ECES 352 Winter 2007Ch. 7 Frequency Response Part 31 Common-Base (CB) Amplifier *DC biasing ● Calculate I C, I B, V CE ● Determine related small signal

Ch. 7 Frequency Response Part 3 1ECES 352 Winter 2007

Common-Base (CB) Amplifier* DC biasing

● Calculate IC, IB, VCE

● Determine related small signal equivalent circuit parameters Transconductance gm

Input resistance rπ

* Midband gain analysis* Low frequency analysis

● Gray-Searle (Short Circuit) Technique Determine pole frequencies

ωPL1, ωPL2, ... ωPLn

● Determine zero frequencies ωZL1, ωZL2, ... ωZLn

* High frequency analysis● Gray-Searle (Open Circuit)

Technique Determine pole frequencies

ωPH1, ωPH2, ... ωPHn

● Determine zero frequencies ωZH1, ωZH2, ... ωZHn

Input at emitter, output at collector.


CB Amplifier - DC Analysis (Same as CE Amplifier)

* GIVEN: Transistor parameters:● Current gain β = 200● Base resistance rx = 65 Ω● Base-emitter voltage VBE,active = 0.7 V● Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K

* Form Thevenin equivalent for base; given VCC = 12.5V● RTh = RB = R1||R2 = 10K||2.5K = 2K● VTh = VBB = VCC R2 / [R1+R2] = 2.5V ● KVL base loop

IB = [VTh-VBE,active] / [RTh+(β +1)RE] IB = 26 μA

* DC collector current IC = β IB

IC = 200(26 μ A) = 5.27 mA* Transconductance gm = IC / VT ; VT = kBT/q = 26 mV

gm = 5.27 mA/26 mV = 206 mA/V * Input resistance rπ =

β / gm = 200/[206 mA/V]= 0.97 K* Check on transistor region of operation

● KVL collector loop● VCE = VCC - IC RC - (β +1) IB RE = 4.4 V

(okay since not close to zero volts).

R1 = 10KR2 = 2.5KRC = 1.2KRE = 0.33K


* Construct small signal ac equivalent circuit (set DC supply to ground)

* Substitute small signal equivalent circuit (hybrid-pi model) for transistor

* Neglect all capacitances● Coupling and emitter bypass capacitors become shorts

at midband frequencies (~ 105 rad/s) Why? Impedances are negligibly small, e.g.

few ohms because CC1, CC2, CE ~ few μF (10-6F)

● Transistor capacitances become open circuits at midband frequencies Why? Impedances are very large, e.g. ~ 10’s M Ω

because Cπ , Cμ ~ pF (10-12 F)

* Calculate small signal voltage gain AVo = Vo /Vs

CB Amplifier - Midband Gain Analysis

10)1)(/10(

115 FsradC

ZC

75

10)1)(/10(

11

pFsradCZC

High and Low Frequency AC Equivalent Circuit


CB Amplifier - Midband Gain Analysis

dBdBA

VVA

K

K

KKK

KK

rRR

rR

V

V

KK

K

rrI

Ir

V

V

KKVmARRgV

RRVg

V

V

V

V

V

V

V

V

V

VA

Vo

Vo

eEs

eE

s

e

xe

CLmCLmo

s

e

e

o

s

oVo

14)20.0log(20

/20.0001.094.0218

001.00050.5

0050.0

0051.033.05

0051.033.0

94.0065.097.0

97.0

)(

21892.1/206

KR

KR

KR

KR

KR

KR

S

E

C

L

5

5.2

10

33.0

2.1

9

2

1

Equivalent resistance re

KKKrr

rg

rr

rg

r

r

rr

rg

r

V

V

I

Vr

r

rgV

rgVI

r

VVgI

EnodeatKCL

I

Vr

x

m

x

m

x

m

e

e

ee

mme

me

e

ee

0051.02001

97.0065.0

11

11

11

0

Ve

+_

Iπ

rr

r

V

V

rr

V

r

VI

xe

x

e

re

βIπ

Voltage gain is less than one !


What Happened to the CB Amplifier’s Midband Gain?

* Source resistance Rs = 5K is killing the gain.

● Why? Rs >> re = 0.0051 K so

Ve/Vs<<1

* Need to use a different signal source with a very low source resistance Rs , i.e. ~ few ohms

* Why is re so low?

● Vs drives formation of Ve

● Ve creates Vπ across rπ

● Vπ turns on dependent current source

● Get large Ie for small Ve so re =Ve/Ie is very small.

dBdBA

VVA

and

KK

K

KKK

KK

rRR

rR

V

V

RFor

ceresislowwithsourcesignalNew

K

K

KKK

KK

rRR

rR

V

V

VVA

Vo

Vo

eEs

eE

s

e

s

eEs

eE

s

e

Vo

2.40)5.102log(20)(

/5.1025.094.0218

5.0005.0005.0

0050.0

0051.033.0005.0

0051.033.0

5

tan

001.00050.5

0050.0

0051.033.05

0051.033.0

/20.0001.094.0218

Ve

+_

re

Voltage gain is now much bigger than one !


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

* Draw low frequency AC circuit● Substitute AC equivalent circuit for transistor

(hybrid-pi for bipolar transistor) ● Include coupling and base capacitors CC1, CC2, CB

● Ignore (remove) all transistor capacitances Cπ , Cμ

* Turn off signal source, i.e. set Vs= 0

● Keep source resistance RS in circuit (do not remove)

* Consider the circuit one capacitor Cx at a time ● Replace all other capacitors with short circuits● Solve remaining circuit for equivalent resistance Rx seen

by the selected capacitor● Calculate pole frequency using● Repeat process for each capacitor finding equivalent

resistance seen and the corresponding pole frequency

* Determine the dominant (largest) pole frequency* Calculate the final low pole frequency using

xxPx CR

1

xx

PnPPPxLP CR

1...21


Common Base - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique

* Base capacitor CB = 12 μF

sradxCR

xKFRC

KKK

KKKKK

RRrgrrRR

RRrgrV

RRgr

V

rR

RRgr

VRRVgIVV

r

VIce

V

Vr

rV

V

I

VR

RrRR

BxCPL

xCB

SEmxBxC

SEm

SEm

i

SEmSEmi

iiii

ixBxC

B

B

B

B

/83102.1

11

sec102.10.112

0.103.22

005.033.020197.0065.02

1

1

11

11

sin/

21

2

Vi

+

_RxCB

Ri

Iπ

Low Frequency AC Equivalent Circuit

Vπ

Vx

Vo

Ix



Input coupling capacitor CC1 = 2 μF

sradxxRC

xKFRC

KKK

KKKK

rg

rrRRR

rg

rr

Irg

rrI

I

Vr

rrIV

IrgVgII

I

VrrRR

I

VR

C

C

C

C

xCCPL

xCC

m

xEsxC

m

x

m

x

e

ee

xe

mme

e

eeeEs

x

xxC

/100.5sec100.2

11

sec100.2010.02

010.00051.0005.0

201

97.0065.033.0005.0

1

11

1

45

12

51

1

1

1

1

Ve

Ie

Iπ

Ve

+_

re

Vπ

Vo

Ix Rs

Vx



srad

PLPLPLPL

/116,5033000,5083

321

* Output coupling capacitor CC2 = 3 μF

sradFKCR

KKKRRR

CCPL

CLC

/3332.10

11

2.102.19

223

2

* Low 3dB frequency

Vo

RL

VX

RC

Dominant low frequency pole is due to CC1 !


* What are the zeros for the CB 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 RB in parallel with CB

* Impedance given by

* When Z’B , Iπ 0, so gmVπ 0, so Vo 0

* Z’B when s = - 1 / RBCB so pole for CB is at

Common Base - Low Frequency Zeros

BB

BB

B

BBB

BCBB

CBB

CsR

RZ

R

CsRsC

RZRZ

ZRZ

B

B

1

11111

'

'

'

sradFKCR BB

ZL /42122

113

sss

ssF

sss

s

sss

ssssss

ssssF

L

ZL

PLPLPL

ZLZLZL

PLPLPL

ZLZLZLL

000,501

331

831

421

)(

000,501

331

831

10101

111

111)(

3

321

321

321

321

Iπ


Common Base - Low Frequency Poles and ZerosMagnitude Bode Plot

2222

3

2

1

321

000,501log10

331log10

831log10

421log102.40)(

2.405.102log20)(

000,50

33

83

42,0,0

000,501

331

831

421

5.102000,50

133

183

1

421

5.102)(

dBdBA

dBdBA

jjj

j

sss

sA

Mo

PL

PL

PL

ZLZLZL


Common Base - Low Frequency Poles and ZerosPhase Shift Bode Plot

000,50tan

33tan

83tan

42tan)(

000,50

33

83

42,0,0

000,501

331

831

421

5.102000,50

133

183

1

421

5.102)(

1111

3

2

1

321

PL

PL

PL

ZLZLZL

jjj

j

sss

sA


Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique

* Draw high frequency AC equivalent circuit● Substitute AC equivalent circuit for transistor

(hybrid-pi model for transistor with Cπ, Cμ)

● Consider coupling and emitter bypass capacitors CC1, CC2, CB as shorts

● Turn off signal source, i.e. set Vs = 0

● Keep source resistance RS in circuit

● Neglect transistor’s output resistance ro

* Consider the circuit one capacitor Cx at a time

● Replace all other transistor capacitors with open circuits

● Solve remaining circuit for equivalent resistance Rx seen by the selected capacitor

● Calculate pole frequency using● Repeat process for each capacitor

* Calculate the final high frequency pole using

xxPHx CR

1

xxPHnPHPHPxPH

CR

11...

1111

21

1


Common Base - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique

High frequency AC equivalent circuit

NOTE: We neglect rx here since the base is grounded. This simplifies our analysis,but doesn’t change theresults appreciably.



* Equivalent circuit for Ze

11

11

1

1

1

1

1

1

1

1

1

1

1

r

rg

rr

where

ZrZSo

sCrg

r

sCrg

r

sCr

rgsC

r

rgV

V

I

VZ

sCr

rgV

gsCr

VVg

sC

V

r

VI

givesEnodeatKCLVVI

VZ

me

Cee

m

m

mme

e

e

ee

me

memee

e

ee

ee

Ze

Ze

Replace thiswith this.

Ve

+_

Parallel combination of a resistor and capacitor.


* Pole frequency for Cπ =17pF


sradxsxpFCR

KKKKR

KK

rg

rr

RRrR

xCPH

xC

me

sEexC

/105.2101.4

1

174.2

11

4.20024.0005.033.00048.0

8.40048.02001

97.0

1

10111

Turn off signal source when finding resistance seen by capacitor.



* Equivalent circuit for Capacitor Cμ = 1.3 pF

* Pole frequency for Cμ =1.3pF

sradxsxpFKCR

KKKR

RRR

xCPH

xC

LCxC

/101.7104.1

1

3.105.1

11

05.192.1

892

sradxsx

xxCRCR

PH

xCxCPH

/109.61044.1

1

104.1101.4

11

89

911

* High 3 dB frequency

Dominant high frequency pole is due to Cμ !

Rs || RE || r π

= 0


Common Base - High Frequency Zeros

* What are the high frequency zeros for the CB amplifier?

* Voltage gain can be written as

* When Vo/Vπ 0, we have found a zero.

* For Cμ , we get Vo 0 when ω since the output will be shorted to ground thru Cμ .

* Similarly,we get a zero from Cπ when when ω since ZC π = 1/sCπ 0, so the voltage Vπ 0.

* Both Cπ and Cμ give high frequency zeros at ω !

s

o

s

oV V

V

V

V

V

VA


Common Base - High Frequency Poles and ZerosMagnitude

2

8

2

10

82

101

2

1

810810

101.71log10

105.21log102.40)(

2.405.102log20)(

101.7105.2

101.71

105.21

15.102)(

101.71

105.21

115.102)(

xxdBdBA

dBdBA

xx

xj

xj

A

x

s

x

s

ss

sA

Mo

PH

PH

ZH

ZH


Common Base - High Frequency Poles and ZerosPhase Shift

8

110

1

82

101

2

1

810810

101.7tan

105.2tan)(

101.7105.2

101.71

105.21

15.102)(

101.71

105.21

115.102)(

xx

xx

xj

xj

A

xs

xs

ss

sA

PH

PH

ZH

ZH


Comparison of CB to CE Amplifier

CE (with RS = 5K) 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

7.27)6.24log(20

/6.2412.094.0218

sradFK

CRRrr

R

sradFKCRR

sradFKCrrRR

sradFKCR

EBsx

E

PL

CCLPL

CxBSPL

EEZPZPZP

/342,512016.0

1

1

1

/3332.10

11

/8827.5

11

/2521233.0

110

3

22

11

321

sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmA

C

g

SBxLC

mLC

PH

SBx

PH

mZHZH

/109.53.1130

1

111

1

/100.11759.0

11

/106.13.1

/206,

6

2

81

1121

Note: CB amplifier has much better high frequency performance!


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

Note: CB amplifier has much better high frequency performance!


Conclusions* Voltage gain

● Can get good voltage gain from both CE and CB amplifiers.● Low frequency performance similar for both amplifiers.● CB amplifier gives better high frequency performance !

CE amplifier has dominant pole at 5.0x107 rad/s. CB amplifier has dominant pole at 7.1x108 rad/s.

* Bandwidth approximately 14 X larger!* Miller Effect multiplication of C by the gain is avoided in

CB configuration.

* Current gain● For CE amplifier, current gain is high AI = Ic / Ib

● For CB amplifier, current gain is low AI = Ic / Ie (close to one)!● Frequency dependence of current gain similar to voltage gain.

* Input and output impedances are different for the two amplifiers! ● CB amplifier has especially low input resistance.

Documents

ECES 352 Winter 2007Ch. 7 Frequency Response Part 31 Common-Base (CB) Amplifier *DC biasing ● Calculate I C, I B, V CE ● Determine related small signal