Collection of Solved Feedback Amplifier Problems


 Damian Mills
 1 years ago
 Views:
Transcription
1 c Copyright 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 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 smallsignal analysis. SeriesShunt Example Figure (a) shows the ac signal circuit of a seriesshunt 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
2 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 signalflow graph simplifies the solution. Figure 2 shows the signalflow 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: Signalflow 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
3 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 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. SeriesShunt Example 2 A seriesshunt 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
4 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 Figure 4: Signalflow 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 R 3 b 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
5 SeriesShunt Example 3 A seriesshunt 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 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 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
6 Figure 6: Signalflow 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 R 3 b R 3 + R 4 Notice that the product Ab ispositive.thismustbetrueforthefeedbacktobenegative. Numerical evaluation of the voltage gain yields 42.8 v The resistances R A and R B are given by R B i t SeriesShunt Example 4 R A R MΩ R a R 5k (R 3 + R 4 ) 32.3 Ω A seriesshunt 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
7 Figure 7: (a) Amplifier circuit. (b) Emitter equivalent circuit for calculating i e and i e2. Figure 8: Circuit with feedback removed. 7
8 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 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 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)] kω R B i t R a R 4k (R 2 + R 3 ) Ω 8
9 ShuntShunt Example Figure 0(a) shows the ac signal circuit of a shuntseries 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 m 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 signalflow 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
10 Figure : Signalflow 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 kω r s2 + R 2 R 3 + R 4 b R S Note that Ab is dimensionless and positive. Numerical evaluation yields i +( ) ( 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 Ω ShuntShunt Example 2 A shuntshunt 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 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 shuntshunt 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
11 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 Ω Figure 3: Signalflow graph for the equations.
12 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 kω b R 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 ShuntShunt Example 3 R a v g R b.3 kω i R A R kω R a R R B i t R c 2.22 kω A shuntshunt 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 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
13 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
14 Figure 6: Signalflow 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 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 kω R a R 4
15 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 ShuntShunt Example 4 A shuntshunt 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 m 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
16 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 R 2 Figure 9: Signalflow 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
17 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 SeriesSeries Example R B i t ve i R c Ω A R A Figure 20(a) shows the ac signal circuit of a seriesseries 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
18 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 signalflow graph simplifies the solution. Figure 2 shows the signalflow 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 S r s + R 5 r s2 + R 3 b R 5 000Ω Note that ba is dimensionless. Numerical evaluation yields i d v 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 Ω Because the gate current of M is zero, the input resistance is R A R 50kΩ. 8
19 SeriesSeries Example 2 A seriesseries 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 Ω, R 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
20 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: Signalflow 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 + re S α r 0 e2 re3 0 + R 6k (R + R 5 ) α 20
21 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 The voltage gain is given by i c3 A v v i c3 R 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 )] MΩ R B R Ω SeriesSeries Example 3 A seriesseries 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
22 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: Signalflow 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 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 ms v The voltage gain is given by i e2 A v v i e2 R 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) kω R B (R b R 4 ) kr kω 22
23 SeriesSeries Example 4 A seriesseries 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
24 Figure 28: Signalflow 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 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 ms The voltage gain is given by i c2 A v v i c2 R 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 SeriesSeries Example 5 A seriesseries 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
http://users.ece.gatech.edu/~mleach/ece3050/notes/feedback/fbexamples.pdf
c Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. Feedback Amplifiers CollectionofSolvedProblems A collection of solved
More informationFEEDBACK INTRODUCTION
FEEDBACK INTRODUCTION Most physical systems incorporate some form of feedback. Feedback can be either negative (degenerative) or positive (regenerative). In amplifier design, negative feedback is applied
More informationVoltage Divider Bias
Voltage Divider Bias ENGI 242 ELEC 222 BJT Biasing 3 For the Voltage Divider Bias Configurations Draw Equivalent Input circuit Draw Equivalent Output circuit Write necessary KVL and KCL Equations Determine
More informationThe BJT Differential Amplifier. Basic Circuit. DC Solution
c Copyright 010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The BJT Differential Amplifier Basic Circuit Figure 1 shows the circuit
More informationBJT AC Analysis. by Kenneth A. Kuhn Oct. 20, 2001, rev Aug. 31, 2008
by Kenneth A. Kuhn Oct. 20, 2001, rev Aug. 31, 2008 Introduction This note will discuss AC analysis using the beta, re transistor model shown in Figure 1 for the three types of amplifiers: commonemitter,
More informationTransistor Characteristics and Single Transistor Amplifier Sept. 8, 1997
Physics 623 Transistor Characteristics and Single Transistor Amplifier Sept. 8, 1997 1 Purpose To measure and understand the common emitter transistor characteristic curves. To use the base current gain
More informationFigure 1: Commonbase amplifier.
The CommonBase Amplifier Basic Circuit Fig. 1 shows the circuit diagram of a single stage commonbase amplifier. The object is to solve for the smallsignal voltage gain, input resistance, and output
More informationLab 7: Operational Amplifiers Part I
Lab 7: Operational Amplifiers Part I Objectives The objective of this lab is to study operational amplifier (op amp) and its applications. We will be simulating and building some basic op amp circuits,
More informationBasic Laws Circuit Theorems Methods of Network Analysis NonLinear Devices and Simulation Models
EE Modul 1: Electric Circuits Theory Basic Laws Circuit Theorems Methods of Network Analysis NonLinear Devices and Simulation Models EE Modul 1: Electric Circuits Theory Current, Voltage, Impedance Ohm
More informationVer 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)
Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.
More informationSmall Signal Analysis of a PMOS transistor Consider the following PMOS transistor to be in saturation. Then, 1 2
Small Signal Analysis of a PMOS transistor Consider the following PMOS transistor to be in saturation. Then, 1 I SD = µ pcox( VSG Vtp)^2(1 + VSDλ) 2 From this equation it is evident that I SD is a function
More informationBipolar Junction Transistors
Bipolar Junction Transistors Physical Structure & Symbols NPN Emitter (E) ntype Emitter region ptype Base region ntype Collector region Collector (C) B C Emitterbase junction (EBJ) Base (B) (a) Collectorbase
More informationApplication Note AN1017401. A Low Impedance PIN Diode Driver Circuit with Temperature Compensation
Application Note AN74 A Low mpedance PN Diode Driver Circuit with Temperature Compensation Two Philips BAP5 PN diodes are used in an RF attenuator with a low impedance driver circuit to significantly
More informationSeries and Parallel Circuits
Direct Current (DC) Direct current (DC) is the unidirectional flow of electric charge. The term DC is used to refer to power systems that use refer to the constant (not changing with time), mean (average)
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module 2 DC Circuit Lesson 5 Nodevoltage analysis of resistive circuit in the context of dc voltages and currents Objectives To provide a powerful but simple circuit analysis tool based on Kirchhoff s
More informationCommon Base BJT Amplifier Common Collector BJT Amplifier
Common Base BJT Amplifier Common Collector BJT Amplifier Common Collector (Emitter Follower) Configuration Common Base Configuration Small Signal Analysis Design Example Amplifier Input and Output Impedances
More informationCommonEmitter Amplifier
CommonEmitter Amplifier A. Before We Start As the title of this lab says, this lab is about designing a CommonEmitter Amplifier, and this in this stage of the lab course is premature, in my opinion,
More informationPhysics 623 Transistor Characteristics and Single Transistor Amplifier Sept. 13, 2006
Physics 623 Transistor Characteristics and Single Transistor Amplifier Sept. 13, 2006 1 Purpose To measure and understand the common emitter transistor characteristic curves. To use the base current gain
More informationSuperposition Examples
Superposition Examples The following examples illustrate the proper use of superposition of dependent sources. All superposition equations are written by inspection using voltage division, current division,
More informationLab 4: BJT Amplifiers Part I
Lab 4: BJT Amplifiers Part I Objectives The objective of this lab is to learn how to operate BJT as an amplifying device. Specifically, we will learn the following in this lab: The physical meaning of
More informationCommon Emitter BJT Amplifier Design Current Mirror Design
Common Emitter BJT Amplifier Design Current Mirror Design 1 Some Random Observations Conditions for stabilized voltage source biasing Emitter resistance, R E, is needed. Base voltage source will have finite
More informationSchool of Engineering Department of Electrical and Computer Engineering
1 School of Engineering Department of Electrical and Computer Engineering 332:223 Principles of Electrical Engineering I Laboratory Experiment #4 Title: Operational Amplifiers 1 Introduction Objectives
More informationBJT AC Analysis 1 of 38. The r e Transistor model. Remind Qpoiint re = 26mv/IE
BJT AC Analysis 1 of 38 The r e Transistor model Remind Qpoiint re = 26mv/IE BJT AC Analysis 2 of 38 Three amplifier configurations, Common Emitter Common Collector (Emitter Follower) Common Base BJT
More informationSeries and Parallel Resistive Circuits
Series and Parallel Resistive Circuits The configuration of circuit elements clearly affects the behaviour of a circuit. Resistors connected in series or in parallel are very common in a circuit and act
More informationThe basic cascode amplifier consists of an input commonemitter (CE) configuration driving an output commonbase (CB), as shown above.
Cascode Amplifiers by Dennis L. Feucht Twotransistor combinations, such as the Darlington configuration, provide advantages over singletransistor amplifier stages. Another twotransistor combination
More informationUnit/Standard Number. High School Graduation Years 2010, 2011 and 2012
1 Secondary Task List 100 SAFETY 101 Demonstrate an understanding of State and School safety regulations. 102 Practice safety techniques for electronics work. 103 Demonstrate an understanding of proper
More informationBlock Diagram Reduction
Appendix W Block Diagram Reduction W.3 4Mason s Rule and the SignalFlow Graph A compact alternative notation to the block diagram is given by the signal ow graph introduced Signal ow by S. J. Mason
More information3.4  BJT DIFFERENTIAL AMPLIFIERS
BJT Differential Amplifiers (6/4/00) Page 1 3.4 BJT DIFFERENTIAL AMPLIFIERS INTRODUCTION Objective The objective of this presentation is: 1.) Define and characterize the differential amplifier.) Show the
More informationContent Map For Career & Technology
Content Strand: Applied Academics CTET11 analysis of electronic A. Fractions and decimals B. Powers of 10 and engineering notation C. Formula based problem solutions D. Powers and roots E. Linear equations
More informationIn the previous chapter, it was discussed that a
240 10 Principles of Electronics Single Stage Transistor Amplifiers 10.1 Single Stage Transistor Amplifier 10.2 How Transistor Amplifies? 10.3 Graphical Demonstration of Transistor Amplifier 10.4 Practical
More informationLecture 24: Oscillators. Clapp Oscillator. VFO Startup
Whites, EE 322 Lecture 24 Page 1 of 10 Lecture 24: Oscillators. Clapp Oscillator. VFO Startup Oscillators are circuits that produce periodic output voltages, such as sinusoids. They accomplish this feat
More informationTRANSISTOR AMPLIFIERS AET 8. First Transistor developed at Bell Labs on December 16, 1947
AET 8 First Transistor developed at Bell Labs on December 16, 1947 Objective 1a Identify Bipolar Transistor Amplifier Operating Principles Overview (1) Dynamic Operation (2) Configurations (3) Common Emitter
More informationLaboratory 4: Feedback and Compensation
Laboratory 4: Feedback and Compensation To be performed during Week 9 (Oct. 2024) and Week 10 (Oct. 2731) Due Week 11 (Nov. 37) 1 PreLab This PreLab should be completed before attending your regular
More informationLED level meter driver, 12point, power scale, dot or bar display
LED level meter driver, 12point, power scale, dot or bar display The is a monolithic IC for LED power meter applications. The display level range is 9mVrms to 380mVrms (Typ.) divided into 12 points with
More informationGeneralized Transistor Amplifier
Generalized Transistor Amplifier * Perspectie: look at the arious configurations of bipolar and MOS transistors, for which a smallsignal oltage or current is transformed (e.g., usually amplified  increased
More informationBasic FET Ampli ers 6.0 PREVIEW 6.1 THE MOSFET AMPLIFIER
C H A P T E R 6 Basic FET Ampli ers 6.0 PREVIEW In the last chapter, we described the operation of the FET, in particular the MOSFET, and analyzed and designed the dc response of circuits containing these
More informationLecture 18: Common Emitter Amplifier. Maximum Efficiency of Class A Amplifiers. Transformer Coupled Loads.
Whites, EE 3 Lecture 18 Page 1 of 10 Lecture 18: Common Emitter Amplifier. Maximum Efficiency of Class A Amplifiers. Transformer Coupled Loads. We discussed using transistors as switches in the last lecture.
More informationCurrent mirrors are commonly used for current sources in integrated circuit design. This section covers other current sources that are often seen.
c Coyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Comuter Engineering. Current Sources Current mirrors are commonly used for current sources
More informationAn FET Audio Peak Limiter
1 An FET Audio Peak Limiter W. Marshall Leach, Jr., Professor Georgia Institute of Technology School of Electrical and Computer Engineering Atlanta, Georgia 303320250 USA email: mleach@ee.gatech.edu Copyright
More informationW03 Analysis of DC Circuits. Yrd. Doç. Dr. Aytaç Gören
W03 Analysis of DC Circuits Yrd. Doç. Dr. Aytaç Gören ELK 2018  Contents W01 Basic Concepts in Electronics W02 AC to DC Conversion W03 Analysis of DC Circuits (self and condenser) W04 Transistors and
More informationLABORATORY 2 THE DIFFERENTIAL AMPLIFIER
LABORATORY 2 THE DIFFERENTIAL AMPLIFIER OBJECTIVES 1. To understand how to amplify weak (small) signals in the presence of noise. 1. To understand how a differential amplifier rejects noise and common
More informationChapter 7: AC Transistor Amplifiers
Chapter 7: AC Transistor Amplifiers The transistor amplifiers that we studied in the last chapter have some serious problems for use in AC signals. Their most serious shortcoming is that there is a dead
More informationDiploma in Applied Electronics
DUBLIN INSTITUTE OF TECHNOLOGY KEVIN STREET, DUBLIN 8 Diploma in Applied Electronics YEAR II SUMMER EXAMINATIONS 1999 ELECTRIC CIRCUITS MR. P. Tobin MR. C. Bruce DATE Attempt FIVE questions with a maximum
More informationThe 2N3393 Bipolar Junction Transistor
The 2N3393 Bipolar Junction Transistor CommonEmitter Amplifier Aaron Prust Abstract The bipolar junction transistor (BJT) is a nonlinear electronic device which can be used for amplification and switching.
More informationNotes about Small Signal Model. for EE 40 Intro to Microelectronic Circuits
Notes about Small Signal Model for EE 40 Intro to Microelectronic Circuits 1. Model the MOSFET Transistor For a MOSFET transistor, there are NMOS and PMOS. The examples shown here would be for NMOS. Figure
More informationBasic BJT Amplifiers CHAPTER PREVIEW
CHAPTER Basic BJT Amplifiers In the previous chapter, we described the structure and operation of the bipolar junction transistor, and analyzed and designed the dc response of circuits containing these
More informationProgrammable Single/Dual/Triple Tone Gong SAE 800
Programmable Single/Dual/Triple Tone Gong Preliminary Data SAE 800 Bipolar IC Features Supply voltage range 2.8 V to 18 V Few external components (no electrolytic capacitor) 1 tone, 2 tones, 3 tones
More informationTWO PORT NETWORKS hparameter BJT MODEL
TWO PORT NETWORKS hparameter BJT MODEL The circuit of the basic two port network is shown on the right. Depending on the application, it may be used in a number of different ways to develop different
More informationPhysics 160. Fun with Op Amps. R. Johnson May 13, 2015
Physics 160 Lecture 14 Fun with Op Amps. Johnson May 13, 015 Ideal OpAmp Differential gain, of course. Commonmode gain is ideally zero. Such an ideal opamp of course does not exist, but a first analysis
More informationWHY DIFFERENTIAL? instruments connected to the circuit under test and results in V COMMON.
WHY DIFFERENTIAL? Voltage, The Difference Whether aware of it or not, a person using an oscilloscope to make any voltage measurement is actually making a differential voltage measurement. By definition,
More informationMOSFET DIFFERENTIAL AMPLIFIER (TWOWEEK LAB)
page 1 of 7 MOSFET DIFFERENTIAL AMPLIFIER (TWOWEEK LAB) BACKGROUND The MOSFET is by far the most widely used transistor in both digital and analog circuits, and it is the backbone of modern electronics.
More informationV out. Figure 1: A voltage divider on the left, and potentiometer on the right.
Living with the Lab Fall 202 Voltage Dividers and Potentiometers Gerald Recktenwald v: November 26, 202 gerry@me.pdx.edu Introduction Voltage dividers and potentiometers are passive circuit components
More informationAZ i Z s + Z i. [V s +(V ts + V n + I n Z s )] (1)
Euivalent Noise Input Voltage Figure shows the amplifier noise model with a Thévenin input source, where V s is the source voltage, Z s = s + jx s isthesourceimpedance,v ts is the thermal noise voltage
More informationCHAPTER 10 OPERATIONALAMPLIFIER CIRCUITS
CHAPTER 10 OPERATIONALAMPLIFIER CIRCUITS Chapter Outline 10.1 The TwoStage CMOS Op Amp 10.2 The FoldedCascode CMOS Op Amp 10.3 The 741 OpAmp Circuit 10.4 DC Analysis of the 741 10.5 SmallSignal Analysis
More informationFast analytical techniques for electrical and electronic circuits. Jet Propulsion Laboratory California Institute of Technology
Fast analytical techniques for electrical and electronic circuits Vatché Vorpérian Jet Propulsion Laboratory California Institute of Technology PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE
More informationChapter 5: Analysis of TimeDomain Circuits
Chapter 5: Analysis of TimeDomain Circuits This chapter begins the analysis of circuits containing elements with the ability to store energy: capacitors and inductors. We have already defined each of
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module DC Circuit Lesson 4 Loop Analysis of resistive circuit in the context of dc voltages and currents Objectives Meaning of circuit analysis; distinguish between the terms mesh and loop. To provide
More informationBJT Amplifier Circuits
JT Amplifier ircuits As we have developed different models for D signals (simple largesignal model) and A signals (smallsignal model), analysis of JT circuits follows these steps: D biasing analysis:
More informationRectifier: It is a circuit which employs one or more diodes to convert ac voltage into pulsating dc voltage. We will consider the following circuits:
Rectifier: It is a circuit which employs one or more diodes to convert ac voltage into pulsating dc voltage. We will consider the following circuits: (i) Half wave rectifier. (ii) Full wave rectifier.
More informationBJT Characteristics and Amplifiers
BJT Characteristics and Amplifiers Matthew Beckler beck0778@umn.edu EE2002 Lab Section 003 April 2, 2006 Abstract As a basic component in amplifier design, the properties of the Bipolar Junction Transistor
More informationLecture 21: Junction Field Effect Transistors. Source Follower Amplifier
Whites, EE 322 Lecture 21 Page 1 of 8 Lecture 21: Junction Fiel Effect Transistors. Source Follower Amplifier As mentione in Lecture 16, there are two major families of transistors. We ve worke with BJTs
More informationCOMMONSOURCE JFET AMPLIFIER
EXPERIMENT 04 Objectives: Theory: 1. To evaluate the commonsource amplifier using the small signal equivalent model. 2. To learn what effects the voltage gain. A selfbiased nchannel JFET with an AC
More informationChapter 8 Differential and Multistage Amplifiers. EE 3120 Microelectronics II
1 Chapter 8 Differential and Multistage Amplifiers Operational Amplifier Circuit Components 2 1. Ch 7: Current Mirrors and Biasing 2. Ch 9: Frequency Response 3. Ch 8: ActiveLoaded Differential Pair 4.
More informationElectronics. Discrete assembly of an operational amplifier as a transistor circuit. LD Physics Leaflets P4.2.1.1
Electronics Operational Amplifier Internal design of an operational amplifier LD Physics Leaflets Discrete assembly of an operational amplifier as a transistor circuit P4.2.1.1 Objects of the experiment
More informationJunction Field Effect Transistor (JFET)
Junction Field Effect Transistor (JFET) The single channel junction fieldeffect transistor (JFET) is probably the simplest transistor available. As shown in the schematics below (Figure 6.13 in your text)
More informationA Simple CurrentSense Technique Eliminating a Sense Resistor
INFINITY Application Note AN7 A Simple CurrentSense Technique Eliminating a Sense Resistor Copyright 998 A SIMPE CURRENTSENSE TECHNIQUE EIMINATING A SENSE RESISTOR INTRODUCTION A sense resistor R S,
More informationBasic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati
Basic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati Module: 2 Bipolar Junction Transistors Lecture2 Transistor
More informationLecture 30: Biasing MOSFET Amplifiers. MOSFET Current Mirrors.
Whites, EE 320 Lecture 30 Page 1 of 8 Lecture 30: Biasing MOSFET Amplifiers. MOSFET Current Mirrors. There are two different environments in which MOSFET amplifiers are found, (1) discrete circuits and
More informationFeedback Amplifier Configurations
12 eedback Amplifier Configurations John Choma, Jr. University of Southern California 12.1 ntroduction... 121 12.2 SeriesShunt eedback Amplifier... 121 Circuit Modeling and Analysis eedorward Compensation
More informationELECTRONICS. EE 42/100 Lecture 8: OpAmps. Rev C 2/8/2012 (9:54 AM) Prof. Ali M. Niknejad
A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 8 p. 1/23 EE 42/100 Lecture 8: OpAmps ELECTRONICS Rev C 2/8/2012 (9:54 AM) Prof. Ali M. Niknejad University of California, Berkeley
More informationProblem set #5 EE 221, 09/26/ /03/2002 1
Chapter 3, Problem 42. Problem set #5 EE 221, 09/26/2002 10/03/2002 1 In the circuit of Fig. 3.75, choose v 1 to obtain a current i x of 2 A. Chapter 3, Solution 42. We first simplify as shown, making
More informationMAS.836 HOW TO BIAS AN OPAMP
MAS.836 HOW TO BIAS AN OPAMP OpAmp Circuits: Bias, in an electronic circuit, describes the steady state operating characteristics with no signal being applied. In an opamp circuit, the operating characteristic
More information30. Bode Plots. Introduction
0. Bode Plots Introduction Each of the circuits in this problem set is represented by a magnitude Bode plot. The network function provides a connection between the Bode plot and the circuit. To solve these
More informationMaking Accurate Voltage Noise and Current Noise Measurements on Operational Amplifiers Down to 0.1Hz
Author: Don LaFontaine Making Accurate Voltage Noise and Current Noise Measurements on Operational Amplifiers Down to 0.1Hz Abstract Making accurate voltage and current noise measurements on op amps in
More informationDesigning Stable Compensation Networks for Single Phase Voltage Mode Buck Regulators
Designing Stable Compensation Networks for Single Phase Voltage Mode Buck Regulators Technical Brief December 3 TB47. Author: Doug Mattingly Assumptions This Technical Brief makes the following assumptions:.
More informationOBJECTIVE QUESTIONS IN ANALOG ELECTRONICS
1. The early effect in a bipolar junction transistor is caused by (a) fast turnon (c) large collectorbase reverse bias (b)fast turnoff (d) large emitterbase forward bias 2. MOSFET can be used as a
More informationCHAPTER 28 ELECTRIC CIRCUITS
CHAPTER 8 ELECTRIC CIRCUITS 1. Sketch a circuit diagram for a circuit that includes a resistor R 1 connected to the positive terminal of a battery, a pair of parallel resistors R and R connected to the
More informationObjectives The purpose of this lab is build and analyze Differential amplifiers based on NPN transistors (or NMOS transistors).
1 Lab 03: Differential Amplifiers (BJT) (20 points) NOTE: 1) Please use the basic current mirror from Lab01 for the second part of the lab (Fig. 3). 2) You can use the same chip as the basic current mirror;
More informationElectronics Prof. D.C. Dube Department of Physics Indian Institute of Technology, Delhi
Electronics Prof. D.C. Dube Department of Physics Indian Institute of Technology, Delhi Module No. #06 Power Amplifiers Lecture No. #01 Power Amplifiers (Refer Slide Time: 00:44) We now move to the next
More informationBJT Amplifier Circuits
JT Amplifier ircuits As we have developed different models for D signals (simple largesignal model) and A signals (smallsignal model), analysis of JT circuits follows these steps: D biasing analysis:
More informationA LowCost VCA Limiter
The circuits within this application note feature THAT218x to provide the essential function of voltagecontrolled amplifier (VCA). Since writing this note, THAT has introduced a new dual VCA, as well
More informationVoltage/current converter opamp circuits
Voltage/current converter opamp circuits This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/,
More informationBasic Op Amp Circuits
Basic Op Amp ircuits Manuel Toledo INEL 5205 Instrumentation August 3, 2008 Introduction The operational amplifier (op amp or OA for short) is perhaps the most important building block for the design of
More informationHomework Assignment 03
Question 1 (2 points each unless noted otherwise) Homework Assignment 03 1. A 9V dc power supply generates 10 W in a resistor. What peaktopeak amplitude should an ac source have to generate the same
More informationChapter 11 Current Programmed Control
Chapter 11 Current Programmed Control Buck converter v g i s Q 1 D 1 L i L C v R The peak transistor current replaces the duty cycle as the converter control input. Measure switch current R f i s Clock
More informationLM118/LM218/LM318 Operational Amplifiers
LM118/LM218/LM318 Operational Amplifiers General Description The LM118 series are precision high speed operational amplifiers designed for applications requiring wide bandwidth and high slew rate. They
More informationEE4232 Review of BJTs, JFETs and MOSFETs
EE4232 Review of BJTs, JFETs and MOSFETs 0 A simplified structure of the npn transistor. 1 A simplified structure of the pnp transistor. 2 Current flow in an npn transistor biased to operate in the active
More informationReading: HH Sections 4.11 4.13, 4.19 4.20 (pgs. 189212, 222 224)
6 OP AMPS II 6 Op Amps II In the previous lab, you explored several applications of op amps. In this exercise, you will look at some of their limitations. You will also examine the op amp integrator and
More informationExperiment 8 SeriesParallel Circuits
Experiment 8 SeriesParallel Circuits EL 111  DC Fundamentals By: Walter Banzhaf, E.K. Smith, and Winfield Young University of Hartford Ward College of Technology Objectives: 1. For the student to measure
More informationLecture 060 PushPull Output Stages (1/11/04) Page 0601. ECE 6412  Analog Integrated Circuits and Systems II P.E. Allen  2002
Lecture 060 PushPull Output Stages (1/11/04) Page 0601 LECTURE 060 PUSHPULL OUTPUT STAGES (READING: GHLM 362384, AH 226229) Objective The objective of this presentation is: Show how to design stages that
More informationwww.jameco.com 18008314242
Distributed by: www.jameco.com 18008314242 The content and copyrights of the attached material are the property of its owner. LF411 Low Offset, Low Drift JFET Input Operational Amplifier General Description
More informationDesigning a Poor Man s Square Wave Signal Generator. EE100 Lab: Designing a Poor Man s Square Wave Signal Generator  Theory
EE100 Lab:  Theory 1. Objective The purpose of this laboratory is to introduce nonlinear circuit measurement and analysis. Your measurements will focus mainly on limiters and clamping amplifiers. During
More informationMATERIALS. Multisim screen shots sent to TA.
Page 1/8 Revision 0 9Jun10 OBJECTIVES Learn new Multisim components and instruments. Conduct a Multisim transient analysis. Gain proficiency in the function generator and oscilloscope. MATERIALS Multisim
More informationLecture Notes: ECS 203 Basic Electrical Engineering Semester 1/2010. Dr.Prapun Suksompong 1 June 16, 2010
Sirindhorn International Institute of Technology Thammasat University School of Information, Computer and Communication Technology Lecture Notes: ECS 203 Basic Electrical Engineering Semester 1/2010 Dr.Prapun
More informationBipolar Transistor Amplifiers
Physics 3330 Experiment #7 Fall 2005 Bipolar Transistor Amplifiers Purpose The aim of this experiment is to construct a bipolar transistor amplifier with a voltage gain of minus 25. The amplifier must
More informationPHYS 210 Electronic Circuits and Feedback
PHYS 210 Electronic Circuits and Feedback These notes give a short introduction to analog circuits. If you like to learn more about electronics (a good idea if you are thinking of becoming an experimental
More informationHOW TO USE MULTIMETER. COMPILE BY: Dzulautotech
HOW TO USE MULTIMETER COMPILE BY: Dzulautotech 1. GENERAL Electricity is absolutely necessary for an automobile. It is indispensable when the engine is started, the air fuel mixture is ignited and exploded,
More informationPositive Feedback and Oscillators
Physics 3330 Experiment #6 Fall 1999 Positive Feedback and Oscillators Purpose In this experiment we will study how spontaneous oscillations may be caused by positive feedback. You will construct an active
More informationTransistor Amplifiers
Physics 3330 Experiment #7 Fall 1999 Transistor Amplifiers Purpose The aim of this experiment is to develop a bipolar transistor amplifier with a voltage gain of minus 25. The amplifier must accept input
More informationDifferential Amplifier Offset. Causes of dc voltage and current offset Modeling dc offset R C
ESE39 ntroduction to Microelectronics Differential Amplifier Offset Causes of dc voltage and current offset Modeling dc offset mismatch S mismatch β mismatch transistor mismatch dc offsets in differential
More information