Chap14 Single Transistors Amplifiers

7/27/2019 Chap14 Single Transistors Amplifiers

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Jaeger/Blalock7/1/03

Microelectronic Circuit DesignMcGraw-Hill

Chapter 14

Single-Transistors Amplifiers

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock

Chap 14 - 1


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Chapter Goals

Detailed study of three broad classes of amplifiers

Inverting amplifiers- that provide high voltage gain with a 1800

phase shift, common-emitter and common-source configurations,

Followers- that provide nearly unity gain similar to op amp voltage

follower, common-collector and common-drain configurations, Noninverting amplifiers- that provide high voltage gain with no

phase shift, common-base and common-gate configurations.

Detailed design of voltage gain, input voltage range, current gain, input

and output resistances, coupling and bypass capacitor design and lower

cutoff frequency for each type of amplifier. Understand differences between SPICE ac (small-signal), transient

(large-signal) and transfer function analysis modes.

Chap 14 - 2


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Signal Injection and Extraction: BJT

In forward-active region,

To cause change in current, vBE= vB - vEmust be changed.Base or emitter terminals are used to inject signal because

even if Early voltage is considered, collector voltage has

negligible effect on terminal currents.

Substantial changes in collector or emitter currents cancreate large voltage drops across collector and emitter

resistors and collector or emitter can be used to extract

output. Since iB is a factor ofbFsmaller than iCoriE

currents, base terminal is not used to extract output.

TVBEv

F

SI

Ei

TVBEv

SI

Ci

exp

exp

T

VBEv

FO

SI

Bi exp

b

Chap 14 - 3


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Signal Injection and Extraction: FET

In pinch-off region,

To cause change in current, vGS= vG - vSmust be changed.Gate or source terminals are used to inject signal because

even with channel-length modulation, drain voltage has

negligible effect on terminal currents.

Substantial changes in drain or source currents can createlarge voltage drops across drain and source resistors and

drain or source can be used to extract output. Since iG is

always zero, gate terminal is not used to extract output.

2

2

TN

VGS

vnK

Di

Si

Chap 14 - 4


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Amplifier Families

Constraints for signal injection and extraction yield three families of

amplifiers

Common-Emitter (C-E)/Common- Source (C-S)

Common-Base (C-B)/Common- Gate (C-G) Common-Collector (C-C)/Common- Drain (C-D)

All circuit examples here use the four-resistor bias circuits to establish

Q-point of the various amplifiers

Coupling and bypass capacitors are used to change the ac equivalent

circuits.

Chap 14 - 5


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Inverting Amplifiers: Common-Emitter

and Common-Source Circuits

AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier

Chap 14 - 6


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Inverting Amplifiers: Terminal Voltage

Gain

Using test source vb to drive the base

terminal of the transistor, neglecting

ro,

ER

mg

LR

mg

ACEvt

ERomgo

LR

o

bv

ov

vtA

1

)1(/ bb

b

1obAssuming

For C-S Amplifier, take limit of voltagegain of C-E amplifier as

and

IfREandRSare set to zero, the voltagegain has an upper bound of

IfREandRSare large, the voltage gainhas a lower bound of

r

b rmgo

SR

mg

LR

mg

ACSvt

1

CCVA

CEvt

LRmgA

CSvtA

CEvt

10

DDVA

CSvt

R

LR

ACSvtA

CEvt

R is the unbypassed

resistor in the

emitter or source.

Chap 14 - 7


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Inverting Amplifiers: Input Signal Range

For small-signal operation, magnitude ofvbe developed across r in small-

signal model must be less than 5 mV.

If , vb can be increased beyond 5 mV limit.

In case of FET, magnitude ofvgs must be less than 0.2(VGS- VTN).

Presence ofRSincreases permissible value ofvg

r

ER

ERmg

r

1

bv

ibev

V)1(005.01005.0E

Rmg

o

ER

ERmgb

v

b

1E

Rmg

)(2.01 TN

VGS

V

SRmg

gvgsv

)1)((2.0 SRmgTN

VGS

Vgv

Chap 14 - 8


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Inverting Amplifiers: Condition forgmR

>>1

This condition simplifies the gain expression and is used to stabilize

voltage gain, achieve high input and output resistances and increase

input signal range.

For BJT:

For MOSFET:

V025.0

1

ER

EI

TV

ER

EI

TV

ER

CI

ERmg

2

12

TNV

GSV

SR

DI

TNV

GSV

SR

DI

SR

mg

Chap 14 - 9


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Inverting Amplifiers: Input Resistance

and Overall Voltage GainInput resistance looking into the base

terminal is given by

For C-S Amplifier,

)1(

)1(

ERmgrRCEin

ERor

i

bv

RCEin

b

r

RCSin

RCEinBR

IR

RCEinBRA

CEvt

ivb

vA

CEvt

ivb

v

bvov

ivovA

CEv

Overall voltage gain is

For C-S Amplifier,

GR

IR

GR

ACSvtA

CSv

Chap 14 - 10


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Inverting Amplifiers: Voltage Gain

Calculations (Example)

Problem: Find overall voltage gain.

Given data: Q-point values and values forRI,R1,R2,R3,R7 ,for both

BJT and FET as well as values forREandRS.

Assumptions: Small-signal operating conditions.

Analysis: For C-E Amplifier,

k18k100k22

73

k313k3(101)k160)1(

k104k300k16021

RR

L

R

ERmgrR

CEin

RRB

R

75.5k313

)k18(100

RCEin

LR

oA

CEvt

b61.5

RCEinB

RI

R

RCEinBR

ACEvtA

CEv

Chap 14 - 11


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Inverting Amplifiers: Voltage Gain

Calculations (Example contd.)

Analysis: For C-S Amplifier,

k18k100k2273

k892M2.2.5M121

RRL

R

RRG

R

46.4)k2(mS)491.0(1

)k18(mS)491.0(

1

S

Rm

g

LRmgACSvt

45.4

GRIR

GR

ACS

vtA

CS

v

Chap 14 - 12


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Inverting Amplifiers: Output Resistance

0ixi0i

0ev0)1(

1ev

i)1(ev

o

rth

RE

Ro

ERo

b

b

b

rthR

ev-

i

ButRout = ro whenRE= 0, not infinite.

Now, we also include ro in our analysis.

xixvoutR

b

b

rth

RE

RE

RoorR

rthRER

ER

ERr

thR

oro

1out

xii

xiev

ev)ixi(evrvxv

Chap 14 - 13


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(contd.)

b rmgo Assuming and , with

.for

Finite current gain of BJT places an upper limit on size of output

resistance. rappears in parallel withREifRth is neglected. If we letREbe

infinite, maximum value of output resistance is

Output resistance of C-S amplifier is given by,

thR

ERr )( E

Ror

)(out

)()(1out

ERr

fR

ERr

for

ERrmgorR

1)(

E

Rrmg

oroR )1(out b

S

RmgorR 1out

Chap 14 - 14


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(Example) Problem: Find output resistance.


BJT and FET as well as values forREandRS.

Assumptions: Small-signal operating conditions. Small -signal values

are known.


For C-S Amplifier,

4.55Mk3k2.10k96.1

k3100(1k2191

out

b

rth

RE

RE

RoorR

k442k20.491mS)(1k2231out

SRmgorR

Chap 14 - 15


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Inverting Amplifiers: Current Gain

Terminal current gain is the ratio of the current delivered to the load

resistor to the current being supplied to the base terminal.

oACEit b

ACSit

Chap 14 - 16


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Inverting Amplifiers: Summary

C-E and C-S amplifiers have similar voltage gains.

C-S amplifier provides extremely high input resistance but that of C-E is also

substantial due to the fREterm.

Output resistance of C-E amplifier is much higher than that of C-S amplifier as

f is much larger for BJT than for FET.

Input signal range of C-E amplifier is also higher than that of C-S amplifier.

Current gains of both are identical to those of individual transistors.

Following transformation is used to simplify circuit analysis by absorbingRE

(orRS) into the transistor (For FET, current gain and input resistance are

infinite).

ERmg

mgmg1

' )1(' ERmgrr )1(' ERmgoror

ormgo bb ''' formgf

'''

Chap 14 - 17


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Follower Circuits: Common-Collector

and Common-Drain Amplifiers

AC equivalent for C-C Amplifier AC equivalent for C-D Amplifier

Chap 14 - 18


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Follower Circuits: Terminal Voltage

Gain

Neglecting ro,

LR

mg

LR

mg

ACCvt

L

Ro

r

LR

o

b

v

ov

vtA

1

)1(

)1(

b

b

1obAssuming

For C-S Amplifier, take limit of voltagegain of C-E amplifier as

and

In most C-C and C-D amplifiers,

Output voltage follows input voltage,

hence theses circuits are called

followers. BJT gain is closer to unity

than FET. Mostly,

ro can be neglected as gain


20/59



Follower Circuits: Input Signal Range

For small-signal operation, magnitude ofvbe developed across r in small-

signal model must be less than 5 mV.

If , vb can be increased beyond 5 mV limit.Since only small

portion of input signal appears across base-emitter or gate-source terminals,

followers can be used with relatively large input signals without violating

small-signal limits.

In case of FET, magnitude ofvgs must be less than 0.2(VGS- VTN).

rL

R

LRmg

r

1

bv

ibev

V)1(005.01005.0L

Rmg

o

LR

LRmgb

v

b

1L

Rmg

)(2.01 TN

VGS

V

LRmg

gv

gsv )1)((2.0 L

RmgTNV

GSVgv

Chap 14 - 20


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Follower Circuits: Input Resistance and

Overall Voltage GainInput resistance looking into the base

terminal is given by

For C-S Amplifier,

LRor

i

bv

RCCin

)1( b

r

RCDin

RCCinBR

IR

RCCinBRA

CEvt

ivb

vA

CCvt

ivb

v

bvov

ivovA

CCv


For C-S Amplifier,

GR

IR

GR

ACDvtA

CCv

Chap 14 - 21


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Follower Circuits: Voltage Gain


Problem: Find overall voltage gain.


BJT and FET.


Analysis: For C-C Amplifier,

k117)k2.10(101k2.10)1(

k5.1174

k10421

L

RorRCE

in

RRL

R

RRB

R

b

991.0k117

)k5.11(101)1(

RCCin

LR

oA

CCvt

b956.0

RCCinB

RI

R

RCCinBR

ACCvtA

CCv

Chap 14 - 22


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Follower Circuits: Voltage Gain

Calculations (Example contd.)

Analysis: For C-D Amplifier,

k7.1074

892k21

RRL

R

RRG

R

840.0)k7.10(mS)491.0(1

)k7.10(mS)491.0(

1

L

Rm

g

LRmgACDvt

838.0

GRIR

GR

ACD

vtA

CDv

Chap 14 - 23


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Follower Circuits: Output Resistance

111

xvxviixi

o

thR

o

r

o

rth

R

RCCout

rth

Ro

rth

Ro

bb

b

b

b

1

1

1

o

thR

mgo

thR

mgoR

CCout bb

In case of FET,

Thus equivalent resistance looking into

emitter or source of a transistor is

approximately 1/gm.

mgR

CD

out

1

Current is injected

into emitter of BJT.

1

iiev

o

thR

mgo

b

Current oi coming out of collector mustbe supported by veb = oi/gm, given by

first term. ib =-i/bo+1creates voltage

drop inRth given by second term

Chap 14 - 24


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Follower Circuits: Output Resistance

(Example) Problem: Find output resistance. Given data: Q-point values and values forRI,R1,R2,R4,R7 ,for both

BJT and FET.

Assumptions: Small-signal operating conditions. Small -signal values

are known.

Analysis: For C-C Amplifier,

For C-D Amplifier,

120101

k96.1

mS80.9

990.0

1

1

o

thR

mgR

CCout b

k04.2mS91.4

11

mgR

CDout

Chap 14 - 25


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Follower Circuits: Current Gain


resistor to the current being supplied from the Thevenin source.

1i

1i

oA

CC

itb

ACDit

Chap 14 - 26


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Follower Circuits: Summary

Both C-C and C-D amplifiers have voltage gains approaching unity.

C-D amplifier provides extremely high input resistance because of

infinite resistance looking into gate terminal of FET as compared to C-

C amplifier.

Output resistance of C-C amplifier is much lower than the C-D

amplifier due to higher transconductance of BJT than an FET for given

operating current.

Both C-C and C-D amplifiers can handle relatively large input signal

levels..

Current gains of FET is inherently infinite, whereas that of BJT is

limited by its finite value ofbo.

Chap 14 - 27


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Noninverting Amplifiers: Common-Base

and Common-Gate Circuits

AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier

Chap 14 - 28


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Noninverting Amplifiers: Terminal

Voltage Gain and Input Resistance

Polarities ofvbe and dependentcurrent sourcegmvbe are both

reversed, signal source is

transformed to its Norton

equivalent ro is neglected.

LRmg

ev

ovA

CBvt

For C-S Amplifier, take limit of voltage

gain of C-E amplifier as

r

LRmgA

CGvt

mgo

r

i

ev

RCBin

1

1

b

m

gR

CGin

1

Chap 14 - 29


30/59



Noninverting Amplifiers: Input Signal

Range

For small-signal operation,

In case of FET,

Relative size ofgm andRIdetermine signal-handling limits.

4

4

)4

(1

iv

ebv

RI

R

R

RI

Rmg

V)1(005.0I

Rmgbv

)1)((2.0I

RmgTNV

GSV

iv

)1(ebv

iv

IRmg forRI>>R4.

)1(sgviv

IRmg

Chap 14 - 30


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Jaeger/Blalock7/1/03 Microelectronic Circuit DesignMcGraw-Hill

Noninverting Amplifiers: Overall

Voltage Gain

44

)4

(1

4

4

RI

R

R

RI

RmgL

R

m

g

RCBinR

IR

RCBinR

ACBvt

ivev

evov

ivovA

CBv


For C-S Amplifier,

ForRI>>R4,

4

4

)4(1 RIR

R

RIRmg

LRmg

ACGv

IRmgL

RmgA

CGCBv

1

,

For ,

This is the upper bound on gain.

For ,

1IRmg

LRmgA

CGvt

LRmgA

CBvt

1th

Rmg

IR

LR

A

CG

vtA

CB

vt

ro can be neglected as gain


32/59


Noninverting Amplifiers: Voltage Gain


Problem: Find overall voltage gain. Given data: Q-point values and values forR1,R2,R3,R7 ,for both BJT

and FET,RI=2 k,R4 =12 k.



For C-S Amplifier,

k1873

102/1

RRL

R

mgRCBin176 LRmgACBvt

59.8

4

4

)4

(1

RI

R

R

RI

Rmg

ACBvtA

CBv

k1873 RRLR 84.8

LRmgACSvt

k04.2/1 mgRCGin 11.4

4

4

)4

(1

RI

R

R

RI

Rmg

ACGvtACGv

Chap 14 - 32


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Noninverting Amplifiers: Output

Resistance

b

rth

Rth

RoorR

CBout 1

Desired resistance is that looking intocollector with base grounded and resistorRth

in emitter.

The redrawn equivalent circuit is same as

that for C-E amplifier except resistance in

base is zero and resistance in emitter is

relabeled asRth.

Using b rmgo

)()(1th

Rrfth

RrmgorRCGoutR

CBout

Chap 14 - 33


34/59


Noninverting Amplifiers: Current Gain


resistor to the current being supplied to the base terminal.

1ei1i

oA

CB

it

1ACGit

Chap 14 - 34


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Noninverting Amplifiers: Summary

C-B and C-G amplifiers have similar voltage and current gains.

Numerical differences occur due to difference in parameter values of

BJT and FET at similar operating points.

C-B amplifier can achieve high output resistance due to higher

amplification factor of BJT.

C-B amplifier can more easily reach low levels of output resistance due

to higher transconductance of BJT for a given operating current.

Input signal range of C-G amplifier is inherently larger than that of C-B

amplifier.

Chap 14 - 35


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Selecting Amplifier Configuration

A single-transistor amplifier with a gain of 80 dB and input resistance of 100

k.

Av = 1080/20 = 10,000. For even best BJTs, gain< f= 40VA = 40(150) =

6000 and FET typically has much lower intrinsic gain. Hence such large

gain cant be achieved by single-transistor amplifier. A single-transistor amplifier with gain of 52 dB, input resistance of 250 k.

Av = 1052/20 = 400. Since we need large gain and relatively large input

resistance, we can use C-E amplifier.Av = 20VCC, so, VCC=20 V.

which is small but acceptable.

For FET, even with small gate overdrive, VDD =100 V which is too large

A105105.2

)V025.0(100k250 b

C

I

CI

TVor

Chap 14 - 36


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Selecting Amplifier Configuration

(contd.)

A single-transistor amplifier with gain of 30dB and input resistance of 5M.

Av = 1030/20 = 31.6. Since we need large input resistance and moderate

gain, we can use C-S amplifier.

Input resistance of can be set by our choice of gate-bias resistors. For C-

E amplifier, required high input resistance could be attained but values of

base bias resistor could be a limiting factor.

A single-transistor amplifier with gain of 0dB and input resistance of 20M

with load resistor of 10k. Gain of 0 dB implies a source follower,Rin=boRL=100(10k)=1M, so

BJT cant meet input resistance requirement, A source follower can be

used easily.

30V5.0

DD

V

TNV

GSV

DDV

vA

Chap 14 - 37


38/59


Coupling and Bypass Capacitor Design

Since impedance of a capacitor increases with decreasing frequency,

coupling and bypass capacitors reduce amplifier gain at low

frequencies.

To choose capacitor values, short-circuit time constant method is used:each capacitor is considered separately with all other capacitors

replaced by short circuits.

To neglect a capacitor, the magnitude of capacitive impedance must be

much smaller than the equivalent resistance appearing at its terminals.

Chap 14 - 38


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Coupling and Bypass Capacitor Design:

C-E and C-S Amplifiers

RCEinB

Rin

R For C-E amplifier,

RCEoutRout

R3

For C-S amplifier,

RCSinG

Rin

R RCSoutRout

R3

For coupling capacitorC1,


w is chosen to be lowest frequency

for which midband operation is

needed in given application.

inR

IR

Cw

11

outRR

C

7

13

w

Chap 14 - 39


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C-E and C-S Amplifiers (contd.)

In this case, we can neglect impedancesof capacitors C1 and C3 , the find the

equivalent resistance looking up into

emitter or source of amplifier.

mgE

RR

C

16

1

2w

mgS

RR

C

16

12

w

Chap 14 - 40


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C-E and C-S Amplifiers (Example)

Problem: Choose values of coupling and bypass capacitors. Given data:f= 1000Hz, values of all resistors and input and output

resistances for both C-E and C-S amplifiers.

Analysis:For C-E amplifier:

k1.78 RCEinBRinR

F02.01

1.99nF1

1

w

C

inR

IR

C

F0.682

nF2.67)/1(

6

12

w

Cm

gE

RRC

F015.03

nF31.1

7

13

w

C

outRR

C

For C-S amplifier:

k

892

GRinRpF1800

1178pF

11

C

inR

IR

Cw

F0.562

nF3.55)/1(

6

12

w

Cm

g

S

RRC

F015.03

nF31.1

7

13

w

C

outRR

C

Chap 14 - 41


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C-C and C-D Amplifiers

RCCinB

Rin

R

For C-E amplifier,

RCCoutRout

R4

For C-S amplifier,

R

CS

inGR

inR

R

CD

outR

outR

4



inR

IR

Cw

11

outRR

C

7

13 w

Chap 14 - 42


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C-C and C-D Amplifiers (Example)



Analysis:For C-C amplifier:

k5.95 RCCinBRinR

pF82001

816pF1

1

C

inR

IR

Cw

pF82003

pF795

7

13

C

outRR

Cw

For C-D amplifier:

k892G

R

in

R

pF10001

89pF1

1

Cin

RI

RC

w

pF82003

pF782

7

13

C

outRR

Cw

1204

RCCoutRout

R k74.1

4R

CDoutRout

R

Chap 14 - 43


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7/1/03


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C-B and C-G Amplifiers

RCBin

Rin

R4

For C-E amplifier,

RCBoutRout

R3

For C-S amplifier,

RCGinRinR 4 R

CGoutRoutR 3



inR

IR

Cw

11

outRR

C

7

13 w

Chap 14 - 44


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C-B and C-G Amplifiers (contd.)

In this case, we can neglect impedancesof capacitors C1 and C3 , the find the

equivalent resistance looking up into

emitter or source of amplifier.

21

)4)(1(21

RRRCGeq

IRRorRRRCBeq

b

RCGCB

eq

C,

12

w

Chap 14 - 45


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C-B and C-G Amplifiers (Example)



Analysis: For C-B amplifier:

100k

102k

134

R

CB

inRinR

F82.01

75.8nF1

1

w

C

inR

IR

C

F015.03

nF31.1

7

13

w

C

outRR

C

k9.21M93.3k223

RCBoutRoutR

F0.027nF38.2

)4

)(1(21

1

2

bw

IRR

orRR

C

Chap 14 - 46


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Coupling and Bypass Capacitor Design: C-

B and C-G Amplifiers (Example contd.)

For C-G amplifier:

k74.1k04.2k124

RCGinR

inR

F42.01

42.6nF1

1

w

C

inR

IR

C

F015.03

nF31.1

7

13

w

C

outRR

C

k9.20k410k224

RCGoutRoutR

pF1800pF178

21

12

RR

Cw

Chap 14 - 47


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Lower Cutoff Frequency of an Amplifier

We can choose capacitor values to set the lower cutoff frequency of the

amplifier at desired value.

Pole associated with a capacitor occurs at the frequency at which

capacitive reactance is equal to resistance at the capacitor terminals.

In discussed amplifiers, there are several poles and a bandwidth

shrinkage occurs at low frequencies.

A transfer function with n identical poles at wo is given by

Lower cutoff frequency is higher than frequency corresponding to

individual poles.

n

o

nAjT

22

mid)(

ww

ww

1/122mid)(

no

L

A

LjT

www

Chap 14 - 48


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Dominant Pole Design

Instead of having the lower cutoff frequency set by the interaction of

several poles, it can be set by the pole associated with just one of the

capacitors. The other capacitors can be chosen to have their pole

frequencies much belowfL.

Capacitor associated with emitter or source part of the circuit tends to

be the largest due to low resistance presented by emitter or source

terminal of transistor and is commonly used to setfL.

Values of other capacitors are increased by a factor of 10 to push their

corresponding poles to much lower frequencies.

Chap 14 - 49


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Follower Design Example

Problem: Design an amplifier with given specifications.

Given data:Av>0.95,Rin>20M,Rout 6600 (beyond range of normal

BJTs). So we choose source-follower

configuration.

Chap 14 - 50


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Follower Design Example (contd.)

M22G

Rin

R

Tolerance of 5% is included.

Now,

LRD

InK

LRmgL

RmgL

Rmg

2

19

1995.01

k3k3 SRLR

We chooseRL=1.5 k

mA96.8D

InK

GSV

SR

DI

SSV

TNV

GSVn

K

DI

2

2

We can chooseID = 5

mA withKn =20mA/V2

to give VSS=16.7+ VTN.If we choose a MOSFET

with VTN= 1.5 V,V21.2

GSV

k3S

R

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Follower Design Example (contd.)

VSS=16.7+ VTN. gives a lower bound. Ifwe choose VSS= 20 V,RS=3.56 k .

To choose VDD :

For small-signal operation,

If we choose VDD =1.2 V, small-signal

criteria are fulfilled and MOSFET stays

saturated for all signals.

V5.1

SvTNVSvDDV

TNV

GSV

Sv

GSV

DDv

V7.2)19)(71.0(2.0

1)1(2.0

LRmgL

Rmg

LRmgTN

VGS

Vggv

Coupling capacitors mustnt affectcircuit at frequencies > 50 Hz.

Choose C1 =1500 pF and C2 =10 F.

pF1451

M22

1)Hz50(2

1 C

C

F03.12

k097.3

2)Hz50(2

1

C

C

4.691

mg

SRoutR

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Common-Source Amplifier Design

Example

Problem: Design an amplifier with given specifications.

Given data:Av> 50,Rin= 75,f >500 kHz.

Analysis:Moderate input resistance of C-E and C-S amplifiers can be

limited by reducing sizes of bias resistors (called swamping of

impedance level). JFET is chosen over BJT due to its better signalhandling capability and simple bias circuit design.

To deliver overall gain of 50, amplifier

must deliver a gain of 100.

We can choose VDD =20 V

and VGS- VP= 0.2 V

27575

75 iv

iv

GR

IR

GR

ivgsv

PV

GSV

DDV

vA

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Example (contd.)

k100

k100

DR

DR

LR

ID of 0.2 mA inRD = 100 k needs avoltage drop equal to total power

supply of 20 V. So power supply

voltage must be increased.

For pinch-off region operation,

Choose VDD = 25 V

k9

DIGS

V

SR

VGS- VPmust be small for high gain,

so JFET with largeIDSShas to be

chosen for a reasonableID. Choose

JFET withIDSS= 20 mA and VP=-2 V

mA2.0

2

1

PVGS

V

DSSIDI

k50mg

v2A

LR

V22)2()8.1(8.120

DDV

DDV

PVGSV

SRDIDRDIDDV

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Example (contd.)

Coupling capacitors mustnt affectcircuit at frequencies > 500 kHz.

Choose C1 =0.022 F, C2 =0.0068 F

and C3 =20 pF

nF12.21

150

1)kHz500(2

1 C

C

pF6442

494

2)kHz500(2

1 C

C

4741k1.9 mg

eqR

pF59.13k

2003

)kHz500(2

1

CC

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Common-Source Amplifier with Body

Effect

SR

mbgmgS

R )bsvgsv(oisv

svbsv svgvgsv

thv

)1(1sv

SR

mg

SRmg

mgmbg

gv)1(1

oi

SR

mg

mg

SRmg

LRmgL

R

vtA )1(1gv

oi-

gvov

gsv)1(

gsv)(bsvgsvi

mgmb

gmgmbgmg

S

RmgorRCSout )1(1

RCSin

ACSit

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Common-Drain Amplifier with Body

Effect

gv)1(1

oi

L

Rmgmg

LRmg

LRmgL

R

vtA

)1(1gv

oi

gvov

)1(

1

mgR

CDout

RCDin

ACDit

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Common-Gate Amplifier with Body

Effect

LRmgA

CGvt )1(

)1(

1

mgR

CGin

))(1(1

SR

IRmgorR

CGout

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Results of Body Effect

Gain of C-G amplifier more closely approaches theRL/Rth limit.

Gain of source follower is degraded.

Input resistance of C-G and output resistance of C-D amplifier is

lowered. Output resistance of both C-S and C-G amplifiers is raised.

Body effect increases input signal range.

Ch 14 59

Chap14 Single Transistors Amplifiers

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Transcript of Chap14 Single Transistors Amplifiers