltage gain vo/vs For the circuit in problem 2 use a single resistor to provide voltage shunt feedback to stabilize the gain with feedback at ~40 V/V. What value of feedback resistor should be used and show how you would connect it by adding it to the schematic of problem 2. Page 77