9. Discrete Transistor Amplifiersaries.ucsd.edu/.../ECE65_W13-9-Discrete_Amps-A.pdf · Use a capacitor to separate bias voltage from the signal. Simplified biasing problem. Used in

9. Discrete 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

ECE 65, Winter2013, F. Najmabadi

How to add signal to the bias

F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (2/42)

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 (DC)


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. (We do NOT want to operate the amplifier in these 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

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 almost all circuits, we can use fundamental amplifier configurations,

instead of solving signal circuits.

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

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

We will solve this configuration

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)


)||||( LComi

o RRrgvv

−=

Fundamental CE configuration

Signal input at the base Signal output at the collector No RE

)||( Lomi

o Rrgvv ′−=

LCL RRR ||=′

Discrete CE Amplifier (Ri)


|| πrRR Bi =

| πrRR CEi ==Fundamental CE configuration

πrR =

Discrete CE Amplifier (Ro)


1) Set vsig = 0

|| oCo rRR =

0 =πv

Controlled current source becomes open circuit because gm vπ = 0

2) Replace transistor with its SSM

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

∞→ πr

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)


/)||( 1

)||( oLDSm

LDm

i

o

rRRRgRRg

vv

++−=

Fundamental CS configuration with RS

LDL RRR ||=′

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

/ 1

oLSm

Lm

i

o

rRRgRg

vv

′++′

−=

Discrete CS Amplifier with RS (Ri)


Gi RR =

| / ∞== RSCSiRR ∞=R

Fundamental CS configuration with RS

Discrete CS Amplifier with RS (Ro)


1) Set vsig = 0

Since ig = 0, vg = 0 and gate is grounded


Discrete CS Amplifier with RS (Ro)


[ ] )1( || SmoDo RgrRR +=

Attach vx , compute ix (Ro = vx /ix )

Somo

xy

Somoyx

SySyomoyx

Syogsmyx

Sygs

Rrgrvi

RrgrivRiRirgriv

RirvgivRiv

)1(

])1([

)(

)(

++=

++=

+−−=

+−=

−=KVL:

)]1([||][||

])1([||

)1(

SmoD

x

SomoD

xx

SomoD

xx

Somo

x

D

xy

D

xx

RgrRv

RrgrRvi

RrgrRvi

Rrgrv

Rvi

Rvi

+=

+≈

++=

+++=+=KCL:

By KCL

Discrete CE and CS Amplifiers with RE / RS


[ ] )1( ||

/)||( 1

)||(

SmoDo

Gi

oLDSm

LDm

i

o

RgrRRRR

rRRRgRRg

vv

+==

++−=

i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

+++=

+

++=

+++−=

sigBE

EoCo

oLC

EEBi

EoLCEm

LCm

i

o

RRRrRrRR

rRRRRrRR

rRrRRRgRRg

vv

||1 ||

]/)||[(1 ||

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

π

π

π

β

β

Discrete CB Amplifier


Real Circuit


CB amplifier Input at the gate Output at the drain


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

Discrete CB Amplifier (Gain)


Fundamental CB Configuration

LCL RRR ||=′ Signal input at the emitter Signal output at the collector

)||||( LComi

o RRrgvv

+= )||( Lomi

o Rrgvv ′+=

Discrete CB Amplifier (Ri)


1

)||( || ||

+

+=

om

LCoEi rg

RRrrRR π

1

||| om

LoCBi rg

RrrRR+

′+== π Fundamental

CB configuration

1

|| om

Lo

rgRrrR

+′+

= π

Discrete CB Amplifier (Ro)


1) Set vsig = 0 2) Replace transistor with its SSM

Discrete CB Amplifier (Ro)


{ } )]||||(1[ || sigEmoCo RRrgrRR π+=

Attach vx , compute ix (Ro = vx /ix )

1

1

11

1

1

1

)1(

])1([

)(

)(

||||

Rrgrvi

RrgrivRiRirgriv

RirvgivRiv

rRRR

omo

xy

omoyx

yyomoyx

yomyx

y

sigE

++=

++=

+−−=

+−=

−=

=

π

π

π

KVL:

)]1([||][||

])1([||

)1(

11

1

1

RgrRv

RrgrRvi

RrgrRvi

Rrgrv

Rvi

Rvi

moC

x

omoC

xx

omoC

xx

omo

x

C

xy

C

xx

+=

+≈

++=

+++=+=KCL:

By KCL

Discrete CB and CG Amplifiers

F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (24/42) i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

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

)||( ||||

)||||(

sigEmoCo

om

LCoEi

LComi

o

RRrgrRRrgRRrrRR

RRrgvv

π

π

+=

+

+=

+=

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

)||( ||

)||||(

sigSmoDo

om

LDoSi

LDomi

o

RRgrRRrgRRrRR

RRrgvv

+=

+

+=

+=

∞→ πr

Discrete CD Amplifier (Source Follower)


Real Circuit


CD amplifier Input at the gate Output at the source


Discrete CD Amplifier (Gain)


Fundamental CD Configuration

Signal input at the gate Signal output at the source

)||||(1

)||||( LSom

LSom

i

o

RRrgRRrg

vv

+=

LSL RRR ||=′

)||(1

)||( Lom

Lom

i

o

RrgRrg

vv

′+′

=

Discrete CD Amplifier (Ri)


Gi RR =

| ∞== CDiRR ∞=R

Fundamental CD Configuration

Discrete CD Amplifier (Ro)


1) Set vsig = 0


3) Attach vx , compute ix (Ro = vx /ix )

omSo

omS

xx

o

x

m

x

S

xx

o

xgsm

S

xx

xgs

rgRRrgR

vi

rv

gv

Rvi

rvvg

Rvi

vv

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

/1

=

=

++=

+−=

−=

KCL:

* Because 1/gm << ro

mSo g

RR 1|| ≈

Since ig = 0, vg = 0 and gate is grounded

Discrete CC and CD Amplifiers


|| 1 ||

)||||(1

)||||(

om

So

Gi

LSom

LSom

i

o

rg

RR

RRRRrg

RRrgvv

=

=+

=

i

o

sigi

i

sig

o

vv

RRR

vv

×+

=

[ ]

osigB

Eo

LEoBi

LEom

LEom

i

o

rRRr

RR

RRrrRRRRrg

RRrgvv

|| ||

||

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

)||||(

β

β

π

π

+=

+=+

=

Documents

9. Discrete Transistor Amplifiersaries.ucsd.edu/.../ECE65_W13-9-Discrete_Amps-A.pdf · Use a capacitor to separate bias voltage from the signal. Simplified biasing problem. Used in