FREQUENCY RESPONSE OF AMPLIFIERS · 3 Frequency Response of Amplifiers * In reality, all amplifiers...

1

FREQUENCY RESPONSE OF AMPLIFIERS* Effects of capacitances within transistors and in amplifiers

* Build on previous analysis of amplifiers from EENG 341Transistor DC biasingSmall signal amplification, i.e. voltage and current gainTransistor small signal equivalent circuit

* Use in Bipolar and FET Transistors Amplifiers and their analysis* Build on previous analysis of single time constant circuits

Review simple RC, LC and RLC circuitsRecall frequency dependent impedances for C and LReview frequency dependence in transfer functions

* Magnitude and phase* Examine origins of frequency dependence in amplifier gain

Identify capacitors and their origins; find the dominant CDetermine equivalent R and determine RC time constantUse to describe approximately the amplifier’s frequency behaviorExamine effects of other capacitors

* GOAL: Use results of analysis to modify circuit design to improve performance.

2

Analysis of Amplifier Performance

DC bias or quiescent point

iB iC

vBE vCE

* Previously analyzed DC bias pointAC analysis (midband gain)

* Neglected all capacitances in the transistor and circuit

* Gain at middle frequencies, i.e. not too high or too low in frequency

3

Frequency Response of Amplifiers* In reality, all amplifiers have a limited

range of frequencies of operationCalled the bandwidth of the amplifierFalloff at low frequencies

* At ~ 100 Hz to a few kHz* Due to coupling capacitors at the

input or output, e.g. CC1 or CC2 Falloff at high frequencies

* At ~ 100’s MHz or few GHz* Due to capacitances within the

transistors themselves.

Equivalent circuit for bipolar transistor

20 logT(ω)

ω

Midband Gain

4

Frequency Response of Amplifiers

* First approximation – describe the amplifier’s high and low frequency responses in terms of that of single time constant (STC) circuits

High frequency falloff –Like that of a low pass filter

* Simple RC equivalent circuit * Shunting capacitor shorts signal at the

output at high frequenciesLow frequency falloff

Like that of a high pass filter* Simple RC equivalent circuit* Series capacitor blocks output signal at

low frequencies (acts like open circuit)* Amplifier frequency analysis

Determine equivalent R for each CCompare and find the most important (dominant RC) combination

Find the dominant one (RC)at high frequencies

Find the dominant one (RC)at low frequencies

Low-Pass Network

High-Pass Network

( )⎩⎨⎧

→∞∞→

→==0 (open)

011ωω

ω asasshort

CjsCZC

Vi

Vi

5

Review of Complex Numbers* Complex numbers

General form a + bj wherea = real part, b = imaginary part and

Magnitude of complex numberPhase of complex numberPhasor form

* Complex number mathMultiplication of two complex numbers a + bj = Me jθ and c + dj =Ne jφ

Reciprocal of a complex number

1−=j

θθ

jj eMMebja

ORba

bjba

ababja

bjabja

bjabja

−==+

+−

+=

+−

=−−

+=

+

111

11222222

22 babjaM +=+=

⎟⎠⎞

⎜⎝⎛= −

ab1tanθ

θjMebja =+

( )( )

( )( ) ( )( ) ( )ϕθϕθ +==++

++−=++

jjj MNeNeMedjcbjaOR

adbcjbdacdjcbja )()(

6

Amplifier Transfer Function (Gain) - General Form* A (s) = Gain Function (general form of amplifier transfer function)

AM = midband gain (independent of frequency)FH(s) = high frequency function (acts like low pass filter)

FL(s) = low frequency function (acts like high pass filter)

HH s

sFω/1

1)(+

=

)()()( sFsFAsA LHM=

ssF

LL /1

1)(ω+

=

ωHωL

Magnitude

FH(s)FL(s)

AM

7

Amplifier Transfer Function (Gain) - General Form

)()()( sFsFAsA LHM=

Coupling Capacitors

Transistor’s Capacitors

ωHωL

FH(s)FL(s)

AM

8

Frequency Response of MOSFET vs BJT AmplifiersEquivalent circuit for bipolar transistor Equivalent circuit for MOSFET

Similar equivalent circuits

Corresponding amplifier circuits

Similar frequency performance

Common Source Amplifier

Common Emitter Amplifier

Gain Gain

9

Amplifier Transfer Function (Gain) - General Form

)()()( sFsFAsA LHM=

Coupling Capacitors

Transistor’s Capacitors

ωHωL

Now we consider the low frequency behavior.

10

Summary* Examined origin of falloff in amplifier gain at low and high frequencies.

Degradation in magnitude of the gain.Shift in phase of output relative to input.

* Due to presence of capacitors within the amplifier (Create poles and zeros).Coupling capacitors limit gain at low frequencies.Transistor’s capacitances limit gain at high frequencies.

* Examined and quantified the falloff due to single and multiple poles and zeroes. Bode plots of gain and phase shift with frequency.

* Next, we will apply this method of analysis to transistor amplifiers.Multiple capacitors so multiple RC combinations.Investigate how to determine which capacitors are most important and limit the bandwidth.Examine how to change the amplifier to get better frequency performance.

Higher frequency operation before falloff (improved bandwidth).Better low frequency behavior.

11

Analysis of Bipolar Transistor Amplifiers* Single stage amplifiers

Common Emitter (CE)Common Base (CB)Emitter Follower (EF) (Common Collector)

* DC biasingCalculate IC, IB, VCEDetermine related small signal equivalent circuit parameters

Transconductance gmInput resistance rπ

* Low frequency analysisGray-Searle (Short Circuit) Technique

Determine the pole frequencies ωPL1, ωPL2, ... ωPLnDetermine the zero frequencies ωZL1, ωZL2, ... ωZLn

* High frequency analysisGray-Searle (Open Circuit) Technique

Determine the pole frequencies ωPH1, ωPH2, ... ωPHnDetermine the zero frequencies ωZH1, ωZH2, ... ωZHn

12

CE Amplifier Frequency Analysis - Long and Difficult Way* DC analysis: IC , IB , VCE ; gm , rπ* Draw ac equivalent circuit* Substitute hybrid-pi model for

transistor* Obtain KVL equations (at least one

for each capacitor in circuit (5))* Solve set of 5 simultaneous equations

to obtain voltage gain AV = Vo /Vs* Put expression in standard form for

gain AV(ω) = AVo FH(ω) FL(ω)* Identify midband gain AVo* Determine FH(ω) part and factor to

Determine high frequency poles ωPH1 and ωPH2Determine high frequency zeros ωZH1 and ωZH2

* Determine FL(ω) part and factor toDetermine low frequency poles ωPL1, ωPL2 and ωPL3Determine high frequency zeros ωZL1, ωZL2 and ωZL3

* Is there an easier way ? YES⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=

21

21

11

11)(

PP

ZZH ss

ss

sF

ωω

ωω

( )( )( )( )( )( )

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

=

++++++

=

sss

sss

sssssssF

PPP

ZZZ

PPP

ZZZL

321

321

321

311

111

111

)(

ωωω

ωωωωωωωωω

13

CE Amplifier - Starting Point is DC Analysis

* Q is quiescent point (DC bias point)* Q needs to be in the active region

IC = β IB* If Q is in saturation (VCE < 0.3 V) , then

IC < β IB and there is little or no gain from the transistor amplifier

* If the transistor is in the cutoff mode, there is virtually no IC so there is no gain, i.e. gm 0.

* Q depends on the choice of R1 and R2 since they determine the size of IB.

* Q point determines the size of the small signal parameters

Transconductance gm = IC / VTVT = kBT/q = 26 mVInput resistance rπ = β / gm

Saturationregion

Cutoff region

Activeregion

Q point

14

Example of CE Amplifier - DC Analysis

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

* Form Thevenin equivalent for base; given VCC = 12.5VRTh = RB = R1||R2 = 10K||2.5K = 2KVTh = VBB = VCC R2 / [R1+R2] = 2.5V

DC Base Current (use 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 resistancerπ = β / gm = 200/[206 mA/V]= 0.97 K

* Check on transistor region of operation (Find VCE)KVL collector loopVCE = VCC - IC RC - (β +1) IB RE = 4.4 V

(okay since not close to zero volts, i.e. > 0.2V).

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

15

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

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

* Neglect all capacitancesCoupling and emitter bypass capacitors become shorts at midband frequencies (~ 105 rad/s)

Why? Their impedances are negligibly small, e.g. few ohms because CC1, CC2, CE are large,

e.g.~ few µF (10-6F)

Transistor capacitances become open circuits at midband frequencies

Why? Their impedances are very large, e.g. ~ 10’s M Ω because Cπ , Cµ are very small, e.g. ~ pF(10-12 F)

* Calculate small signal voltage gain AVo = Vo /Vs

CE Amplifier - Midband Gain Analysis

Ω== 10)1)(/10(

1~1 5 FsradCZC µω

Ω== 75 10)1)(/10(1~1

pFsradCZC ω

Hybrid-Pi Model for BJT

16

CE Amplifier - Midband Gain Analysis

( ) ( ) ( )( )

( )[ ]( )[ ]

( )[ ]( )[ ]

( )( )( )( ) dBdBA

VVA

KK

KKKKKKK

RrrRRrr

VV

KKK

rrr

VV

KKVmARRgV

RRVgVV

VV

VV

VV

VVA

Vo

Vo

Bxs

Bx

s

i

xi

CLmCLmo

s

i

i

o

s

oVo

7.27)6.24log(20

/6.2412.094.0218

12.068.568.0

0.2065.097.050.2065.097.0

94.0065.097.0

97.0

21892.1/206

=−=

−=−=

==++

+=

++

+=

=+

=+

=

−=−=−=−

=

==

π

π

π

ππ

π

π

π

π

π

KRKR

KRKR

KRKR

S

E

C

L

55.2

1033.02.1

9

2

1

======

RL||RC

+

-Vi

Io

+

-

Vi

Break voltage gain into a series of voltage ratios

Vs

Vorx

rπ gmVπ

Negative sign means output signal is 180o out of phase with the input signal.

Vπ

17

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

* Draw AC equivalent circuit at low frequency Include coupling and emitter bypass capacitors CC1,CC2, CESubstitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Ignore (remove) all transistor capacitances Cπ , Cµ

* Turn off signal source, i.e. set Vs= 0Keep source resistance RS in circuit (do not remove)

* Analyze the circuit one capacitor Cx at a time Replace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor finding equivalent resistance seen by it and find corresponding pole frequency

* Determine which is dominant (largest) low frequency pole* Calculate the final, low 3dB frequency using

xxPx CR

1=ω

∑ ∑=+== xxPnPPPxLP CR1...21 ωωωωω

18

Common Emitter - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique

( ) ( )

( ) sradFKCR

KKKKKrrRRIVR

CCPL

xBSX

XC

/8827.5

11

7.5)97.0065.025

111

1

===

=++=++==

µω

π

( ) sradFKCR

KKKRRIVR

CCPL

CLX

XC

/3332.10

11

2.102.19

222

2

===

=+=+==

µω

Vo

Vo

VXVX

Vπ

Vπ

rπ

r π

rX

rX

RS RB RC RL

RS

VS gmVπ RC RL

RE

RB

CC1

CE

CC2

* Output coupling capacitor CC2 = 3 µF* Input coupling capacitor CC1 = 2 µF

AC equivalent circuitat low frequency

IX IX

19

Common Emitter - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique

( ) sradFKCR

KKKKK

K

RRrrRR

RRrrrg

R

RRrrrg

RRrrRVI

R

RRrrVrg

RRrrV

RVI

RRrrVrrIV

RRrrVI

RVI

VgIIIIVR

EExPL

BsxEEx

Bsx

m

E

Bsx

m

BsxEx

x

Ex

Bsx

xm

Bsx

x

E

xx

Bsx

x

Bsx

x

E

xe

mexx

xEx

/342,512016.0

11

016.0201

2597.0065.033.0

1

11

111

?

3 ===

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=

+++

+=

+++

+++==

+++

+++=

++−

==

++−

==

−−===

µω

βπ

π

π

π

π

π

π

π

π

π

ππππ

ππ

ππ

sradPLPLPLPL /546353423388321 =++=++= ωωωω

rX

rπ Vπ gmVπ

VX

RS RB

RB

RS rX

rπ Vπ gmVπ RC RL

RE

CC1

CE

CC2

REIe

Ix

Iπ

Emitter bypass capacitor CE = 12 µF

+

VoVS

REx

KCL at node E

Recall β = gmrπ

Low 3db frequency

20

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

* Draw AC equivalent circuit at high frequency Substitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ )Remove coupling and emitter bypass capacitors CC1, CC2, CE (consider as shorts).Turn off signal source, i.e. set Vs = 0Keep 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 circuitsSolve remaining circuit for the equivalent resistance Rx seen by the selected capacitorCalculate the pole frequency usingRepeat process for each capacitor

* Calculate the final high 3dB frequency using

xxPHx CR

1=ω

∑∑=⎥

⎦

⎤⎢⎣

⎡++=⎥

⎦

⎤⎢⎣

⎡=

−−

xxPHnPHPHPxPH

CR

11...1111

21

1

ωωωωω

Cπ comes from Emitter-Base p-n junction.

Cµ comes from Base-Collector p-n junction.

21

Common Emitter - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique

AC equivalent circuit at High frequency

CE shorts REat high frequencies.

Vo

VS

VoVS Vπ

22


* Equivalent circuit for Capacitor Cπ = 17 pF

( ) ( )

( )

sradxxCR

xpFKCR

KKKKKRRrrIVR

xPH

x

SBxX

Xx

/100.1sec100.1

11sec100.11759.0

59.052065.097.0

881

8

===

==

=+=+==

−

−

ππ

ππ

ππ

ω

VoVS Vπ

VX

IX

Note: This frequency is very high due to the very small size of the capacitor.

23


* Equivalent circuit for Capacitor Cµ = 1.3 pF

( ) ( )( )

( ) ( )( )

( )

sradxxxCRCR

isfrequencydBfrequencyhighFinal

sradxxCR

xpFKCRK

KKKKKK

VmAKK

RRrrRR

gRRR

getweIVRforsolvingandVforngSubstituti

RRV

RRgV

RVV

RVVVgI

VVVorVVVRV

RVVgICnodeAt

RRrrIVso

RRrV

rVIBnodeAt

xxPH

xPH

x

SBxLC

mLCx

x

xx

LCx

LCm

L

x

C

xmx

xoox

L

o

C

omx

SBxx

SBxx

/106.5sec107.1100.1

11

3

/109.5sec107.1

11sec107.13.1130

130

52065.097.091

2.11/206192.1

111

1111

0Using

11'

678

672

7

=+

=+

=

===

==

=

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛ +++=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛+++=

=

⎥⎦

⎤⎢⎣

⎡++⎥

⎦

⎤⎢⎣

⎡++−=

−−

−−−=

−==++−

−−−=

⎥⎦

⎤⎢⎣

⎡

++=

++=

−−

−

−

µµππ

µµ

µµ

πµ

µπ

πππ

π

ππ

π

ππ

π

π

π

ω

ω

B/

Need to solve forX

XX I

VR =µ

VXIX

Vπ

Vπ

C

24

Common-Base (CB) Amplifier* DC biasing

Calculate IC, IB, VCEDetermine related small signal equivalent circuit parameters


* Midband gain analysis* Low frequency analysis

Gray-Searle (Short Circuit) Technique

Determine pole frequencies ωPL1, ωPL2, ... ωPLn

Determine zero frequencies ωZL1, ωZL2, ... ωZLn


Determine pole frequencies ωPH1, ωPH2, ... ωPHn

Determine zero frequencies ωZH1, ωZH2, ... ωZHn

Input at emitter, output at collector.

25

CB Amplifier - DC Analysis (Same as CE Amplifier)

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


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




* Check on transistor region of operationKVL collector loopVCE = 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

26

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

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

* Neglect all capacitancesCoupling 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 Ωbecause Cπ , Cµ ~ pF (10-12 F)

* Calculate small signal voltage gain AVo = Vo /Vs

CB Amplifier - Midband Gain Analysis

Ω=== 10)1)(/10(

115 FsradC

ZC µω

Ω=== 75 10)1)(/10(11

pFsradCZC ω

High and Low Frequency AC Equivalent Circuit

27

CB Amplifier - Midband Gain Analysis

( ) ( ) ( )( )

[ ][ ]

[ ][ ]

( )( )( )( ) dBdBA

VVAKK

KKKKK

rRRrR

VV

KKK

rrIIr

VV

KKVmARRgV

RRVgVV

VV

VV

VV

VVA

Vo

Vo

eEs

eE

s

e

xe

CLmCLmo

s

e

e

o

s

oVo

14)20.0log(20

/20.0001.094.0218

001.00050.50050.0

0051.033.050051.033.0

94.0065.097.0

97.0)(

21892.1/206

−==

=−−=

==+

=+

=

−=+

−=

+−=

−=−=−=−

=

==

ππ

πππ

π

π

π

π

π

KRKR

KRKR

KRKR

S

E

C

L

55.2

1033.02.1

9

2

1

======

Equivalent resistance re

KKKrrrg

rrrg

rr

rrrg

rVV

IVr

rrgV

rgVI

rVVgI

EnodeatKCLIVr

x

m

x

m

x

m

e

e

ee

mme

me

e

ee

0051.02001

97.0065.011

11

11

0

=++

=++

=++

=

+⎟⎟⎠

⎞⎜⎜⎝

⎛ +−−=

+−==

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

=++

=

βπ

π

π

π

π

π

π

π

π

π

π

ππ

ππ

π

ππ

Ve+_

Iπ

π

ππ

ππ

ππ

rrr

VV

rrV

rVI

xe

x

e

+−

=

+−

==

re

βIπ

Voltage gain is less than one !

28

What Happened to the CB Amplifier’s Midband Gain?

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

Why? Rs >> re = 0.0051 Kso Ve/Vs

29


* Draw low frequency AC circuitSubstitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Include coupling and base capacitors CC1, CC2, CBIgnore (remove) all transistor capacitances Cπ , Cµ


* Consider the circuit one capacitor Cx at a timeReplace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat 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=ω

∑ ∑=+==xx

PnPPPxLP CR1...21 ωωωωω

30

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

* Base capacitor CB = 12 µF

( )

( )( ) ( )

( )( )( )

( )( )[ ]( )( )[ ]

( )

sradxCR

xKFRC

KKK

KKKKK

RRrgrrRR

RRrgrV

RRgr

VrR

RRgr

VRRVgIVV

rVIce

VVr

rVV

IVR

RrRR

BxCPL

xCB

SEmxBxC

SEm

SEm

i

SEmSEmi

iiii

ixBxC

B

B

B

B

/83102.111

sec102.10.112

0.103.22

005.033.020197.0065.02

1

1

11

11

sin/

21

2

===

==

==

++=

+++=

++=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++=++=

====

+=

−

−

ω

µ

ππ

πππ

ππ

π

πππππ

π

ππ

ππ

πππ

Vi

+

_RxCBRi

Iπ

Low Frequency AC Equivalent Circuit

Vπ

VxVo

Ix

31


Input coupling capacitor CC1 = 2 µF

( )( )

( )( )

( )( )

( )

sradxxRC

xKFRC

KKK

KKKK

rgrrRRR

rgrr

IrgrrI

IVr

rrIVIrgVgII

IVrrRR

IVR

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

20197.0065.033.0005.0

1

11

1

451

2

51

1

1

1

1

===

==

=+=

⎥⎦

⎤⎢⎣

⎡ ++=

⎥⎦

⎤⎢⎣

⎡

++

+=

++

=+−

+−==

+−=+−=−−=

=+==

−

−

ω

µ

π

π

π

π

ππ

ππ

ππ

ππππ

Ve

Ie

Iπ

Ve+_

re

Vπ

Vo

Ix Rs

Vx

32


srad

PLPLPLPL

/116,5033000,5083

321

=++=

++= ωωωω

* Output coupling capacitor CC2 = 3 µF

( ) sradFKCR

KKKRRR

CCPL

CLC

/3332.10

112.102.19

223

2

===

=+=+=

µω

* Low 3dB frequency

VoRL

VX

RC

Dominant low frequency pole is due to CC1 !

33


* Draw high frequency AC equivalent circuitSubstitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ) Consider coupling and emitter bypass capacitors CC1, CC2, CB as shortsTurn off signal source, i.e. set Vs = 0Keep 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 circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor

* Calculate the final high frequency pole using

xxPHx CR

1=ω

∑∑=⎥

⎦

⎤⎢⎣

⎡++=⎥

⎦

⎤⎢⎣

⎡=

−−

xxPHnPHPHPxPH

CR

11...1111

21

1

ωωωωω

34

Common Base - Analysis of High Frequency PolesGray-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.

35


* Equivalent circuit for Ze

βππ

ππ

π

ππ

π

ππ

ππ

π

π

ππ

π

ππ

π

π

π

π

π

+=

+=

=

⎟⎠⎞⎜

⎝⎛

⎥⎦

⎤⎢⎣

⎡+

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞⎜

⎝⎛

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎥⎦

⎤⎢⎣

⎡+

+=

⎥⎦

⎤⎢⎣

⎡+

+==

⎥⎦

⎤⎢⎣

⎡+

+=

⎥⎦

⎤⎢⎣

⎡++=−

⎟⎠⎞⎜

⎝⎛

+=

−==

11

11

11

1

1

1

11

1

1

11

rrg

rr

where

ZrZSo

sCrgr

sCrgr

sCr

rgsCr

rgV

VIVZ

sCr

rgV

gsCr

VVg

sC

VrVI

givesEnodeatKCLVVIVZ

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.

36

* Pole frequency for Cπ =17pF


( ) sradxsxpFCR

KKKKR

KKrg

rr

RRrR

xCPH

xC

me

sEexC

/105.2101.4

1174.211

4.20024.0005.033.00048.0

8.40048.02001

97.01

10111 ==Ω

==

Ω===

Ω==+

=+

=

=

−π

π

π

π

π

π

ω

Turn off signal source when finding resistance seen by capacitor.

37


* Equivalent circuit for Capacitor Cµ = 1.3 pF

* Pole frequency for Cµ =1.3pF

( ) sradxsxpFKCR

KKKR

RRR

xCPH

xC

LCxC

/101.7104.1

13.105.1

11

05.192.1

892 ====

==

=

−µµ

µ

µ

ω

sradxsx

xxCRCR

PH

xCxCPH

/109.61044.11

104.1101.411

89

911

==

+=

+=

−

−−

ω

ωµπ µπ

* High 3 dB frequency

Dominant high frequency pole is due to Cµ !

Rs || RE || r π

= 0

38

Comparison of CB to CE AmplifierCE (with RS = 5K) CB (with RS = 5Ω)

Midband Gain

Low Frequency Poles and Zeros

High Frequency Poles and Zeroes

( ) [ ][ ]( )( )( )

( ) dBdBAVVA

rrr

rRRrR

RRgVV

VV

VVA

Vo

Vo

xeEs

eECLm

s

e

e

oVo

2.40

/4.1025.094.0218

=

+=−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

+−==

π

ππ

π

[ ] ( )

( ) ( ) sradxpFKCRR

sradxpFCRRr

LCPH

sEePH

ZHZH

/101.73.105.1

11

/105.2174.211

,

82

101

21

===

=Ω

==

∞=∞=

µ

π

ω

ω

ωω

( )

( )( )[ ]{ } ( )

( )

( ) ( ) sradFKCRR

sradxKF

rgrrRRC

sradKFCRRrgrrR

sradFKCR

CCLPL

mx

EsC

PL

BSEmxBPL

BBZPZPZP

/3332.10

11

/100.5010.02

1

1

1

/83112

11

1

/42122110

23

4

1

2

1

321

==+

=

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡++

+

=

==+++

=

=====

µω

µω

µω

µωωω

ππ

ππ

( )[ ] ( )[ ]( )[ ]( )( )( )

( ) dBdBAVVA

RrrRRrr

rrrRRg

VV

VV

VV

VVA

Vo

Vo

Bxs

Bx

xCLm

s

i

i

o

s

oVo

7.27)6.24log(20

/6.2412.094.0218

=−=

−=−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+⎥⎦

⎤⎢⎣

⎡

+−===

π

π

π

ππ

π

( )

( )[ ] ( )

( ) ( )

( ) sradFKC

RRrrR

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

==

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛

+++

=

==+

=

==++

=

=====

µ

β

ω

µω

µω

µωωω

π

π

( )[ ] ( )

( ) ( )( )

( ) sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmACg

SBxLC

mLC

PH

SBxPH

mZHZH

/109.53.1130

1

111

1

/100.11759.0

11

/106.13.1

/206,

6

2

81

1121

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=

==+

=

===∞=

µπ

ππ

µ

ω

ω

ωω

Note: CB amplifier has much better high frequency performance!

39

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

Midband Gain



( ) [ ][ ]( )( )( )

( ) dBdBAVVA

rrr

rRRrR

RRgVV

VV

VVA

Vo

Vo

xeEs

eECLm

s

e

e

oVo

2.40

/4.1025.094.0218

=

+=−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

+−==

π

ππ

π

[ ] ( )

( ) ( ) sradxpFKCRR

sradxpFCRRr

LCPH

sEePH

ZHZH

/101.73.105.1

11

/105.2174.211

,

82

101

21

===

=Ω

==

∞=∞=

µ

π

ω

ω

ωω

( )

( )( )[ ]{ } ( )

( )

( ) ( ) sradFKCRR

sradxKF

rgrrRRC

sradKFCRRrgrrR

sradFKCR

CCLPL

mx

EsC

PL

BSEmxBPL

BBZPZPZP

/3332.10

11

/100.5010.02

1

1

1

/83112

11

1

/42122110

23

4

1

2

1

321

==+

=

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡++

+

=

==+++

=

=====

µω

µω

µω

µωωω

ππ

ππ

( )[ ] ( )[ ]( )[ ]( )( )( )

( ) dBdBAVVA

RrrRRrr

rrrRRg

VV

VV

VV

VVA

Vo

Vo

Bxs

Bx

xCLm

s

i

i

o

s

oVo

6.45)191log(20

/19193.094.0218

=−=

−=−=⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++

⎥⎦

⎤⎢⎣

⎡+

−===π

π

π

ππ

π

( )

( )[ ] ( )

( ) ( )

( ) sradxFKC

RRrrR

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

==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=

==+

=

==++

=

=====

µ

β

ω

µω

µω

µωωω

π

π

( )[ ] ( )

( ) ( )( )

( ) sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmACg

SBxLC

mLC

PH

SBxPH

mZHZH

/100.53.14.15

1

111

1

/100.917065.0

11

/106.13.1

/206,

7

2

81

1121

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=

==+

=

===∞=

µπ

ππ

µ

ω

ω

ωω

Note: CB amplifier has much better high frequency performance!

40

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 Cµ by the gain is avoided in

CB configuration.

* Current gainFor CE amplifier, current gain is high AI = Ic / IbFor 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.

41

Emitter-Follower (EF) Amplifier* DC biasing

Calculate IC, IB, VCEDetermine related small signal equivalent circuit parameters


* Midband gain analysis* Low frequency analysis

Gray-Searle (Short Circuit) Technique

Determine pole frequencies ωPL1, ωPL2, ... ωPLn

Determine zero frequencies ωZL1, ωZL2, ... ωZLn


Determine pole frequencies ωPH1, ωPH2, ... ωPHn

Determine zero frequencies ωZH1, ωZH2, ... ωZHn

High and Low Frequency AC Equivalent Circuit

42

EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier)* GIVEN: Transistor parameters:

Current gain β = 200Base resistance rx = 65 ΩBase-emitter voltage VBE,active = 0.7 VResistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K


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




* Check on transistor region of operationKVL collector loopVCE = 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.

43

EF Amplifier - Midband Gain Analysis

( )( ) ( ) ( )

( )( )

( )( ) 998.0905.1

9.165065.02005.0

65065.02

999.065065.0

65

015.0661

1

1

1

6697.020133.091

==++

+=

+++

=

=+

=+

=

=+

=+

=+

=

==+

=⎥⎦

⎤⎢⎣

⎡+

=

==

KK

KKKKKKK

RrRRRrR

VV

KKK

RrR

VV

VVVV

VVV

KKK

rrgRR

V

VgrVRR

VV

VV

VV

VV

VV

VVA

ixBS

ixB

s

b

ix

i

b

i

ooi

mEL

mELo

s

b

b

i

i

o

s

oVo

π

π

ππ

π

π

π

ππ

π

π

π

π

Ω=====

==

5005.05.2

1033.0

09

2

1

KRKR

KRKR

KRKR

S

E

C

L

Equivalent input resistance Ri

( ) KKVVr

rV

VVIVR ooii 6566197.01 =+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==π

π

ππ

π

π

( )( )( )( )( ) dBdBA

A

Vo

Vo1.0987.0log20)(

987.0998.0999.0015.066−==

==

Vi

+

_

RiVb

+

_

Iπ

KVmAg

r

VmAmVmA

VIg

m

T

Cm

97.0/206

200

/2062627.5

===

===

βπ

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!

44


* Draw low frequency AC circuitSubstitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Include coupling capacitors CC1, CC2Ignore (remove) all transistor capacitances Cπ , Cµ


* Consider the circuit one capacitor Cx at a time Replace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rxseen by the selected capacitorCalculate 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=ω

∑ ∑=+== xxPnPPPxLP CR1...21 ωωωωω

45

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


( )

( )( ) ( )( )[ ]( )( )

( )

( )

( )

( )

sradxRC

xKFRC

KKKKK

RrRRR

KKKK

RRrgrIVR

RRrgrIRRVgIrIV

IVRRrRR

IVR

xCCPL

xCC

ixBsxC

LEmi

i

LEmLEmi

iiixBs

x

xxC

/256sec109.3

11sec109.395.12

95.165065.02005.0

65933.0)201(97.0

1

1

3111

311

1

1

===

==

=++=

++=

=+=

++==

++=++=

=++==

−

−

ω

µ

πππ

πππππππ

π

Ri

Vi

Iπ

IX

46

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

srad

PLPLPL

/29337256

21

=+=

+= ωωω


( )[ ][ ]

( )

( ) sradFKCR

KKKKrRRR

KKKKK

rgRRrr

rgIRRrrI

r

RRrrIV

rgIVgII

IVrrRRR

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

===

=+=+=

=++

=

+

++=

+−

++−=

++−=

+−=−−=

=+=

µω

π

π

ππ

ππ

ππ

ππππ

* Low 3 dB frequency

Iπ

Ve

Ie IXre

So dominant low frequency pole is due to CC1 !

47


* Draw high frequency AC equivalent circuitSubstitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ) Consider coupling and emitter bypass capacitors CC1 and CC2 as shortsTurn off signal source, i.e. set Vs = 0Keep 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 circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor

* Calculate the final high frequency pole using

xxPHx CR

1=ω

∑∑=⎥

⎦

⎤⎢⎣

⎡++=⎥

⎦

⎤⎢⎣

⎡=

−−

xxPHnPHPHPxPH

CR

11...1111

21

1

ωωωωω

48

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

* Redrawn High Frequency Equivalent Circuit

sourcecurrent dependent to

due resistance equivalent'

'

=−

=

==

πVgVRwhere

RRRIVZ

m

oS

SLEe

oeq

Ie

( ) ( ){ }

( ){ }

( ) ⎥⎦

⎤⎢⎣

⎡ +=

⎥⎦

⎤⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎦

⎤⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

⎛−=+=

=

+−=−=

+=+==

+=⎭⎬⎫

⎩⎨⎧

++=

=−++

π

π

π

π

π

ππ

πππππ

π

ππππ

ππππ

π

ygyRRZ

gyy

RR

gyg

RRRRRRRZ

findcanweRRRZSince

gyRRgVg

VRso

sCrZrz

ydefinewewhere

gyRRgsCr

RRVV

RRVVgVsC

rVEatKCL

mLEeq

mLE

m

m

LELESLEeq

SLEeq

mLEmm

oS

C

mLEmLEo

LE

om

1

11111

1

1111

1

0

'

'

'

Zeq

EIe

zπ =1/yπ

49


( ) ( ) ( )( )

( )

( )

( )

( )

( )( )[ ] ( )[ ]

( ) ( )[ ]( )

( )LEmC

CLEmLEb

b

LEmLEm

LE

LEm

LEb

LEm

LE

LEmLE

mLEeqb

b

RRgCssC

Zwhere

ZRRgrRRZ

isZso

RRgCs

RRgr

RR

RRgr

CsrRRZ

rCsrsC

ryforngSubstituti

RRgy

RR

yRRg

RRyy

gyRRy

ZZ

ZisgroundandBbetweenimpedanceTotal

+

==

++=

++

+

+=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

⎥⎦

⎤⎢⎣

⎡ ++=

+=+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

++=+⎥

⎦

⎤⎢⎣

⎡ +=+=

1

1'

11

111

1

1

11

11

1

1

111

'

'

''

'

'

'

'

π

π

π

π

π

ππ

π

πππ

ππ

π

πππ

π

π

Modified Equivalent Circuit

Replace thiswith this.

ZB’

ZB’

zπ =1/yπ

Looks like a resistor in parallel with a capacitor.

50


( )[ ] [ ]( )( )( )[ ] [ ]

[ ] [ ]

( )( )( )

sradx

sxpFKKKVmA

pFKRRg

CRCR

KKK

KKKKKKKVmAK

RRRRrRRgrR

PH

LEmxC

xCPH

LEBSxLEmxC

/100.1

1086.91

255.0386.01

933.0/206117386.0

1

1

11

386.039.06.64

933.02005.0065.0933.0/206197.0

1

101

11

'''

1

'

=

==

⎥⎦

⎤⎢⎣

⎡

+

=

⎥⎦

⎤⎢⎣

⎡

+

==

==

+++=

+++=

−

ω

ωπ

πππ

ππ

RxCπ

* Pole frequency for Cπ =17 pF

51


( )[ ]( ) [ ]( )( )( )[ ] ( )( ) [ ]

( )

( )sradx

sxpFKCR

KKKK

KKKKKKKVmAK

RRrRRRRgrR

PH

xCPH

BSxLELEmxC

/101.1

101.91

3.107.011

07.007.032.06.64

2005.0065.0933.0933.0/206197.0

1

102

112

=

===

=+=

+++=

+++=

−

ω

ωµµ

πµ

* Pole frequency for Cµ =1.3 pF

52


* Alternative Analysis for Pole Due to Cπ

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )[ ] ( )

( )( ) ( )

( )( ) ( )

( )( )

( ) sradxpFKCR

KKKKKK

KKKKKK

RRrRRrgRRrRR

rIIr

IVR

soRRrRRrg

RRrRRII

RRrRRIRRrRRrgI

RRrIIVRRIrgII

sorIVandIIIBut

RRrIIVRRVgI

givesloopbasearoundKVLIIr

IV

IVRandVVNote

xCH

BSxLEm

BSxLE

xx

xxC

BSxLEm

BSxLE

x

BSxLExBSxLEm

BSxxLEmx

xe

BSxxLEme

xxx

xxCx

/100.117006.0

11is frequency pole theSo

006.02005.0065.0933.0201

2005.0065.0933.097.0

1

1

1get wegRearrangin

0

0

:

101P ===

=++

++=

+++++

===

+++++

=

++=+++

=+−+−+−−

=−=

=+−+−+−

====

ππ

ππ

πππ

π

π

ππ

πππππ

ππππ

πππ

πππππ

ω

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.

53

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

( )( ) ( )

( )( )

( )[ ] [ ] [ ]

[ ] [ ] ( )( )

( )( ) ( )( )[ ] ( )( )[ ][ ] ( )[ ][ ] [ ]

( ) sradxpFKCR

KKK

KKKKKK

RRrgrRRr

RRrgrRRrRRrgrRRr

RRrIVR

RRrgrrVRRr

rRRrIV

V

RRrrVRRrIRRrIIV

RRrgrrVV

sorRR

rgVRRVgIVV

xCH

LEmBsx

LEmBsxLEm

Bsx

Bsx

x

xxC

LEmxBsxBsxxx

BsxBsxxBsxxx

LEmx

LEmLEmx

/101.13.107.0

11is Cfor frequency pole theSo

07.00.6507.0

933.0)201(97.02005.0065.0

1

111

1

11

get wegRearrangin

11

get wefor ngSubstituti

writealsocan We

1

11

102P ===

==

++=

+++=

+++

+

=

+++

+

+==

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++−+=

+−+=+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++=

⎥⎦

⎤⎢⎣

⎡++=++=

µµ

µ

ππ

ππππ

µ

ππ

π

π

π

π

ππ

ππ

ππ

ππππππ

ω

VxIx

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.

54

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

CE EF

Midband Gain



( )[ ] ( )[ ]( )[ ]( )( )( )

( ) dBdBAVVA

RrrRRrr

rrrRRg

VV

VV

VV

VVA

Vo

Vo

Bxs

Bx

xCLm

s

i

i

o

s

oVo

6.45)191log(20

/19193.094.0218

=−=

−=−=⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++

⎥⎦

⎤⎢⎣

⎡+

−===π

π

π

ππ

π

( )

( )[ ] ( )

( ) ( )

( ) sradxFKC

RRrrR

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

==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=

==+

=

==++

=

=====

µ

β

ω

µω

µω

µωωω

π

π

( )[ ] ( )

( ) ( )( )

( ) sradxpFK

CRRrrRR

gRR

sradxpFKCRRrr

sradxpF

VmACg

SBxLC

mLC

PH

SBxPH

mZHZH

/100.53.14.15

1

111

1

/100.917065.0

11

/106.13.1

/206

7

2

81

1121

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=

==+

=

====∞

µπ

ππ

µ

ω

ω

ωω

( )[ ][ ]

( )( )

( )( )( )( ) dBdBA

VVA

RrRRRrR

RrR

RRrRR

VV

VV

VV

VVA

Vo

Vo

ixBS

ixB

ix

i

LE

EL

s

b

b

i

i

oVo

1.0

/987.0)998.0(999.0015.066

−=

==

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛

++==

π

π

π

( )[ ] ( )

( )[ ] ( ) sradKFCrRR

sradKFRrRRC

CeELPL

ixBsCPL

ZPZP

/3793

11

/25695.12

110

22

11

21

==+

=

==++

=

==

µω

µω

ωω

( )

( )

( )[ ] [ ]

( ) sradxpFK

CRRrRRRRgr

sradxpFK

RRgCR

sradxpFKCr

rg

BSxLELEmPH

LEmxC

PH

mZHZH

/101.13.107.0

1

11

/100.126.0386.0

1

1

1

/102.11797.0

20111

10

2

10

'

1

1021

==

+++=

==

⎥⎦

⎤⎢⎣

⎡

+

=

==⎟⎟⎠

⎞⎜⎜⎝

⎛ +==∞

µπ

ππ

ππ

π

ω

ω

ωω

Better low frequency response !

Much better high frequency response !

55

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 Cµ by the gain is avoided in EF.

* Current gainFor CE amplifier, current gain is high β = Ic/IbFor EF amplifier, current gain is also high Ie/Ib = β +1 !Frequency dependence of current gain similar to voltage gain.

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

56

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.

pFCCpFCC

rr xx

29.13

0100

21

21

21

21

====

≈===

µµ

ππ

ββ

EF CE

57

Cascade Amplifier - DC analysis

{ }

( )

( )

( )

.

7.86.3)101(

7.087.31

87.3)3.4(899

8999.8)101(1

9.83.4110050

7.05

]1[

50100100

510200100

12

2

22222

112

22

111

1

2

12111111

1

211

21

21

okayisanalysiseapproximatsoII

AK

VVI

RIVVVKARIV

IandVcalculateNow

AAIIThen

AKK

VVI

ionapproximatfirstaasINeglectingRIIRIVV

QBaseKVL

KKKRRR

VVKKV

RRRV

EB

B

EBBEB

EEB

BB

BE

B

B

EBBThBBETh

Th

CCTh

58

Cascade Amplifier - Midband Gain Analysis

( ) ( )[ ]

( )( )

KKKK

rRrgr

IrRIrgrI

R

Em

Emi

178)9.23.4)(101(9.2

1

1

21111

1

2111111

=+=

++=

++=

πππ

π

πππππ

[ ]( )

( )

( )

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

=+

=+

=

=+

=++

=

==⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

−=−

=

==

KKKKKKK

RRRRRRR

VV

KKKK

rRIrgrIrI

VV

KKVmArRg

rVrRVgI

VV

VRRVg

VV

VV

VV

VV

VV

VVA

BS

i

s

i

Emi

EmEm

CLmo

s

i

i

o

s

oVo

πππππ

πππ

πππ

πππ

π

π

π

π

π

π

π

π

π

( ) ( )( )( ) 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

59

Cascade Amplifier - Low Frequency Poles and Zeroes

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

Three low frequency polesEquivalent resistance may depend on rπ for both transistors.

* Find three low frequency zeroes.

( )( )( )( )( )( )

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

=

++++++

=

sss

sss

sssssssF

LPLPLP

LZLZLZ

LPLPLP

LZLZLZL

321

321

321

321

111

111

)(

ωωω

ωωωωωωωωω

60

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


( )( ) ( )( )[ ]( )( )

( )

( )

sradxRC

xKFRC

KKKK

RRRR

KKKK

rRrgrIVR

rRrgrIrRVgIrIVIVRRRR

IVR

xCCPL

xCC

iBsxC

Emi

i

EmEmi

iiiBs

x

xxC

/23sec103.4

11sec103.40.431

0.43178504

1789.23.4)101(9.2

1

1

211

1

211

1

21111

2111112111111

1

===

==

=+=

+=

=+=

++==

++=++=

=+==

−

−

ω

µ

ππππ

πππππππππ

π

Ri

Vi

Iπ1

IX

RE1 rπ2

rπ1 Vπ1

Vπ2+

_

RE1 rπ2

61


( ) sradFKCR

KKKRRR

CCPL

CLC

/1251811

844

222

2

===

=+=+=

µω

Vo

Vo

VX

Vπ2

rX2

RC RL

gm2Vπ2 RC RL

RE2 CE

CC2


rπ2

62


( ) sradFKCR

KKKrRR

KKKK

r

KKKrg

RrrgI

RrII

Vr

rgrRr

rgIrRrI

IVr

rRIVR

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

7.39.21

')1()'(

1)1()(

3

22

2

11

1

111

11

1

11

22

112

222

1122

2

22

22

===

===

=+

=

=+

=

++

=+−+−

==

++

=+−+−

==

==

µω

π

π

ππ

ππ

π

π

ππ

ππ

sradPLPLPLPL /88273412523321 =++=++= ωωωω

Emitter bypass capacitor CE = 47 µF

RE1

r π1 Vπ1

Ie1

Iπ1

Iπ2

VX

Ix

gm2Vπ2rπ2 Vπ2

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 SS7.3

' 21=

=

63

Comparison of Cascade to CE Amplifier

CE* Cascade (EF+CE)

Midband Gain



( )( )( )( )( ) ( ) dBdBA

VVAVV

VV

VV

VVA

Vo

Vo

S

i

i

oVo

6.352.60log20

/2.6091.0016.08.6068

1

1

2

2

=−=

−=−=

= ππ

π

π

( )

( )

( )

( ) sradxpFK

sradxpFK

sradxpFK

sradxpFK

sradxsradx

PH

PH

PH

PH

ZHZH

ZHZH

/105.2221

,/100.1152063.0

1

,/104.126.3

1

,/100.89.1309.0

1

/107.1,/105.2

,,

84

83

82

81

104

93

21

==

==

==

==

==

∞=∞=

ω

ω

ω

ω

ωω

ωω

( )

( )( )[ ]{ } ( )

( ) ( )

( ) sradKFCrR

sradFKCRR

sradKFCrRrgrRR

sradFKCR

EeEPL

CCLPL

CEmBSPL

EEZPZPZP

/73403.047

1)(

1

/1251811

/234311

11

/9.5476.3110

223

22

1221111

2321

===

==+

=

==+++

=

=====

µω

µω

µω

µωωω

πππ

( )( )( ) dBdBA

VVAVV

VV

VVA

Vo

Vo

s

o

s

oVo

5.29)30log(20

/3037.02.81

2

2

=−=

−=−=

== ππ

( )

[ ] ( )

( ) ( )

( ) sradFKC

RRrR

sradFKCRR

sradFKCrRR

sradFKCR

EBs

E

PL

CCLPL

CBSPL

EEZPZPZP

/59147036.0

1

1

1

/1251811

/15714.4

11

/9.5476.3110

22

3

22

121

2321

==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

==+

=

==+

=

=====

µ

β

ω

µω

µω

µωωω

π

π

[ ] ( )

( ) ( )( )

( ) sradxpFK

CRRrRR

gRR

sradxpFKCRRr

sradxpF

VmACg

SBLC

mLC

PH

SBPH

mZHZH

/100.42125

1

111

1

/108.49.135.1

11

/100.22

/6.40,

6

222

2

7

221

10

2

221

==

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=

===

===∞=

µπ

ππ

µ

ω

ω

ωω

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

25 X improvement in bandwidth !

2 X improvement in voltage gain !

64

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 CESo 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 Ei1789.23.4)101(9.2

1 2111=+=

++= ππ β

( )( )

KKKKrRrR

KKKrg

RrrgI

RrIIVr

EeX

m

S

m

S

e

ee

063.09.23.4065.0

065.0101

7.39.21

'1

'

211

11

1

111

11

1

11

===

=+

=

++

=+−+−

==

π

π

π

ππ

ππ

Pole for Capacitor CT = 152 pF

( )amplifier. CE for the /100.4 versus

cascade for the /100.1152063.0

1

6

83

sradx

sradxpFKPH

==ω( )KKKRRR

pFKVmApFpFRgCCC

CLL

LmT

244'

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

=++=++= µπ

re1

Ri1Ri2

65

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

pFpFpFKKVmApFpFRRgCCCsorg

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

130221813.1172.19/20613.1171

by limited is eperformanc frequency high theamplifier CEFor

8

111

68

1111

==

=++=++=++=

Ω=++

=

==

=++=++=++=

−

ω

ω

µπ

π

π

µπ

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

Large Miller Effect

Small Miller Effect

66

Other Examples of Multistage AmplifiersCE CE EF EF

Darlington Pair

67

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

Amplifier with FETs and Bipolar Transistors

68

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 input1/2 of signal at output1/2 of source resistance

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

* Advantage: Rejects common noise pickup on inputNo coupling capacitors so can operate down to zero frequency.

Vo+ _

69

Differential Amplifier Analysis

( )( )

dBdBAKKK

KKVmA

Rrr

rV

RVgVV

V

V

VV

V

VA

Vo

Sx

Cms

o

s

os

oVo

1.54509log20)(

5095.2065.097.0

97.09/206

22

2

2

2

==

−=⎟⎠

⎞⎜⎝

⎛++

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛===

π

π

π

ππ

π

Midband Gain

Low Frequency Poles and Zeros* 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

( )[ ] ( )[ ]

( )[ ] ( )[ ] ( )( )sradx

x

pFKpFpFKKK

CRgCRrr CmsxPH

/101.5sec1095.1

1

27870.01

3.1201172/5065.097.01

12/1

67 ==

=++

=

+++=

−

µππω

High frequency performance is very similar to CE amplifier.

VoVo /2

Vo /2

70

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 highfrequency 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.

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