single transistor amp

8/3/2019 Single Transistor Amp

1/31

SJSU EE223 by Koorosh Aflatooni 1

Single Transistor and MultipleTransistor Amplifiers


2/31


Overview Introduction

Modeling

Single Transistor amplifiers

Common emitter/source

Common base/gate

Common collector/drain

Common emitter/source with degeneration

Multiple transistor amplifiers

Common collector-common emitter

Common collector-common collector

Cascode

Simple cascode

Active cascode

Differential pairs

DC transfer of common emitter/source pairs

DC transfer of common emitter/source pairs with degeneration

Small signal characteristics

Device mismatch


3/31


Modeling

Two port

network

i1

v1

i2

v2 Two port modeling

Express the relation

between input and output

Superposition of the each

source contribution2221212

2121111

vyvyi

vyvyi

+=

+=

y11

y12v2

y21v1

y22


4/31


Modeling (cont.)

Feedback

Bilateral

Unilateral => y12=0 Other terms

Short-circuit transonductance => Gm=y21 Input impedance => Zi=1/y11

Output impedance => Zo=1/y22 Knowing any two parameters leads to

the third parameter

The key is to understand the effect ofloading on performance

Zi Gmv1

Zo

i1

v1

i2

v2

+

_Zi avv1

Zo

i1

v1

i2

v2omiv ZGvva == =0

1

2

2

Norton

to

Thevenin


5/31


Common Emitter Large signal

Collector current is relatedto base current

Output voltage is defined byconsidering load line

Small Signal

Input resistance

Transconductance

Output resistance

Open circuit voltage gain

==

T

i

F

S

B

CB

V

VI

I

II exp

=

T

iSCCCo

VVIRVV exp

m

oi

grR

==

mm gG =

oCo rRR ||=

( )Comv Rrga ||=


6/31


Common Source Large signal

Output voltage related toinput

Transition from cutoff =>active => triode

Small signal

Input resistance

Transconductance

Output resistance


The maximum voltage gainfor CS is proportional to1/ID, in contrast to BJT thatis independent of current

( )22

tiDox

DDo VVRL

WCVV =

iR

mm gG =

oDo rRR ||=

( )Domv Rrga ||=


7/31


Common Base Small signal

Modifying -model to T model todecouple the dependent currentsource between input-output ports

Input resistance

Transconductance

Output resistance


Compare to common emitter, the inputresistance is reduced by (1+b) as wellthe the current gain

ei rR =

r

r

gG

b

mm

+=

1

Co RR =

Cmv Rga =


8/31


Common Gate Large signal

Not much interesting

Small signal Input resistance

Transconductance

Output resistance


mbm

igg

R+

=1

mbmm ggG +=

Do RR =

( )Dmbmv

Rgga +=


9/31


Common Gate (cont.)

Considering a case with ro =>

bilateral because of feedbackprovided by output => inputresistance depends on outputload

Small signal Input resistance

Transconductance Ro has no effect since this is

measured with output shorted

Output resistance

ombm

LDoi

rgg

RRrR

)(1

||

+++

=

( )( )Sombmo RrggRR += ||

mbmm ggG +=


10/31


Common Collector

Emitter follower

Ideally base-emitter voltage remains

constant, independent of collector voltage In reality it is not quite constant

It is not unilateral

Small signal ( model)

Input resistance

Voltage gain

Output resistance

( )( )oLoi rRrR ||1++=

o

o

So rRrR ||

1++=

( )( )oLoS

v

rR

rRa

||11

1

++

+=


11/31


Example

Calculate the input resistance, output resistance,

and voltage gain of the emitter follower. AssumeRS=RL=1k, =100, rb=0, ro =>, Io=100A.


12/31


Common Drain

Source follower

Ideally, source follows gatevoltage

In reality, it deviates due tobody effect and channelmodulation effect

Small signal

Input resistance

Voltage gain

Depends on body effect

Output resistance

=iR

( )L

oombm

omv

R

rrgg

rga

+++=

1

Lo

mbm

o

Rrgg

R11

1

+++

=


13/31


Common Emitter with Emitter

Degeneration Adding the resistance to

emitter: Reduces the

transconductance, andincreases output/inputresistances

Small signal

Input resistance

Transconductance

Output resistance

( )

+++

+++=

ECo

o

Co

EoiRRr

Rr

RrR1

1

+++

=

omo

Em

oo

E

mm

rgRg

rR

gG11

11

1

( ) ( )[ ]EmoEo RrgrRrR ||1|| ++=


14/31


Common Source with Source

Degeneration

Small signal

Input resistance

Transconductance

Output resistance

=iR

( ) Smbmm

mRgg

gG

++=

1

( )[ ]SmbmoSo RggrRR +++= 1


15/31


Single Transistor Summary

Moderate to

high

LowAi ~ 1Av > 1Common-Base

LowHighAi > 1Av > 1Emitter-Follower

Moderate to

high

ModerateAi > 1Av > 1Common-Emitter

Moderate to

high

LowAi ~ 1Av > 1Common-Gate

Low-Av ~ 1Source-Follower

Moderate to

high

-Av > 1Common Source

Output

resistance

Input

resistance

Current gainVoltage gainConfiguration


16/31


Multiple Transistor Amplifiers In many applications performance of a single stage amplifier is not

sufficient to meet various requirements

Need to combine multiple transistors to achieve voltage, current,input/output impedance adjustments

In general, the overall voltage/current gain is not simply the product ofall the stages, but it is a function of loading (input/output resistance) ofeach stage => need specific analysis

Some popular combinations Common collector- common emitter & Common collector- common

collector

Cascode

Super source follower

Differential pair

Stage 1Av1

Stage 2Av2

Stage nAvn

Ri1Ro1 RinRi2

RonRo2


17/31


Common Collector- Common

Emitter Goal:

To achieve higher input resistance and gain

Operation principle: Ibias provides the DC biasing

Q2 appears as load on emitter of Q1 => inputresistance increases; also gives two stages ofcurrent gain

Consider a combined transistor Small circuit analysis

Input resistance

Transconductance

Current gain

Output resistance

( )( )ooi rrrR ||1 21 ++=

++

=

2

1

2

)1(1

1

r

rgG

o

mm

( )1+= ooc

2oo rR =


18/31


Darlington Configuration

Similar to:

cc-cc: as discussed cc-ce: but in Darlington

collector of Q1 givesfeedback path=>

reduction of outputresistance & increase ofinput capacitance

BiCMOS version finds

many applications High input resistance

Large transconductance


19/31


Cascode Configuration

Bipolar version

Common emitter- common base

Small circuit analysis Input resistance

Transconductance

Voltage gain

Output resistance

1rRi =

1mm gG =

ovA =

+

+=

o

om

omoo rg

rgrR

12

122

1

1


20/31


Cascode Configuration (cont.)

MOSFET version:

Common source- common gate

Output resistance can be tuned

=> limited by power supply

voltage and signal swing

Small circuit analysis Input resistance

Transconductance

Output resistance

iR

1mm gG

( ) 2122 oombmo rrggR +=


21/31


Cascode Configuration (cont.)

Active cascode

Using an amplifier to toprovide negative feedback and

increases the output

resistance

Only works at frequencies

amplifier has gain


22/31


Super Source Follower

Goal:

Reduce output resistance ofsource follower => useful if you

need to drive a resistive load

Small signal Output resistance

( ) 1211

11

ommbm

orggg

R+

=


23/31


Differential Pair

Goal: to eliminate the common sources (e.g.,

noise sources) and amplify differential inputsignal

Analysis

Large signalBipolar: Linear region 26mV around zero

MOSFET:

Small signal

ff


24/31


BJT Differential Pair

Large Signal Analysis Assuming

Rtrail very large and ro can be ignored

Steps:

Write KCL for input signals to emitter of transistors

Relate Ic1 and Ic2 to Itrail

Relate output voltages to input voltages Highlights

Useful range ~ Linear range ~


25/31


MOSFET Differential Pair

Large Signal Analysis Assuming

Rtrail very large and ro can be ignored

Steps: Write KCL for input signals to emitter of

transistors

Relate Id1 and Id2 to Itrail

Relate output voltages to input voltages Highlights

Useful range ~ Linear range

Voltage gain

How to increase the useful range? W/L

Over-drive

1

2

L

Wk

ITrail

'

2

DTrailRIk'


26/31


Example

Compare the forward transconductance of a

MOSFET differential gain against a bipolardifferential gain? (assume ITrail=500A andk=100A/V2, W/L=1,F=1)

21

)/('4

4'

2id

Trailid

Traild V

LWkIV

LWkII +=

L

WIk

V

Ig TrailV

id

dm id 4

'|(max)

0

1 =

==

+

=

T

i

TrailF

c

V

v

I

I1

1

exp1

T

TrailFV

i

cm

V

I

V

Ig

i 2|(max)

01

1

1 =

==

VAgm /35(max) = VAgm/9766(max) =

MOSFETBJT

Diff ti l P i


27/31


Differential Pair

Small Signal Analysis

Vid/2

-Vid/2

-Vic

Breaking analysis to:

Differential mode

Common mode

Ideally we like Adm-cm=0

and Acm-dm=0, in realitythey are not

Common mode rejection

ratios (CMRR)

Other important ratios

iccmidcmdmoc

icdmcmiddmod

vAvAv

vAvAv

+=+=

cm

dm

A

ACMRR =

dmcm

dm

A

A

cmdm

dm

A

A

Diff ti l P i


28/31


Differential Pair

Small Signal Analysis (cont.)21

In a balanced differential

pair, increase if current inpath 1, means current inpath 2 decreases by sameamount

Voltage across Rtrail stay

constantDropping Rtrail makes no

difference

Voltage gain

The gmb has no effect sincesource to ground stays at aconstant potential

( )omdm rRgA =

Diff ti l P i


29/31


Differential Pair

Small Signal Analysis (cont.) Due to symmetry, we could

assume no current flowsbetween two sections Breaking the circuit into two

sections

Each of these sections present

a degenerate source followerconfiguration

Common mode gain

In case ro>0

Note: increase of Rtrail leads to

increase of CMRR

1 2

( )Smbm

mm

Rgg

gG

++=

1

Degenerate Source

follower

( ) TrailmbmDm

DmcmRgg

RgRGA

21 ++

==

( )( )[ ]{ }TrailmbmoTrailD

Trailmbm

mDmcm RggrRR

Rgg

gRGA 212

21+++

++

==

( ) Trailmbm RggCMRR ++ 21


30/31


Example

Find the differential-mode gain, common-mode gain,

and differential-mode input resistance for a bipolar

differential pair? (assume ITrail=20A, RTrail=10M,RC=100k, VEE=VCC=5V, =150, and neglect rb, ro,and r.

7810020

=

== K

V

ARgA

T

Cmdm

005.01

11

=

++

= C

o

Trailm

mcm R

Rg

gA


31/31


Summary

Review of various single and two state

amplifiers, including differential pairs

End of chapter problems: 3-2, 3-4, 3-7, 3-9,3-14, 3-16, 3-24

single transistor amp

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