ECES 352 Winter 2007Ch. 7 Frequency Response Part 51 Comparison of Amplifier Configurations Midband...

Ch. 7 Frequency Response Part 5 1ECES 352 Winter 2007

Comparison of Amplifier Configurations

Midband Characteristics*

• These are approximate expressions neglecting the effects of the biasing resistors R1 and R2 and the source resistance RS.

J. Millman and A. Grabel, Microelectronics, 2nd Ed., McGraw Hill, NY (1987), p. 420.


Characteristics of Amplifier Configurations

Current gain is large ( β) for CE and EF, but < 1 for CB.

Voltage gain is large for CE and CB, but < 1 for EF.

Input resistance is • Very small (few Ωs) for CB, • Medium (few KΩs) for CE, but • Very large (~ 10’s of KΩs) for EF.

Output resistance is • Very small (few Ωs) for EF, • Very large (~ 100’s of KΩs) for CE and CB.


Numerical Comparison of Amplifier Configurationsfor the Same Transistor and DC Biasing

• These are approximate expressions neglecting the effects of the biasing resistors R1 and R2 and the source resistance RS.


Comparison of CB to CE Amplifier (with same Rs = 5 Ω)

CE (with RS = 5 Ω) CB (with RS = 5Ω)

Midband Gain

Low Frequency Poles and Zeros

High Frequency Poles and Zeroes

dBdBA

VVA

rr

r

rRR

rRRRg

V

V

V

V

V

VA

Vo

Vo

xeEs

eECLm

s

e

e

oVo

2.40

/4.1025.094.0218

sradxpFKCRR

sradxpFCRRr

LCPH

sEePH

ZHZH

/101.73.105.1

11

/105.2174.2

11

,

82

101

21

sradFKCRR

sradxKF

rg

rrRRC

sradKFCRRrgrrR

sradFKCR

CCLPL

m

xEsC

PL

BSEmxBPL

BBZPZPZP

/3332.10

11

/100.5010.02

1

1

1

/83112

1

1

1

/42122

110

23

4

1

2

1

321

dBdBA

VVA

RrrR

Rrr

rr

rRRg

V

V

V

V

V

V

V

VA

Vo

Vo

Bxs

Bx

xCLm

s

i

i

o

s

oVo

6.45)191log(20

/19193.094.0218

sradxFK

CRRrr

R

sradFKCRR

sradFKCrrRR

sradFKCR

EBsx

E

PL

CCLPL

CxBSPL

EEZPZPZP

/107.112005.0

1

1

1

/3332.10

11

/71427.0

11

/2521233.0

110

43

22

11

321

sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmA

C

g

SBxLC

mLC

PH

SBx

PH

mZHZH

/100.53.14.15

1

111

1

/100.917065.0

11

/106.13.1

/206,

7

2

81

1121


Comparison of EF to CE Amplifier (For RS = 5Ω )

CE EF

Midband Gain



dBdBA

VVA

RrrR

Rrr

rr

rRRg

V

V

V

V

V

V

V

VA

Vo

Vo

Bxs

Bx

xCLm

s

i

i

o

s

oVo

6.45)191log(20

/19193.094.0218

sradxFK

CRRrr

R

sradFKCRR

sradFKCrrRR

sradFKCR

EBsx

E

PL

CCLPL

CxBSPL

EEZPZPZP

/107.112005.0

1

1

1

/3332.10

11

/71427.0

11

/2521233.0

110

43

22

11

321

sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmA

C

g

SBxLC

mLC

PH

SBxPH

mZHZH

/100.53.14.15

1

111

1

/100.917065.0

11

/106.13.1

/206

7

2

81

1121

dBdBA

VVA

RrRR

RrR

Rr

R

RRr

RR

V

V

V

V

V

V

V

VA

Vo

Vo

ixBS

ixB

ix

i

LE

EL

s

b

b

i

i

oVo

1.0

/987.0)998.0(999.0015.066

sradKFCrRR

sradKFRrRRC

CeELPL

ixBsCPL

ZPZP

/3793

11

/25695.12

11

0

22

11

21

sradxpFK

CRRrRRRRgr

sradxpFK

RRgC

R

sradxpFKCr

rg

BSxLELEmPH

LEmxC

PH

mZHZH

/101.13.107.0

1

1

1

/100.126.0386.0

1

1

1

/102.11797.0

20111

10

2

10

'

1

1021


Comparison of Amplifier Configurations

Midband Gain and High and Low Frequency Performance

CE CB EF

Midband Voltage Gain -191 V/V +102 V/V +0.987 V/V45.6dB 40.2dB - 0.1dB

Low 3dB Frequency 1.7x104 rad/s 5.0x104 rad/s 2.6x102 rad/s

High 3dB Frequency 5.0x107 rad/s 7.1x108 rad/s 1.0x1010 rad/s

• Results for all three amplifiers with the smaller (5Ω) source resistance RS.

RS= 5 Ω


Cascade Amplifier

* Emitter Follower + Common Emitter (EF+CE)

* Voltage gain from CE stage, gain of one for EF.

* Low output resistance from EF provides a low source resistance for CE amplifier so good matching of output of EF to input of CE amplifier

* High frequency response (3dB frequency) for Cascade Amplifier is improved over CE amplifier.

pFCC

pFCC

rr xx

2

9.13

0

100

21

21

21

21

EF CE


Cascade Amplifier - DC analysis

.

7.86.3)101(

7.087.3

1

87.3)3.4(899

8999.8)101(1

9.83.4110050

7.05

]1[

50100100

510200

100

12

2

22222

112

22

111

1

2

12111111

1

211

21

21

okayisanalysiseapproximatsoII

AK

VVI

RIVV

VKARIV

IandVcalculateNow

AAII

Then

AKK

VVI

ionapproximatfirstaasINeglecting

RIIRIVV

QBaseKVL

KKKRRR

VVK

KV

RR

RV

EB

B

EBBEB

EEB

BB

BE

B

B

EBBThBBETh

Th

CCTh

KVmAg

r

V

mA

V

A

V

I

V

Ig

KVmAg

r

V

mA

V

A

V

I

V

Ig

m

T

B

T

Cm

m

T

B

T

Cm

9.2/0.34

100

0.340256.0

)7.8(100

9.2/8.34

100

8.340256.0

)9.8(100

2

22

2222

1

11

1111

10021

Small Signal Parameters

IE1

IB2

IRE1

IB1


Cascade Amplifier - Midband Gain Analysis

KKKK

rRrgr

I

rRIrgrIR

Em

Emi

178)9.23.4)(101(9.2

1

1

21111

1

2111111

91.0178)100100(4

178)100100(

)(

)(

016.0)9.23.4)(101(9.2

9.2

)(1

8.60)9.23.4(1.351

68

21

21

2111111

111

21111

21111

1

2

2

22

2

1

1

2

2

KKKK

KKK

RRRR

RRR

V

V

KKK

K

rRIrgrI

rI

V

V

KKV

mArRg

rV

rRVgI

V

V

V

RRVg

V

V

V

V

V

V

V

V

V

V

V

VA

BS

i

s

i

Emi

EmEm

CLmo

s

i

i

o

s

oVo

dBdBA

VVA

Vo

Vo

6.35)2.60log(20

/2.6091.0016.0)8.60(68

Vπ2

+

_

+

_Vπ1

+

Vi

_

Note: Voltage gain is nearly equal to that of the CE stage, e.g. – 68 !

Note: rx1 = rx2 = 0 so equivalent circuit is simplified.Iπ1

Ri


Cascade Amplifier - Low Frequency Poles and Zeroes

* Use Gray-Searle (Short Circuit) Technique to find the poles.

● Three low frequency poles● Equivalent resistance may

depend on rπ for both transistors.

* Find three low frequency zeroes.

sss

sss

sss

ssssF

LPLPLP

LZLZLZ

LPLPLP

LZLZLZL

321

321

321

321

111

111

)(


Cascade Amplifier - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique

Input coupling capacitor CC1 = 1 μF

sradxRC

xKFRC

KKKK

RRRR

KKKK

rRrgrI

VR

rRrgrIrRVgIrIV

I

VRRRR

I

VR

xCCPL

xCC

iBsxC

Emi

i

EmEmi

iiiBs

x

xxC

/23sec103.4

11

sec103.40.431

0.43178504

1789.23.4)101(9.2

1

1

211

1

211

1

21111

2111112111111

1

Ri

Vi

Iπ1

IX

RE1rπ2

rπ1 Vπ1

Vπ2

+

_

RE1 rπ2



sradFKCR

KKKRRR

CCPL

CLC

/12518

11

844

222

2

Vo

Vo

VX

Vπ2

rX2

RC RL

gm2Vπ2 RCRL

RE2 CE

CC2

* Output coupling capacitor CC2 = 1 μF

rπ2



sradFKCR

KKKrRR

KKKK

r

KKK

rg

Rr

rgI

RrI

I

Vr

rg

rRr

rgI

rRrI

I

Vr

rRI

VR

EExPL

eEEx

e

m

S

m

S

e

Ee

m

eE

m

eE

e

Ee

eEx

xEx

/73447029.0

11

029.0029.06.3

029.0101

065.03.49.2

065.0101

7.39.2

1

'

)1(

)'(

1)1(

)(

3

22

2

11

1

111

11

1

11

22

112

222

1122

2

22

22

sradPLPLPLPL /88273412523321

Emitter bypass capacitor CE = 47 μF

RE1

r π1 Vπ1

Ie1

Iπ1

Iπ2

VX

Ix

gm2Vπ2rπ2Vπ2

RE2

IE2

VE1

re1

Ie2

VE2

re2

Low 3 dB Frequency

The pole for CE is the largest and therefore themost important in determining the low 3 dB frequency.

gm1Vπ1

K

RRRR SS

7.3

' 21


* What are the zeros for the Cascade amplifier?

* For CC1 and CC2 , we get zeros at ω = 0 since ZC = 1 / jωC and these capacitors are in the signal line, i.e. ZC at ω = 0 so Vo 0.

* Consider RE in parallel with CE

* Impedance given by

* When Z’E , Iπ 0, so gmVπ 0, so Vo 0

* Z’E when s = - 1 / RE2CE so pole for CE is at

Cascade Amplifier - Low Frequency Zeros

sss

ssF

sss

s

sss

ssssss

ssssF

L

ZL

PLPLPL

ZLZLZL

PLPLPL

ZLZLZLL

7341

1251

231

9.51

)(

7341

1251

231

10101

111

111)(

3

321

321

321

321

EE

EE

E

EEE

ECEE

CEE

CsR

RZ

R

CsRsC

RZRZ

ZRZ

E

E

2

2'

2

22'

2'

1

11111

sradFKCR EE

ZL /9.5476.3

11

23


Cascade Amplifier - High Frequency Poles and Zeroes

HPHPHPHP

HZHZHZHZH

ssss

ssss

sF

4321

4321

1111

1111

)(

* Use Gray-Searle (Open Circuit) Technique to find the poles.

● Four high frequency poles● Equivalent resistance may

depend on rπ for both transistors.

* Find four high frequency zeroes.

High Frequency Equivalent Circuit


Cascade Amplifier - High Frequency Poles

KRRRR SS 7.3' 21

KK

K

KK

K

KKKK

rgr

rR

r

RrRR

I

V

rRRIrgr

rR

r

RV

rRIrgr

VR

r

VIV

So

IrgIIVgIIandr

VI

But

rRIRIIV

loopbasearoundKVL

mES

ESX

X

ESXmES

X

EXmX

SX

XX

XmXmeX

EeSXX

09.02.58

3.5

)101(9.2

9.26.3

9.2

7.31/9.26.37.3

1'

1/'

'1'

1

/IVfor solving and gRearrangin

0)(1'

1

0)(')(

111

21

121

21111

21

1

XX

211111

11111111

1

2111

VX

Ix

Iπ1

Ie1

Ix- Iπ1

+

_

Pole for Capacitor Cπ1 = 13.9 pF

sradxpFKPH /100.8

9.1309.0

1 81



K

RRRR SS

7.3

' 21

K

KKKK

rRrgrR

rRrgrR

rRrgrR

I

V

rRrgrR

VIV

I

R

VIIorRIIV

Also

rRrgrI

rRIrgrIV

So

rgIVgII

EmS

EmS

EmS

X

X

EmS

XXX

S

XXSXX

Em

EmX

mme

6.3

9.23.4)101(9.27.3

)(1'

)(1'

)(1'

)(1'

/IVfor solving and gRearrangin

and abovein for ngSubstituti

''

)(1

)(1

1

21111

21111

21111

21111

XX

1

11

211111

2111111

1111111

VX

Ix Iπ1

Ie1

+

_

Pole for Capacitor Cμ1 = 2 pF

sradxpFKPH /104.1

26.3

1 82

Ix- Iπ1



KRRRR SS 7.3' 21 Simplified Equivalent Circuit

222

222212

152}2)/34(1{29.13

'1

CC

pFKVmApFpF

RgCCCCC LmT

Using Miller’s Theorem, replace Cμ2 by two capacitors.

CLmCLmo RRg

V

RRVg

V

V

V

VK 2

2

22

21

2

22

2222

222221

11

11

1)1(

CRRg

CK

CC

CRRgCKCC

CLm

CLm



KRRRR SS 7.3' 21

KKKKrRrR

KKK

rg

Rr

rgI

RrI

I

Vr

EeX

m

S

m

S

e

ee

063.09.23.4065.0

065.0101

7.39.2

1

'

1

'

211

11

1

111

11

1

11

Vπ1

IxIe1

+

_

Pole for Capacitor CT = 152 pF

sradxpFKPH /100.1

152063.0

1 83

VX

re1

Iπ1

Ve1

Pole for Output Capacitor Cμ2 = 2 pF

sradxpFKCR

KKKR

RRRR

xCPH

xC

LCLxC

/105.222

11

244

'

8

24

gm2Vπ2 VX+_


Cascade Amplifier - High Frequency Zeroes

sradx

KpFrC

rg

so

rC

rgs

or

rsCrgII

VsCVgII

mZH

m

me

me

/105.2)9.2(9.13

1011

1

01

0

9

11

113

11

11

111111

111111

Ie1

* When does Vo = 0?

* When ω → ∞, ZCμ1→ 0, so signal shorted to ground. ωZH1= ∞.

* When ω → ∞, ZCπ2→ 0, so rπ2 shorted, so Vπ2 = 0. ωZH2= ∞.

* For Cπ1 , we get a zero when Ie1 = 0.

Ie1


Cascade Amplifier - High Frequency Zeroes

sradxpF

VmA

C

g

so

C

gs

or

VgVsCVVsCI

mZH

m

moC

/107.12

/34 10

2

24

2

2

2222222

I Cμ2

* When does Cμ2 produce a zero, i.e. make Vo = 0?

* For Cμ2 , we get a zero when IRL’ = 0, or ICμ2 = gm2Vπ2 , i.e. the output load resistance RL’ is starved of any current.

IRL’= 0

Zero for Output Capacitor Cμ2 = 2 pF



8888

109

8888

109

4321

4321

104.11

105.21

1011

1081

107.11

105.21

)(

104.11

105.21

1011

1081

107.11

105.2111

1111

1111

)(

xs

xs

xs

xs

xs

xs

sF

xs

xs

xs

xs

xs

xsss

ssss

ssss

sF

H

HPHPHPHP

HZHZHZHZH


Comparison of Cascade to CE Amplifier

CE* Cascade (EF+CE)

Midband Gain



dBdBA

VVA

V

V

V

V

V

V

V

VA

Vo

Vo

S

i

i

oVo

6.352.60log20

/2.6091.0016.08.6068

1

1

2

2

sradxpFK

sradxpFK

sradxpFK

sradxpFK

sradxsradx

PH

PH

PH

PH

ZHZH

ZHZH

/105.222

1

,/100.1152063.0

1

,/104.126.3

1

,/100.89.1309.0

1

/107.1,/105.2

,,

84

83

82

81

104

93

21

sradKFCrR

sradFKCRR

sradKFCrRrgrRR

sradFKCR

EeEPL

CCLPL

CEmBSPL

EEZPZPZP

/73403.047

1

)(

1

/12518

11

/23431

1

1

1

/9.5476.3

110

223

22

1221111

2321

dBdBA

VVA

V

V

V

V

V

VA

Vo

Vo

s

o

s

oVo

5.29)30log(20

/3037.02.81

2

2

sradFK

CRRr

R

sradFKCRR

sradFKCrRR

sradFKCR

EBs

E

PL

CCLPL

CBSPL

EEZPZPZP

/59147036.0

1

1

1

/12518

11

/15714.4

11

/9.5476.3

110

22

3

22

121

2321

sradxpFK

CRRrRR

gRR

sradxpFKCRRr

sradxpF

VmA

C

g

SBLC

mLC

PH

SB

PH

mZHZH

/100.42125

1

111

1

/108.49.135.1

11

/100.22

/6.40,

6

222

2

7

22

1

10

2

221

* CE stage with same transistor, biasing resistors, source resistance and load as cascade.

25 X improvement in bandwidth !

2 X improvement in voltage gain !


Comparison of Cascade to CE Amplifier

* Why the better voltage gain for the cascade?● Emitter follower gives no voltage gain!● Cascade has better matching with source than CE.

Cascade amplifier has an input resistance that is higher due to EF first stage.

Versus Ri2 = rπ2 = 2.5 K for CE

So less loss in voltage divider term (Vi / Vs ) with the source resistance.

* 0.91 for cascade vs 0.37 for CE.

* Why better bandwidth?

● Low output resistance re1 of EF stage gives smaller effective source resistance for CE stage and higher frequency for dominant pole due to CT (including Cμ2)

KKKK

rRrR Ei

1789.23.4)101(9.2

1 2111

KKKKrRrR

KKK

rg

Rr

rgI

RrI

I

Vr

EeX

m

S

m

S

e

ee

063.09.23.4065.0

065.0101

7.39.2

1

'

1

'

211

11

1

111

11

1

11

Pole for Capacitor CT = 152 pF

amplifier. CE for the /100.4 versus

cascade for the /100.1152063.0

1

6

83

sradx

sradxpFKPH

KKKRRR

pFKVmApFpFRgCCC

CLL

LmT

244'

152}2)/34(1{29.13'1 222

re1

Ri1

Ri2


Another Useful Amplifier – Cascode (CE+CB) Amplifier

* Common Emitter + Common Base (CE + CB) configuration

* Voltage gain from both stages* Low input resistance from second CB stage

provides first stage CE with low load resistance so Miller Effect multiplication of Cμ1 is much smaller.

* High frequency response dramatically improved (3 dB frequency increased).

● Bandwidth is much improved (~130 X).

sradxpFKK

pFpFpFKKVmApFpFRRgCCCso

rg

rrR

bygivenisRwhereamplifierCBofRbecomesamplifierCEstagefirstforRamplifiercascodeFor

sradxx

pFpFpFKKVmApFpFRRgCCC

PHin

CLmin

m

xi

iiL

PHin

CLmin

/103.76.292.1

1

6.19)11(3.1172.1005.0/20613.1171

51

,

/106.5sec108.1

1

30221813.1172.19/20613.1171

by limited is eperformanc frequency high theamplifier CEFor

8

111

68

1111

Bandwidth is improved by a factor of 130X over that for the CE amplifier !

Large Miller Effect

Small Miller Effect


Example of Cascode (CE +CB) Amplifier

http://www.freescale.com ECTW Conf. Proceedings 2003.


Other Examples of Multistage Amplifiers

CE CE EF EF

Darlington Pair


Other Examples of Multistage AmplifiersPush – Pull Amplifier Amplifier with Npn and Pnp Transistors

Amplifier with FETs and Bipolar Transistors


Differential Amplifier

* Similar to CE amplifier, but two CE’s operated in parallel

* Signal applied between two equivalent inputs instead of between one input and ground

* Common emitter resistor or current source used

* Current shared or switched between two transistors (they compete)

* Analyze using equivalent half-circuit● 1/2 of signal at input● 1/2 of signal at output● 1/2 of source resistance

* Gain and frequency response similar to CE amplifier for high frequencies

* Advantage: ● Rejects common noise pickup on input● No coupling capacitors so can operate

down to zero frequency.

Vo

+ _


Differential Amplifier Analysis

dBdBA

KKK

KKVmA

Rrr

r

V

RVgVV

V

V

V

VV

V

A

Vo

Sx

Cm

s

o

s

o

s

o

Vo

1.54509log20)(

5095.2065.097.0

97.09/206

22

2

2

2

Midband Gain


* Direct coupled so no coupling capacitors and no emitter bypass capacitor* No low frequency poles and zeros* Flat (frequency independent) gain down to zero frequency

High Frequency Poles and ZerosDominant pole using Miller’s Thoerem

sradxx

pFKpFpFKKK

CRgCRrr CmsxPH

/101.5sec1095.1

1

27870.0

1

3.1201172/5065.097.0

1

12/

1

67

High frequency performance is very similar to CE amplifier.

Vo

Vo /2

Vo /2


Summary* In this chapter we have shown how to analyze the high and low frequency

dependence of the gain for an amplifier.● Analyzed the effects of the coupling capacitors on the low frequency response

Found the expressions for the corresponding poles and zeros. Demonstrated Bode plots of magnitude and phase.

● Analyzed the effects of the capacitances within the transistor on the high frequency response.

Found the expressions for the corresponding poles and zeros. Demonstrated Bode plots of the magnitude and phase.

* Analyzed the high and low frequency performance of the three bipolar transistor amplifiers: common emitter, common base and emitter follower.

● Found the expressions for the corresponding poles and zeros.● Demonstrated Bode plots of the magnitude and phase.

* Demonstrated how to find the expressions for the gain and the high and low frequency poles and zeros for multistage amplifiers.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 51 Comparison of Amplifier Configurations Midband...

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