Feedback Ampliﬁers - Georgia Institute of …brewer.ece.gatech.edu/ece4043/Lecture_Notes/duck...feedback ampliﬁers are shown in Fig. 1. Depending on the type of feedback, the variables

c° Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School ofElectrical and Computer Engineering.

Feedback Amplifiers

Collection of Solved Problems

A collection of solved feedback amplifier problems can be found at the below link. The solutionsare based on the use of the Mason Flow Graph described below.

http://users.ece.gatech.edu/~mleach/ece3050/notes/feedback/FBExamples.pdf

Basic Description of Feedback

A feedback amplifier is one in which the output signal is sampled and fed back to the input to forman error signal that drives the amplifier. The basic block diagrams of non-inverting and invertingfeedback amplifiers are shown in Fig. 1. Depending on the type of feedback, the variables x, y,and z are voltages or currents. The diagram in Fig. 1(a) represents a non-inverting amplifier. Thesumming junction at its input subtracts the feedback signal from the input signal to form the errorsignal z = x − by which drives the amplifier. If the amplifier has an inverting gain, the feedbacksignal must be added to the input signal in order for the feedback to be negative. This is illustratedin Fig. 1(b). The summing junction at the input adds the feedback signal to the input signal toform the error signal z = x+ by. In both diagrams, the gain around the loop is negative and equalto −bA, where both A and b are positive real constants. Because the loop-gain is negative, thefeedback is said to be negative. If the gain around the loop is positive, the amplifier is said to havepositive feedback which causes it to be unstable.

Figure 1: Feedback amplifier block diagrams. (a) Non-inverting. (b) Inverting.

In the non-inverting amplifier of Fig. 1(a), the error signal is given by z = x− by. The outputsignal can be written.

y = Az = A (x− by) (1)

This can be solved for the gain to obtain

y

x=

A

1 + bA(2)

We see that the effect of the feedback is to reduce the gain by the factor (1 + bA). This factor iscalled the “amount of feedback”. It is often specified in dB by the relation 20 log |1 + bA|.

In the inverting amplifier of Fig. 1(b), the error signal is given by z = x + by. When x goespositive, y goes negative, so that the error signal represents a difference signal. The output signalcan be written

y = −Az = −A (x+ by) (3)

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This can be solved for the gain to obtain

y

x=−A1 + bA

(4)

We see that the amount of feedback for the inverting amplifier is the same as for the non-invertingamplifier.

If A is large enough so that bA >> 1, the gain of the non- inverting amplifier given by Eq. (2)can be approximated by

y

x' A

bA=1

b(5)

The gain of the inverting amplifier given by Eq. (4) can be approximated by

y

x' −A

bA= −1

b(6)

These are important results, for they show that the gain is set by the feedback network and not bythe amplifier. In practice, this means that an amplifier without feedback can be designed withouttoo much consideration of what its gain will be as long as the gain is high enough. When feedbackis added, the gain can be reduced to any desired value by the feedback network.

The product bA in Eqs. (2) and (4) must be dimensionless. Thus if A is a voltage gain (voltagein-voltage out) or a current gain (current in-current out), then b must be dimensionless. If A isa transconductance gain (voltage in-current out), b must have the units of ohms (Ω). If A is atransresistance gain (current in-voltage out), b must have the units siemens (f). How these aredetermined is illustrated below.

We have assumed so far that the gains A and b are positive real constants. In general, thegains are phasor functions of frequency. This leads to a stability problem in feedback amplifiers.As frequency is increased, |A| must eventually decrease because no amplifier can have an infinitebandwidth. The decrease in |A| is accompanied with a phase shift so that bA can be equal to anegative real number at some frequency. Suppose that bA = −1 at some frequency. Eqs. (3) and(4) show that the gain becomes infinite at that frequency. An amplifier with an infinite gain at anyfrequency can put out a signal at that frequency with no input signal. In this case, the amplifieris said to oscillate. It can be shown that an amplifier will oscillate if |bA| ≥ 1 at any frequency forwhich bA is a negative real number, i.e. the phase of bA is ±180.

In the block diagrams of Fig. 1, the input and output variables can be modeled as either avoltage or a current. It follows that there are four combinations of inputs and outputs that representthe possible types of feedback. These are summarized in Table 1 where the names for each aregiven. These names come from the way that the feedback network connects between the input andoutput stages. This is explained in the following for each type of feedback.

Table 1. The Four Types of Feedback

NameInput

Variable xOutputVariable y

ErrorVariable z

ForwardGain A

FeedbackFactor b

Series-Shunt Voltage v Voltage v Voltage v Voltage Gain DimensionlessShunt-Shunt Current i Voltage v Current i Transresistance siemens (f)Series-Series Voltage v Current i Voltage v Transconductance ohms (Ω)Shunt-Series Current i Current i Current i Current Gain Dimensionless

The Mason Signal-Flow Graph

The analysis of feedback amplifiers is facilitated by the use of the Mason signal-flow graph. Asignal-flow graph is a graphical representation of a set of linear equations which can be used to

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write by inspection the solution to the set of equations. For example, consider the set of equations

x2 = Ax1 +Bx2 +Cx5 (7)

x3 = Dx1 +Ex2 (8)

x4 = Fx3 +Gx5 (9)

x5 = Hx4 (10)

x6 = Ix3 (11)

where x1 through x6 are variables and A through I are constants. These equations can be rep-resented graphically as shown in Fig. 2. The graph has a node for each variable with branchesconnecting the nodes labeled with the constants A through I. The node labeled x1 is called a sourcenode because it has only outgoing branches. The node labeled x6 is called a sink node because ithas only incoming branches. The path from x1 to x2 to x3 to x6 is called a forward path becauseit originates at a source node and terminates at a non-source node and along which no node isencountered twice. The path gain for this forward path is AEI. The path from x2 to x3 to x4 tox5 is called a feedback path because it originates and terminates on the same node and along whichno node is encountered more than once. The loop gain for this feedback path is EFHC.

Figure 2: Flow graph for the equations.

Mason’s formula can be used to calculate the transmission gain from a source node to anynon-source node in a flow graph. The formula is

T =1

∆

Xk

Pk∆k (12)

where Pk is the gain of the kth forward path, ∆ is the graph determinant, and∆k is the determinantwith the kth forward path erased. The determinant is given by

∆ = 1− (sum of all loop gains)

+

µsum of the gain products of all possiblecombinations of two non-touching loops

¶−µ

sum of the gain products of all possiblecombinations of three non-touching loops

¶+

µsum of the gain products of all possiblecombinations of four non-touching loops

¶− · · · (13)

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For the flow graph in Fig. 2, the objective is to solve for the gain from node x1 to node x6.There are two forward paths from x1 to x6 and three loops. Two of the loops do not touch eachother. Thus the product of these two loop gains appears in the expression for ∆. The path gainsand the determinant are given by

P1 = AEI (14)

P2 = DI (15)

∆ = 1− (B + CEFH +GH) +B ×GH (16)

Path P1 touches two loops while path P2 touches one loop. The determinants with each path erasedare given by

∆1 = 1−GH (17)

∆2 = 1− (B +GH) +B ×GH (18)

Thus the overall gain from x1 to x6 is given by

x6x1=

AEI × (1−GH) +DI × [1− (B +GH) +B ×GH]

1− (B + CEFH +GH) +B ×GH(19)

This equation can also be obtained by algebraic solution of the equations in Eqs. (7) through (11).

Review of Background Theory

This section summarizes several BJT small-signal ac equivalent circuits which are used to writethe circuit equations in the following sections. Fig. 3(a) shows an npn BJT with Thévenin sourcesconnected to its base and emitter and a load resistor connected to its collector. First we definethe emitter intrinsic resistance re, the collector-emitter resistance r0, the resistance r0e, and thetransconductance Gm. These are given by

re =VTIE

(20)

r0 =VCE + VA

IC(21)

r0e =Rtb + rx1 + β

+ re (22)

Gm =a

r0e +Rte(23)

where VT is the thermal voltage, IE is the emitter bias current, VCE is the quiescent collector-emittervoltage, VA is the Early voltage, IC is the quiescent collector current, rx is the base spreadingresistance, β is the base-collector current gain, and α is the emitter-collector current gain. Thelatter two are related by α = β/ (1 + β) and β = α/ (1− α).

The small-signal ac Thévenin equivalent circuit seen looking into the base is shown in Fig. 3(b),where vb(oc) and rib are given by

vb(oc) = vter0 +Rtc

Rte + r0 +Rtc(24)

rib = rx + (1 + β) re +Rte(1 + β) r0 +Rtc

r0 +Rte +Rtc(25)

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Figure 3: Small-signal equivalent circuits.

The small-signal ac Thévenin equivalent circuit seen looking into the emitter is shown in Fig. 3(c),where ve(oc) and rie are given by

ve(oc) = vtbr0 +Rtc/ (1 + β)

r0e + r0 +Rtc/ (1 + β)(26)

rie = r0er0 +Rtc

rie + r0 +Rtc/ (1 + β)(27)

The small-signal ac Norton equivalent circuit seen looking into the collector is shown in Fig. 3(d),where ic(sc) and ric are given by

ic(sc) = Gmbvtb −Gmevte (28)

ric =r0 + r0ekRte

1−GmRte(29)

The transconductances Gmb and Gme are given by

Gmb = Gmr0 −Rte/β

r0 + r0ekRte(30)

Gme = Gmr0 + r0e/α

r0 + r0ekRte(31)

The r0 Approximations

In some of the examples that follow, the analysis is simplified by making use of the so-called r0approximations. That is, we assume that r0 → ∞ in all equations except in the one for ric. Thisassumption makes Gmb, Gme, rib, and rie independent of r0. In addition, it makes Gmb = Gme sothat we can denote Gmb = Gme = Gm. In this case,

ic(sc) = Gm (vtb − vte) =α

r0e +Rte(vtb − vte) (32)

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vb(oc) = vte (33)

rib = rx + (1 + β) (re +Rte) (34)

ve(oc) = vtb (35)

rie = r0e (36)

ib =ic(sc)

β(37)

ie =ic(sc)

α(38)

If vte = 0 and Rre = 0, r0 appears as a resistor from the collector to ground. If Rtc = 0, r0 appearsas a resistor from emitter to ground. In either case, r0 can be easily included in the analysis bytreating it as an external resistor from either the emitter or the collector to ground.

Series-Shunt Feedback

A series-shunt feedback amplifier is a non-inverting amplifier in which the input signal x is a voltageand the output signal y is a voltage. If the input source is a current source, it must be convertedinto a Thévenin source for the gain to be in the form of Eq. (2). Because the input is a voltageand the output is a voltage, the gain A represents a dimensionless voltage gain. Because bA mustbe dimensionless, the feedback factor is also dimensionless. Two examples are given below. Thefirst is an op-amp example. The second is a BJT example.

Op Amp Example

Fig. 4(a) shows an op amp with a feedback network consisting of a voltage divider connectedbetween its output and inverting input. The input signal is connected to the non-inverting input.Because the feedback does not connect to the same terminal as the input signal, the summing isseries. The feedback network connects in shunt with the output node, thus the sampling is shunt.

To analyze the circuit, we replace the circuit seen looking out of the op-amp inverting inputwith a Thévenin equivalent circuit with respect to vo and the circuit seen looking into the feedbacknetwork from the vo node with a Thévenin equivalent circuit with respect to i1. We replace theop amp with a simple controlled source model which models the differential input resistance, theopen-loop voltage gain, and the output resistance. A test source it is added at the output in orderto calculate the output resistance. The circuit is shown in Fig. 4(b), where Rid is the differentialinput resistance, A0 is the open-loop gain, R0 is the output resistance of the op amp. The feedbackfactor b is given by

b =R1

R1 +RF(39)

The error signal z in Fig. 4(b) is a voltage which we denote by ve. It is the difference betweenthe two voltage sources in the input circuit and is given by

ve = vs − bvo (40)

By voltage division, the voltage vi which controls the op amp output voltage is

vi = veRid

RS +Rid +R1kRF(41)

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Figure 4: (a) Series-shunt op-amp example. (b) Circuit with feedback removed.

Signal tracing shows that the negative feedback has the effect of making the current i1 smaller. Forthis reason, we will neglect the i1R1 source in the output circuit in calculating vo. By superposition,it follows that vo can be written

vo = A0vi(R1 +RF ) kRL

R0 + (R1 +RF ) kRL+ itR0k (R1 +RF ) kRL (42)

To calculate the input resistance, we need the current i1. It is given by

i1 =viRid

(43)

To simplify the equations, let us define

k1 =Rid

RS +Rid +R1kRF(44)

k2 =(R1 +RF ) kRL

R0 + (R1 +RF ) kRL(45)

Req = R0k (R1 +RF ) kRL (46)

The circuit equations can be rewrittenvi = k1ve (47)

vo = k2A0vi + itReq (48)

The flow graph for these equations is shown in Fig. 5. The determinant is given by

∆ = 1− [k1k2A0 (−b)] = 1 + bk1k2A0 = 1 + bA (49)

From the flow graph, the voltage gain is given by

vovs=1

∆k1k2A0 =

k1k2A01 + bk1k2A0

=A

1 + bA(50)

It follows that A is given byA = k1k2A0 (51)

This is the gain from vs to vo with b = 0. If bAÀ 1, the gain approaches

vovs→ A

bA=1

b= 1 +

RF

R1(52)

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Figure 5: Flow graph for the series-shunt op-amp example.

This is the familiar formula for the gain of the non-inverting op amp.From the flow graph, the output resistance is given by

rout =voit=1

∆Req =

R0k (R1 +RF ) kRL

1 + bA(53)

Similarly, the input resistance is given by

rin =

µi1vs

¶−1=

µ1

∆

k1Rid

¶−1= (1 + bA) (RS +Rid +R1kRF ) (54)

Note that the voltage gain and the output resistance are decreased by the feedback. The inputresistance is increased by the feedback. These are properties of the series-shunt topology.

Transistor Example

The ac signal circuit of an example BJT series-shunt feedback amplifier is shown in Fig. 6. Weassume that the dc solutions to the circuit are known. The feedback network is in the form of avoltage divider and consists of resistors RF1 and RF2. Because the input to the feedback networkconnects to the vo node, the amplifier is said to employ shunt sampling at the output. The outputof the feedback network connects to the emitter of Q1. Because this is not the node to which vsconnects, the circuit is said to have series summing at the input. The following analysis assumesthe r0 approximations for Q1. That is, we neglect r0 in all equations except when calculating ric1.

Figure 6: Series-shunt amplifier.

In order for the amplifier to have negative feedback, the voltage gain from the emitter of Q1 tothe collector of Q2 must be inverting. When the feedback signal is applied to its emitter, Q1 is a

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common-base stage which has a non-inverting voltage gain. Q2 is a common-emitter stage whichhas an inverting gain. Thus the amplifier has an inverting voltage gain from the emitter of Q1 tothe collector of Q2 so that the feedback is negative.

To remove the feedback, we replace the circuit seen looking out of the emitter of Q1 with aThévenin equivalent circuit with respect to vo. The circuit seen looking into RF1 from the vo nodeis replaced with a Thévenin equivalent circuit with respect to ie1. Fig. 7 shows the circuit. A atest current source is connected to the vo node to calculate rout. The Thévenin voltage looking outof the emitter of Q1 is given by

vte1 = voR1

R1 +RF= bvo (55)

where b is the feedback factor given by

b =R1

R1 +RF(56)

Figure 7: Circuit with feedback removed.

For the circuit of Fig. 7, the error voltage ve is given by

ve = vs − bvo (57)

Signal tracing shows that the negative feedback has the effect of making the current ie1 smaller.For this reason, we will neglect it in calculating vo. To circuit equations are

ic1(sc) = Gm1 (vs − bvo) = Gm1ve (58)

vtb2 = −ic1(sc)ric1kRC1 (59)

Rtb2 = ric1kRC1 (60)

ic2(sc) = −Gmb2vtb2 (61)

vo =£ic2(sc) + it

¤× ric2kRC2k (RF1 +RF2) (62)

To solve for rin, we need ib1. If we use the r0 approximations, it is given by

ib1 =ic1(sc)

β1(63)

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To simplify the flow graph, let us define

Req1 = ric1kRC1 (64)

Req2 = ric2kRC2k (RF1 +RF2) (65)

The flow graph for the equations is shown in Fig. 8. The determinant is given by

∆ = 1− [Gm1 (−Req1) (−Gm2)Req2 (−b)] = 1 + bA (66)

where A is given byA = Gm1Req1Gm2Req2 (67)

This is the gain from vs to vo with b = 0. From the flow graph, the voltage gain is given by

vovs=1

∆Gm1Req1Gm2Req2 =

A

1 + bA(68)

The output resistance is given by

rout =voit=1

∆Req2 =

Req2

1 + bA(69)

Figure 8: Flow graph for the series-shunt example.

The input resistance is given by rin = vs/ib1. Because vs is an independent variable, it can beused to solve for ib1/vs not vs/ib1. Thus the input resistance can be written

rin =

µib1vs

¶−1=

µ1

∆

Gm1

β1

¶−1= ∆

β1Gm1

(70)

But β1/Gm1 is given by

β1Gm1

= β1r0e1 +Rte1

α1= RS + rx1 + (1 + β1) (re1 +Rte1) (71)

Thus rin is given byrin = (1 + bA) [RS + rx1 + (1 + β1) (re1 +Rte1)] (72)

Summary of the Effects of Series-Shunt Feedback

From the examples above, it can be seen that the voltage gain is divided by the factor (1 + bA).The input resistance is multiplied by the factor (1 + bA). And the output resistance is divided bythe factor (1 + bA).

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Shunt-Shunt Feedback

A shunt-shunt feedback amplifier is an inverting amplifier in which the input signal x is a currentand the output signal y is a voltage. If the input source is a voltage source, it must be convertedinto a Norton source for the gain to be in the form of Eq. (4). Because the input is a current andthe output is a voltage, the gain A represents a transresistance with the units Ω. Because bA mustbe dimensionless, the feedback factor has the units of f. Two examples are given below. The firstis an op-amp example. The second is a BJT example.

Op Amp Example

Fig. 9(a) shows an op amp with a feedback network consisting of a resistor connected between itsoutput and its inverting input. The input signal is connected through a resistor to the invertinginput. Because the feedback connects to the same terminal as the input signal, the summing isshunt. The feedback network connects in shunt with the output node, thus the sampling is shunt.

To analyze the circuit, we replace the circuit seen looking out of the vi node through RS with aNorton equivalent circuit with respect to vs and the circuit seen looking out of the vi node throughRF with a Norton equivalent circuit with respect to vo. This must always be done with the shunt-shunt amplifier in order for the gain with feedback to be of the form −A/ (1 + bA), where A and bare positive and the − sign is necessary because the circuit has an inverting gain. In addition, wereplace the circuit seen looking out of the vo node through RF with a Thévenin equivalent circuitwith respect to vi.

The circuit with feedback removed is shown in Fig. 9(b), where r0in is the input resistance seenby the source current is. The Norton current seen looking out of the vi node is represented by thebvo source, where b is the feedback factor. The current is and the feedback factor are given by

is =vsRS

(73)

b =1

RF(74)

The error current ie is the total current delivered to the vi node. It is given by

ie = is + bvo (75)

Because vo is negative when is is positive, the current bvo subtracts from is to cause ie to bedecreased.

Figure 9: (a) Shunt-shunt op amp circuit. (b) Circuit with feedback removed.

Fig. 10 shows the circuit with the op amp replaced with a controlled source model which modelsthe differential input resistance Rid, the open-loop voltage gain A0, and the output resistance R0.

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A test source it is added at the output in order to calculate the output resistance. The voltage vicontrols the op-amp output voltage. It is given by

vi = (is + bvo)RSkRF kRid = ieRSkRFkRid (76)

Signal tracing shows that the negative feedback has the effect of making the voltage vi smaller. Forthis reason, we will neglect the vi source in calculating vo. It follows that vo is given by

vo = −A0viRFkRL

R0 +RFkRL+ itR0kRFkRL (77)

Figure 10: Shunt-shunt circuit with the op amp replaced with a controlled source model.


Req1 = RSkRFkRid (78)

Req2 = R0kRFkRL (79)

k =RFkRL

R0 +RFkRL(80)

The circuit equations can be rewritten

ie = is + bvo (81)

vi = ieReq1 (82)

vo = −kA0vi + itReq2 (83)


∆ = 1− [Req1 (−A0k) b] = 1 + bReq1A0k (84)

From the flow graph, the transresistance gain is given by

vois=1

∆(−Req1A0k) =

−Req1A0k

1 + bReq1A0k=−A1 + bA

(85)

It follows that A is given byA = Req1A0k (86)

If bAÀ 1, the transresistance gain approaches

vois→ −A

bA= −1

b= −RF (87)

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Figure 11: Flow graph for the shunt-shunt amplifier.

We note that A is the negative of the gain from is to vo calculated with b = 0. Also, −bA is theloop gain in the flow graph.

From the flow graph, the output resistance is given by

rout =voit=1

∆Req2 =

ROkRFkRL

1 + bA(88)

Similarly, the input resistance is given by

r0in =viis=1

∆Req1 =

RSkRF kRid

1 + bA(89)

We note that Req2 is the output resistance with the vi source zeroed at the output. Also, Req1 isthe input resistance with the bvo source zeroed at the input.

The voltage gain of the original circuit in Fig. 9(a) is given by

vovs=

isvs

vois=

1

RS

−A1 + bA

' 1

RS

−AbA

= −RF

RS(90)

where the approximation holds for bA À 1. This is the familiar gain expression for the invertingop amp amplifier. The input resistance is obtained from r0in with the relation

rin = RS +

µ1

r0in− 1

RS

¶−1(91)

Transistor Example

Fig. 12 shows the ac signal circuit of an example BJT shunt-shunt feedback amplifier. The feedbacknetwork consists of the resistor RF which connects between the output and input nodes. BecauseRF connects to the output node, the amplifier is said to have shunt sampling. Because the currentfed back through RF to the input node combines in parallel with the source current, the circuit issaid to have shunt summing. Thus the amplifier is said to have shunt-shunt feedback. In order forthe feedback to be negative, the voltage gain from vi to vo must be inverting. Q1 is a common-emitter stage which has an inverting gain. Q2 is a common-collector stage which has a non-invertinggain. Thus the overall voltage gain is inverting so that the feedback is negative.

Fig. 13 shows the equivalent circuit with feedback removed. The circuits seen looking out ofthe vi node through RS and through RF , respectively, are converted into Norton equivalent circuitswith respect to vs and vo. The source current is and the feedback factor b are given by

is =vsRS

(92)

b =1

RF(93)

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Figure 12: Example BJT shunt-shunt amplifier.

Figure 13: Shunt-shunt amplifier with feedback removed.

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The feedback network at the output is modeled by a Thévenin equivalent circuit with respect tovi. The external current source it is added to the circuit so that the output resistance can becalculated. Signal tracing shows that the input voltage vi is reduced by the feedback. Therefore,we neglect the effect of the vi controlled source in the output circuit when calculating vo.

The circuit equations arevtb1 = ieRSkRF = ieReq1 (94)

ie = is + bvo (95)

vi = ieRSkRFkrib1 = ieReq2 (96)

ic1(sc) = Gmb1vtb1 (97)

vtb2 = −ic1(sc)ric1kRC1 = −ic1(sc)Req3 (98)

ve2(oc) =r02

r0e2 + r02vtb2 = k1vtb2 (99)

vo =RE2kRF

rie2 +RE2kRFve2(oc) + it × rie2kRE2kRF = k2ve2(oc) + itReq4 (100)

The flow graph for the equations is shown in Fig. 14. The determinant is given by

∆ = 1− [Req1Gmb1 (−Req3) k1k2b] = 1 + bReq1Gmb1Req3k1k2 (101)

The transresistance gain is given by

vois=1

∆Req1Gmb1 (−Req3) k1k2 =

−Req1Gmb1Req3k1k21 + bReq1Gmb1Req3k1k2

=−A1 + bA

(102)

It follows that A is given byA = Req1Gmb1Req3k1k2 (103)

The input and output resistances are given by

r0in =viis=

Req2

∆=

R1kRFkrib1∆

(104)

rout =voit=

Req4

∆=

rie2kRE2kRF

∆(105)

It can be seen from these expressions that the transresistance gain, the input resistance, and theoutput resistance are all decreased by a factor equal to the amount of feedback.

Figure 14: Flow graph for the shunt-shunt amplifier.

The voltage gain of the original circuit in Fig. 12 is given by

vovs=

isvs

vois=

1

RS

−A1 + bA

' 1

RS

−AbA

= −RF

RS(106)

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where the approximation holds for bA À 1. This is the familiar gain expression for the invertingop amp amplifier. The input resistance is obtained from r0in with the relation

rin = RS +

µ1

r0in− 1

RS

¶−1(107)

Summary of the Effects of Shunt-Shunt Feedback

Notice from the examples above that the transresistance gain and the voltage gain are divided bythe factor (1 + bA). The input resistance r0in is divided by the factor (1 + bA). And the outputresistance is divided by the factor (1 + bA).

Series-Series Feedback

A series-series feedback amplifier is a non-inverting amplifier in which the input signal x is a voltageand the output signal y is a current. If the input source is a current source, it must be convertedinto a Thévenin source for the gain to be in the form of Eq. (2). Because the input is a voltage andthe output is a current, the gain A represents a transconductance with the units f. Because bAmust be dimensionless, the feedback factor has the units of Ω. An op-amp example is given below.

Fig. 15(a) shows an op-amp circuit in which a resistor R1 is in series with the load resistorRL. The voltage across R1 is fed back into the inverting op-amp input. The voltage across R1 isproportional to the output current io. This is said to be series sampling at the output. Becausethe feedback does not connect to the same op-amp input as the source, the circuit is said to haveseries summing. Thus the circuit is called a series-series feedback amplifier.

Figure 15: (a) Example series-series feedback amplifier. (b) Circuit with feedback removed.

To remove the feedback, the circuit seen looking out of the op-amp inverting input is replacedwith a Thévenin equivalent circuit with respect to io. The circuit seen looking below RL is replacedwith a Thévenin equivalent circuit with respect to i1. The circuit with feedback removed is shown inFig. 15(b), where the op amp is replaced with a controlled source model that models its differentialinput resistance Rid, its gain A0, and its output resistance R0. The feedback factor b is given by

b = R1

A test voltage source vt is added in series with RL to calculate the output resistance. We definethe resistance rout as the effective series resistance in the output circuit, including RL. It is givenby rout = vt/io. The output resistance seen by RL is given by r0out = rout − RL. Because RL isfloating, it is not possible to label rout on the diagram.

16

Signal tracing shows that the negative feedback has the effect of reducing i1. For this reason,the i1R1 source in the output circuit will be neglected in solving for io. We can write the followingequations

ve = vs − bio (108)

vi =Rid

RS +Rid +R1ve (109)

io =A0vi

R0 +RL +R1+

vtR0 +RL +R1

(110)

i1 =viRid

(111)


k =Rid

RS +Rid +R1(112)

Req = R0 +RL +R1 (113)

The circuit equations can be rewrittenvi = kve (114)

io =A0viReq

+vtReq

(115)


∆ = 1−∙kA0Req

(−b)¸= 1 + bk

A0Req

= 1 + bA (116)

From the flow graph, the transconductance gain is given by

iovs=1

∆kA0Req

=kA0/Req

1 + bkA0/Req=

A

1 + bA(117)

It follows that A is given by

A = kA0Req

(118)

This is the gain from vs to io with b = 0. If bAÀ 1, the gain approaches

iovs→ A

bA=1

b=

1

R1(119)

This is the gain obtained by assuming the op amp is ideal. Because it has negative feedback, thereis a virtual short between its two inputs so that the voltage at the upper node of R1 is vs and thecurrent through R1 is vs/R1. Thus the output current is io = vs/R1.

The input resistance rin and the output resistance rout are given by

rin =

µi1vs

¶−1=

µ1

∆

k

Rid

¶−1= ∆

Rid

k= (1 + bA) (RS +Rid +R1) (120)

rout =

µiovt

¶−1=

µ1

∆

1

Req

¶−1= (1 + bA) (R0 +RL +R1) (121)

Note that the effect of the the feedback is to reduce the gain, increase the input resistance andincrease the output resistance. In the case bA→∞, the output resistance becomes infinite and theload resistor RL is driven by an ideal current source.

17

Figure 16: Flow graph for the series-series feedback amplifier.

Shunt-Series Feedback

A shunt-series feedback amplifier is an inverting amplifier in which the input signal x is a voltageand the output signal y is a current. If the input source is a voltage source, it must be convertedinto a Norton source for the gain to be in the form of Eq. (4). Because the input is a current andthe output is a current, the gain A represents a dimensionless current gain. Because bA must bedimensionless, the feedback factor is dimensionless. An op-amp example is given below.

Fig. 17(a) shows an op-amp circuit in which a resistor R1 connects from the lower node of theload resistor RL to ground. The voltage across R1 causes a current to flow through the feedbackresistor RF into the inverting op-amp input. The current through RF is proportional to the outputcurrent io. This is said to be series sampling at the output. Because the feedback connects to thesame op-amp input as the source, the circuit is said to have shunt summing at the input. Thus thecircuit is called a shunt-series feedback amplifier.

Figure 17: (a) Shunt-series amplifier example. (b) Circuit with feedback removed.

To analyze the circuit, we replace the circuit seen looking out of the vi node through RS with aNorton equivalent circuit with respect to vs and the circuit seen looking out of the vi node throughRF with a Norton equivalent circuit with respect to io. This must always be done with the shunt-series amplifier in order for the gain with feedback to be of the form −A/ (1 + bA), where A andb are positive and the − sign is necessary because the circuit has an inverting gain. In addition,we replace the circuit seen looking through RF into the vi node with a Thévenin equivalent circuitwith respect to vi. The circuit with feedback removed is shown in Fig. 17(b), where the op ampis replaced with a controlled source model that models its differential input resistance Rid, its gainA0, and its output resistance R0. The feedback factor b is is a current divider ratio given by

b =R1

R1 +RF(122)

18

A test voltage source vt is added in series with RL to calculate the output resistance. We definethe resistance rout as the effective series resistance in the output circuit, including RL. It is givenby rout = vt/io. The output resistance seen by RL is given by r0out = rout − RL. Because RL isfloating, it is not possible to label rout on the diagram.

Signal tracing shows that the negative feedback has the effect of reducing vi. For this reason,the vi source in the output circuit will be neglected in solving for io. The error current ie is thesum of the two current sources in the input circuit. We can write the following equations

ie = is + bio (123)

vi = ieRSk (R1 +RF ) kRid = ieReq1 (124)

io =−A0vi

R0 +RL +R1kRF+

vtR0 +RL +R1kRF

=−A0viReq2

+vt

Req2(125)


∆ = 1−∙Req1

µ−A0Req2

¶b

¸= 1 + bReq1

A0Req2

= 1 + bA (126)

From the flow graph, the current gain is given by

iois=1

∆Req1

µ−A0Req2

¶=−Req1A0/Req2

1 + bReq1A0/Req2=−A1 + bA

(127)

It follows that A is given by

A = Req1A0Req2

(128)

This is the negative of the gain from is to io with b = 0. If bAÀ 1, the gain approaches

iois→ −A

bA= −1

b= −

µ1 +

RF

R1

¶(129)

This is the gain obtained by assuming the op amp is ideal. In this case, there is a virtual groundat its inverting input which causes ie = is + bio = 0. Solution for io yields io = −is/b.

Figure 18: Flow graph for the shunt-series amplifier.

The input resistance r0in and the output resistance rout are given by

r0in =viis=1

∆Req1 =

RSk (R1 +RF ) kRid

1 + bA(130)

rout =

µiovt

¶−1=

µ1

∆

1

Req2

¶−1= (1 + bA) (R0 +RL +R1kRF ) (131)

19

Note that the effect of the the feedback is to reduce the gain, decrease the input resistance andincrease the output resistance. In the case bA→∞, the output resistance becomes infinite and theload resistor RL is driven by an ideal current source. The gain from vs to io in the original circuit,the input resistance rin, and the output resistance r0out seen by RL are given by

iovs=

isvs

iois=

1

RS

−A1 + bA

=1

RS

−A1 + bA

(132)

rin = RS +

µ1

r0in− 1

RS

¶−1(133)

r0out = rout −RL (134)

20

Stop HereFigure 10 shows the circuit diagram of a series-series feedback amplifier, where the bias circuits

have been omitted for simplicity. The feedback network consists of resistors RF1 and RF2. Thisnetwork samples the voltage at the collector of Q2 and feeds a voltage back into the emitter ofQ1. The output voltage from the circuit is the voltage at the emitter of Q1. Thus the feedbacknetwork does not sample an output voltage. Instead, it samples a voltage which is proportional tothe output current, i.e. it samples a voltage proportional to ic2 = α2io. Because io is the currentthrough the load resistor RE2, it follows that the feedback is proportional to the load current. Thisis called series sampling. The input summing is identical to that for the circuit of Fig. 4 which isseries summing. Therefore, the circuit is a series-series feedback amplifier.

XXXExp17f10.wmf f10Series-series feedback amplifier.To analyze the circuit, first remove the feedback. The circuit with feedback removed is shown

in Fig. 11. An external voltage source vt is included in series with RE2 in order to calculate theoutput resistance rout. Notice that rout is labeled looking into RE2 from signal ground. To simplifythe analysis, it will be assumed that r0 =∞ for both BJTs. The Thévenin equivalent circuit seenlooking into the emitter of Q2 consists of the voltage source vtb2 = −ic1RC1 in series with theresistor rie2. To solve for io/vi, rin, and rout, we can write the following equations:

io = ie2 =vt − vtb2rie2 +RE2

=vt + ic1RC1

rie2 +RE2(135)

ic1 = Gm1 (vtb1 − vte1) = Gm1

µvi − ic2

RC2

RC2 +RF1 +RF2RF2

¶(136)

ic2 = α2io ib1 =ic1β1

(137)

where Rtb1 = R1, Rte1 = RF2k (RF1 +RC2), and

rie2 = re2 +rx2 +Rtb2

1 + β2= re2 +

rx2 +RC1

1 + β2(138)

XXXExp17f11.wmf f11Series-series amplifier with feedback removed.The flow graph for the equations is given in Fig. 12. The determinant is given by

∆ = 1− RC1

rie2 +RE2× α2 ×

−Gm1RC2RF2

RC2 +RF1 +RF2(139)

The transconductance gain is given by

iovi=1

∆×Gm1 ×

RC1

rie2 +RE2(140)

The output resistance is given by

rout =

µiovt

¶−1= ∆× (rie2 +RE2) (141)

The input resistance is given by

rin =

µib1vi

¶−1= ∆× β1

Gm1= ∆× (R1 + rib1) (142)

where the alternate relation Gm1 = β1/ (R1 + rib1) has been used in the expression for rin andrib1 = rx1 + (1 + β1) (re1 +Rtb1). It is seen from these relations that the transconductance gain is

21

reduced by the amount of feedback, whereas both the output resistance and the input resistanceare increased by the amount of feedback. These are all properties of series-series feedback.

XXXExp17f12.wmf f12Signal flow graph for the series-series amplifier.It follows from Eqs. 85 and 86 that A and b are given by

A = Gm1 ×RC1

rie2 +RE2(143)

b = α2 ×RC2RF2

RC2 +RF1 +RF2(144)

Eq. (5) can be used to write the approximate transconductance gain as

iovi' 1

b=1

α2× RC2 +RF1 +RF2

RC2RF2(145)

Shunt-Series Feedback

XXXExp17f13.wmf f13Shunt-series amplifier circuit.Figure 13 shows the circuit diagram of a shunt-series feedback amplifier, where the bias circuits

have been omitted for simplicity. The feedback network consists of resistor RF . This resistorsamples the voltage at the emitter of Q2 and feeds a current back into the base of Q1. The outputcurrent from the circuit is the collector current in Q2. The feedback network samples a voltagewhich is proportional to the output current, i.e. it samples a voltage proportional to ie2 = io/α2.Because io is the current through the load resistor RC2, it follows that the feedback is proportionalto the load current. This is called series sampling. The input summing is identical to that for thecircuit of Fig. 7 which is shunt summing. Therefore, the circuit is a shunt-series feedback amplifier.

XXXExp17f14.wmf f14Shunt-series amplifier with feedback removed.To solve for the load current and input resistance, it will be assumed that r0 = ∞ for both

transistors. This simplifying assumption forces the output resistance to be infinite. A separatesolution will be given to solve for the output resistance. To analyze the circuit, first remove thefeedback. The circuit with feedback removed is shown in Fig. 14. The Thévenin voltage seenlooking out of the base of Q2 is vtb2 = −ic1RC1. To solve for io/ii and rin, we can write

io = Gm2 (vtb2 − vte2) = Gm2

µ−ic1RC1 − vb1

RE2

RF +RE2

¶(146)

ic1 = Gm1vb1 (147)

vb1 = ie [R1k (RF +RE2) krib1] (148)

ie = ii + ie2RE2

RF +RE2(149)

where ie is the error current, Rtb1 = 0, Rte1 = RE1, Rtb2 = RC1, Rte2 = RE2kRF , and

rib1 = rx1 + (1 + β1) (re1 +RE1) (150)

XXXExp17f15.wmf f15Flow graph for the shunt series amplifier.The flow graph for the equations is given in Fig. 15. The determinant of the graph is given by

∆ = 1−∙Gm1 (−Gm2RC1)−

Gm2RE2

RF +RE2

¸1

α2× RE2

RF +RE2[R1k (RF +RE2) krib1] (151)

22

The current gain is given by

ioii=1

∆[R1k (RF +RE2) krib1]

∙Gm1 (−Gm2RC1)−

Gm2RE2

RF +RE2

¸(152)

The input resistance is given by

rin =vb1ii=1

∆[R1k (RF +RE2) krib1] (153)

Thus both the current gain and the input resistance are decreased by the amount of feedback.These are properties of the shunt-series topology.

It follows from Eqs. (103) and (??) that A and b are given by

A = [R1k (RF +RE2) krib1]∙Gm1 (−Gm2RC1)−

Gm2RE2

RF +RE2

¸(154)

b = − 1α2× RE2

RF +RE2(155)

Note that b is negative. Because A is also negative, the product bA is positive. If this product isnot positive, the amplifier would have positive feedback rather than negative feedback. Eq. (5) canbe used to write the approximate current gain as

ioii' 1

b= −α2

µ1 +

RF

RE2

¶(156)

XXXExp17f16.wmf f16Circuit for calculating rout.Because it is assumed that r02 =∞ in the above analysis, it follows that rout =∞. To obtain

a finite rout, r02 must be added from the collector to the emitter of Q2 as shown in the circuitwith feedback removed in Fig. 16. Note the use of the primes to denote the collector and emittercurrents of Q2 that are inside the resistor r02. In the circuit, the source ii is set to zero and acurrent source it is added in series with the collector of Q2. Because ii = 0, the circuit seen lookinginto RF from the emitter of Q2 can be represented by the resistance RF +R1krib1 to ground. Theoutput resistance calculated from this circuit is the output resistance seen by the load resistor RC2

in Fig. (14). The analysis can be simplified by replacing the circuit seen looking out of the base ofQ2 by a Thévenin equivalent circuit. This is shown in Fig. 17.

XXXExp17f17.wmf f17Circuit for calculating rout after making a Thévenin equivalent circuitlooking out of the base of Q2.

The transresistance Rm is given by

Rm1 =vtb2ie2

(157)

=vtb2ic1

× ic1vb1

× vb1ie2

= (−RC1)×Gm1 ×∙

RE2

RF +RE2×R1k (RF +RE2) krib1

¸where Rtb1 = 0 and Rte1 = RE1. For this circuit

vc2 = i02r02 + ie2Rte2 (158)

i02 = it − i0c2 (159)

i0c2 = Gm2 (−Rmie2 − i02Rte2) (160)

23

ie2 = ic2 + ib2 = ic2 +i0c2β2

(161)

where Rte2 = [RF +R1krib1] kRE2 and ic2 = it.XXXExp17f18.wmf f18Flow graph for calculating rout.The flow-graph for these equations is shown in Fig. 18. There are two loops. The determinant

is given by

∆ = 1−∙Gm2Rte2 −Gm2Rm

1

β2

¸(162)

There are four forward paths from it to vc2, two of which touch both loops, one which touches onlyone loop, and one which touches only the other loop. The output resistance is given by

rout =vcit=1

∆

∙r02∆1 + r02Gm2Rm +Rte2∆2 −Gm2R

2te2

1

β2

¸(163)

where ∆1 = 1+Gm2Rm/β2 and ∆2 = 1−Gm2Rte2. It is straightforward to show that (122) reducesto

rout =r02 (1 +Gm2Rm/α2) + (rie2kRte2)

1−Gm2 (Rte2 −Rm/β2)(164)

where rie2 = re2 + (Rc1 + rx2) / (1 + β2).

Design Examples

Series-Shunt

Figure 19 shows the circuit diagram of a practical series-shunt amplifier. The circuit is to bedesigned for the following specifications: voltage gain vo/vi = 10, RL = Rin ' 10 kΩ, IC1 = 0.5mA, IC2 = 3 mA, V + = −V − = 15 V. It may be assumed that VBE1 = VEB2 = 0.65 V andβ1 = β2 =∞ for the calculations.


Let R8 = 100 Ω. The voltage across R7 is IE2R8 + VEB2 = 0.95 V. Thus R7 = 0.95/0.5 = 1.9kΩ. Let VE1 = −5 V, VC2 = 0 V, and R5 = 20 kΩ. The current through R5 is 5/20 = 0.25mA. The current through R4 is IE1 + 0.25 = 0.75 mA. The voltage across R4 is 10 V. ThusR4 = 10/0.75 = 13.3 kΩ. The current through R6 is 3− 0.25 = 2.75 mA. The voltage across R6 is

24

15 V. Thus R6 = 15/2.75 = 5.45 kΩ. The base voltage of Q1 is VB1 = −5+0.65 = −4.35 V. R1 andR2 set VB1 and Rin. The design equations are VB1 = 15 (R2 −R1) / (R2 +R1) and R1kR2 = 10kΩ. Solution yields R1 = 15.5 kΩ and R2 = 28.3 kΩ. If the open-loop gain of the amplifier is verylarge, the closed-loop gain is given by 1 + R5/ (R3kR4). Let us use this gain expression to designfor a gain of 10

√2, i.e. 3 dB higher than the specified value, in hopes that the actual gain will

be close to 10. It follows that 1 + R5/ (R3kR4) = 10√2. Solution for R3 yields R3 = 1.72 kΩ. If

the circuit oscillates, a small capacitor, typically 10 pF can be connected from collector to base ofQ2. For acceptable low frequency response above the frequency f , the capacitors should satisfyC1 > 1/ (2πf Rin), C2 > 1/ (2πf RL), and C3 > 1/ [2π (R3 +R4kR5kre1)], where re1 = VT/IE1.For f = 10 Hz, acceptable values are C1 = C2 = 10 μF and C3 = 22 μF.

Once the circuit is assembled, R3 can be tweaked to fine tune the gain. Either R1, R2, or bothcan be tweaked to adjust the dc bias voltage at the collector of Q2.

Shunt-Shunt

Figure 20 shows the circuit diagram of a practical shunt-shunt amplifier. The circuit is to bedesigned for the following specifications: voltage gain vo/vi = −10, Rin = 10 kΩ, RL = 100 kΩ,IC1 = 0.5 mA, IC2 = 4 mA, V + = −V − = 15 V. It may be assumed that VBE1 = VBE2 = 0.65 Vand β1 = β2 =∞ for the calculations. Because vi = i1Rin, it follows that the transresistance gainis vo/i1 = −100 kΩ.

Figure 20: Shunt-shunt amplifier.

Let us choose the dc voltage at the emitter of Q1 to be −10 V and the dc voltage at theemitter of Q2 to be +5 V. The current through R5 and R8 is 0.5 mA, thus R5 = (−10 + 15) /0.5m= 10 kΩ and R8 = (15− 5.65) /0.5m = 18 kΩ. The current through R7 is 4 mA, thus R7 =(5 + 15) /4m = 5 kΩ. Let us pick the current through R2 and R3 to be 0.05 mA. If β1min = 100,this current is 10 times IB1max. Thus R2 = (15 + 9.35) /0.05m = 487 kΩ. Let us choose the 5%value R2 = 470 kΩ. It follows that the current through R2 is (15 + 9.35) /470m = 0.0518 mA.Thus R3 = (−9.35 + 15) /0.0518m = 109 kΩ. Let us choose the 5% value R3 = 110 kΩ. Thesmaller the value of R4, the higher the open-loop gain. To prevent possible oscillation problems,the value of R3 should not be too small. Let us pick R3 = 100 Ω. If the open-loop gain is veryhigh, the input resistance is Rin ' R1. Thus let us choose R1 = 10 kΩ to meet the input resistancespecification. If the open-loop gain is infinite, the closed-loop voltage gain is vo/vi = −R6/R1. Fora gain of −10, we must have R6 = 100 kΩ. To correct for a non-infinite open-loop gain, R6 should

25

be larger than this. Let us choose the value R6 = 120 kΩ, which is 20% larger. For the capacitors,let C1 = C2 = C4 = 1μF and C3 = 100 μF.

Series-Series

Figure 21 shows the circuit diagram of a series-series amplifier. The circuit is to be designed forthe following specifications: transconductance gain ie2/vg = 2 mA/V, Rin ' 10 kΩ, RL = 10 kΩ,IC1 = 0.5 mA, IC2 = 4 mA, V + = −V − = 15 V. It may be assumed that VBE1 = VBE2 = 0.65 Vand β1 = β2 =∞ for the calculations.


Let us choose the following dc voltages: VE1 = VC2 = −10 V and VC1 = 2 V. The currentthrough R6 and R9 is 4 mA, thus R6 = (−10 + 15) /4m = 1.3 kΩ and R9 = (15− 2.65) /4m= 3 kΩ. The current through R4 and R8 is 0.5 mA, thus R4 = (−10 + 15) /0.5m = 10 kΩ andR8 = (15− 2) /0.5m = 26 kΩ. Because VB1 = −10 + 0.65 = −9.35 V, we have 30R2/ (R1 +R2)−15 = −9.35. Also, R1R2/ (R1 +R2) = 10 kΩ. These equations can be solved for R1 to obtain 53kΩ. This is not a standard 5% value. Let us pick R1 = 56 kΩ. For VB1 = −9.35 V, it followsthat R2 = 13 kΩ. Let us pick R5 = 1.3 kΩ. If the open-loop gain is infinite, the transconductancegain would be given by ie2/vi = (R6 +R5 +R3kR4) / (R6 ×R3kR4) = 0.002. For the values for R5and R6, it follows from this equation that R3kR4 = 1625 Ω, thus R3 = 1.94 kΩ. To account for anon-infinite open-loop gain, the value R3 = 1.3 kΩ which is about 33% smaller than the 1.94 kΩvalue.

Shunt-Series

Figure 22 shows the circuit diagram of a shunt-series amplifier. The circuit is to be designed forthe following specifications: current gain ic2/i1 = −10, Rin ' 10 kΩ, RL = 10 kΩ, IC1 = 0.5 mA,IC2 = 4 mA, V + = −V − = 15 V, VC2 = +5 V, VC1 = −5 V, and VB1 = VE2 = −10 V. It may beassumed that VBE1 = VBE2 = 0.65 V and β1 = β2 =∞ for the calculations.

Both R7 and R9 conduct a current of 4 mA, thus R7 = (−5− 0.65 + 15) /4m = 2.34 kΩand R9 = (15− 5) /4m = 2.5 kΩ. Both R5 and R8 conduct a current of 0.5 mA, thus R5 =(−10− 0.65 + 15) /0.5m = 8.7 kΩ and R8 = (15 + 5) /0.5m = 40 kΩ. Let us take the currentthrough R2 to be 0.05 mA, i.e. about 10 times the maximum expected value of IB1 for βmin = 100.Thus we have R2 = (15 + 10) /0.05m = 500 kΩ. Also, R3 = (−10 + 15) /0.05m = 100 kΩ. If

26

Figure 22: Shunt-series amplifier.

we assume that the open-loop gain is infinite, the input resistance is R1 and the current gain isic2/i1 = − (1 +R6/R7). Thus, we pick R1 = 10 kΩ and (1 +R6/1.25k) = 10. Solution for R6yields R6 = 21.1 kΩ. We will pick the standard value R6 = 22 kΩ. To complete the design, we pickC1 = C2 = 10 μF, C3 = 100 μF, C4 = 10 μF, and R4 = 100 Ω.

27

Series Shunt Amplifier

Q

-

+

v g2R

V

C 3

R 3

GR 1CR 1

v i

V

1

4R 5R

C

R 6

2

7R

Q

R

2

8

LR

v o

Rp A B, ( )A B⋅

A B+:=

VT 25.9mV:=

VBE 0.65V:=

VEB 0.65V:=

RL 10kΩ:=

Vplus 15V:= Vminus 15− V:= Rg 0:= R1 28kΩ:= R2 15kΩ:=

R3 1.8kΩ:= R4 13kΩ:= R5 20kΩ:= R6 5.1kΩ:= R7 1.8kΩ:= R8 100Ω:=

VB1 VplusR2 R1−

R1 R2+⋅ 4.535− V=:= VE1 VB1 VBE− 5.185− V=:=

I1

I2

⎛⎜⎜⎝

⎞⎟⎟⎠

Rp R4 R5 R6+, ( )R7

R6R4 R5+ R6+

R4⋅

R8−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1−

VE1 Vminus−

VEB

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅0.53

3.034⎛⎜⎝

⎞⎟⎠

mA⋅=:=

VE1 I1 Rp R4 R5 R6+, ( )⋅ I2R6

R4 R5+ R6+R4⋅

⎛⎜⎝

⎞⎟⎠

⋅+ Vminus+ 5.185− V=:=

VC1 Vplus I1 R7⋅− 14.047V=:= VB2 VC1 14.047V=:= VE2 VB2 VEB+ 14.697V=:=

VC2 I2 Rp R6 R4 R5+, ( )⋅ I1R4

R4 R5+ R6+⋅ R6⋅+ Vminus+ 0.678− V=:=

re1VTI1

:= re2VTI2

:= Gm21

re2 R8+:=Gm1

1re1 Rp R5 Rp R3 R4, ( ), ( )+

:=

A Gm1 Gm2 R7⋅ Rp Rp R6 RL, ( ) R5 Rp R3 R4, ( )+, ( )⋅Rp R3 R4, ( ) Rp R6 RL, ( )⋅

Rp R3 R4, ( ) R5+ Rp R6 RL, ( )++

⎛⎜⎝

⎞⎟⎠

⋅ 32.128=:=

bRp R3 R4, ( )

R5 Rp R3 R4, ( )+0.073=:=

1b

13.65= AvA

1 b A⋅+9.58=:=

Put cap to ground from emitter of Q2

Gm21

re2:=

A Gm1 Gm2 R7⋅ Rp Rp R6 RL, ( ) R5 Rp R3 R4, ( )+, ( )⋅Rp R3 R4, ( ) Rp R6 RL, ( )⋅

Rp R3 R4, ( ) R5+ Rp R6 RL, ( )++

⎛⎜⎝

⎞⎟⎠

⋅ 406.789=:=

bRp R3 R4, ( )

R5 Rp R3 R4, ( )+0.073=:= Av

A1 b A⋅+

13.206=:=

0

0

1

1

2

2

3

3

4

4

5

5

6

6

A A

B B

C C

D D

E E

VCC15V

VEE

-15V

V11 Vpk 1kHz 0°

Q1

Q2

R128.3kΩ

C1

10uF

R215.5kΩ

R31.72kΩ

R413.3kΩ

R5

20kΩ R65.45kΩ

R71.9kΩ

RL10kΩ

C2

10uF

C322uF

R8100 Ω

VEE

-15V VEE

-15V VEE

-15V

VCC15V

VCC15V

Out

Shunt-Shunt Amplifier

Vplus 15V:= Vminus 15− V:= β 10200:= RG 0Ω:= R1 10kΩ:=

R2 487kΩ:= R3 109kΩ:= R4 100Ω:= R5 10kΩ:= R6 120kΩ:=

R7 5kΩ:= R8 18kΩ:= RL 10kΩ:= VBE 0.65V:= VT 25.9mV:=

VB1R3 R2−

R2 R3+Vplus⋅ 9.513− V=:= VE1 VB1 VBE− 10.163− V=:=

IE1VE1 Vminus−

R50.484 mA⋅=:= IC1 IE1:= VC1 Vplus IC1 R8⋅− 6.294V=:=

VB2 VC1:= VE2 VB2 VBE− 5.644V=:= IE2VE2 Vminus−

R74.129 mA⋅=:=

re1VTIE1

53.55Ω=:= re2VTIE2

6.273Ω=:= Rp A B, ( )A B⋅

A B+:=

Gm11

re1 Rp R4 R5, ( )+:= Gm2

1re2 Rp R7 Rp RL R6, ( ), ( )+

:=

R23 Rp R2 R3, ( ):= R1 Rp R1 R23, ( ):= RE2 Rp R7 RL, ( ) 3.333 kΩ⋅=:=

A Rp R1 R6, ( ) Gm1 R8−( )⋅Rp RE2 R6, ( )

re2 Rp RE2 R6, ( )+⋅

Rp re2 RE2, ( )R6 Rp re2 RE2, ( )+

+⎡⎢⎣

⎤⎥⎦

⋅ 9.849− 105× Ω=:=

b1−

R6:=

1b

120− kΩ⋅=

ArA

1 A b⋅+106.967− kΩ⋅=:=

0

0

1

1

2

2

3

3

4

4

5

5

6

6

A A

B B

C C

D D

E E

VCC15V

VEE

-15V

Vi1 Vpk 1kHz 0°

Q1

R2487kΩ

C1

1uF

R3109kΩ

R4100 Ω

R510kΩ

R6

120kΩ

R75kΩ

R818kΩ

RL10kΩ

C2

1uF

C3100uF

VEE

-15V VEE

-15V VEE

-15V

VCC15V VCC

15V

Out

R1

10kΩ

Q2

C4

1uF

Series Series Feedback

Vplus 15V:= Vminus 15− V:= VBE 0.65V:= VEB 0.65V:= VT 25.9mV:=

β 10307:= Rp A B, ( )

A B⋅A B+

:= R1 53kΩ:= R2 13kΩ:= R3 1.94kΩ:=

R4 10kΩ:= R5 1.3kΩ:= R6 1.25kΩ:= R7 26kΩ:= R8 3.0875kΩ:=

RL 10kΩ:= VB1 VplusR2 R1−

R2 R1+⋅ 9.091− V=:= VE1 VB1 VBE− 9.741− V=:=

I1

I2

⎛⎜⎜⎝

⎞⎟⎟⎠

R4 R5 R6+( )⋅

R4 R5+ R6+

R7

R4 R6⋅

R4 R5+ R6+

R8−

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

1−

VE1 Vminus−

VEB

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅0.525

4.209⎛⎜⎝

⎞⎟⎠

mA⋅=:=

re1VTI1

:= re2VTI2

:= RF2 Rp R3 R4, ( ):= RF1 R5:= RC2 R6:=

Rte1 Rp RF2 RF1 RC2+, ( ):= Rte2 Rp R8 RL, ( ):= RC1 R7:=

Gm11

re1 Rte1+:= Gm2

1re2 Rte2+

:=

A Gm1RC1

re2 Rte2+⋅ 0.011

1Ω

=:= bRC2 RF2⋅

RC2 RF1+ RF2+486.489Ω=:=

1b

2.056 10 3−×

1Ω

= AmA

1 b A⋅+1.72 10 3−

×1Ω

=:=

0

0

1

1

2

2

3

3

4

4

5

5

6

6

A A

B B

C C

D D

E E

VCC15V

VEE

-15V

V11 Vpk 1kHz 0°

Q1

Q2

R153kΩ

C1

10uF

R213kΩ

R31.94kΩ R4

10kΩ

R5

1.3kΩ

R61.25kΩ

R726kΩ

RL10kΩ

C2

10uF

C322uF

R83.0875kΩ

VEE

-15V VEE

-15V VEE

-15V

VCC15V

VCC15V

Out

Shunt Series

Vplus 15V:= Vminus 15− V:= R1 10kΩ:= R2 500kΩ:= R3 100kΩ:=

R4 100Ω:= R5 8.7kΩ:= R6 21.0375kΩ:= R7 2.3375kΩ:= R8 40kΩ:=

R9 2.5kΩ:= RL 10kΩ:= VBE 0.65V:= VT 25.9mV:=

Rp A B, ( )A B⋅

A B+:= VB1 Vplus

R3 R2−

R3 R2+⋅ 10− V=:= VE1 VB1 VBE− 10.65− V=:=

RG 0Ω:= I1VE1 Vminus−

R50.5 mA⋅=:= VC1 Vplus I1 R8⋅− 5− V=:=

VE2 VC1 VBE− 5.65− V=:=

I2VE2 Vminus−

R74 mA⋅=:= β 10307

:=

re1VTI1

:= re2VTI2

:= R67 Rp R6 R7, ( ):=

Gm11

re1 Rp R4 R5, ( )+:= Gm2

1re2 Rp R7 R6, ( )+

:= RF R6:= RE2 R67:=

RC1 R8:=

R23 Rp R2 R3, ( ):= R1 Rp R1 R23, ( ):=

A Rp R1 RF RE2+, ( ) Gm1 Gm2− RC1⋅( )⋅Gm2 RE2⋅

RF RE2+−

⎡⎢⎣

⎤⎥⎦

⋅ 810.856−=:=

bRE2

RF RE2+− 0.091−=:=

1b

11−=

AiA

1 b A⋅+10.853−=:=

0

0

1

1

2

2

3

3

4

4

5

5

6

6

A A

B B

C C

D D

E E

VCC15V

VEE

-15V

Vi1 Vpk 1kHz 0°

Q1

R2500kΩ

C1

10uF

R3100kΩ

R4100 Ω

R58.7kΩ

R6

21.0375kΩ

R72.3375kΩ

R840kΩ

RL10kΩ

C2

10uF

C3100uF

VEE

-15V VEE

-15V VEE

-15V

VCC15V

VCC

15V

Out

R1

10kΩ

ip

Q2

C4

10uF

b1

R92.5kΩ

c2

Feedback Ampliﬁers - Georgia Institute of …brewer.ece.gatech.edu/ece4043/Lecture_Notes/duck...feedback ampliﬁers are shown in Fig. 1. Depending on the type of feedback, the variables

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