Lecture 27Lecture 27 Amplifier Configurations Reading: CE...

Lecture 27Lecture 27

Amplifier Configurations

Reading:

CE/CS: Jaeger 13.6, 13.9, 13.10, 13.11

CC/CD: Jaeger 14.1, 14.3

CB/CG: Jaeger 14 1 14 4CB/CG: Jaeger 14.1, 14.4

and Notes

ECE 3040 - Dr. Alan DoolittleGeorgia Tech


Voltage Amplifier: Voltage input and Voltage output

ifier

Inpu

t si

stan

ce

rce

Inpu

t si

stan

ce

Res

ista

nce

fier

Out

put

sist

ance

Am

pli

Res

Sour Res

Loa

d R

Am

plif

Res

Source LoadAmplifier•Any signal source has a finite “source resistance”, RS .•The amplifier is often asked to drive current into a load of finite

The controlled source is a Voltage-controlled-Voltage Source

A O Ci i V l G i b f d b l i

impedance, RL (examples: 8 ohm speaker, 50 ohm transmission line, etc…)


Av=Open Circuit Voltage Gain can be found by applying a voltage source with Rs=0, and measuring the open circuit output

voltage(no load or RL=infinity)


Why is the input and output resistance important?

•Only the voltage vin is amplified to Avvin.

•Since Rs and Rin form a voltage divider that determines vin, you want Rin as large as possible (for a voltage amplifier) for maximum voltage gain.large as possible (for a voltage amplifier) for maximum voltage gain.

•Since RL and Rout form a voltage divider that determines vout, you want Routas small as possible (for a voltage amplifier) for maximum voltage gain.



Current Amplifier: Current input and Current output

The controlled source is a Current-controlled-Current SourceThe controlled source is a Current controlled Current Source

Ai=Short Circuit Current Gain can be found by applying a current source with Rs= infinity, and measuring the short circuit output

current (No Load or R 0)current (No Load or RL=0)

•Only the current iin is amplified to Aiiin.•Since Rs and Rin form a current divider that determines iin, you want Rin as small as

ibl (f t lifi ) f i t i


possible (for a current amplifier) for maximum current gain.•Since RL and Rout form a current divider that determines iout, you want Rout as large as possible (for a current amplifier) for maximum current gain.


Transconductance Amplifier: Voltage input and Current output

The controlled source is a Voltage-controlled-Current Source

Gm=Transconductance Gain can be found by applying a voltage source with Rs=0, and measuring the short circuit output current

(No Load or RL=0)(No Load or RL 0)

•Only the voltage vin is amplified to iout=Gmvin.•Since Rs and Rin form a voltage divider that determines vin, you want Rin as large as possible for maximum transconductance gain


possible for maximum transconductance gain.•Since RL and Rout form a current divider that determines iout, you want Rout as large as possible for maximum transconductance gain.


Transresistance Amplifier: Current input and Voltage output

The controlled source is a Current-controlled-Voltage Source

Rm=Transresistance Gain can be found by applying a current source with Rs=infinity, and measuring the open circuit output

voltage (RL=infinity)voltage (RL infinity)•Only the current iin is amplified to vout=Rmiin•Since Rs and Rin form a current divider that determines iin, you want Rin as small as possible for maximum transresistance gain.


p g•Since RL and Rout form a voltage divider that determines vout, you want Rout as small as possible for maximum transresistance gain.


Input Resistance

With the load resistance attached…

l i l d h i /iApply a test input voltage and measure the input current, Rin=vt/it

Or

Apply a test input current and measure the input voltage R v /iApply a test input current and measure the input voltage, Rin= vt/it

Output Resistance

With all input voltage sources shorted and all input current sources opened…

Apply a test voltage to the output and measure the output current, Rout=vt/it

Or

Apply a test current to the output and measure the output voltage, Rout= vt/it


Final Summary of Transistor Amplifier Analysis1.) a.) Determine DC operating point. Make sure the transistors are biased into active mode (forward active for BJTs and Saturation for MOSFET. Do not (confuse the two terms as saturation means a completely different thing for a BJT)and b.) calculate small signal parameters gm, r, ro etc…2.) Convert to the AC only model.

•DC Voltage sources are replaced with shorts to ground•DC Current sources are replaced with open circuits•Large capacitors are replaced with short circuits•Large inductors are replaced with open circuitsLarge inductors are replaced with open circuits

3.) Use a Thevenin circuit where necessary on each leg of transistor

4.) Replace transistor with small signal model5.) Simplify the circuit as much as necessary and solve for gain.6.) Solve for Input Resistance: With the load resistance attached… a.) Apply a test input voltage and measure the input current, Rin=vt/it or b.) Apply a test input current and measure the input voltage R = v /imeasure the input voltage, Rin= vt/it

7.) Solve for Output Resistance: With all input voltage sources shorted and all input current sources opened… a.) Apply a test voltage to the output and measure the output current R =v /i or b ) Apply a test current to the output and measure the output


current, Rout=vt/it or b.) Apply a test current to the output and measure the output voltage, Rout= vt/it

Transistor Amplifier Configurations

Common Emitter and Common Source

Res

ista

nce

t nce

Res

ista

nce

t nce

Inpu

t R

Out

put

Res

ista

n

Inpu

t R

Out

put

Res

ista

n

O ll A lifi C fi ti

Can be modeled as a current amplifier, IC=IB, or a transconductance amplifier , iC=KvBE

Modeled as transconductance amplifier, iDS=KvGS

Overall Amplifier Configuration•Emitter/Source is neither an input nor an output•Input is between base-emitter or gate-sourceO t t i b t ll t itt d d i


•Output is between collector-emitter and drain-source•Is a transconductance amplifier (see small signal models we have used in previous examples)


Common Emitter and Common SourcePortion due to bias circuitry

Portion due to transistor

Portion due to bias circuitry

Portion due to transistor

r gmvor gmvgsR

in r Rc

Rou

t

r is replaced with an open circuit for the MOSFET casecircuit for the MOSFET case

Previously, we have analyzed voltage gain. Now let us look at the amplifier input and output resistance (these are small signal

MOSFET for the 1||2BJT for the ||1||2 RRRinorrRRRin

parameters):


MOSFET BJT for the || orRrRout co


Summary of Common Emitter and Common Source Characteristics

•Very Large Voltage Gain These propertiesVery Large Voltage Gain

•Inverting Voltage Gain (due to –gmro)

•High Input Impedance

These properties make the CE/CS configuration very good for high gain•High Input Impedance

•High Output Impedance

good for high gain stages of amplifiers.

Now let us consider the other two configurations of transistor amplifiers:transistor amplifiers:

•Common Gate/Common Base

C D i /C C ll t


•Common Drain/Common Collector

Transistor Amplifier ConfigurationsCommon Collector and Common Drain

DC Ci itDC Circuit

B

E

CG

S

D

E

Collector (or Drain) is neither an input or output

Input is Base (or Gate)

Output is Emitter (or Source)


p ( )

Transistor Amplifier ConfigurationsCommon Collector

DC Ci it t d t AC E i l t ( d d)DC Circuit converted to AC Equivalent (reduced)

DCDC Circuit

Note the extra ground due to C2

ACib CB

AC Circuit

ib ror

E


Transistor Amplifier ConfigurationsCommon Collector


AC ib

ib

rr

CB

Circuit b ror

E

ib CB

AC Circuit (reduced)

ib

b

r

C

E ibE

vo

7||||41 RrRiv

b


7||||41

7||||41RrRrRiv

RrRiv

oothbth

oboo

Transistor Amplifier ConfigurationsCommon CollectorAC V lt G iAC Voltage Gain

RRvv

RrRrRivRrRiv

h

oothbth

oboo

1||27||||41

7||||41

RRRRRA

RrRrRiRrRi

RsRRRR

vv

vvA

RsRRvv

Lo

oothb

obo

s

th

th

ov

sth

7||||4Rh11||2

7||||417||||41

1||21||2

1||2

RrRRrRRsRR

A

o

oLoth

Lov

1gby r denominato andnumerator gmultiplyin

7||||4Rwhere11||2

||

m

L

rgRRg

RgRsRR

RRA

o

mL

o

thm

Lmv

111||2

1||2

RRg

RgRsRR

RRA

oLo

thm

Lmv

11||2

1||2


VV

RgRgA

Lm

Lmv 1

1

1Rg and RsR1||R2for But Lm

Gain is positive and ~1

Transistor Amplifier ConfigurationsCommon Drain Conversion from DC to AC Equivalent Circuit

DC Ci iCircuit

AC Circuit

ro

DG

R7gmvGS

CircuitS


Transistor Amplifier ConfigurationsCommon Emitter and Common Source


AC DG

g vCircuit

ro

S

gmvGSR7

DG

AC Circuit (reduced)

DG


Transistor Amplifier ConfigurationsCommon Drain AC Voltage Gain

RrRgvRrRvgvv

RrRvgv

omGSoGSmGSth

oGSmo

7||||417||||47||||4

RrRgRRvvv

RsRRRRvv sth

7||||41||2

1||21||2

RRRgRRA

RrRgRrRg

RsRRRR

vv

vv

vv

A

Lm

om

om

s

th

th

GS

GS

ov

7||||4Rh1||2

7||||417||||4

1||21||2

RrRRg

gRsRR

A oLm

Lmv 7||||4Rwhere

11||2||

L

VV

RgRg

A Lmv 1

1

1RgandRsR1||R2for But Lm

Gain is positive and ~1


Rg Lm1

Transistor Amplifier ConfigurationsCommon Collector/Drain Input Resistance

Input Resistance: With the load resistance attached apply a test input voltage and measure the input current, Rin=vx/i (where i=ix)

x Rrv

R 1 LoBJTin Rri

R 1,

Resistance in the emitter circuit is “multiplied” by transistor to increase the input resistance


0,xx

MOSFETinv

iv

R

Transistor Amplifier ConfigurationsCommon Collector Output Resistance

ix

RL

Output Resistance: With all input voltage sources shorted and all input current sources opened, apply a test voltage to the output and measure the

Rv

Rrv

Rrv

rvi

Rrv

Rviii

L

x

th

xo

th

x

o

xo

th

x

L

xox

current sources opened, apply a test voltage to the output and measure the output current, Rout=vx/ix

4||1

1

11

, RrRwhere

RRrivR oL

Lth

x

xBJTout


1o

Two resistors in parallel: RL, and Resistance in the base circuit is “multiplied” by transistor to decrease the output resistance

Transistor Amplifier ConfigurationsCommon Drain Output Resistance

gmvGS

ix

RL

Output Resistance: With all input voltage sources shorted and all input current sources opened apply a test voltage to the output and measure the

Rvvg

rvvgi x

xmx

GSmx

current sources opened, apply a test voltage to the output and measure the output current, Rout=vx/i

4||11

, RrRwhere

Rgi

vR

Rr

oL

Lm

x

xMOSFETout

Lo


L

Two resistors in parallel: RL and inverse transconductance


Summary of Common Collector and Common Drain CharacteristicsSummary of Common Collector and Common Drain Characteristics

Th ti k th•Unity Voltage Gain

•Non-Inverting Voltage Gain

These properties make the CC/CD configuration very good for impedance transformation I ENon Inverting Voltage Gain

•Very High Input Impedance

L O t t I d

transformation, I.E. “buffering” high impedances to low impedances. CC/CD •Low Output Impedance peda ces. CC/Cconfigurations are good for output stages of amplifiers due to their very low output impedance, I.E., very little voltage drop in the output


resistance of the amp.

Transistor Amplifier ConfigurationsCommon Collector/Drain Current Gain

Rth

ith

th

vth (o+1)ithi

oBJTi iiA , 1

th

MOSFETi

th

iiA

i

,


th

Note: since R7 was originally defined as the load, the current gain should actually be (+1) (R4||ro)/(R4||ro+R7) using a current divider.

Transistor Amplifier ConfigurationsCommon Base and Common Gate

DC Ci itDC Circuit

Base (or Gate) is neither an input or output

Input is Emitter (or Source)

Output is Collector (or Drain)


p ( )

Transistor Amplifier ConfigurationsCommon Base


DC Circuit

Note: Jaeger let’s ro go to infinity which

AC Circuit

g v ror

CBmakes the math dramatically easier

vo

gmv or

E


Transistor Amplifier ConfigurationsCommon Base AC Equivalent (reduced)

AC vo

gmv ror

CBv

Circuit o

E

v

rogmvror gmv

v

vo

AC Circuit ( d d) r

v

VRvV 866704

(reduced)

ECE 3040 - Dr. Alan DoolittleGeorgia TechKRRR

VRR

vV

sth

ss

sth

73.14||

8667.04

Transistor Amplifier ConfigurationsCommon Base Voltage Gain

iC

rogmv

vo

C

i

i

r

v

iB

iE

riRivrv

rvv

vgi

iivgiii

BthEth

o

omE

BmBCE

KRRR

VRR

RvV

sth

ss

sth

73.14||

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4

also

rrv

Rrv

rvv

vgv tho

omth

BthEth

1,

R

rR

rg

vvsoRr

vvvgRiv L

o

L

om

oLo

omLCo

1

1

1

,

vRrv

r

R

rR

rg

vv

vgv th

L

o

L

om

mth

1

1


rro


1

1

1

RR

rg

RR

rg

vv Lo

m

Lo

m

,1111

ThusRrr

rR

gvvvRrv

rrR

vgv tho

L

o

L

mththo

L

o

L

mth

44

1

1

1

1

RRRR

Rr

g

vv

vv

vv

As

LL

om

s

th

th

ov

111

1

1

Rrr

R

rR

rg

g

r

tho

L

o

L

om

m

o

4

RgvvvRRandrfor so


1

rR

Rg

Rgvv

vv

vv

Ath

thm

Lm

s

th

th

ov


ththm

Lmv

rR

Rg

RgA

1

mth

Lmv

rgR

RgA

11

m

m

m

mmth

Lmv

gg

rrgrg

gR

RgA

11

mm

mth

Lmv

rgrg

gR

RgA

11

o

o

mth

Lmv eL

gR

RgA

arg110


mth

Lmv gR

RgA

1Gain is positive and can be large

Transistor Amplifier ConfigurationsCommon Base Input Resistance

rogmv

iC

i

r

vvx

vo

iB

-ix=iE

vx v vxx

Input Resistance: With the load resistance attached apply a test input l d h i R /ivoltage and measure the input current, Rin=vx/i

om

o Rr

gd

vvvi

beforeFrom

1,

xEx

L

o

L

oo

o

omE

vvandii

R

rR

vvandrr

vgi

1

1,


L

o

L

om

xox

o

oxxmx R

rR

rg

vvandrv

rvv

vgi

1

1

,

gm

1

Transistor Amplifier ConfigurationsCommon Base Input Resistance

r

vr

R

rR

rvv

vgi x

o

L

o

L

oxx

xmx

1

rLetting

rgm

1m

x

xBJTin

o

giv

R

rLetting

11

,

rr

R

rR

r

gvio

L

o

L

o

mxx1

11

mBJTin rg

rR

r

1,

iv

R xBJTin

11

, o

o

o

o

r

11

rr

R

rR

rg

g

i

L

o

L

om

m

x

11

1

1 mm

o

gg

Input Resistance is very small!


rro Input Resistance is very small!

Transistor Amplifier ConfigurationsCommon Base Output Resistance

v

Replace RL by a voltage source, vx

ror gmv

v

iCi

oib rorv

iB

iix

vx

vr

RthiE

ve

Result follows exactly after discussion in Jaeger pages 668-670 and 683-684


Result follows exactly after discussion in Jaeger, pages 668-670, and 683-684.

Transistor Amplifier ConfigurationsCommon Base Output Resistance

eobox

erx

vriivvv

i rr

iCiB

v

th

thxthxe Rr

RriRriv

but

||

, oib

Rth

rorv

iE

ixvx

ve

vr

th

thxb

RrR

thusRr

Rii

,

ththx

th

thxo

th

thxooxx

RrrRrvR

RrRrir

RrRiriv

tho

thoo

xBJTout Rr

rRr

ri

R

,

rge

rger

ECE 3040 - Dr. Alan DoolittleGeorgia Tech Very

Lar

Eve

n L

ar Output Resistance is HUGE!

Transistor Amplifier ConfigurationsCommon Base Current Gain

1 ol

i ii

A


thi

Transistor Amplifier ConfigurationsCommon Gate Solution

The Common Gate solution can be found by recognizing that the following translations can be made in our small signal model:signal model:

r

oo 1

11

,,

thm

LmMOSFETv

ththm

LmBJTv Rg

RgA

rR

Rg

RgA

mMOSFETin

m

BJTin gR

rg

R 11

1,,

ththmoMOSFEToutth

tho

th

thmo

th

tho

th

thooBJTout RRgrR

RrRr

rRr

Rrgr

RrRr

rRr

RrR

1,,


11 ,, oBJTioBJTi AA


Summary of Common Base and Common Gate CharacteristicsSummary of Common Base and Common Gate CharacteristicsThe input and output impedances are the opposite of what is

•High Voltage Gain

•Non-Inverting Voltage Gain

opposite of what is typically needed for a voltage amplifier. Thus, Common Emitter/Source Non Inverting Voltage Gain

•Very Low Input Impedance

V Hi h O t t I d

amplifiers are normally used instead of Common Base/Gate. The input and

•Very High Output Impedance output impedances are useful for current amplifiers but the current

i i t b t it Thgain is at best unity. Thus a current buffer is one useful application for the Common Base/Gate


Common Base/Gate

Multistage Amplifier Configurations

You can combine or Cascade configurations to produce “High Performance” amplifiersYou can combine or Cascade configurations to produce High Performance amplifiers with High input impedance, low output impedance and huge voltage gains.

21

2

,

1,

vv

vv

vv

vv

vv

A o

inputths

inputth

s

ov

v1v2

vvo

des H

igh

peda

nce,

ly

hig

h ga

in

des H

igh

peda

nce,

a

nd

he

gain

vi

ous

des L

ow

mpe

danc

e,

ut

s pos

itive

stag

e


CS

prov

idIn

put I

mp

Mod

erat

elne

gativ

e g

CE

pro

vid

Inpu

t Im

phi

gh g

ain,

corr

ects

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vst

age

CC

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ain,

bm

aint

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gain

from

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For AC-Coupled amplifiers (capacitors between stages) the DC solution reduces to threeFor AC-Coupled amplifiers (capacitors between stages), the DC solution reduces to three parallel and independent circuits!



For AC-Coupled amplifiers (capacitors between stages) the AC solution reduces to threeFor AC-Coupled amplifiers (capacitors between stages), the AC solution reduces to three circuits, each of which has a load dependent on the input resistance of the next stage! Continued….



Continued (For AC-Coupled amplifiers (capacitors between stages) the AC solutionContinued….(For AC-Coupled amplifiers (capacitors between stages), the AC solution reduces to three circuits, each of which has a load dependent on the input resistance of the next stage!)


Multistage Amplifier ConfigurationsContinued...

)||||(

)||||(1

412222

3

211111

21

inQom

omthGS

G

s

th

RRrgvv

rRrgvv

vv

RRR

vv

)||(1

)1||||(

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o

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Rrgvv

vvv

RrRrgvvv

We just found this!

Note: an exact solution would have RL3||ro3

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)||(1

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3333

3

3333

4

Lomoo

Lomo

R

Rrgr

vvv

Rrgr

vv

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3333

3333

3Lom

Lomo

Rrg

Rrgr

Rrgr

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)||(11

)||()1||||()||||()1(

3333

3333

3312222111132

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21

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Lom

Lomomo

ths

th

s

ov

Rrgr

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RrRrgrRrgvv

vv

vv

vv

vv

vvA


AC-Coupled amplifiers (capacitors between stages) have one major limitation They doAC-Coupled amplifiers (capacitors between stages), have one major limitation. They do not amplify low frequencies or DC voltages. To accomplish this, we must DC-Couplethe stages as shown. Note: for this to be a DC coupled amp, C1 and C6 should also be replaced as shorts.

Since the bias here is usually (2/3)V (V =15V in this example) it is easier to use a


Since the bias here is usually ~(2/3)Vcc (Vcc =15V in this example) , it is easier to use a PNP for the second stage so that VEB+IERE2 ~ (2/3)Vcc

Lecture 27Lecture 27 Amplifier Configurations Reading: CE...

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Transcript of Lecture 27Lecture 27 Amplifier Configurations Reading: CE...