Transistor Amplifiers - University of California, San …aries.ucsd.edu/NAJMABADI/CLASS/ECE65/12-W/Slides/ECE65_W...A Transistor can be configured to act as a resistor for small signals!

Transistor Amplifiers

Lecture notes: Sec. 6

Sedra & Smith (6th Ed): Sec. 5.6, 5.8, 6.6 & 6.8 Sedra & Smith (5th Ed): Sec. 4.6, 4.8, 5.6 & 5.8

F. Najmabadi, ECE65, Winter 2012

How to add signal to the bias


Bias & Signal vGS = VGS + vgs

Bias & Signal vDS = VDS + vds

1. Direct Coupling Use bias with 2 voltage supplies

o For the first stage, bias such that VGS = 0

o For follow-up stages, match bias voltages between stages

Difficult biasing problem Used in ICs Amplifies “DC” signals!

2. Capacitive Coupling Use a capacitor to separate bias

voltage from the signal. Simplified biasing problem. Used in discrete circuits Only amplifies “AC” signals

Capacitive coupling is based on the fact that capacitors appear as open circuit in bias


At a high enough frequency, Zc = 1/ (ωC), becomes small (effectively, capacitors become short circuit). o Mid-band parameters of an Amplifier.*

At low frequencies, Zc cannot be ignored. As Zc depends on frequency, amplifier is NOT linear (for an arbitrary signal) for these low frequencies. o Capacitors introduce a lower cut-off frequency for an amplifier (i.e., amplifier

should be operated above this frequency).

In ECE102, you will see that transistor amplifiers also have an “upper” cut-off frequency

Real Circuit Bias Circuit Signal Circuit

What are amplifier parameters?


:Gain Voltage i

ov v

vA =

∞→

=LRi

ovo v

vA :Gain loop-Open

:ResistanceInput i

ii i

vR =

0

: Amplifier of ResistanceOutput →

=ivo

oo i

vR

Output resistance is the Thevenin resistance between the output terminals!

0

:circuit theof ResistanceOutput →

=sigvo

oout i

vR

Observations on the amplifier parameters


Avo is the maximum possible gain of the amplifier.

Value of Ro is important. o For Ro << RL , Av ≈ Avo

o For Ro = RL , Av = 0.5 Avo o For Ro >> RL , Av ≈ 0

Prefer “small” Ro

vooL

L

i

ov A

RRR

vvA

+==

Value of Ri is important. o For Rsig << Ri , vi ≈ vsig

o For Rsig = Ri , vi = 0.5 vsig o For Rsig >> Ri , vi ≈ 0

Prefer “large” Ri

sigi

i

sig

i

RRR

vv

+=

vsigi

i

i

o

sig

i

sig

o ARR

Rvv

vv

vvA

+=×==

:Gain Overall

How to Solve Amplifier Circuits 1. Find Bias and Signal Circuits.

2. Bias circuit (signal = 0): o Capacitors are open circuit. o Use transistor large-signal model to find the bias point. o Use bias parameters to find small-signal parameters (rπ , gm , ro ).

3. Signal Circuit (IVS becomes short, ICS becomes open circuit): o Assume capacitors are short to find mid-band amplifier parameters. o Replace diodes and/or transistors with their small-signal model. o Solve for mid-band amplifier parameters (Av , Ri , Ro ).

• For most circuits, we can use fundamental amplifier configurations, elementary R forms instead of solving signal circuits.

o Include impedance of capacitors to find the lower cut-off frequency of the amplifier.



Example 1: Draw the small-signal equivalent of the circuit below (assume capacitors are short for small signal).

IVS → 0 R remains Caps short

Ground at the bottom

Replace MOS with its small signal model



Flip PMOS

IVS → 0 Caps short Ground at the bottom

(100k || 33k = 24.8 k)




ICS → 0 (This makes ICS an open circuit)

IVS → 0 Caps short



Basic MOS Amplifier Configurations

We are considering only signal circuit here!

Possible MOS amplifier configurations


Same as Common Gate (vi does not change)

Common-Source Common-Gate Common-Drain

Common-Source with Rs Not Useful

PMOS configurations are the same as those of NMOS


Common-Source Common-Gate Common-Drain

Since PMOS has the same signal model, configurations and results are exactly the same

Common Source Configuration (Gain)


Signal Circuit:

Signal Circuit with MOS SSM:

omvo

Lomi

ov

Logsmo

rgA

RrgvvA

Rrvgv

−=

′−==

′−=

)||(

)||(

Relevant circuit for Gain calculation

By KCL

Common Source Configuration (Ri)


0

∞==

=

i

ii

i

ivR

i


Relevant circuit for Ri calculation

Common Source Configuration (Ro)


oo rR = Current source becomes open circuit


Relevant circuit for Ro calculation (set vi = 0)

Common Source with Source Resistor


0 ∞==⇒=i

iii i

vRiInput Resistance

Small Signal Circuit:


Common Source with Source Resistor (Gain)


omvo

oLSm

Lmv

LSomo

Lom

i

ov

rgArRRg

RgA

RRrgrRrg

vvA

−=

′++′

−≈

′+++′

−==

/1

)1(

0)(

0)(

=−+−

+′

=−−−

+

−=

Simo

So

L

o

Simo

oS

S

S

Sigs

vvgr

vvRv

vvgr

vvRv

vvv

Node voltage method:

Node vS

Node vo

Relevant circuit for Gain calculation

Common Source with Source Resistor (Ro)


)1( Somoo RrgrR ++=

Somo

x

S

S

Smo

xS

S

S

Sgs

Rrgrv

Rv

vgr

vvRv

vv

)1(

0)(

++=

=−−−

+

−=


Node vS

set vi = 0 Attach vx and compute ix

Ro = vx /ix

Somo

x

S

Sx Rrgr

vRvi

)1( ++==

Somox

x

o Rrgrvi

R )1(11++

=≡

Common Gate Configuration


omvo

Lomv

Loo

om

i

ov

rgARrgA

Rrr

rgvvA

≈

′≈

′+==

)||(

)||(1

io

om

Lo

o

imo

io

L

o

igs

vr

rgRr

v

vgr

vvRv

vv

+=

′

=−+−

+′

−=

1||

0)(


Node vo

Gain

Common Gate Configuration (Ri and Ro)


om

L

mi

om

Lo

i

ii

rgR

gR

rgRr

ivR

′+≈

+′+

==

1

1

)()1(

)(

Loiomi

Liogsmii

RrirgvRirvgiv′+=+

′++=KVL:

oo rR =

Current source becomes open circuit

Output Resistance (set vi = 0)

Input Resistance

Common Drain Configuration (Source Follower)


11

)||(1

)||(

≈+

=

′+′

=

om

omvo

Lom

Lomv

rgrgA

RrgRrgA

omLo

oim

oimo

o

L

o

oigs

vgRr

vvg

vvgrv

Rv

vvv

+′

=

=−−+′

−=

||

0)(


Node vo

Gain

Common Drain Configuration (Source Follower)


m

x

o

xgsm

o

xx g

vrvvg

rvi

/1+=−=

Input Resistance

Output Resistance (set vi = 0)

1||1 m

om

o gr

gR ≈=

0=ii

∞==i

ii i

vR

MOS Basic Amplifier Configurations (PMOS circuits are identical)


Common Source with RS oLSm

Lmv rRRg

RgA/1 ′++

′−=

Common Drain/Source Follower

)||(1)||(

Lom

Lomv Rrg

RrgA′+

′=

Common Source )||( Lomv RrgA ′−=

Common Gate )||( Lomv RrgA ′=

MOS Elementary R forms

A Transistor can be configured to act as a resistor for small signals!


oo rR =Set vi = 0, current source becomes open circuit

Ex: Output resistance of a CS Amplifier

or

If we connect any two terminals of a MOS, we get a two-terminal device. o For Small Signals, this two terminal device can be replaced with its

Thevenin equivalent circuit. o As there is NO independent sources present, the Thevenin

equivalent circuit is reduced to a resistor.

Notation: ro is the small-signal resistance between the point and ground

Transistor can be configured to act as a resistor for small signals!


But, MOS should be in saturation for small signal model to work! o Connection between MOS terminals are, therefore, made through

ground for small signals. o In fact, one or both MOS terminals have to be connected to bias power

supplies to ensure that MOS is in saturation

Small Signal Circuit Real Circuit

A)

No Small Signal circuit MOS is NOT in saturation

B)

MOS Elementary R forms (PMOS circuits are identical)


∞

Input resistance of CS Amp

Output resistance of CS Amp with Rs

)1()1(Rgr

RRgr

mo

mo

+≈++

Input resistance of CG Amp

om

o

rgRr

++

1

Diode-connected Transistor Always in saturation! m

om g

rg

1||1≈

Above configurations are for Small Signal. Typically one or both “signal” grounds are actually connected to bias voltage sources to ensure that MOS is in saturation!

or


Basic BJT Amplifier Configurations

We are considering only signal circuit here!

Possible BJT amplifier configurations


Same as Common Base (vi does not change)

Common-Emitter Common-Base Common-Collector

Common-Emitter with RE Not Useful

Common Emitter Configuration (Gain)


Signal Circuit:


omvo

Lomi

ov

Lomo

rgA

RrgvvA

Rrvgv

−=

′−==

′−=

)||(

)||( π

By KCL

Common Emitter Configuration (Ri)


π

π

rivR

rvi

i

ii

ii

==

=


Relevant circuit for Ri calculation

Common Emitter Configuration (Ro)


oo rR = Current source becomes open circuit


Relevant circuit for Ro calculation (set vi = 0)

BJT Basic Amplifier Configurations (PNP circuits are identical)


Common Emitter with RE

Common Collector/ Emitter Follower

)||(1)||(

Lom

Lomv Rrg

RrgA′+

′=

Common Emitter )||( Lomv RrgA ′−=

Common Base )||( Lomv RrgA ′=

)/1)(/(1 πrRrRRgRgA

EoLEm

Lmv +′++

′−=

BJT Elementary R forms (PNP circuits are identical)


om

o

rgRr

++

≈1

or

)1( Rgr mo +≈

βπ Rr +

≈Rr )1( βπ ++


Discrete Amplifier Configurations

We focus on biasing with Emitter/Source degeneration!

Emitter-degeneration bias circuits


Bias with one power supply

(voltage divider)

EEBEBBBB RIVRIV ++=

Bias with two power supplies

EEBEBBEE RIVRIV ++=

Emitter-degeneration bias circuits have similar signal circuits


Bias with one power supply (voltage divider)

Bias with two power supplies

The same circuit for

21 || BBB RRR =

Signal Circuits

By-pass capacitors


Basic CE Configuration There is no RE in the basic Common-Emitter configuration.

However, RE is necessary for bias in discrete circuits. Use a by-pass capacitor

Real Circuit Bias Circuit: Cap is open, RE stabilizes bias

Signal Circuit: Capacitor shorts RE

Discrete Common-Emitter Amplifier


Standard Bias Circuit:* Caps are open circuit

Real Circuit

CE amplifier: Input at the base Output at the collector

* Bias calculations are NOT done here as we have done them before.

Signal circuit of the discrete CE Amplifier


Real Circuit

Short caps Zero bias supplies

Rearrange

Discrete CE Amplifier (Gain)


i

o

sigi

i

sig

o

LComi

o

vv

RRR

vv

RRrgvv

×+

=

−=

)||||(

Basic CE configuration

Signal input at the base Signal output at the collector No RE LCL RRR ||=′

Discrete CE Amplifier (Ri)


Replace transistor with its equivalent resistance

Elementary R form

|| πrRR Bi =

πrR =

πrR =

Discrete CE Amplifier (Ro)


Set vsig = 0 Replace transistor with its equivalent resistance

Elementary R form

|| oCo rRR =

orR =

Discrete CE and CS Amplifiers


||

)||||(

oDo

Gi

LDomi

o

rRRRR

RRrgvv

==

−=

i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

|| ||

)||||(

oCo

Bi

LComi

o

rRRrRR

RRrgvv

==

−=

π

Discrete CS Amplifier with RS


Real Circuit

Signal Circuit Short caps Zero bias supplies

CS amplifier with RS Input at the gate Output at the drain

Bias Circuit Caps open

Discrete CS Amplifier with RS (Gain)


i

o

sigi

i

sig

o

oLDSm

LDm

i

o

vv

RRR

vv

rRRRgRRg

vv

×+

=

++−=

/)||( 1

)||(

Basic CS configuration with RS

LDL RRR ||=′

Signal input at the gate Signal output at the drain RS !

Discrete CS Amplifier with RS (Ri)


Gi RR =


Elementary R form

∞= R

Discrete CS Amplifier with RS (Ro)


[ ] )1( || SSmoDo RRgrRR ++=

Set vsig = 0 Replace transistor with its equivalent resistance Since ig = 0, Rsig and RG can be removed (vg = 0)

Elementary R Configuration

SSmo RRgrR ++= )1(

SR

SSmo RRgrR ++= )1(

Discrete CE and CS Amplifiers with RE / RS


[ ] )1( ||

/)||( 1

)||(

SSmoDo

Gi

oLDSm

LDm

i

o

RRgrRRRR

rRRRgRRg

vv

++==

++−=

i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

[ ]

+++≈

++≈+

−≈

+++−=

sigBE

EoCo

EBi

Em

LDm

i

o

EoLDEm

LDm

i

o

RRRrRrRR

RrRRRgRRg

vv

rRrRRRgRRg

vv

||1 ||

)1( || 1

)||(

)/1](/)||[(1)||(

π

π

π

β

β

Discrete CB Amplifier


Real Circuit




Capacitor CB is necessary. Otherwise, Amp gain drops substantially.

Discrete CB Amplifier (Gain)

Basic CB form

LCL RRR ||=′

Signal input at the source Signal output at the drain

i

o

sigi

i

sig

o

LComi

o

vv

RRR

vv

RRrgvv

×+

=

+=

)||||(

om

LCo

rgRRrR

++

=1

)||(

Discrete CB Amplifier (Ri)

1

)||( ||

+

+=

om

LCoEi rg

RRrRR



LC RR ||

Discrete CB Amplifier (Ro)

{ } )]||(1[ || sigEmoCo RRgrRR +=

Set vsig = 0 Replace transistor with its equivalent resistance


)]||(1[ sigEmo RRgrR +≈

Discrete CB and CG Amplifiers

F. Najmabadi, ECE65, Winter 2012 i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

{ } )]||(1[ || 1

)||( ||

)||||(

sigEmoCo

om

LCoEi

LComi

o

RRgrRRrgRRrRR

RRrgvv

+=

+

+=

+=

{ } )]||(1[ || 1

)||( ||

)||||(

sigSmoDo

om

LDoSi

LDomi

o

RRgrRRrgRRrRR

RRrgvv

+=

+

+=

+=

Discrete CD Amplifier (Source Follower)


Real Circuit




Discrete CD Amplifier (Gain)


Basic CD form

Signal input at the gate Signal output at the source

i

o

sigi

i

sig

o

LSom

LSom

i

o

vv

RRR

vv

RRrgRRrg

vv

×+

=

+=

)||||(1

)||||(

LSL RRR ||=′

Discrete CD Amplifier (Ri)




∞

Gi RR =

Discrete CD Amplifier (Ro)


1 || m

So gRR =

R


Set vsig = 0 Replace transistor with its equivalent resistance Since ig = 0, Rsig and RG can be removed (vg = 0)

Rm

om g

rg

1||1≈

Discrete CC and CD Amplifiers


1 ||

)||||(1

)||||(

mSo

Gi

LSom

LSom

i

o

gRR

RRRRrg

RRrgvv

=

=+

=

i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

[ ]

β

β

π

π

sigBo

LEoBi

LEom

LEom

i

o

RRrR

RRrrRRRRrg

RRrgvv

||

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

)||||(

+≈

++=+

=

Impact of Coupling and Bypass Capacitors (1)


11

1

1

)(1

/11

)/(1

csigip

psigi

i

sig

i

csigi

i

sig

i

CRR

jRRR

vv

CjRRR

vv

+=

−×

+=

++=

ω

ωω

ω

High Pass filter with pole at ωp1

ωω /11

1pi

o

sigi

i

sig

o

jvv

RRR

vv

−××

+=

Impact of Coupling and Bypass Capacitors (2)


Each capacitor introduces a pole!

Poles can be found by inspection: 1) Set vsig = 0 2) Consider each capacitor separately

(i.e., assume all others are short). 3) Find R, the total resistance seen

between capacitor terminals 4) Pole is given by

1 21

cp CR

fπ

=

The lower cut-off frequency of amplifiers can be found from ... 21 ++≈ ppp fff

Documents

Transistor Amplifiers - University of California, San …aries.ucsd.edu/NAJMABADI/CLASS/ECE65/12-W/Slides/ECE65_W...A Transistor can be configured to act as a resistor for small signals!