# part5

```Part IV
Lectures 19 &amp; 20
Multistage and Feedback
Amplifiers
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 1 of 11
Multistage and Compound Amplifiers
Basic Definitions:
Amplifiers that create voltage, current, and/or power gain through the use of two or
more stages (devices) are called multistage (compound) amplifiers.
The circuitry used to connect the output of one stage of a multistage amplifier to
the input of the next stage is called the coupling method. In general, there are three
coupling methods: RC coupling, direct coupling, and transformer coupling.
A popular connection of amplifier stages is the cascade connection. Basically, a
cascade connection is a series connection with the output of one stage then applied as
input to the second stage. A combination of FET and/or BJT stages can be used to
provide high gain and high input impedance, as demonstrated by the following
example.
Example 19-1:
Calculate the voltage gain, output voltage, input impedance, and output impedance for
the cascade amplifier of Fig. 19-1. Calculate the output voltage resulting if a 10-kΩ
load is connected to the output.
Fig. 19-1
Solution:
For the FET amplifier (stage1),
VGS = −1.9V , I D = 2.8mA , and
gm =
2 I DSS
VP
⎛ VGS
⎜⎜1 −
⎝ VP
⎞ 2(10m) ⎛ − 1.9 ⎞
⎟⎟ =
⎟ = 2.6mS .
⎜1 −
−4 ⎝
−4 ⎠
⎠
University of Technology
Electrical and Electronic Engineering Department
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Multistage and Compound Amplifiers
Lecture Nineteen - Page 2 of 11
For the BJT amplifier (stage2),
VB = 4.7V , VE = 4.0V , VC = 11.2V , I E = 4.0mA , and
0.026 26m
re =
=
= 6.5Ω .
IE
4m
Since ri ( stage2) = 15k 4.7 k 200(6.5) = 953.6Ω , the gain of stage1 (when loaded by
stage2) is:
Av1 = − g m [RD ri ( stage 2)] = −2.6m(2.4k 953.6 ) = −1.77
The voltage gain of stage2 is:
R
2.2k
Av 2 = − C = −
= −338.46
re
6.5
The overall voltage gain is then:
Av = Av1 Av 2 = (−1.77)(−338.46) = 599.1
The output voltage is then:
Vo = AvVi = (599.1)(1mV ) ≈ 0.6V .
The input impedance of the amplifier is that of stage1,
Z i = RG = 3.3MΩ .
While the output impedance of the amplifier is that of stage2,
Z o = RC = 2.2kΩ .
If a 10-kΩ load is connected to the output, the resulting voltage across the load is:
R ⋅V
(10k )(0.6)
VL = L o =
= 0.49V .
RL + Z o 10k + 2.2k
When amplifier stages are cascaded to form a multistage amplifier, the dominant
frequency response is determined by the response of the individual stages. In general,
there are two cases to consider:
Different Cutoff Frequencies:
e When the lower-cutoff frequency, fL, of each amplifier stage is different, the
dominant lower-cutoff frequency, f L′ , equals the cutoff frequency of the stage with
highest fL.
e When the higher-cutoff frequency, fH, of each amplifier stage is different, the
dominant higher-cutoff frequency, f H′ , equals the cutoff frequency of the stage with
lowest fH.
e The overall bandwidth of a multistage amplifier is the difference between the
dominant lower-cutoff frequency and the dominant higher-cutoff frequency.
BW = f H′ − f L′
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 3 of 11
Equal Cutoff Frequencies:
e When the lower-cutoff frequencies of each stage in a multistage amplifier are all the
same, the dominant lower-cutoff frequency is increased by a factor of 1
as shown by the following formula:
fL
f L′ =
21 / n − 1
n: the number of stages in the multistage amplifier.
21/ n − 1
e When the higher-cutoff frequencies of each stage are the same, the dominant higher-
cutoff frequency is reduced by a factor of
formula:
21 / n − 1 as shown by the following
f H′ = f H 21 / n − 1
Example 19-2:
Fig. 19-2 shows an amplifier consisting of a common-emitter stage driving an emitterfollower (a common-collector) stage. The transistors have the following parameter
values: Q1: re1 = 15 Ω, β1 = 180, ro1 ≈ ∞, and Q2: re2 = 25 Ω, β2 = 100, ro2 ≈ ∞.
Find Avs = Vo/Vs, and fL.
+ 18V
150kΩ
CS
100Ω 6 μF
+
Vs
−
4.7 kΩ
CI
68kΩ
CC
0.4μF
39kΩ
1.5kΩ CE
40 μF
47 kΩ
10kΩ
20 μF
+
50Ω Vo
Fig. 19-2
Solution:
r ′ = RE RL = 10k 50 ≈ 50Ω ,
ri ( stage2) = R1 R2 β 2 (re 2 + r ′) = 68k 47k (100)(25 + 50) = 5.9kΩ ,
Av1 = −
RC ri ( stage2)
4.7 k 5.9k
=−
= −174.4
re1
15
−
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 4 of 11
68k 47 k 4.7 k
≈ 40Ω ,
100
β2
r′
50
Av 2 =
=
= 0.435
rs′ + re 2 + r ′ 40 + 25 + 50
rs′ =
R1 R2 RC
=
ri ( stage1) = R1 R2 β1re1 = 150k 39k (180)(15) = 2.48kΩ ,
ri ( stage1)
2.48k
Avs =
⋅ Av1 ⋅ Av 2 =
(−174.4)(0.435) = −72.9
ri ( stage1) + Rs
2.48k + 100
f LS =
1
1
=
= 10.3Hz .
2π [ RS + ri ( stage1)]C S 2π (100 + 2.48k )6 μ
150k 39k 100 ⎞
R R R ⎞
⎛
⎛
⎟⎟ ≈ 15Ω ,
Re1 = RE ⎜⎜ re1 + 1 2 S ⎟⎟ = 1.5k ⎜⎜15 +
β1
180
⎝
⎠
⎝
⎠
1
1
f LE =
=
= 265.3Hz .
2πRe1C E 2π (15)40μ
f LI =
1
1
=
= 37.5 Hz .
2π [ RC + ri ( stage2)]C I 2π (4.7 + 5.9k )0.4 μ
Re 2 = RE (re 2 + rs′) = 10k (25 + 40) ≈ 65Ω ,
1
1
f LC =
=
= 69.3Hz .
2π [ RL + Re 2 ]CC 2π (50 + 65k )20 μ
f L = Max.[ f LS , f LE , f LI , f L C ] = 265.3Hz .
Cascode Amplifiers:
A cascode connection has one transistor one top of (in series with) another. Fig. 19-3a
shows a cascode configuration with common-emitter (CE) stage feeding a commonbase (CB) stage. This arrangement is designed to provide a high input impedance with
low voltage gain to insure that the input miller capacitance is at a minimum with the
CB stage providing good high-frequency operation. A practical BJT version of a
cascode amplifier is provided in Fig. 19-3b.
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 5 of 11
(a)
(b)
Fig. 19-3
Example 19-3:
Calculate the voltage gain for the cascode amplifier of Fig. 19-3b.
Solution:
From dc analysis,
VB1 = 4.9V , VB 2 = 10.8V , and I E1 ≈ I C1 = I E 2 ≈ I C 2 = 3.8mA .
The dynamic resistance of each transistor is then:
0.026 26m
=
= 6.8Ω .
re1 ≈ re 2 =
3.8m
IE
The voltage gain of stage1 (common-emitter) is approximately:
r
Av1 = − e 2 = −1 .
re1
The voltage gain of stage2 (common-base) is:
R
1.8k
Av 2 = C =
= 265 .
6.8
re 2
Resulting in an overall cascode amplifier gain of
Av = Av1 Av 2 = (−1)(265) = −265 .
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 6 of 11
Example 19-4:
The silicon transistors in Fig. 19-4 have the following parameter values:
Q1: β1 = 100, ro1 ≈ ∞, Cbc = 4 pF, Cbe = 10 pF and Q2: α2 ≈ 1, ro2 ≈ ∞.
Find IC1, IC2, VC1, VC2, Avs, and fHi1.
Fig. 19-4
Solution:
10k (12)
= 2.8V , VE1 = VB1 − 0.7 = 2.8 − 0.7 = 2.1V ,
10k + 33k
V
2.1
10k (12)
I C 1 ≈ I E1 = E1 =
= 2.1mA = I E 2 ≈ I C 2 , VB 2 =
= 6V ,
RE1 1k
10k + 10k
VC1 = VE 2 = VB 2 − 0.7 = 6 − 0.7 = 5.3V ,
VC 2 = VCC − I C 2 RC 2 = 12 − 2.1m(2k ) = 7.8V .
VB1 =
0.026 26m
=
= 12.4Ω ,
2.1m
I E1
2k 10k
R R
r
Av1 = − e 2 ≈ −1 , Av 2 = C L =
= 134.4,
12.4
re1
re 2
ri ( stage1) = 33k 10k 100(12.4) = 1.07 kΩ ,
ri ( stage1)
1.07 k
Avs =
⋅ Av1 ⋅ Av 2 =
(−1)(134.4) = −123 .
ri ( stage1) + RS
1.07k + 100
I E1 ≈ I E 2 =&gt; re 2 = re1 =
RTH i1 = ri ( stage1) RS = 1.07 k 100 = 91.5Ω ,
Ci1 = Cbe + C Mi = Cbe + Cbc (1 − Av1 ) = 10 p + 4 p(2) = 18 pF ,
1
1
f Hi1 =
=
= 96.7 MHz .
2πRTHi1Ci1 2π (91.5)(18 p )
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Electrical and Electronic Engineering Department
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Multistage and Compound Amplifiers
Lecture Nineteen - Page 7 of 11
Darlington Amplifiers:
A very popular connection of two BJTs for operation as one &quot;superbeta&quot; transistor is
the Darlington connection as shown in Fig. 19-5. The main feature of the Darlington
connection is that the composite transistor acts as a single unit with current gain that is
the product of the current gains of the individual transistors. That is
β D = β1 β 2 = β 2
Fig. 19-5
Example 19-5:
For the Darlington emitter-follower circuit shown in Fig. 19-6a, determine IB, IC, IE, VE,
VB, VC, Zb, Zi, Zo, Ai, and Av. Use βD = 8000, VBE = 1.6 V, and ri = 5 kΩ.
(a)
(b)
Fig. 19-6
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Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 8 of 11
Solution:
From dc analysis:
V − VBE
18 − 1.6
=
I B = CC
= 2.56 μA ,
RB + β D RE 3.3M + 8000(390)
I E = ( β D + 1) I B ≈ β D I B = I C = 8000(2.56μ ) = 20.48mA ,
VE = I E RE = 20.48m(390) = 8V ,
VB = VE + VBE = 8 + 1.6 = 9.6V , and
VC = VCC = 18V .
From ac equivalent circuit of Fig. 19-6b:
Z b = ri + β D RE = 5k + 8000(390) = 3.125MΩ ,
Z i = RB Z b = 3.3M 3.125M = 1.605MΩ ,
r
r
Z o = RE ri i ≈ i
[Derive]
βD
βD
5k
= 0.625 ,
8000
β R
8000(3.3M )
Ai = D B =
= 4109 , and
RB + Z b 3.3M + 3.125M
RE + β D RE
390 + (8000)(390)
Av =
=
= 0.998 ≈ 1 .
ri + ( RE + β D RE ) 5k + [390 + (8000)(390)]
=
Feedback Pair Amplifiers:
The feedback pair connection (see Fig. 19-7) is a two-transistor circuit that operates
like the Darlington circuit. Notice that the feedback pair uses a pnp transistor driving an
npn transistor, the two devices acting effectively much like one pnp transistor. As with
a Darlington connection, the feedback pair provides very high current gain (the product
of the transistor current gains). A typical application uses a Darlington connection and a
feedback pair connection to provide complementary transistor operation.
Fig. 19-7
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 9 of 11
Example 19-6:
For the feedback pair connection amplifier circuit shown in Fig. 19-8:
1. Calculate the dc bias currents and voltages to provide Vo(dc) at one-half the
supply voltage.
2. Calculate the ac circuit values of Zi, Zo, Ai, and Av. Assume that ri1 = 3 kΩ.
Fig. 19-8
Solution:
From the Q1 bias-emitter loop, on obtains
VCC − I C RC − VEB1 − I B1 RB = 0 , VCC − β1 β 2 I B1 RC − VEB1 − I B1 RB = 0 =&gt;
V − VEB1
18 − 0.7
I B1 = CC
=
= 4.45μA .
RB − β1 β 2 RC 2 M + (140)(180)(75)
The base Q2 current is then
I B 2 = I C1 = β1 I B1 = 140(4.45μ ) = 0.623mA ,
resulting in a Q2 collector current of
I C 2 = β 2 I B 2 = 180(0.623m) = 112.1mA ,
and the current through RC is then
I C = I E1 + I C 2 = 0.623m + 112.1m ≈ I C 2 = 112.1mA .
The dc voltage at the output is thus
Vo (dc) = VCC − I C RC = 18 − 112.1m(75) = 9.6V ,
and
Vi (dc) = Vo (dc) − VEB1 = 9.6 − 0.7 = 8.9V .
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 10 of 11
From the ac equivalent circuits of Fig. 19-9a and b:
Z i ≈ RB (ri1 + β1β 2 RC )
= 2 M [3k + (140)(180)] = 974kΩ .
Z o = RC ri1 (ri1 β1 ) [ri1 ( β1 β 2 )] ≅
=
Ai =
ri1
β1 β 2
=
[Derive]
3k
= 0.12Ω .
(140)(180)
β1 β 2 R B
[Derive]
RB + Z i
(140)(180)(2 M )
=
= 16950 .
2 M + 974k
Av =
[Derive]
β1 β 2 RC
β1 β 2 RC + ri1
[Derive]
(140)(180)(75)
= 0.9984 ≈ 1 .
(140)(180)(75) + 3k
(a)
(b)
Fig. 19-9
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Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Multistage and Compound Amplifiers
Lecture Nineteen - Page 11 of 11
Exercises:
1. The transistors in Fig. 19-10 have the following parameter values: Q1: gm1 = 4 mS,
rd1 ≈ ∞, and Q2: re2 = 30 Ω, ro2 ≈ ∞. Find Avs.
Fig.19-10
2. Fig. 19-11 shows a common-emitter stage driving a Darlington pair connected as an
emitter follower. The β-values for the silicon transistors are β1 = 200, β2 = 100, and
β3 = 100. Find Avs.
Fig. 19-11
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Feedback Amplifiers
Lecture Twenty - Page 1 of 13
Feedback Amplifiers
Feedback Concepts:
A typical feedback connection is shown in Fig. 20-1. The input signal, Vs, is applied to
a mixer network, where it is combined with a feedback signal, Vf. The difference of
these signals, Vi, is then the input voltage to the amplifier. A portion of the amplifier
output (sampled signal), Vo, is connected to the feedback network (β), which provides a
reduced portion of the output as feedback signal to the input mixer network.
If the feedback signal is of opposite polarity to the input signal, as shown in
Fig. 20-1, negative feedback results. While negative feedback results in reduced overall
voltage gain, a number of improvements are obtained, among them being:
1.
2.
3.
4.
5.
6.
Higher input impedance.
Lower output impedance.
Better stabilized voltage gain.
Improved frequency response.
Reduced noise.
More linear operation.
Fig. 20-1
Feedback Connection Types:
There are four basic ways of connecting the feedback signal. Both voltage and current
can be fed back to the input either in series or parallel. Specifically, there can be:
1. Voltage-series feedback (Fig. 20-2a).
2. Voltage-shunt feedback (Fig. 20-2b).
3. Current-series feedback (Fig. 20-2c).
4. Current-shunt feedback (Fig. 20-2d).
In the list above, voltage refers to connecting the output voltage as input to the
feedback network; current refers to tapping off some output current through the
feedback network. Series refers to connecting the feedback signal in series with the
input signal voltage; shunt refers to connecting the feedback signal in shunt (parallel)
with an input current source.
Generally, series feedback connections tend to increase the input resistance,
while shunt feedback connections tend to decrease the input resistance. Voltage
feedback tends to decrease the output impedance, while current feedback tends to
increase the output impedance.
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 2 of 13
(a) Voltage-Series Feedback.
(b) Voltage-Shunt Feedback.
(c) Current-Series Feedback.
(d) Current-Shunt Feedback.
Fig. 20-2
Gain with Feedback:
The gain without feedback, A, is that of the amplifier stage. With feedback, β, the
overall gain of the circuit is reduced by a factor (1 + βA), as detailed below. A summary
of the gain, feedback factor, and gain with feedback of Fig. 20-2 is provided for
reference in Table 20-1.
Table 20-1
Feedback types
Parameters
Voltage-Series Voltage-Shunt Current-Series Current-Shunt
Gain without feedback A
Feedback
β
Gain with feedback
Af
Vo
Vi
Vf
Vo
Ii
If
Io
Vi
Vf
Io
Ii
If
Vo
Vo
Vs
Vo
Vo
Is
Io
Io
Vs
Io
Io
Is
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Voltage-Series Feedback (Voltage Amplifier):
From Fig. 20-2a and Table 20-1;
V
Vo
Vo
AVi
,
Af = o =
=
=
Vs Vi + V f Vi + βVo Vi + βAVi
Avf =
Vo
Av
=
Vs 1 + β v Av
Voltage-Shunt Feedback (Transresistance Amplifier):
From Fig. 20-2b and Table 20-1;
V
Vo
Vo
AI i
=
=
,
Af = o =
I s Ii + I f
I i + β Vo I i + β AI i
Azf =
Vo
Az
=
I s 1 + β g Az
Current-Series Feedback (Transconductance Amplifier):
From Fig. 20-2c and Table 20-1;
I
Io
Io
AVi
=
=
,
Af = o =
Vs Vi + V f Vi + βI o Vi + βAVi
Agf =
Ag
Io
=
Vs 1 + β z Ag
Current-Shunt Feedback (Current Amplifier):
From Fig. 20-2d and Table 20-1;
I
Io
Io
AI i
Af = o =
=
=
I s Ii + I f
I i + β I o I i + β AI i
Aif =
Io
Ai
=
I s 1 + β i Ai
Feedback Amplifiers
Lecture Twenty - Page 3 of 13
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 4 of 13
Input Impedance with Feedback:
The input impedance for the connections of Fig. 20-2 are dependent on whether series
or shunt feedback is used. For series feedback, the input impedance is increased, while
shunt feedback decreases the input impedance.
Series Feedback:
From Fig. 20-3 with voltage-series feedback;
Vi + V f Vi + βVo Vi + βAVi
V
Z if = s =
=
=
= Z i + ( βA) Z i ,
Ii
Ii
Ii
Ii
Z if = Z i (1 + β A)
The input impedance with series feedback is seen to be the value of the input
impedance without feedback multiplied by the factor (1 + βA) and applies to both
voltage-series (Fig. 20-2a) and current-series (Fig. 20-2c) configurations.
Fig, 20-3
Shunt Feedback:
From Fig. 20-4 with voltage-shunt feedback;
V
Vi
Vi
Vi I i
,
Z if = i =
=
=
I s Ii + I f
I i + β Vo I i I i + β Vo I i
Z if =
Zi
1 + βA
This reduced input impedance applies to the voltage-shunt connection of
Fig. 20-2b and the current-shunt connection of Fig. 18.2d.
University of Technology
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Feedback Amplifiers
Lecture Twenty - Page 5 of 13
Fig, 20-4
Output Impedance with Feedback:
The output impedance for the connections of Fig. 20-2 are dependent on whether
voltage or current feedback is used. For voltage feedback, the output impedance is
decreased, while current feedback increases the output impedance.
Voltage Feedback:
For the voltage-series feedback circuit of Fig. 20-3, 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
V = IZ o + AVi ,
Vi = −V f for Vs = 0 ,
V = IZ o − AV f = IZ o − A( βV ) =&gt; V + βAV = IZ o ,
Z of =
Zo
V
=
I 1 + βA
The above equation shows that with voltage feedback the output impedance is
reduced from that without feedback by the factor (1 + βA).
Current Feedback:
From Fig. 20-5 with current-series feedback;
Vi = V f for Vs = 0 ,
I=
Z of
V
V
V
− AVi =
− AV f =
− AβI =&gt; Z o (1 + βA) I = V ,
Zo
Zo
Zo
V
= = Z o (1 + βA)
I
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 6 of 13
Fig, 20-5
A summary of the effect of feedback on input and output impedance is provided
in Table 20-2.
Table 20-2
Impedances
Z if
Input
Output
Feedback types
Voltage-Series Voltage-Shunt Current-Series Current-Shunt
Z i (1 + βA)
Zi
1 + βA
Z i (1 + βA)
Zi
1 + βA
(increased)
(decreased)
(increased)
(decreased)
Zo
1 + βA
Zo
1 + βA
Z o (1 + βA)
Z o (1 + βA)
(decreased)
(decreased)
(increased)
(increased)
Z of
Gain Stability (Sensitivity and Desensitivity) with Feedback:
The fractional change in amplification with feedback divided by the fractional change
without feedback is called the sensitivity of the gain. If the equation Af = A/(1+ βA) is
differentiated with respect to A, the absolute value of resulting equation is
dA f
A
=
1 dA
1 + βA A
Hence the sensitivity is 1/|1+ βA|. This shows that magnitude of the relative change in
gain with feedback is reduced by the |1+ βA| compared to that without feedback. The
reciprocal of sensitivity is called the desensitivity D, or
D = 1 + βA
The fractional change in gain without feedback is divided by the desensitivity D when
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 7 of 13
In particular, if |βA| &gt;&gt; 1, then
Af =
A
A
1
≈
=
1 + βA βA β
and the gain may be made to depend entirely on the feedback network. The worst
offenders with respect to stability are usually the active devices (transistors) involved.
If the feedback network contains only stable passive elements, the improvement in
stability may indeed be pronounced.
Bandwidth with Feedback:
Fig. 20-6 shows that the amplifier with negative feedback has more bandwidth (Bf) than
the amplifier without feedback (B). The feedback amplifier has a higher upper 3-dB
frequency and smaller lower 3-dB frequency.
Fig. 20-6
Method of Analysis of A Feedback Amplifier:
It is desirable to separate the feedback amplifier into two blocks, the basic amplifier A
and the feedback network β, because with a knowledge of A and β, we can calculate the
important parameters of the feedback amplifier, namely, Af, Zif, and Zof. The basic
amplifier configuration without feedback but taking the loading of the β network into
account is obtained by applying the following rules:
To find the input circuit:
1. Set Vo = 0 for voltage feedback (sampling). In other words, short the output node.
2. Set Io = 0 for current feedback (sampling). In other words, open the output loop.
To find the output circuit:
1. Set Vi = 0 for shunt feedback. In other words, short the input node.
2. Set Ii = 0 for series feedback. In other words, open the input loop.
These procedures ensure that the feedback is reduced to zero without altering the
University of Technology
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Feedback Amplifiers
Lecture Twenty - Page 8 of 13
The complete analysis of a feedback amplifier is obtained by carrying out the
following steps:
1. Identify the topology;
(a) Is the sampled signal Xo a voltage or a current? In other words, is Xo taken at
the output node or from the output loop?
(b) Is the feedback signal Xf a voltage or a current? In other words, is Xf applied
in series or in shunt with the external excitation?
2. Draw the basic amplifier circuit without feedback, following the rules listed above.
3. Use a Thevenin's source if Xf is a voltage and a Norton's source if Xf is a current.
4. Replace each active device by the proper model.
5. Indicate Xf and Xo on the circuit obtained by carrying out steps 2, 3, and 4.
Evaluate β = Xf / Xo.
6. Evaluate A by applying KVL and KCL to the equivalent circuit obtained after
step 4.
7. From A and β, find D = 1 + βA, Af, Zif, and Zof.
Table 20-3 summarizes the above procedure and should be referred to when
carrying out the analyses of the feedback circuits discussed in the following examples.
Table 20-3
Feedback topology
Parameters
Amplifier type
(1)
(2)
(3)
(4)
Voltage-Series
Voltage-Shunt
Current-Series
Current-Shunt
Voltage
Transresistance
Transconductance
Current
Sampled signal Xo
Voltage (Shunt) Voltage (Shunt)
Current (Series)
Current (Series)
Feedback signal Xf
Voltage (Series) Current (Shunt)
Voltage (Series)
Current (Shunt)
Vo = 0
Ii = 0
Vo = 0
Vi = 0
Io = 0
Ii = 0
Io = 0
Vi = 0
Signal source
Thevenin
Norton
Thevenin
Norton
A = Xo Xi
β = X f Xo
Av = Vo Vi
β v = V f Vo
Az = Vo I i
β g = I f Vo
Ag = I o Vi
Ai = I o I i
βi = I f Io
D = 1 + βA
1 + β v Av
1 + β g Az
Av D
Zi D
Zo D
Az D
Zi D
Zo D
To find input loop, set
To find output loop, set
Af
Zif
Zof
β z = V f Io
1 + β z Ag
Ag D
Zi D
Zo D
1 + β i Ai
Ai D
Zi D
Zo D
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 9 of 13
Example 20-1 (Voltage-Series Feedback):
Calculate Avf, Zif, and Zof for the amplifier of Fig. 20-7(a). Assume RS = 0, hfe = 50,
hie = 1.1 kΩ, hre = hoe = 0, and identical transistors.
+ 25V
+ 25V
10kΩ
150kΩ
47 kΩ 4.7 kΩ
Vo
Vs
Q1
47 kΩ
Q2
33kΩ
4.7 kΩ
Z if
150kΩ
10kΩ
Vo
Vs
Z of
R2 = 4.7kΩ
47 kΩ 4.7kΩ
Q1
47 kΩ
4.7 kΩ
Q2
4.7 kΩ
33kΩ
R2
4.7 kΩ
R1 V f
+
−
R1
R1 = 0.1kΩ
R2
(a)
(b)
Fig. 20-7
Solution:
Referring to the first topology (voltage-series) in Table 20-3 and from the amplifier
circuit without feedback of Fig. 20-7(b);
RL1 = 10k 47 k 33k 1.1k = 942Ω , RE1 = R1 R2 = 0.1k 4.7 k = 98Ω ,
− h fe RL1
− 50(942)
Av1 =
=
= −7.8
hie + h fe RE1 1.1k + 50(98)
RL 2 = 4.7 k (4.7 k + 0.1k ) = 2.37 kΩ , Av 2 =
− h fe RL 2
hie
=
− 50(2.37 k )
= −108 .
1.1k
Vf
Vo
R1
0.1k
1
= Av1 Av 2 = (−7.8)(−108) = 842 , β v =
=
=
=
,
Vi
Vo R1 + R2 0.1k + 4.7 k 48
A
842
842
D = 1 + β v Av = 1 +
= 18.5 , Avf = v =
= 45.5
48
D 18.5
Av =
Z i = hie + h fe RE1 = 1.1k + 50(98) = 6kΩ , Z if = Z i D = 6k (18.5) = 111kΩ .
Z o = RL 2 = 2.37 kΩ , Z of =
Z o 2.37 k
=
= 128Ω .
D
18.5
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 10 of 13
Example 20-2 (Voltage-Shunt Feedback):
The amplifier of Fig. 20-8(a) has the following parameters: RC = 4 kΩ, RF = 40 kΩ,
RS = 10 kΩ, hie = 1.1 kΩ, hfe = 50, and hre = hoe = 0. Find Avf, the impedance seen
by the voltage source, and Zof.
RF
+ VCC
+ VCC
RC
RC
RS
Vs
Vo
Vo
RB
Z of
V
Is = s
RS
Z if
If
RS
(a)
RF
RF
(b)
Fig. 20-8
Solution:
Referring to the second topology (voltage-shunt) in Table 20-3 and from the amplifier
circuit without feedback of Fig. 20-8(b);
RL = RC RF = 4k 40k = 3.64kΩ , RB = RS RF = 10k 40k = 8kΩ ,
− h fe I b RL − h fe RL RB (−50)(3.64k )(8k )
V
V
Az = o = o =
=
=
= −160kΩ ,
Ii
Is
Is
RB + hie
8k + 1.1k
I
1
1
βg = f = −
=−
= −0.025mS , D = 1 + β g Az = 1 + (0.025m)(160k ) = 5 ,
Vo
RF
40k
Azf − 32k
V
V
A
− 160k
Azf = z =
= −32kΩ , Avf = o = o =
=
= −3.2
Vs I s R S
RS
10k
D
5
Z i = RB hie = 8k 1.1k = 967Ω , Z if =
Z i 967
=
= 193Ω ,
D
5
(10k )(193)
= 197Ω ,
10k − 193
= 10k + 197 = 10.2kΩ .
Z if = Z bf RS ⇒ Z bf =
Z sf = RS + Z bf
Z o = RL = 3.64kΩ , Z of =
Z o 3.64k
=
= 728Ω .
D
5
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 11 of 13
Example 20-3 (Current-Series Feedback):
For the amplifier of Fig. 20-9(a), write suitable mathematical expressions to determine
Avf and Zif.
+ VCC
RC
RS
Vs
Io
Vo
RS
Z if
Vs
RE
+
Vo RC
−
RE
RE
+V
(a)
f
−
(b)
Fig. 20-9
Solution:
Referring to the third topology (current-series) in Table 20-3 and from the amplifier
circuit without feedback of Fig. 20-9(b);
− h fe
I o − h fe I b
=
=
Vi
Vs
RS + hie + RE
V
−I R
β z = f = o E = − RE
Io
Io
h fe RE
RS + hie + (1 + h fe ) RE
=
D = 1 + β z Ag = 1 +
RS + hie + RE
RS + hie + RE
Ag
− h fe
Agf =
=
D RS + hie + (1 + h fe ) RE
Ag =
Avf =
− h fe RC
I o RC
= Agf RC =
Vs
RS + hie + (1 + h fe ) RE
Z i = RS + hie + RE
Z if = Z i D = RS + hie + (1 + h fe ) RE
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 12 of 13
Example 20-4 (Current-Shunt Feedback):
The amplifier of Fig. 20-10(a) has the following parameters: RC1 = 3 kΩ, RC2 = 0.5 kΩ,
RE2 = 50 Ω, RF = RS = 1.2 kΩ, hie = 1.1 kΩ, hfe = 50, and hre = hoe = 0. Find Avf and
the impedance seen by the voltage source.
+ VCC
+ VCC
RC1
RC1
Vo
RS
Vs
RC 2
Q1
Z if
RF
Io
RC 2
Vo
R
Q2
Q1
Is
RE 2
RS
Q2
RF
RE 2
(a)
RF
If
RE 2
(b)
Fig. 20-10
Solution:
Referring to the fourth topology (current-shunt) in Table 20-3 and from the amplifier
circuit without feedback of Fig. 20-10(b);
Ri 2 = hie + h fe ( RE 2 RF ) = 1.1k + 50(50 1.2k ) = 3.5kΩ ,
R = R S ( R F + R E ) = 1 .2 k (1 .2 k + 50 ) = 1 .2 k 1 .25 k = 612 Ω ,
I
− RC1
− I c 2 − I c 2 I b 2 I c1 I b1
R
Ai = o =
⋅ h fe ⋅
=
⋅
⋅
⋅
= −h fe ⋅
Ii
Is
I b 2 I c1 I b1 I s
RC1 + Ri 2
R + hie
⎛ − 3k ⎞
⎛ 612 ⎞
= (−50)⎜
⎟(50)⎜
⎟ = (−50)(−0.462)(50)(0.358) = 413 ,
⎝ 3k + 3.5k ⎠
⎝ 612 + 1.1k ⎠
I
RE 2
50
50
βi = f =
=
=
= 0.04 ,
I o RF + RE 2 1.2k + 50 1250
D = 1 + β i Ai = 1 + (0.04)(413) = 17.5 ,
V
− I c 2 RC 2 Aif RC 2 (23.6)(0.5k )
A
413
=
=
= 9.8
Aif = i =
= 23.6 , Avf = o =
Vs
I s RS
RS
1.2k
D 17.5
Z i = R hie = 612 1.1k = 393Ω , Z if =
(1.2k )(23)
= 23.5Ω ,
1.2k − 23
= 1.2k + 23.5 = 1.22kΩ .
Z if = Z bf RS ⇒ Z bf =
Z sf = RS + Z bf
Z i 393
=
= 23Ω ,
D 17.5
University of Technology
Electrical and Electronic Engineering Department
Second Year, Electronics I, 2009 - 2010
Feedback Amplifiers
Lecture Twenty - Page 13 of 13
Exercises:
1. For each amplifier in Fig. 20-11, write a mathematical expression to calculate Avf .
+ VCC
RC
RS
+ VDD
Vs
RE
Vs
Vo
+
Vo
+
RS
−
−
(a)
(b)
+ VDD
+ VDD
RD
R1
+
R2
Vs
RC
Vo
+
Vo RL
+
−
Vs
−
RS
−
(c)
(d)
Fig. 20-11
2. The amplifier circuit of Fig. 20-9(a) is to have an overall transconductance
gain of –1 mS, a voltage gain of –4, and a desensitivity of 50. If RS = 1 kΩ,
hfe = 150, and ro is negligible. Find RE, RC, Zif, and ICQ.
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