Collection of Solved Feedback Amplifier Problems

c Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. Collection of Solved Feedback Amplifier Problems This document contains a collection of solved feedback amplifier problems involving one or more active devices. The solutions make use of a graphical tool for solving simultaneous equations that is called the Mason Flow Graph (also called the Signal Flow Graph). When set up properly, the graph can be used to obtain by inspection the gain of a feedback amplifier, its input resistance, and its output resistance without solving simultaneous equations. Some background on how the equations are written and how the flow graph is used to solve them can be found at http://users.ece.gatech.edu/~mleach/ece3050/notes/feedback/fdbkamps.pdf The gain of a feedback amplifier is usually written in the form A ( + ba), wherea is the gain with feedback removed and b is the feedback factor. In order for this equation to apply to the four types of feedback amplifiers, the input and output variables must be chosen correctly. For amplifiers that employ series summing at the input (alson called voltage summing), the input variable must be a voltage. In this case, the source is modeled as a Thévenin equivalent circuit. For amplifiers that employ shunt summing at the input (also called current summing), the input variable must be a current. In this case, the source is modeled as a Norton equivalent circuit. When the output sampling is in shunt with the load (also called voltage sampling), the output variable must be a voltage. When the output sampling is in series with the load (also called current sampling), the output variable must be a current. These conventions are followed in the following examples. The quantity Ab is called the loop gain. For the feedback to be negative, the algebraic sign of Ab must be positive. If Ab is negative the feedback is positive and the amplifier is unstable. Thus if A is positive, b must also be positive. If A is negative, b must be negative. The quantity ( + Ab) is called the amount of feedback. It is often expressed in db with the relation 20 log ( + Ab). For series summing at the input, the expression for the input resistance is of the form R IN ( + ba), where R IN is the input resistance without feedback. For shunt summing at the input, the expression for the input resistance is of the form R IN ( + ba). For shunt sampling at the output, the expression for the output resistance is of the form R O ( + ba), wherer O is the output resistance without feedback. To calculate this in the examples, a test current source is added in shunt with the load. For series sampling at the output, the expression for the output resistance is of the form R O ( + ba). To calculate this in the examples, a test voltage source is added in series with the load. Most texts neglect the feedforward gain through the feedback network in calculating the forward gain A. When the flow graph is used for the analysis, this feedforward gain can easily be included in the analysis without complicating the solution. This is done in all of the examples here. The dc bias sources in the examples are not shown. It is assumed that the solutions for the dc voltages and currents in the circuits are known. In addition, it is assumed that any dc coupling capacitors in the circuits are ac short circuits for the small-signal analysis. Series-Shunt Example Figure (a) shows the ac signal circuit of a series-shunt feedback amplifier. The input variable is v and the output variable is. The input signal is applied to the gate of M and the feedback signal is applied to the source of M. Fig. (b) shows the circuit with feedback removed. A test current source i t is added in shunt with the output to calculate the output resistance R B. The feedback at the source of M is modeled by a Thévenin equivalent circuit. The feedback factor or feedback ratio b is the coefficient of in this source, i.e. b R / (R + R 3 ). The circuit values are g m 0.00 S, r s g m R 3 9kΩ, R 4 kω, andr 5 00 kω. The following equations can be written for the circuit with feedback removed: kω, r 0, R kω, R 2 0kΩ, i d G m v a G m R v a v v ts v ts r s + R kr 3 R + R 3 i d2 g m v tg2

Figure : (a) Amplifier circuit. (b) Circuit with feedback removed. R R 4 v tg2 i d R 2 i d2 R C + i t R C + i d R D R C R 4 k (R + R 3 ) R D R + R 3 + R 4 The voltage v a is the error voltage. The negative feedback tends to reduce v a,making v a 0 as the amount of feedback becomes infinite. When this is the case, setting v a 0yields the voltage gain /v b +R 3 /R. Although the equations can be solved algebraically, the signal-flow graph simplifies the solution. Figure 2 shows the signal-flow graph for the equations. The determinant of the graph is given by G m [ R 2 ( g m2 ) R C + R D ] R R + R 3 ( ) Figure 2: Signal-flow graph for the equations. The voltage gain /v is calculated with i t 0.Itisgivenby G m [ R 2 ( g m2 ) R C + R D ] v (R 2 g m2 R C + R D ) r s + R kr 3 + (R 2 g m2 R C + R D ) R r s + R kr 3 R + R 3 2

This is of the form where A A v +Ab r s + R kr 3 (R 2 g m2 R C + R D )4.83 R b 0. R + R 3 Note that Ab is dimensionless. Numerical evaluation yields 4.83 v +0.483 3.26 The output resistance R B is calculated with v 0.Itisgivenby R B i t R C R C +Ab 63 Ω Note that the feedback tends to decrease R B. Because the gate current of M is zero, the input resistance is R A R 5 00 kω. Series-Shunt Example 2 A series-shunt feedback BJT amplifier is shown in Fig. 3(a). A test current source is added to the output to solve for the output resistance. Solve for the voltage gain /v, the input resistance R A, and the output resistance R B. Assume β 00, r π 0kΩ, α β/( + β), g m β/r π, r e α/g m, r 0, r x 0, R kω, R 2 kω, R 3 2kΩ, R 4 4kΩ, andr 5 0kΩ. The circuit with feedback removed is shown in Fig. 3(b). The circuit seen looking out of the emitter of Q is replaced with a Thévenin equivalent circuit made with respect to. A test current source i t is added to the output to solve for the output resistance. Figure 3: (a) Amplifier circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write R 3 i e G v a v a v G R 3 + R 4 re 0 + R re 0 R 3kR 4 +β + r e i c αi e i b i c β v tb2 i c R 2 i e2 G 2 v tb2 G 2 r 0 e2 r 0 e2 R 2 +β + r e 3

R 3 R 5 i c2 αi e2 i c2 R a + i t R a + i e R b R a R 5 k (R 3 + R 4 ) R b R 3 + R 4 + R 5 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 4. The determinant is G [α R 2 G 2 α R a + R b ] R 3 R 3 + R 4 +G [α R 2 G 2 α R a + R b ] R 3 R 3 + R 4 9.09 Figure 4: Signal-flow graph for the equations. The voltage gain is G [α R 2 G 2 α R a + R b ] v G [α R 2 G 2 α R a + R b ] +G [α R 2 G 2 α R a + R b ] R 3 R 3 + R 4 This is of the form where A i +Ab A G [α R 2 G 2 α R a + R b ] re 0 + R α R 2 R 3 R 5 3kR 4 re2 0 α R 5 k (R 3 + R 4 )+ R 3 + R 4 + R 5 24.27 R 3 b 0.333 R 3 + R 4 Notice that the product Ab ispositive.thismustbetrueforthefeedbacktobenegative. Numerical evaluation of the voltage gain yields The resistances R A and R B are given by v A 2.67 R A ib v G α/β βr0 e α R B i t β α (r0 e + R 3 kr 4 ).32 MΩ R a R 5k (R 3 + R 4 ) 42.5 Ω 4

Series-Shunt Example 3 A series-shunt feedback BJT amplifier is shown in Fig. 5(a). Solve for the voltage gain /v, the input resistance R A, and the output resistance R B.ForJ,assumeg m 0.003 S, andr 0. ForQ 2,assume β 2 00, r π2 2.5kΩ, α 2 β 2 / ( + β 2 ), g m2 β 2 /r π2, r e2 α 2 /g m2, r 02, r x2 0. The circuit elements are R MΩ. R 2 0kΩ, R 3 kω, R 4 20kΩ, andr 5 0kΩ. The circuit with feedback removed is shown in Fig. 5(b). The circuit seen looking out of the source of J is replaced with a Thévenin equivalent circuit made with respect to. A test current source i t is added in shunt with the output to solve for the output resistance. Figure 5: (a) Amplifier circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write v a v R 3 R 3 + R 4 i d G v a G r s + R 3 kr 4 v tb2 i d R 2 i e2 G 2 v tb2 G 2 r 0 e2 r 0 e2 R 2 +β 2 + r e2 R 3 R 5 i c2 α 2 i e2 i c2 R a + i t R a + i d R b R a R 5 k (R 3 + R 4 ) R b R 3 + R 4 + R 5 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 6. The determinant is The voltage gain is G [ R 2 G 2 α R a + R b ] R 3 R 3 + R 4 +G [R 3 G 2 α R a + R b ] R 3 R 3 + R 4 2.08 G [ R 2 G 2 α 2 R a + R b ] v G [R 2 G 2 α 2 R a + R b ] +G [R 3 G 2 α R a + R b ] R 3 R 3 + R 4 5

Figure 6: Signal-flow graph for the equations. This is of the form where A i A +Ab A G [R 2 G 2 α 2 R a + R b ] R 2 R 3 R 4 r s + R 3 kr 4 re 0 α R 5 k (R 3 + R 4 )+ R 3 + R 4 + R 5 42.8 R 3 b 0.0467 R 3 + R 4 Notice that the product Ab ispositive.thismustbetrueforthefeedbacktobenegative. Numerical evaluation of the voltage gain yields 42.8 v + 42.8 0.0476 20 The resistances R A and R B are given by R B i t Series-Shunt Example 4 R A R MΩ R a R 5k (R 3 + R 4 ) 32.3 Ω A series-shunt feedback BJT amplifier is shown in Fig. 7(a). A test current source is added to the output to solve for the output resistance. Solve for the voltage gain /v, the input resistance R A, and the output resistance R B. Assume β 50, r π 2.5kΩ, α β/( + β), g m β/r π, r e α/g m, r 0, r x 0, R kω, R 2 00 Ω, R 3 9.9kΩ, R 4 0kΩ, andr 5 0kΩ. The circuit with feedback removed is shown in Fig. 8. The circuit seen looking out of the base of Q 2 is a Thévenin equivalent circuit made with respect to the voltage. A test current source i t is added in shunt with the output to solve for the output resistance. The emitter equivalent circuit for calculating i e and i e2 is shown in Fig. 7(b). For this circuit and the circuit with feedback removed, we can write i e G v e v e v R 2 R 2 + R 3 G i b i c β r e + r 0 e2 v tb3 i c R i e3 G 2 v tb3 G 2 r 0 e3 r 0 e2 R 2kR 3 +β + r e i c αi e r 0 e3 R +β + r e R 2 R 4 i c3 αi e3 i c3 R a + i t R a i b2 R b R a R 4 k (R 2 + R 3 ) R b R 2 + R 3 + R 4 6

Figure 7: (a) Amplifier circuit. (b) Emitter equivalent circuit for calculating i e and i e2. Figure 8: Circuit with feedback removed. 7

The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 9. The determinant is G α R G 2 α R a R b R 2 +β R 2 + R 3 +G α R G 2 α R a R b R 2 +β R 2 + R 3 8.004 The voltage gain is This is of the form where v v Figure 9: Flow graph for the equations. G α R G 2 α R a R b +β G α R G 2 α R a R b +G α R G 2 α R a A i +Ab R b +β +β R 2 R 2 + R 3 A G α R G 2 α R a R b +β r e + re2 0 α R re3 0 α R 4 k (R 2 + R 3 ) R 2 R 4 +β R 2 + R 3 + R 4 700.4 R 2 b 0.0 R 2 + R 3 Notice that the product Ab ispositive.thismustbetrueforthefeedbacktobenegative. Numerical evaluation of the voltage gain yields A v 87.5 The resistances R A and R B are given by ib G α/β R A R 5 k R 5 k R 5 k [ ( + β) (r e + r 0 e2)] 8.032 kω R B i t R a R 4k (R 2 + R 3 ) 624.7 Ω 8

Shunt-Shunt Example Figure 0(a) shows the ac signal circuit of a shunt-series feedback amplifier. The input variable is v and the output variable is. The input signal and the feedback signal are applied to the base of Q. A test current source i t is added in shunt with the output to calculate the output resistance R B.Fortheanalysis to follow convention, the input source consisting of v in series with R must be converted into a Norton equivalent. This circuit is the current i v /R in parallel with the resistor R. Fig. 0(b) shows the circuit with feedback removed and the source replaced with the Norton equivalent. The feedback at the base of Q is modeled by a Norton equivalent circuit /R 4 in parallel with the resistor R 4. The feedback factor or feedback ratio b is the negative of the coefficient of in this source, i.e. b R4. The circuit values are β 00, g m 0.05 S, r x 0, r ib β /g m 2kΩ, r 0, g m2 0.00 S, r s2 gm2 kω, r 02, R kω, R 2 kω, R 3 0kΩ, andr 4 0kΩ. Figure 0: (a) Amplifier circuit. (b) Circuit with feedback removed. The following equations can be written for the circuit with feedback removed: v b i a R b i a i + R 4 R b R kr 4 kr ib i c g m v b i d i c R 2 r s + R 2 i d R c + i t R c + v b R 3 R 3 + R 4 R c R 3 kr 4 The current i a is the error current. The negative feedback tends to reduce i a,making i a 0 as the amount of feedback becomes infinite. When this is the case, setting i a 0yields the current gain /i R 4. Although the equations can be solved algebraically, the signal-flow graph simplifies the solution. Figure shows the flow graph for the equations. The determinant of the graph is given by R b g m R 2 r s + R 2 R c + R 3 R 3 + R 4 The transresistance gain is calculated with i t 0.Itisgivenby R b g m R 2 R c + R 3 r s2 + R 2 R 3 + R 4 i + R 4 g m R 2 r s2 + R 2 (R 3 kr 4 )+ R 3 (R kr 4 kr ib ) R 3 + R 4 (R kr 4 kr ib ) g m R 2 (R 3 kr 4 )+ R 3 r s2 + R 2 R 3 + R 4 9 R 4

Figure : Signal-flow graph for the equations. This is of the form A i +Ab where A (R kr 4 kr ib ) g m R 2 (R 3 kr 4 )+ R 3 77.8 kω r s2 + R 2 R 3 + R 4 b R 4 0 4 S Note that Ab is dimensionless and positive. Numerical evaluation yields 77.8 0 3 i +( 77.8 0 3 ) ( 0 4 8.86 kω ) The voltage gain is given by i 8.86 v i v i R The resistance R a is calculated with i t 0.Itisgivenby R a v b i R b R kr 4 kr ib 7.7 Ω +Ab Note that the feedback tends to decrease R a.theresistancer A is calculated as follows: R A R + R a The resistance R B is calculated with i 0.Itisgivenby R B i t R.077 kω R c R 3kR 4 +Ab 569.4 Ω Shunt-Shunt Example 2 A shunt-shunt feedback JFET amplifier is shown in Fig. 2(a). Solve for the voltage gain /v, the input resistance R A, and the output resistance R B.Assumeg m 0.005 S, r s gm 200 Ω, r 0, R 3kΩ, R 2 7kΩ, R 3 kω, R 4 0kΩ. The circuit with feedback removed is shown in Fig. 2(b) In this circuit, the source is replaced by a Norton equivalent circuit consisting of a current i v /R in parallel with the resistor R. This is necessary for the feedback analysis to conform to convention for shunt-shunt feedback.. The circuit seen looking up into R 2 is replaced with a Norton equivalent circuit made with respect to.a test current source i t is added in shunt with the output to solve for the output resistance. 0

Figure 2: (a) Amplifier circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write v g i R b + R 2 R b R b R kr 2 i d G m v g G m i d R c + i t R c + v g R 4 R 2 + R 4 R c R 2 kr 4 r s + R 3 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 3. The determinant is R b G m R c + R 4 R 2 + R 4 R 2 +R b G m R c R 4 R 2 + R 4 R 2.853 Ω Figure 3: Signal-flow graph for the equations.

The transresistance gain is i R b G m R c + R 4 R 2 + R 4 R b G m R c R 4 R 2 + R 4 +R b G m R c + R 4 R 2 + R 4 R 2 This is of the form where A i +Ab A R b G m R c R 4 R 2 + R 4 (R kr 2 ) G m R 2 kr 4 R 4 R 2 + R 4 5.97 kω b R 2 0.429 ms Note that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation yields A 3.22 kω i The voltage gain is given by i A v i v.074 R The resistances R a, R A,andR B are given by Shunt-Shunt Example 3 R a v g R b.3 kω i R A R + 4.82 kω R a R R B i t R c 2.22 kω A shunt-shunt feedback BJT amplifier is shown in Fig. 4. The input variable is the v and the output variable is the voltage. The feedback resistor is R 2. The summing at the input is shunt because the input through R and the feedback through R 2 connect in shunt to the same node, i.e. the v b node. The output sampling is shunt because R 2 connects to the output node. Solve for the voltage gain /v, the input resistance R A, and the output resistance R B. Assume β 00, r π 2.5kΩ, g m β/r π, α β/( + β), r e α/g m, r 0, r x 0, V T 25mV. The resistor values are R kω, R 2 20kΩ, R 3 500 Ω, R 4 kω, andr 5 5kΩ. The circuit with feedback removed is shown in Fig. 5. A test current source i t isaddedinshuntwith the output to solve for the output resistance R B. In the circuit, the source is replaced by a Norton equivalent circuit consisting of a current i v R 2

Figure 4: Amplifier circuit. Figure 5: Circuit with feedback removed. in parallel with the resistor R. This is necessary for the feedback analysis to conform to convention for shunt summing. The circuit seen looking into R 2 from the collector of Q 3 is replaced with a Thevenin equivalent circuit made with respect with v b. For the circuit with feedback removed, we can write i e i + R 2 v b i e R b R b R kr 2 kr π i c g m v b v b2 i c R c R c R 3 kr π v b3 i c2 R d R d R 4 kr π i c3 g m v b3 ( i c3 + i t ) R e + v b R 5 R 2 + R 5 R e R 2 kr 5 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 6. The determinant is R b g m R c g m R d g m R e + R 5 R 2 + R 5 R 2 +R b g m R c g m R d g m R e R 5 R 2 + R 5 R 2 354.7 3

Figure 6: Signal-flow graph for the equations. The transresistance gain is R b g m R c g m R d g m R e + R 2 R 2 + R 5 i R b g m R c g m R d g m R e R 2 R 2 + R 5 +R b g m R c g m R d g m R e R 2 R 2 + R 5 R 2 R b g m R c g m R d g m R e R 2 R 2 + R 5 + R b g m R c g m R d g m R e R 2 R 2 + R 5 R 2 This is of the form i e2 v A +Ab where A and b are given by A R b g m R c g m R d g m R e R 5 R 2 + R 5 R kr 2 kr π g m R 3 kr π g m R 4 kr π g m R 2 kr 5 R 5 R 2 + R 5 7.073 MΩ b 50 S R 2 Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transresistance gain yields A 9.94 kω i The resistances R a and R A are R a v b R c i.945 Ω R A R +.002 kω R a R 4

The resistance R B is The voltage gain is v b i R B i t vb i R e.28 Ω A 9.9 R A Shunt-Shunt Example 4 A shunt-shunt feedback BJT amplifier is shown in Fig. 7. The input variable is the v and the output variable is the voltage. The feedback resistor is R 2. The summing at the input is shunt because the input through R and the feedback through R 2 connect in shunt to the same node, i.e. the v e node. The output sampling is shunt because R 2 connects to the output node. Solve for the voltage gain /v, the input resistance R A, and the output resistance R B. For Q and Q 2,assumeβ 00, r π 2.5kΩ, g m β/r π, α β/( + β), r e α/g m, r 0, r x 0, V T 25mV.ForJ 3,assumeg m3 0.00 S and r 03. The resistor values are R kω, R 2 00 kω, R 3 0Ω, R 4 30kΩ, andr 5 0kΩ. Figure 7: Amplifier circuit. The circuit with feedback removed is shown in Fig. 8. A test current source i t isaddedinshuntwith the output to solve for the output resistance R B. In the circuit, the source is replaced by a Norton equivalent circuit consisting of a current i v R in parallel with the resistor R. This is necessary for the feedback analysis to conform to convention for shunt summing. The circuit seen looking into R 2 from the collector of Q 3 is replaced with a Thévenin equivalent circuit made with respect with v e. Figure 8: Circuit with feedback removed. 5

For the circuit with feedback removed, we can write i e i + R 2 v e i e R b R b R kr 2 kr e i c g m v e v tg3 i c R 3 i d3 g m3 v tg3 v tb2 i d3 R 4 i e2 G v tb2 G re2 0 + R 2kR 5 r 0 e2 R 4 + r e2 R 5 ( i e2 + i t ) R c + v e R c R 2 kr 5 R 2 + R 5 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 9. The determinant is R b g m R 3 g m2 R 4 G R c + R 5 R 2 + R 5 +R b g m R 3 g m2 R 4 G R c R 5 R 2 + R 5 R 2 3.0 R 2 Figure 9: Signal-flow graph for the equations. The transresistance gain is R b g m R 3 g m2 R 4 G R c + R 2 R 2 + R 5 i R b g m R 3 g m2 R 4 G R c R 2 R 2 + R 5 +R b g m R 3 g m2 R 4 G R c R 2 R 2 + R 5 R 2 R b g m R 3 g m2 R 4 G R c R 2 R 2 + R 5 + R b g m R 3 g m2 R 4 G R c R 2 R 2 + R 5 R 2 R 2 + R 5 R 2 This is of the form i e2 A v +Ab where A and b are given by A R b g m R 3 g m2 R 4 G R c R 2 R 2 + R 5 R kr 2 kr e g m R 3 g m2 R 4 re2 0 + R R 2 kr 5 R 5 2kR 5 R 2 + R 5.2MΩ 6

b 0 S R 2 Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transresistance gain yields The resistances R a and R A are The resistance R B is The voltage gain is A 99. kω i R a v e R b i 0.24 Ω R A R + kω R a R v e i Series-Series Example R B i t ve i R c 80.49 Ω A 99.09 R A Figure 20(a) shows the ac signal circuit of a series-series feedback amplifier. The input variable is v and the output variable is i d2. The input signal is applied to the gate of M and the feedback signal is applied to thesourceofm. Fig. 20(b) shows the circuit with feedback removed. A test voltage source v t is added in series with the output to calculate the output resistance R b. The feedback at the source of M is modeled by a Thévenin equivalent circuit. The feedback factor or feedback ratio b is the coefficient of i d2 in this source, i.e. b R 5. The circuit values are g m 0.00 S, r s gm kω, r 0, R 50kΩ, R 2 0kΩ, R 3 kω, R 4 9kΩ, andr 5 kω. Figure 20: (a) Amplifier circuit. (b) Circuit with feedback removed. 7

The following equations can be written for the circuit with feedback removed: i d G m v a G m i d2 G m2 v b G m2 r s + R 5 v a v v ts v ts i d2 R 5 r s2 + R 3 v b v t v tg2 v tg2 i d R 2 The voltage v a is the error voltage. The negative feedback tends to reduce v a,making v a 0 as the amount of feedback becomes infinite. When this is the case, setting v a 0yields the transconductance gain i d2 /v b R5. Although the equations can be solved algebraically, the signal-flow graph simplifies the solution. Figure 2 shows the signal-flow graph for the equations. The determinant of the graph is given by G m ( R 2 ) ( ) G m2 R 5 ( ) Figure 2: Flow graph for the equations. The transconductance gain i d2 /v is calculated with v t 0.Itisgivenby i d2 G m ( R 2 ) ( ) G m2 v R 2 r s + R 5 r s + R 5 + R 2 R 5 r s + R 5 r s + R 5 This is of the form i d2 A v +Ab where A G m ( R 2 ) ( ) G m2 R 2 2.5 0 3 S r s + R 5 r s2 + R 3 b R 5 000Ω Note that ba is dimensionless. Numerical evaluation yields i d2 2.5 0 3 v + 000 2.5 0 3 7.24 0 4 S The resistance R b is calculated with v 0.Itisgivenby id2 Gm2 R b (+ba)(r s2 + R 3 )7kΩ v t Note that the feedback tends to increase R b.theresistancer B is calculated as follows: R B (R b R 3 ) kr 3 857. Ω Because the gate current of M is zero, the input resistance is R A R 50kΩ. 8

Series-Series Example 2 A series-series feedback BJT amplifier is shown in Fig. 22. The input variable is the voltage v and the output variable is the voltage. The feedback is from i e2 to the emitter of Q. Because the feedback does not connect to the input node, the input summing is series. The output sampling is series because the feedback is proportional to the current that flows in series with the output rather than the output voltage. Solve for the transconductance gain i c3 /v, the voltage gain /v, the input resistance R A, and the output resistance R B. Assume β 00, I C 0.6mA, I C2 ma, I C3 4mA, α β/( + β), g m I C /V T, r e αv T /I C, r 0, r x 0, V T 25mV, R 00 Ω, R 2 9kΩ, R 3 5kΩ, R 4 600 Ω, R 5 640 Ω, and R 6 00 Ω. The circuit with feedback removed is shown in Fig. 23. Figure 22: Amplifier circuit. The circuit looking out of the emitter of Q is a Thévenin equivalent made with respect to the current i e3. The output current is proportional to this current, i.e. i c3 αi e3. Because r 0 for Q 3, the feedback does not affect the output resistance seen looking down through R 4 because it is infinite. For a finite r 0,a testvoltagesourcecanbeaddedinserieswithr 4 to solve for this resistance. It would be found that a finite r 0 for Q 3 considerably complicates the circuit equations and the flow graph. Figure 23: Circuit with feedback removed. 9

For the circuit with feedback removed, we can write R 6 R v e v i e3 i e G v e G R 6 + R 5 + R r e + R k (R 5 + R 6 ) i b i c β v tb2 i c R 2 i e2 G 2 v tb2 G 2 r 0 e2 i c2 αi e2 v tb3 i c2 R 3 i e3 G 3 v tb3 k i e G 3 R R 6 i c αi e re2 0 R 2 +β + r e re3 0 + R 6k (R + R 5 ) k R + R 5 + re3 0 kr 6 R 6 + re3 0 i c3 αi e3 i c2 R 4 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 24. The determinant is ½ ¾ R 6 R G [(α R 2 G 2 α R 3 G 3 ) k ] R 6 + R 5 + R R 6 R +G (α R 2 G 2 α R a k ) R 6 + R 5 + R 25.5 r e + R k (R 5 + R 6 ) Figure 24: Signal-flow graph for the circuit. The transconductance gain is i c3 G (α R 2 G 2 α R 3 G 3 k ) α v G (α R 2 G 2 α R 3 G 3 k ) α R 6 R +[G (α R 2 G 2 α R a k )] R 6 + R 5 + R G [(α R 2 G 2 α R 3 G 3 ) k ] α R 6 R +[G (α R 2 G 2 α R a k ) α] R 6 + R 5 + R α This is of the form i c3 A v +Ab where A and b are given by A G [(α R 2 G 2 α R 3 G 3 ) k ] α α R 2 α R 3 R R 6 R + R 5 + re3 0 kr 6 R 6 + re3 0 20.83 S α r 0 e2 re3 0 + R 6k (R + R 5 ) α 20

R 6 R b R 6 + R 5 + R α 2.02 Ω Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields i c3 v A 0.083 The voltage gain is given by i c3 A v v i c3 R 4 49.7 The resistances R A and R B are given by R A ib v G α/β ( + β) G ( + β) [r e + R k (R 5 + R 6 )] 3.285 MΩ R B R 4 600 Ω Series-Series Example 3 A series-series feedback BJT amplifier is shown in Fig. 25(a). The input variable is the voltage v and the output variable is the voltage. The feedback is from i e2 to i c2 to the emitter of Q. Because the feedback does not connect to the input node, the input summing is series. Because the feedback does not sample the output voltage, the sampling is series. That is, the feedback network samples the current in series with the outpu. Solve for the transconductance gain i e2 /v, the voltage gain /v, the input resistance R A,and the output resistance R B. Assume β 00, r π 2.5kΩ, α β/( + β), r e α/g m, r 0, r x 0, V T 25mV, R 00 Ω, R 2 kω, R 3 20kΩ, andr 4 0kΩ. Figure 25: (a) Amplifier circuit. (b) Circuit with feedback removed. The circuit with feedback removed is shown in Fig. 25(b). The circuit seen looking out of the emitter of Q is replaced with a Thévenin equivalent circuit made with respect with i c2. The output current i e2 is proportional to this current, i.e. i e2 αi c2. A test voltage source v t is added in series with the output to solve for the output resistance. The resistance seen by the test source is labeled R b. For the circuit with feedback removed, we can write v e v i c2 R 2 i e G v e G i c αi e i b i c r e + R + R 2 β 2

v tb2 i c R 3 i e2 G 2 (v t v tb2 ) G 2 re2 0 + R re2 0 R 3 4 +β + r e i c2 αi e2 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 26. The determinant is (G α R 3 G 2 α R 2 ) +G α R 2 G 2 α R 2.8 The transconductance gain is Figure 26: Signal-flow graph for the equations. This is of the form where A and b are given by i e2 G α R 3 G 2 v (G α R 3 G 2 ) +(G α R 3 G 2 ) α R 2 i e2 v A +Ab A G α R 3 G 2 α R 3 r e + R + R 2 re2 0 + R 4 0.97 ms b αr 2.98 kω Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields i e2 A 0.325 ms v The voltage gain is given by i e2 A v v i e2 R 4 3.25 The resistances R A and R B are given by ib G α/β R A ( + β)(r e + R + R 2 ) 602 kω v ie2 Gm2 R b (R 4 + r 0 v t e2) 28.68 kω R B (R b R 4 ) kr 4 6.53 kω 22

Series-Series Example 4 A series-series feedback BJT amplifier is shown in Fig. 27(a). The input variable is the voltage v and the output variable is the current i c2. The feedback is from i c2 to i e2 to the gate of J. The input summing is series because the feedback does not connect to the same node that the source connects. The output sampling is series because the feedback is proportional to the output current i c2. Solve for the transconductance gain i c2 /v, the voltage gain /v, the input resistance R A, and the output resistance R B.ForJ, assume that g m 0.00 S, r s gm 000 Ω, andr 0. ForQ 2,assumeβ 2 00, r π2 2.5kΩ, α 2 β 2 / ( + β 2 ), r e2 α 2 /g m2, r 02, r x2 0, V T 25mV. The resistor values are R kω, R 2 0kΩ, R 3 kω, R 4 0kΩ, R 5 kω, andr 6 0kΩ. Figure 27: (a) Amplifier circuit. (b) Circuit with feedback removed. The circuit with feedback removed is shown in Fig. 27(b). The circuit seen looking out of the emitter of Q is replaced with a Thévenin equivalent circuit made with respect with i e2. The output current is proportional to this current, i.e. i c2 α 2 i e2. Because r 02, the feedback does not affect the output resistance seen looking down through R 6 because it is infinite. For a finite r 02, a test voltage source can be added in series with R 6 to solve for this resistance. It would be found that a finite r 02 considerably complicates the circuit equations and the flow graph. For the circuit with feedback removed, we can write i d g m v e v e i e2 R 5 R 3 R 3 + R 4 + R 5 v v tb2 i d R 2 i e2 G v tb2 G r 0 e2 + R 5k (R 3 + R 4 ) r 0 e2 R 2 +β 2 + r e2 i c2 α 2 i e2 The equations can be solved algebraically or by a flow graph. The flow graph for the equations is shown in Fig. 28. The determinant is R 5 R 3 g m R 2 G R 3 + R 4 + R 5 R 5 R 3 +g m R 2 G R 3 + R 4 + R 5.80 23

Figure 28: Signal-flow graph for the equations. The transconductance gain is This is of the form where A and b are given by i e2 g m R 2 G α 2 v g m R 2 G α 2 R 5 R 3 +g m R 2 G R 3 + R 4 + R 5 (g m R 2 G α 2 ) R 5 R 3 +(g m R 2 G α 2 ) R 3 + R 4 + R 5 α 2 i e2 v A +Ab A g m R 2 G α 2 g m R 2 re2 0 + R 5k (R 3 + R 4 ) α 2 9.56 ms R 5 R 3 b 84.7 Ω R 3 + R 4 + R 5 α 2 Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields i c2 v A 5.284 ms The voltage gain is given by i c2 A v v i c2 R 6 52.84 The resistances R A and R B are given by R A R k id v ³ gm R k R k 643 Ω g m R B R 6 0kΩ Series-Series Example 5 A series-series feedback BJT amplifier is shown in Fig. 29. The input variable is the current i and the output variable is the current i e2. The feedback path is the path from i e2 to i c2 to i e3 to i c3 to the emitter of Q. The input summing is series because the feedback does not connect to the input node. The output 24