Eeg381 electronics iii chapter 2 - feedback amplifiers

Prof Fawzy IbrahimEEG381 Ch2 Feedback1 of 39
EEG381 Electronics III
CHAPTER 2
Feedback Amplifiers
[[SedraSedra ((Ch.Ch. 88),), BoylestadBoylestad ((Ch.Ch. 1717)])]
Prof. Hassan Elghitani Prof. Fawzy Ibrahim
Electronics and Communication Department
Misr International University (MIU)
Prof Hassan Elghitani

Chapter Contents
2.1 Feedback Concepts
2.2The Four Basic Feedback Topologies
12.2.1 Series - Shunt Feedback or Voltage Amplifiers
2.2.2 shunt-series feedback or Current Amplifiers
2.2.3 Series-Series Feedback or Transconductance Amplifiers
2.2.4 Shunt - Shunt Feedback or Transresistance Amplifiers
2.2.5 Summary of Feedback Topologies
2.3 Negative Feedback Voltage Amplifiers
2.3.1 Gain Calculation
2.3.2 Bandwidth Extension
2.3.3 Input and output Impedance
2.3.4 Noise Reduction
2.3.5 Advantages and Disadvantages of negative feedback

Feedback
Positive
Regenerative
“Oscillators”
Negative
Degenerative
“Amplifiers”
Prof Fawzy IbrahimEEG381 Ch2 Feedback3 of 39 Prof Hassan Elghitani

• Most physical systems incorporate some form of feedback. Feedback has
been mentioned previously. In particular, feedback was used in op-amp
circuits as described in in chapter 1.
• A typical feedback connection is shown in Fig. 2.1. The input signal, Vs, is
applied to a mixer network, where it is combined with a feedback signal, Vf.
The sum or difference of these signals, Vi, is then the input voltage to the
amplifier. A portion of the amplifier output, Vo, is connected to the feedback
network (), which provides a reduced portion of the output as feedback
signal to the input mixer network.
Fig. 2.1 Block diagram (a) Positive feedback (b) Negative feedback amplifier.
(a) (b)

1.1 Feedback Concepts (Continued)
• Depending on the relative polarity of the signal being fed back into a circuit,
one may have negative or positive feedback.
• Negative feedback results in decreased voltage gain, for which a number of
circuit features are improved as summarized below.
• Positive feedback drives a circuit into oscillation as in various types of
oscillator circuits that will be discussed in The next chapters.
• Negative feedback results in reduced overall voltage gain, a number of
improvements are obtained, among them being:
1. Better stabilized voltage gain or desensitize the gain: that is, make the
value of the gain less sensitive to variations in the value of circuit
components, such as might be caused by changes in temperature.
2. Improved frequency response or extend the bandwidth of the amplifier.
3. Control the input and output impedances or higher input impedance and
lower output impedance.
4. Reduce the effect of noise: that is, minimize the contribution to the output of
unwanted electric signals generated, either by the circuit components
themselves, or by extraneous interference.
5. More linear operation or Reduce nonlinear distortion: that is, make the
output proportional to the input (in other words, make the gain constant,
independent of signal level).

1. Series-Shunt Feedback (Voltage amplifiers)
i/p mixing o/p sampling
2. Shunt- Series Feedback (Current amplifiers)
3. Series-Series Feedback (Transconductance amplifiers)
4. Shunt-Shunt Feedback (Transresistance amplifiers)
2.2 The Four Basic Feedback Topologies
Prof Fawzy IbrahimEEG381 Ch2 Feedback6 of 39 Prof Hassan Elghitani

2.2 The Four Basic Feedback Topologies
• Based on the quantity to be amplified (voltage or current) and on the
desired form of output (voltage or current), amplifiers can be classified into
four categories. These categories Classified as follows:
2.2.1 Series - Shunt Feedback or Voltage Amplifiers
• Voltage amplifiers are intended to amplify an input voltage signal and
provide an output voltage signal. The voltage amplifier is essentially a
voltage-controlled voltage source. A suitable feedback topology is the
voltage-mixing (series connection at the input ) and voltage sampling
(parallel or shunt connection at the output) as shown in Fig. 2.2.
• Because of the Thevenin representation of the source, the feedback signal
Vf should be a voltage that can be mixed with the source voltage in series.
Fig. 2.2 Series - shunt feedback or voltage amplifier topology.
• This topology not only stabilizes the
voltage gain but also results in a
higher input resistance (intuitively, a
result of the series connection at the
input) and a lower output resistance
(intuitively, a result of the parallel
connection at the output), which are
desirable properties for a voltage
amplifier.

2.2 The Four Basic Feedback Topologies (Continued)
2.2.2 shunt-series feedback or Current Amplifiers
• The input signal in a current amplifier is essentially a current, and thus the
signal source is most conveniently represented by its Norton equivalent.
• The output quantity of interest is current; hence the feedback network
should sample the output current.
• The feedback signal should be in current form so that it may be mixed in
shunt with the source current.
• Thus the feedback topology suitable for a current amplifier is the current-
mixing current-sampling topology, illustrated in Fig. 2.3.
Fig. 2.3 shunt-series feedback or Current amplifier topology.
• Because of the parallel (or shunt)
connection at the input, and the
series connection at the output, this
feedback topology is also known as
shunt series feedback.
• This topology not only stabilizes the
current gain but also results in a
lower input resistance, and a higher
output resistance, both desirable
properties for a current amplifier.

2.2.3 Series-Series Feedback or Transconductance Amplifiers
• In transconductance amplifiers the input signal is a voltage and the output
signal is a current.
• It follows that the appropriate feedback topology is the voltage-mixing
current-sampling topology, illustrated in Fig. 2.4.
Fig. 2.4 Series-series feedback or transconductance amplifier topology.

2.2.4 Shunt - Shunt Feedback or Transresistance Amplifiers
• In Transresistance amplifiers the input signal is current and the output
signal is voltage.
• It follows that the appropriate feedback topology is of the current - mixing
voltage-sampling type, shown in Fig. 2.5. The presence of the parallel (or
shunt) connection at both the input and the output makes this feedback
topology also known as shunt - shunt feedback.
Fig. 2.5 Shunt - shunt feedback or Transresistance amplifier topology.

2.2.5 Summary of Feedback Topologies
• A summary of the gain, feedback factor, and gain with feedback of Figs. 2.2
to 2,5 is provided for reference in Table 2.1.

• Fig. 2.6 shows the basic structure or signal-flow diagram of a negative
feedback amplifier. Ao is the Gain of the open-loop amplifier and  is the
feedback factor.
• Thus its output voltage Vo is related to the input voltage Vi by:
• The output Vo is fed to the load as well as to a feedback network, which
produces a sample of the output.
Fig. 2.6 General structure of negative feedback amplifier
ioo VAV  (2.1)

2.3 Negative Feedback Voltage Amplifiers (Continued)
• This sample Vf is related to Vo by the feedback factor  as:
• The feedback signal Vf subtracted from the source signal Vs, which is the
input to the complete feedback amplifier, to produce the signal Vi which is
the input to the basic amplifier which is given by:
• Here we note that it is this subtraction that makes the feedback negative. In
essence, negative feedback reduces the signal that appears at the input of
the basic amplifier.
• The gain of the feedback amplifier can be obtained by combining Eqs. (2.1)
through (2.3) as:
• The quantity Af is called the feedback gain or closed loop gain, a name that
follows from Fig. 2.6, A0 is the open loop gain or gain without feedback,
T = A0 is the loop gain, and (1+ A0) is the amount of feedback.
T
A
A
A
V
V
A o
o
o
s
o
f




11 
of VV  (2.2)
fsi VVV  (2.3)
(2.4)

• For the feedback to be negative, the loop gain Ao should be positive; that
is, the feedback signal Vf should have the same sign as Vs .
• Thus this resulting in a smaller difference signal Vi. Equation (2.4) indicates
• that for positive Ao the gain-with-feedback, Af will be smaller than the open-
loop gain Ao by the quantity 1+ Ao, which is called the amount of feedback.
• In other words, the overall gain will have very little dependence on the gain
of the basic amplifier, Ao, a desirable property because the gain Ao is
usually a function of many manufacturing and application parameters, some
of which might have wide tolerances.
• Therefore, the closed-loop gain is almost entirely determined by the
feedback elements.
• Equations (2.1) through (2.3) can be combined to obtain the following
expression for the feedback signal Vf and the output of comparison circuit or
mixer, Vf as:
s
o
s
o
o
s
o
f V
T
A
V
A
A
V
V
A




11



ss
os
o
i V
T
V
AV
V
V




1
1
1
1

(2.5)
(2.6)

Gain Desensitivity
• The gain of the closed-loop amplifier is less sensitive to variation of the gain
of the basic amplifier this property can be analytically established as follows:
Assume that  is constant. Taking differentials of both sides of Eq. (2.4)
results in:
Dividing Eq. (2.7) by Eq. (2.4) yields:
• which says that the percentage change in Af (due to variations in some
circuit parameter) is smaller than the percentage change in Ao by the
amount of feedback.
• For this reason the amount of feedback, (1 +  Ao) or (1 + T), is also known
as the Desensitivity factor.
(2.7)
(2.8)
222
)1()1()1(
)()1(
T
dA
A
dA
A
dAAdAA
dA o
o
o
o
oooo
f









o
o
o
o
of
f
A
dA
TA
dA
AA
dA
)1(
1
)1(
1






Example 2.1:
For the Op Amp circuit shown in Fig. 2.7, if the op amp has infinite input
resistance (Ri = ), zero output resistance (Ro = o) and the open-loop
voltage gain Ao = 100 .
(a) Find an expression for the feedback factor .
(b) Find the ratio R2/R1 to obtain a closed loop voltage gain Af of 10.
(c) What is the amount of feedback in decibels?
(d) If the source voltage, Vs = 1 V, find the output voltage, Vo, the feedback
voltage Vf, and the input voltage Vi.
(e) If Ao decreases by 20%, what is the corresponding decrease in Af ?
Solution
(a) The feedback factor,  is given by:
(b) From (2.4), closed loop voltage gain Af is:
Fig. 2.7 An Op Amp circuit
21
1
RR
R
V
V
o
f


T
A
A
A
V
V
A o
o
o
s
o
f




11 

Example 2.1 Solution
When Af = 10, then
(c) The amount of feedback =1 +Ao = 1 + 100 x 9x10-2 = 10 = 20 dB.
(d) The output voltage, Vo = Af x Vs = 10 x 1 = 10 V.
The feedback voltage Vf =  x Vo = 9x10-2 x 10 = 0.9 V.
The input voltage, Vi = Vs - Vf = 1 – 0.9 V = 0.1 V
(b) From (2.8), the percentage change in Af is given by:
2
109
1001
100
10 


 x

11.101
1
1
2
21
1





R
R
RR
R
%44.2
80
%20
)801091(
1
)1(
1
2





 
xxA
dA
AA
dA
o
o
of
f


• Consider an amplifier whose high-frequency response is characterized by a
single pole. Its gain at mid and high frequencies can be expressed as :
where Ao denotes the midband gain and b is the upper 3-dB
frequency. Application of negative feedback, with a frequency-independent
factor , around this amplifier results in a closed-loop gain Af(s) given by:
• Substituting for A(s) from Eq. (2.9) results, after a little manipulation, in:
• Thus the feedback amplifier will have a midband gain of:
• and an upper 3-dB frequency bf given by:
• It follows that the upper 3-dB frequency is increased by a factor equal to the
amount of feedback while midband gain is decreased by the same factor.
Cascading amplifier is the solution to increase the voltage gain as in Ch. 1.
(2.9)
(2.10)
b
o
s
A
sA
/1
)(


)(1
)(
)(1
)(
)(
sT
sA
sA
sA
sAf





)1(/1
)1/(
)1(/1
)1/(
)(
Ts
TA
As
AA
sA
b
o
ob
oo
f








(2.11)
)1/()1/( TAAAA ooof   (2.12a)
)1()1( TA bobbf   (2.12b)

Example 2.2:
Consider the noninverting op-amp circuit of Fig.2.8. Let the open-loop gain,
Ao has a low-frequency value of 104 and a uniform 20 dB/decade rolloff at
high frequencies with a 3-dB frequency, fb of 100 Hz. If R1 = 1 kΩ and R2 =
9 kΩ, for the closed-loop amplifier find:
(a) The low-frequency gain Af.
(b) The upper 3-dB frequency, fbf .
Solution
(a) From Eq. (2.12a) Af , is given by:
(b) From Eq. (2.12a) fbf ) is given by:
Fig. 2.8 An Op Amp circuit
VVTAAAA ooof /99.91001/10)1/()1/( 4
 
1.0
91
1
21
1





RR
R

kHzxTfAff bobbf 1.1001001100)1()1(  
1001)1()1(  TAFBofamountThe o

2.3.3 Input and output Impedances
Input Impedance
• Consider the series-shunt feedback amplifier connection shown in Fig. 2.9,
the input impedance can be determined as follows:
• The input impedance with series feedback is increased and has the value of
the input impedance without feedback multiplied by a factor equal to the
amount of feedback (1 +  Ao) = (1 + T).
(2.13)
i
ios
i
os
i
fs
i
i
i
Z
VAV
Z
VV
Z
VV
Z
V
I
 





iioiisiosii ZIAZIVVAVZI  
)1()1( TZAZZAZ
I
V
Z ioiioi
i
s
if  

Fig. 2.9 Voltage amplifier feedback connection model

Output Impedance
• The series-shunt feedback amplifier connection shown in Fig. 2.9, provides
sufficient circuit detail to determine the output impedance with feedback.
The output impedance is determined by applying a voltage, V, resulting in a
current, I, with Vs shorted out (Vs = 0). The voltage V is then given by:
For Vs = 0
so that
or
• allows solving for the output resistance with feedback:
• The output impedance with series feedback is reduced and has the value of
the output impedance without feedback divided by a factor equal to the
amount of feedback (1 +  Ao) or (1 +T).
(2.14)
ioo VAIZV 
)1()1( T
Z
A
Z
I
V
Z o
o
o
of





fi VV 
)( VAIZVAIZV ooioo 
oo IZVAV  )(

Example 2.3:
Determine the voltage gain, Af, breakdown frequency, fbf, input impedance
Zi and output impedance, Zo with feedback for voltage amplifier, shown in
Fig. 2.9, having Ao = 100, fb = 200 Hz, Ri = 10 kΩ and Ro = 20 kΩ for:
(a)  = 0.1 (b)  = 0.5
Solution
Using Eqs. (2.12), (2.13), and (2.14), we obtain
(a)
(b)
VVAAA oof /09.911/100)1/(   kHzxAobbf 2.211200)1(  
 kxkAZZ oiif 1101110)1(  



 k
k
A
Z
Z
o
o
of 82.1
11
20
)1( 
11)1001.01()1(  xAFBofamountThe o
51)1005.01()1(  xAFBofamountThe o
VVAAA oof /96.151/100)1/(   kHzxAobbf 2.1051200)1(  
 kxkAZZ oiif 5105110)1(  



 16.292
51
20
)1(
k
A
Z
Z
o
o
of


2.3.4 Noise Reduction
• Negative feedback can be employed to reduce the noise or interference in
an amplifier or, more precisely, to increase the ratio of signal to noise.
• From Fig. 2.10 (a),the output voltage Vo is related to the input voltage Vi by:
where D is the distortion generated by the amplifier.
• With the negative feedback amplifier of Fig. 2.10 (b), when Vs =0, to study
the effect of noise, the output voltage Vo is related to the input voltage Vi by:
• Feedback distortion < open loop distortion
Fig. 2.10 Voltage amplifier noise model: (a) Open loop; (b) With feedback
DVAV ioo 
(2.15)
(a) (b)
)1()1( T
D
A
D
dDdADVAd
o
oio






The distortion with feedback is reduced and
has the value of distortion without feedback,
D, divided by a factor equal to the amount of
feedback (1 +  Ao) or (1 + T).

1. Negative feedback results in reduced overall voltage gain, the solution is to
do cascading of amplifiers.
Eq. (2.4):
2. Better stabilized voltage gain or desensitize the gain: that is, make the
value of the gain less sensitive to variations in the value of circuit
components..
Eq. (2.8):
3. Improved frequency response or extend the bandwidth of the amplifier.
Eq. (2.12):
3. Higher input impedance and Lower output impedance.
Eq. (2.13): Eq. (2.14):
4. Reduce the effect of noise: that is, minimize the contribution to the output of
unwanted electric signals generated, either by the circuit components
themselves, or by extraneous interference.
Eq. (2.15):
2.3.5 Advantages and Disadvantages of negative feedback
T
A
A
A
V
V
A o
o
o
s
o
f




11 
o
o
o
o
of
f
A
dA
TA
dA
AA
dA
)1(
1
)1(
1





)1( obbf A 
)1()1( TZAZZ ioiif  
)1()1( T
Z
A
Z
Z o
o
o
of





)1()1( T
D
A
D
d
o 





• Fig. 2.11 illustrates two Common Emitter (CE) connected in cascade.
• The input and output of the overall amplifier are ac coupled through capacitors
C1 and C5.
• Bypass capacitors C2 and C4 are used to obtain maximum voltage gain from the
two inverting amplifiers.
• Interstage coupling capacitor C3 transfers the ac signals between the amplifiers
but provides isolation at dc. Thus, the individual Q-points of the transistors are
not affected by connecting the stages together.
• Fig. 2.12 gives the dc equivalent circuit for the amplifiers in which the capacitors
have been removed.
• The amplifier is characterized to determine its voltage gain, Av = Ao, input
resistance Ri and output resistance Ro by using the small signal models of the
transistors Q1 and Q2 by applying exactly the same procedure as discussed in
Electronics II Course.
Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps26 of 39
2.4 Feedback in Multistage BJT Amplifiers
2.4.1 Two Stage ac-Coupled Amplifier Structures

2.4 Feedback in Multistage BJT Amplifiers (Continued)
Fig. 2.11 Two stage cascaded common-emitter amplifiers.

Fig. 2.12 dc equivalent circuit for the two stage cascaded common-emitter amplifiers:
(a) dc circuit; (b) Thevenin equivalent for each section.
(a) (b)

Fig. 2.13 shows the small signal model for the amplifier circuits. We note the
following:
1. The input resistance of the amplifier, Ri is equal to the input resistance of the first
stage.
2. The load resistance of the first stage (RL1) is equal to the input resistance of the
second stage, Rin2.
3. The output resistance of the amplifier, Rout is equal to the output resistance of the
second stage.
4. The output voltage of the first stage (vo1) is the input to the second stage.
5. The total voltage gain is given by:
Therefore, the overall voltage gain is equal to the product of the gains of
individual single transistor amplifier stages.
2.4.2 Analysis of two stage (CE) Amplifier
21
1
1
vv
o
o
s
o
s
o
vo xAA
v
v
x
v
v
v
v
AA 
(2.16)
(2.18)
11 // rRR Bi 
22 // oCo rRR  (2.17)

Fig. 2.13 Small signal equivalent circuit for two Stage ac-Coupled Amplifiers
211 // RRRB 
T
C
m
V
I
g 1
1 
11
1
mB
T
gI
V
r

 
C
A
o
I
V
r 1
1 
Where:
432 // RRRB 
T
C
m
V
I
g 2
2 
22
2
mB
T
gI
V
r

 
C
A
o
I
V
r 2
2 
and
11 // rRR Bi 
22 // oCo rRR 

• Fig. 2.14 illustrates two Common Emitter (CE) connected in cascade. The
resistors Rx and Ry represent the feedback loop that take part of the output
voltage, Vo and feed it back to the input.
• The input voltage vi is given by:
• The feedback voltage vf is related to the output voltage vo by:
• In order to have a negative feedback, vf must be phase with vs and vo, therefore
two stages or any other even numbed stages are used in cascade since each
stage cause 180o shift. If it is a single stage of CE amplifier (or odd number of
stages), it would be positive feedback.
• Since  is defined, we use Equs. (2.4) and (2.13) to calculate the feedback gain,
Af, input resistance Ri and the output resistance Ri as already explained in this
chapter.
2.4.3 Analysis of two stage (CE) Amplifier with negative Feedback
fsebbei vvvvvv  111
yx
x
o
f
o
yx
x
of
RR
R
v
v
v
RR
R
vv



 
(2.19)
(2.20)

Fig. 2.14 Two stage cascaded common-emitter amplifiers with negative feedback.

Example 2.4
For the two stage identical common-emitter amplifiers shown in Fig. 2.14, let
Vcc = 9V, R1 = R3 = 27 kΩ, R2 = R4 = 15 kΩ, RE1 = RE2 =1.2 kΩ, Rc1 = Rc2 = 2.2
kΩ. The transistors have  = 100 and VA = 100 V. If the amplifier operates
between a source for which Rs = 10 kΩ and a load of RL = 2 kΩ, For the
amplifier without feedback (neglect Rx and Ry) do:
(a) Perform the dc analysis to calculate the currents IE1, IB1, IC1, IE2, IB2 and IC2.
(b) Determine the transistors small signal model parameters r1, gm1, ro1, r1, gm2
and ro2.
(c) Replace the transistors with their models and find the overall values of:
(i) The input resistance Ri . (ii) the output resistance Ro .
(iii) voltage gain Ao = AV = v0/vs (in ratio and dB).
(d) If Rx = 1 kΩ and Ry = 9 kΩ, calculate the feedback factor , amplifier gain Af,
input resistance Rif and output resistance Rof.
Solution:
(a) For dc analysis C1 through C5 are open circuit or of infinite impedance, so the
resulting circuit is shown in Fig. 2.15 (a). When the voltage divider is replaced
with its Thevenin equivalent the equivalent circuit is shown in Fig. 2.15(b).

Example 2.4 Solution:
(a) (b)
Fig. 2.15 Two Stage Amplifier: (a) dc equivalent circuit ; (b) Thevenin equivalent circuits
.

Since the amplifiers are typical, VBB and RB are given by:
Since the amplifiers are typical, the currents IE, IB and IC are given by:
V
x
RR
RV
VV CC
BBBB 21.3
1527
159
21
2
21 




  mA
RR
VV
II
BB
E
BEBB
EE 94.1
101
64.92.1
7.021.3
)1(
21 
















 k
x
RR
RR
RRRR BB 64.9
1527
1527
//
21
21
2121
A
I
II E
BB 

19
101
94.1
1
21 


mAxIII ECC 92.194.199.021  
99.0
101
100
1







(b) The transistors small signal model parameters r, gm and ro. are given by:
The small signal models for both transistors are shown in Fig. 2.15 (c).
(c) From in Fig. 2.15 (c), we can determine the following:
(i) The input resistance Ri = RB1 // r 1 = 9.64 // 1.3 = 1.15 kΩ.
(ii) the output resistance Ro. = Rout1. = RC2 // r 02 = 2.21 // 39.1 = 2.11 kΩ.
The input resistance of the second stage or the load of the first stage is:
Rinput2 = RL1 = RB2 // r 2 = 9.64 // 1.3 = 1.15 kΩ.
VmA
xV
I
gg
T
C
mm /8.76
1025
92.1
3
1
21  
 
k
xgI
V
rr
mB
T
3.1
108.76
100
3
11
21


 
k
xI
V
rr
C
A
oo 1.52
1092.1
100
3
1
21

Example 2.4
Solution:
Fig. 2.15 (c) The small signal models for Two stage ac-coupled amplifier
VmAgg mm /8.7621 
 krr 3.121 
 krr oo 1.5221

(c) From in Fig. 2.15 (c), we can determine the following:
(iii) The overall voltage gain using Eq. (3.1) is as follows:
Then
dBVV
x
v
vRRg
x
RR
R
v
v
x
v
v
v
v
A
in
inLoutm
insig
in
in
o
sig
in
sig
o
v
41.15/896.5
)15.1//11.2(8.76
15.110
15.1)//( 1111
1






21
1
1
vv
o
o
sig
o
sig
o
v xAA
v
v
x
v
v
v
v
A 
dBVV
v
vRRg
v
v
A
o
oLoutm
o
o
v 94.37/86.78)2//11.2(8.76
)//(
1
1
1
2 
dBVVxxAAA vvv 35.53/88.46486.78895.521 

(d) From Equ. (2.20) the feedback factor  is given by:
From Equ. (2.4), the amplifier feedback gain Af, is given by:
From Equ. (2.13) The amplifier feedback input resistance Rif, is given by:
From Equ. (2.8) The amplifier feedback output resistance Rof is given by:
1.0
91
1






kk
k
RR
R
yx
x

79.9
488.47
88.464
11





T
A
A
A
A o
o
o
f

 kTRARR ioiif 6.54)488.47(15.1)1()1( 




 4.44
488.47
11.2
)1()1( T
R
A
R
R o
o
o
of

488.47)88.4641.01()1()1(  xTAFBofamountThe o

Eeg381 electronics iii chapter 2 - feedback amplifiers

More Related Content

What's hot

Similar to Eeg381 electronics iii chapter 2 - feedback amplifiers

Recently uploaded

In this document

Eeg381 electronics iii chapter 2 - feedback amplifiers