320 Lecture 31

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Whites, EE 320 Lecture 31 Page 1 of 5

2009 Keith W. Whites

Lecture 31: Common Source Amplifier.

Weve studied MOSFET small-signal equivalent models and the

biasing of MOSFET amplifiers in the previous three lectures.Well now apply those skills by looking closely at three basic

MOSFET amplifier types:

1.Common source amplifiers, including configurations with a

source resistor (called source degeneration).

2.Common gate amplifiers.

3.

Common drain (or source follower) amplifiers.

All of these amplifier types are appropriate for discrete

component designs. In the case of IC amplifiers, well also show

corresponding designs implemented in CMOS, beginning in

Lecture 33.

Common Source Small-Signal Amplifier

This type of amplifier is shown in Fig. 4.43a, as biased by a

combination of voltage and current sources:

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(Fig. 4.43a)

Assuming sufficiently large values for the coupling capacitors(CC1 and CC2) and the bypass capacitor (CS) so that their

reactances are very small at the frequency of operation the

equivalent small-signal circuit for this amplifier is shown in Fig.

4.43(b):

(Fig. 4.43b)

The text mentions performing the small-signal analysis directly

on the amplifier circuit, as illustrated in Fig. 4.43(c). We do notrecommend this approach. It is better to take the time and

construct the small-signal equivalent circuit, as were doing

here.

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Small-Signal Amplifier Characteristics

As we did when studying BJT amplifiers, well calculate thefollowing quantities for this MOSFET common source

amplifier: Rin, Av, Gv, Gi, and Rout.

Input resistance, Rin. From the small-signal circuit above, and

noting that 0g

i = , then

in GR R= (4.78),(1)

Partial small-signal voltage gain, Av. From the output side of

the small-signal circuit

( )|| ||o m gs o D Lv g v r R R= (2)

while at the input,

i gs

v v= (3)

Substituting (3) into (2), we find that the partialvoltage gain

as

( )|| ||ov m o D Li

vA g r R R

v = (4.80),(4)

Overall small-signal voltage gain, Gv. As we did with BJT

amplifiers, we can derive an expression for Gvin terms of Av.

By definition,

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sig sig sig

v

o i o iv v

i

A

v v v vG A

v v v v

=

= = (5)

Applying voltage division at the input of the small-signalequivalent circuit,

insig sig

(1)in sig sig

Gi

G

RRv v v

R R R R= =

+ +

(6)

Substituting (6) into (5) and using (4) we find

( )sig

|| ||Gv m o D L

G

RG g r R R

R R

=

+

(4.82),(7)

Overall small-signal current gain, Gi. Using current division

at the output in the small-signal model above||

||o D

o m gs

o D L

r Ri g v

r R R

=

+

(8)

while at the input,gs i G

v i R= (9)

Substituting (9) into (8) we find that the overall small-signal

current gain is||

||o o D

i m G

i o D L

i r RG g R

i r R R

=

+

(10)

Notice that asG

R in this last expression,i

G . Is this

a problem? (To help answer this question, notice that as

GR , 0

ii .) Also, what does

iG as

GR mean

for the overall small-signal outputpowerof this amplifier?

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Output resistance, Rout. To calculate the output resistance, we

first set sig 0v = , which also means that 0m gsg v = . The input

impedance of the dependent current source is infinite.

Consequently,

out ||o DR r R= (4.85),(11)

Summary

In summary, we find for the CS small-signal amplifier that it hasa

oHigh input resistance [see (1)].

oRelatively high small-signal voltage gain [see (7)].

oVery high small-signal current gain [see (10)].

oRelatively high output resistance [see (11)].

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320 Lecture 31