eces 352 winter 2007ch. 7 frequency response part 41 emitter-follower (ef) amplifier *dc biasing ●...

Ch. 7 Frequency Response Part 4 1ECES 352 Winter 2007

Emitter-Follower (EF) 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

High and Low Frequency AC Equivalent Circuit


EF Amplifier - DC Analysis (Nearly the 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 - (β +1) IB RE = 10.8 V (was 4.4 V for CE

amplifier) (okay since not close to zero volts).

R1 = 10KR2 = 2.5KRC = 0 KRE = 0.33K Note: Only difference here from CE case is VCE is larger

since RC was left out here in EF amplifier.


EF Amplifier - Midband Gain Analysis

998.0

905.1

9.1

65065.02005.0

65065.02

999.065065.0

65

015.0661

1

1

1

6697.0

20133.09

1

K

K

KKKK

KKK

RrRR

RrR

V

V

KK

K

Rr

R

V

V

VVVV

V

V

V

KKK

r

rgRR

V

VgrV

RR

V

V

V

V

V

V

V

V

V

V

V

VA

ixBS

ixB

s

b

ix

i

b

i

ooi

mEL

mEL

o

s

b

b

i

i

o

s

oVo

5005.0

5.2

10

33.0

0

9

2

1

KR

KR

KR

KR

KR

KR

S

E

C

L

Equivalent input resistance Ri

KKV

Vr

r

V

VV

I

VR ooii 6566197.01

dBdBA

A

Vo

Vo

1.0987.0log20)(

987.0998.0999.0015.066

Vi

+

_

Ri

Vb

+

_

Iπ

KVmAg

r

VmAmV

mA

V

Ig

m

T

Cm

97.0/206

200

/20626

27.5

DC analysis is nearly the same!IB , IC and gm are all the same. Only VCE is different since RC=0.

VO

NOTE: Voltage gain is only ~1!This is a characteristic of the EF amplifier!Cannot get voltage gain >1 for this amplifier!


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 capacitors CC1, CC2

● 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 corresponding pole frequency

* Calculate the final low 3 dB frequency using

xxPx CR

1

xx

PnPPPxLP CR

1...21


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

Input coupling capacitor CC1 = 2 μF

srad

xRC

xKFRC

KKKKK

RrRRR

KKKK

RRrgrI

VR

RRrgrIRRVgIrIV

I

VRRrRR

I

VR

xCCPL

xCC

ixBsxC

LEmi

i

LEmLEmi

iiixBs

x

xxC

/256sec109.3

11

sec109.395.12

95.165065.02005.0

65933.0)201(97.0

1

1

311

1

311

1

1

Ri

Vi

Iπ

IX


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

srad

PLPLPL

/29337256

21

* Output coupling capacitor CC2 = 3 μF

sradFKCR

KKKKrRRR

KKKKK

rg

RRrr

rgI

RRrrIr

RRrrIV

rgIVgII

I

VrrRRR

CxCPL

eELxC

m

SBx

m

SBxe

SBxe

mme

e

eeeELxC

/373005.9

11

005.9005.033.09

005.0201

005.02065.097.0

11

1

222

2

2

* Low 3 dB frequency

Iπ

Ve

Ie IX

re

So dominant low frequency pole is due to CC1 !


Emitter Follower - Low Frequency Zeros

* What are the zeros for the EF amplifier?

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

ss

sF

ss

ss

ssss

sssF

L

PLPL

ZLZL

PLPL

ZLZLL

371

2561

1)(

371

2561

0101

11

11)(

21

21

21

21


Emitter Follower - Low Frequency Poles and ZerosMagnitude Bode Plot

22

2

1

2

1

371log10

2561log101.0)(

1.0998.0log20)(

37256

0

0

371

2561

1998.0

371

2561

1998.0)(

dBdBA

dBdBA

jjss

A

Mo

PL

PL

ZL

ZL


Emitter Follower - Low Frequency Poles and ZerosPhase Shift Bode Plot

37tan

256tan)(

37256

0

0

371

2561

1998.0

371

2561

1998.0)(

11

2

1

1

1

PL

PL

ZL

ZL

jjss

A


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 and CC2 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


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

* Redrawn High Frequency Equivalent Circuit

sourcecurrent dependent to

due resistance equivalent'

'

Vg

VRwhere

RRRI

VZ

m

oS

SLEe

oeq

Ie

y

gyRRZ

gy

y

RR

gy

g

RRRRRRRZ

findcanweRRRZSince

gyRRgVg

VRso

sCrZrz

ydefinewewhere

gyRRgsCr

RRV

V

RR

VVgVsC

r

VEatKCL

mLEeq

mLE

m

m

LELESLEeq

SLEeq

mLEmm

oS

C

mLEmLEo

LE

om

1

11111

1

1111

1

0

'

'

'

Zeq

EIe

zπ =1/yπ



LEm

C

CLEmLEb

b

LEmLEm

LE

LEm

LEb

LEm

LE

LEmLE

mLEeqb

b

RRgC

ssCZwhere

ZRRgrRRZ

isZso

RRg

Cs

RRgr

RR

RRg

rCsr

RRZ

r

CsrsC

ryforngSubstituti

RRg

yRR

y

RRgRR

yy

gyRR

yZZ

ZisgroundandBbetweenimpedanceTotal

1

1

'

1

1

111

1

1

1

1

11

1

1

111

'

'

''

'

'

'

'

Modified Equivalent Circuit

Replace thiswith this.

ZB’

ZB’

zπ =1/yπ

Looks like a resistor in parallel with a capacitor.



sradx

sxpFK

KKVmApF

KRRg

CR

CR

KKK

KKKKKKKVmAK

RRRRrRRgrR

PH

LEmxC

xCPH

LEBSxLEmxC

/100.1

1086.9

1

255.0386.0

1

933.0/206117

386.0

1

1

11

386.039.06.64

933.02005.0065.0933.0/206197.0

1

101

11

'''

1

'

RxCπ

* Pole frequency for Cπ =17 pF



sradx

sxpFKCR

KKKK

KKKKKKKVmAK

RRrRRRRgrR

PH

xCPH

BSxLELEmxC

/101.1

101.9

1

3.107.0

11

07.007.032.06.64

2005.0065.0933.0933.0/206197.0

1

102

112

* Pole frequency for Cμ =1.3 pF



* Alternative Analysis for Pole Due to Cπ

sradxpFKCR

KKKKKK

KKKKKK

RRrRRrg

RRrRRr

I

Ir

I

VR

soRRrRRrg

RRrRR

I

I

RRrRRIRRrRRrgI

RRrIIVRRIrgII

sorIVandIIIBut

RRrIIVRRVgI

givesloopbasearoundKVL

I

Ir

I

V

I

VRandVVNote

xCH

BSxLEm

BSxLE

xx

xxC

BSxLEm

BSxLE

x

BSxLExBSxLEm

BSxxLEmx

xe

BSxxLEme

xxx

xxCx

/100.117006.0

11

is frequency pole theSo

006.02005.0065.0933.0201

2005.0065.0933.097.0

1

1

1

get wegRearrangin

0

0

:

101P

EIe

Vx

Ix

Iπ

Ix-Iπ

Ie+gmVπ

We get the same result here for the high frequency pole associated with Cπ as we did using the equivalent circuit transformation.


Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique* Alternative Analysis for Pole Due to Cµ

sradxpFKCR

KKK

KKKKKK

RRrgrRRr

RRrgrRRrRRrgr

RRr

RRr

I

VR

RRrgr

rVRRr

rRRrIV

V

RRrr

VRRrIRRrIIV

RRrgr

rVV

sor

RRrgVRRVgIVV

xCH

LEmBsx

LEmBsxLEm

Bsx

Bsx

x

xxC

LEmxBsxBsxxx

BsxBsxxBsxxx

LEmx

LEmLEmx

/101.13.107.0

11

is Cfor frequency pole theSo

07.00.6507.0

933.0)201(97.02005.0065.0

1

111

1

11

get wegRearrangin

1

1

get wefor ngSubstituti

writealsocan We

1

11

102P

Vx

Ix

Iπ

Ix-Iπ

Iπ+gmVπ

E

We get the same result here for the high frequency pole associated with Cµ as we did using the equivalent circuit transformation.


Emitter Follower - High Frequency Zeros

* What are the high frequency zeros for the EF 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 node B’ will be shorted to ground and Vπ = 0 .

* Similarly, we get a zero from Cπ when yπ + gm = 0 since we showed earlier that

* Also, can see this from

s

o

s

oV V

V

V

V

V

VA

sradxpFKCr

rg

bygiveniszeroourso

Cr

rgs

getwesforSolving

gsCr

gy

andsCr

y

gy

mZH

m

mm

m

/102.11797.0

20111

11

01

1Using

0

102

mLEo gyRR

V

V

0,01

1

0

omm

LE

om

LE

om

VthengygsCr

If

RR

VgsC

rV

RR

VVgVsC

r

VEatKCL


Emitter Follower - High Frequency Poles and ZerosMagnitude

2

10

2

10

2

10

102

101

102

1

1010

10

1010

10

101.11log10

100.11log10

102.11log101.0)(

1.0987.0log20)(

/101.1

/100.1

/102.1

101.11

100.11

102.11

987.0)(

101.11

100.11

1102.1

1987.0)(

xxxdBA

dBdBA

sradx

sradx

sradx

xj

xj

xj

A

x

s

x

s

s

x

s

sA

V

Vo

PH

PH

ZH

ZH


Emitter Follower - High Frequency Poles and ZerosPhase Shift

10

110

110

1

102

101

102

1

1010

10

1010

10

101.1tan

100.1tan

102.1tan)(

/101.1

/100.1

/102.1

101.11

100.11

102.11

987.0)(

101.11

100.11

1102.1

1987.0)(

xxx

sradx

sradx

sradx

xj

xj

xj

A

xs

xs

sxs

sA

PH

PH

ZH

ZH


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

CE EF

Midband Gain

Low Frequency Poles and Zeros

High Frequency Poles and Zeroes

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

Better low frequency response !

Much better high frequency response !


Conclusions* Voltage gain

● Can get good voltage gain from CE but NOT from EF amplifier (AV 1).

● Low frequency performance better for EF amplifier.● EF amplifier gives much better high frequency performance!

CE amplifier has dominant pole at 5.0x107 rad/s. EF amplifier has dominant pole at 1.0x1010 rad/s.

* Bandwidth approximately 200 X larger!

* Miller Effect multiplication of C by the gain is avoided in EF.

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

● For EF amplifier, current gain is also high Ie/Ib = +1 !

● Frequency dependence of current gain similar to voltage gain.

* Input and output impedances are different for the two amplifiers!

eces 352 winter 2007ch. 7 frequency response part 41 emitter-follower (ef) amplifier *dc biasing ●...

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