part iv lectures 19 & 20 multistage and feedback...

Part IV Lectures 19 & 20

Multistage and Feedback Amplifiers

University of Technology Multistage and Compound Amplifiers Electrical and Electronic Engineering Department Lecture Nineteen - Page 1 of 11

Second Year, Electronics I, 2009 - 2010 Dr. Ahmed Saadoon Ezzulddin

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. Cascade Amplifiers: 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), , , and VVGS 9.1−= mAI D 8.2=

mSmVV

VIg

P

GS

P

DSSm 6.2

49.11

4)10(212

=⎟⎠⎞

⎜⎝⎛

−−

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−= .



For the BJT amplifier (stage2), , VVB 7.4= VVE 0.4= , VVC 2.11= , mAI E 0.4= , and

Ω=== 5.64

26026.0mm

Ir

Ee .

Since Ω== 6.953)5.6(2007.415)2( kkstageri , the gain of stage1 (when loaded by stage2) is: [ ] ( ) 77.16.9534.26.2)2(1 −=−=−= kmstagerRgA iDmv The voltage gain of stage2 is:

46.3385.6

2.22 −=−=−=

krRA

e

Cv

The overall voltage gain is then: 1.599)46.338)(77.1(21 =−−== vvv AAA The output voltage is then: . VmVVAV ivo 6.0)1)(1.599( ≈==The input impedance of the amplifier is that of stage1, Ω== MRZ Gi 3.3 . While the output impedance of the amplifier is that of stage2, Ω== kRZ Co 2.2 . If a 10-kΩ load is connected to the output, the resulting voltage across the load is:

Vkk

kZRVRV

oL

oLL 49.0

2.210)6.0)(10(=

+=

+⋅

= .

Frequency Response of Cascade Amplifiers: 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:

When the lower-cutoff frequency, fL, of each amplifier stage is different, the dominant lower-cutoff frequency, Lf ′ , equals the cutoff frequency of the stage with highest fL.

When the higher-cutoff frequency, fH, of each amplifier stage is different, the dominant higher-cutoff frequency, Hf ′ , equals the cutoff frequency of the stage with lowest fH.

The overall bandwidth of a multistage amplifier is the difference between the dominant lower-cutoff frequency and the dominant higher-cutoff frequency.

LH ffBW ′−′=



+

−sV

SC

V18+

IC

CC

+

−oV

EC

Ωk68

Ωk47Ωk10 Ω50

Ωk7.4Ωk150

Ωk39Ωk5.1

Ω100 Fμ6

Fμ40

Fμ20

Fμ4.0

Equal Cutoff Frequencies:

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 121 /1 −n as shown by the following formula:

12 /1 −

=′nL

Lff n: the number of stages in the multistage amplifier.

When the higher-cutoff frequencies of each stage are the same, the dominant higher-cutoff frequency is reduced by a factor of 12 /1 −n as shown by the following formula:

12 /1 −=′ nHH ff

Example 19-2: Fig. 19-2 shows an amplifier consisting of a common-emitter stage driving an emitter-follower (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.

Fig. 19-2 Solution: Ω≈==′ 505010kRRr LE , Ω=+=′+= kkkrrRRstager ei 9.5)5025)(100(4768)()2( 2221 β ,

4.17415

9.57.4)2(

11 −=−=−=

kkrstagerR

Ae

iCv



Ω≈==′ 40100

7.44768

2

21 kkkRRRr C

s β,

435.0502540

502

2 =++

=′++′

′=

rrrrAes

v

Ω=== kkkrRRstager ei 48.2)15)(180(39150)1( 1121 β ,

9.72)435.0)(4.174(10048.2

48.2)1(

)1(21 −=−

+=⋅⋅

+=

kkAA

RstagerstagerA vv

si

ivs

HzkCstagerR

fSiS

LS3.10

6)48.2100(21

)]1([21

=+

=+

=μππ

.

Ω≈⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 15

18010039150

155.11

2111

kkk

RRRrRR SeEe β

,

HzCR

fEe

LE3.265

40)15(21

21

1===

μππ.

HzkCstagerR

fIiC

LI5.37

4.0)9.57.4(21

)]2([21

=+

=+

=μππ

.

Ω≈+=′+= 65)4025(10)( 22 krrRR seEe ,

HzkCRR

fCeL

LC3.69

20)6550(21

][21

2=

+=

+=

μππ.

HzffffMaxf

CIES LLLLL 3.265],,,.[ == . 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 common-base (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.



(a) (b)

Fig. 19-3 Example 19-3: Calculate the voltage gain for the cascode amplifier of Fig. 19-3b. Solution: From dc analysis, , , and VVB 9.41 = VVB 8.102 = mAIIII CECE 8.32211 =≈=≈ . The dynamic resistance of each transistor is then:

Ω===≈ 8.68.3

26026.021 m

mI

rrE

ee .

The voltage gain of stage1 (common-emitter) is approximately:

11

21 −=−=

e

ev r

rA .

The voltage gain of stage2 (common-base) is:

2658.6

8.12

2 ===k

rRA

e

Cv .

Resulting in an overall cascode amplifier gain of 265)265)(1(21 −=−== vvv AAA .



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:

Vkk

kVB 8.23310

)12(101 =

+= , VVV BE 1.27.08.27.011 =−=−= ,

221

111 1.2

11.2

CEE

EEC IImA

kRVII ≈====≈ , V

kkkVB 6

1010)12(10

2 =+

= ,

VVVV BEC 3.57.067.0221 =−=−== , VkmRIVV CCCCC 8.7)2(1.212222 =−=−= .

=> 21 EE II ≈ Ω==== 4.121.2

26026.01

12 mm

Irr

Eee ,

11

21 −≈−=

e

ev r

rA , ,4.1344.12

102

22 ===

kkr

RRA

e

LCv

Ω== kkkstageri 07.1)4.12(1001033)1( ,

123)4.134)(1(10007.1

07.1)1(

)1(21 −=−

+=⋅⋅

+=

kkAA

RstagerstagerA vv

Si

ivs .

Ω=== 5.9110007.1)1(

1kRstagerR SiTHi

, pFppACCCCC vbcbeMibei 18)2(410)1( 11 =+=−+=+= ,

MHzpCR

fiTHi

Hi 7.96)18)(5.91(2

12

111

1 ===ππ

.



Darlington Amplifiers: A very popular connection of two BJTs for operation as one "superbeta" 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 2

21 ββββ ==D

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



Solution: From dc analysis:

AMRR

VVIEDB

BECCB μ

β56.2

)390(80003.36.118

=+−

=+−

= ,

mAIIII CBDBDE 48.20)56.2(8000)1( ===≈+= μββ , , VmRIV EEE 8)390(48.20 === , and VVVV BEEB 6.96.18 =+=+= . VVV CCC 18== From ac equivalent circuit of Fig. 19-6b: Ω=+=+= MkRrZ EDib 125.3)390(80005β , Ω=== MMMZRZ bBi 605.1125.33.3 ,

D

i

D

iiEo

rrrRZββ

≈= [Derive]

625.080005

==k ,

4109125.33.3

)3.3(8000=

+=

+=

MMM

ZRRA

bB

BDi

β , and

1998.0)]390)(8000(390[5

)390)(8000(390)(

≈=++

+=

+++

=kRRr

RRAEDEi

EDEv β

β .

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



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 , 0 011121 =11 =−−− BBEBCCCC RIVRIV −−− BBEBCBCC RIVRIV ββ =>

AMRR

VVICB

EBCCB μ

ββ45.4

)75)(180)(140(27.018

21

11 =

+−

=−−

= .

The base Q2 current is then mAIII BCB 623.0)45.4(1401112 ==== μβ , resulting in a Q2 collector current of mAmII BC 1.112)623.0(180222 === β , and the current through RC is then mAImmIII CCEC 1.1121.112623.0 221 =≈+=+= . The dc voltage at the output is thus VmRIVdcV CCCCo 6.9)75(1.11218)( =−=−= , and VVdcVdcV EBoi 9.87.06.9)()( 1 =−=−= .



From the ac equivalent circuits of Fig. 19-9a and b: )( 211 CiBi RrRZ ββ+≈ [Derive] [ ] Ω=+= kkM 974)180)(140(32 .

21

1211111 ])([)(

βββββ i

iiiCorrrrRZ ≅= [Derive]

Ω== 12.0)180)(140(

3k .

iB

Bi ZR

RA+

= 21ββ [Derive]

169509742

)2)(180)(140(=

+=

kMM .

121

21

iC

Cv rR

RA+

=ββββ

[Derive]

19984.03)75)(180)(140(

)75)(180)(140(≈=

+=

k.

(a)

(b)

Fig. 19-9



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

University of Technology Feedback Amplifiers Electrical and Electronic Engineering Department 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. Higher input impedance. 2. Lower output impedance. 3. Better stabilized voltage gain. 4. Improved frequency response. 5. Reduced noise. 6. 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.



(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 i

o

VV

i

o

IV

i

o

VI

i

o

II

Feedback β o

f

VV

o

f

VI

o

f

IV

o

f

II

Gain with feedback Af s

o

VV

s

o

IV

s

o

VI

s

o

II



Voltage-Series Feedback (Voltage Amplifier): From Fig. 20-2a and Table 20-1;

ii

i

oi

o

fi

o

s

of AVV

AVVV

VVV

VVVA

ββ +=

+=

+== ,

vv

v

s

ovf A

AVVA

β+==

1

Voltage-Shunt Feedback (Transresistance Amplifier): From Fig. 20-2b and Table 20-1;

ii

i

oi

o

fi

o

s

of AII

AIVI

VII

VIVA

ββ +=

+=

+== ,

zg

z

s

ozf A

AIVA

β+==

1

Current-Series Feedback (Transconductance Amplifier): From Fig. 20-2c and Table 20-1;

ii

i

oi

o

fi

o

s

of AVV

AVIV

IVV

IVIA

ββ +=

+=

+== ,

gz

g

s

ogf A

AVIA

β+==

1

Current-Shunt Feedback (Current Amplifier): From Fig. 20-2d and Table 20-1;

ii

i

oi

o

fi

o

s

of AII

AIII

III

IIIA

ββ +=

+=

+==

ii

i

s

oif A

AIIA

β+==

1



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;

iii

ii

i

oi

i

fi

i

sif ZAZ

IAVV

IVV

IVV

IVZ )(βββ

+=+

=+

=+

== ,

)1( AZZ iif β+= 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;

ioii

ii

oi

i

fi

i

s

iif IVII

IVVI

VII

VIVZ

ββ +=

+=

+== ,

A

ZZ iif β+=

1

This reduced input impedance applies to the voltage-shunt connection of Fig. 20-2b and the current-shunt connection of Fig. 18.2d.



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 , io AVIZV += for , fi VV −= 0=sV )( VAIZAVIZV ofo β−=−= => oIZAVV =+ β ,

A

ZIVZ o

of β+==

1

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; for fi VV = 0=sV ,

IAZVAV

ZVAV

ZVI

of

oi

oβ−=−=−= => VIAZo =+ )1( β ,

)1( AZIVZ oof β+==



Fig, 20-5 A summary of the effect of feedback on input and output impedance is provided in Table 20-2.

Table 20-2

Feedback types Impedances Voltage-Series Voltage-Shunt Current-Series Current-Shunt

)1( AZi β+ A

Zi

β+1 )1( AZi β+

AZi

β+1 Input ifZ

(increased) (decreased) (increased) (decreased)

AZo

β+1

AZo

β+1 )1( AZo β+ )1( AZo β+ Output ofZ

(decreased) (decreased) (increased) (increased) 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

A

dAAA

dAf

β+=

11

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

AD β+=1

The fractional change in gain without feedback is divided by the desensitivity D when feedback is added.



In particular, if |βA| >> 1, then

βββ1

1=≈

+=

AA

AAAf

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 loading on the basic amplifier.



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 (1) (2) (3) (4) Parameters

Voltage-Series Voltage-Shunt Current-Series Current-Shunt

Amplifier type 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)

To find input loop, set 0=oV 0=oV 0=oI 0=oI To find output loop, set 0=iI 0=iV 0=iI 0=iV Signal source Thevenin Norton Thevenin Norton

io XXA = iov VVA = ioz IVA = iog VIA = ioi IIA =

of XX=β ofv VV=β ofg VI=β ofz IV=β ofi II=β AD β+=1 vv Aβ+1 zg Aβ+1 gz Aβ+1 ii Aβ+1

Af DAv DAz DAg DAi Zif DZi DZi DZi DZi Zof DZo DZo DZo DZo



sV

V25+

Ωk7.4Ωk10Ωk150

oV

Ωk47Ωk7.4

Ω= kR 1.01

Ω= kR 7.42

Ωk7.4

Ωk331Q 2Q

Ωk47

ifZ

ofZ sV

V25+

Ωk7.4Ωk10Ωk150

oV

Ωk47Ωk7.4

1R

Ωk7.4

Ωk331Q 2Q

Ωk47

2R

1R

2R+

−fV

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. (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); Ω== 9421.13347101 kkkkRL , Ω=== 987.41.0211 kkRRRE ,

8.7)98(501.1

)942(501

11 −=

+−

=+

−=

kRhhRh

AEfeie

Lfev

Ω=+= kkkkRL 37.2)1.07.4(7.42 , 1081.1

)37.2(5022 −=

−=

−=

kk

hRh

Aie

Lfev .

842)108)(8.7(21 =−−=== vvi

ov AA

VVA ,

481

7.41.01.0

21

1 =+

=+

==kk

kRR

RVV

o

fvβ ,

5.1848

84211 =+=+= vv AD β , 5.455.18

842===

DAA v

vf

Ω=+=+= kkRhhZ Efeiei 6)98(501.11 , Ω=== kkDZZ iif 111)5.18(6 .

, Ω== kRZ Lo 37.22 Ω=== 1285.18

37.2 kDZZ o

of .



oV

SRFR

CRCCV+

FRS

ss R

VI =fI

BRoV

ifZ

ofZSR

FRCR

CCV+

sV

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. (a) (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); Ω=== kkkRRR FCL 64.3404 , Ω=== kkkRRR FSB 84010 ,

Ω−=+

−=

+

−=

−=== k

kkkk

hRRRh

IRIh

IV

IVA

ieB

BLfe

s

Lbfe

s

o

i

oz 160

1.18)8)(64.3)(50( ,

mSkRV

I

Fo

fg 025.0

4011

−=−=−==β , 5)160)(025.0(11 =+=+= kmAD zgβ ,

Ω−=−

== kkDAA z

zf 325

160 , 2.31032

−=−

====kk

RA

RIV

VVA

S

zf

Ss

o

s

ovf

Ω=== 9671.18 kkhRZ ieBi , Ω=== 1935

967DZZ i

if ,

Ω=−

=⇒= 19719310

)193)(10(kkZRZZ bfSbfif ,

Ω=+=+= kkZRZ bfSsf 2.1019710 .

Ω== kRZ Lo 64.3 , Ω=== 728564.3 k

DZZ o

of .



oV

ifZ

SR

CRCCV+

sVER

SR

sVER ER

CR

fV

oI

+ −

+

−oV

Example 20-3 (Current-Series Feedback): For the amplifier of Fig. 20-9(a), write suitable mathematical expressions to determine Avf and Zif. (a) (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);

EieS

fe

s

bfe

i

og RhR

hV

IhVIA

++

−=

−==

Eo

Eo

o

fz R

IRI

IV

−=−

==β

EieS

EfeieS

EieS

Efegz RhR

RhhRRhR

RhAD

++

+++=

+++=+=

)1(11 β

EfeieS

feggf RhhR

hDA

A)1( +++

−==

EfeieS

CfeCgf

s

Covf RhhR

RhRA

VRIA

)1( +++

−===

EieSi RhRZ ++= EfeieSiif RhhRDZZ )1( +++==



ifZ

SR

1CR

CCV+

sV

2CR

2ER

oV

FR1Q 2Q

SR

1CR

CCV+

2CR

2ER

oV

1Q 2Q

FR

2ERsI

R

FR fI

oI

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. (a) (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); Ω=+=+= kkkRRhhR FEfeiei 5.3)2.150(501.1)( 22 , Ω==+=+= 61225.12.1)502.1(2.1)( kkkkRRRR EFS ,

ie

feiC

Cfe

s

b

b

c

c

b

b

c

s

c

i

oi hR

RhRR

RhII

II

II

II

II

IIA

+⋅⋅

+−

⋅−=⋅⋅⋅−

=−

==21

11

1

1

1

2

2

22

413)358.0)(50)(462.0)(50(1.1612

612)50(5.33

3)50( =−−=⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+−−=

kkkk ,

04.01250

50502.1

502

2 ==+

=+

==kRR

RII

EF

E

o

fiβ ,

5.17)413)(04.0(11 =+=+= ii AD β ,

6.235.17

413===

DAA i

if , 8.92.1

)5.0)(6.23(222 ===−

==k

kRRA

RIRI

VVA

S

Cif

Ss

Cc

s

ovf

Ω=== 3931.1612 khRZ iei , Ω=== 235.17

393DZZ i

if ,

Ω=−

=⇒= 5.23232.1

)23)(2.1(kkZRZZ bfSbfif ,

Ω=+=+= kkZRZ bfSsf 22.15.232.1 .



Exercises:

1. For each amplifier in Fig. 20-11, write a mathematical expression to calculate Avf .

ER

CCV+

+

−sV

oV

CR

SR

SR

DDV+

oV+

−sV

(a) (b)

DR

1R

2RLR

+

−oV

+

−sV

DDV+

SR

DDV+

+

−sV

oVCR

(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.

part iv lectures 19 & 20 multistage and feedback...

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