320 Lecture 34

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Whites, EE 320 Lecture 34 Page 1 of 9

© 2009 Keith W. Whites

Lecture 34: MOSFET Common GateAmplifier.

We’ll continue our discussion of discrete MOSFET amplifierswe began with the common source amplifier in Lectures 31 and

32.

Here we’ll cover the common gate amplifier , which is shown in

Fig. 4.45. It has a grounded gate terminal, a signal input at the

source terminal, and the output taken at the drain.

(Fig. 4.45a)

Small-Signal Amplifier Characteristics

As we’ve done with previous amplifiers in this course, we’ll

calculate the following small-signal quantities for this MOSFET

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common gate amplifier: Rin, Av, Avo, Gv, Gi, Ais, and Rout. To

begin, we construct the small-signal equivalent circuit:

(Fig. 4.45b)

The T model was used since we ignored r o while Rsig appears in

series with 1/gm.

•

Input resistance, Rin. Because the gate is grounded, we can seedirectly from this small-signal equivalent circuit that

in

1

m

Rg

= (4.91),(1)

Actually, this result may not be that readily apparent to you

since while the gate is grounded, the current in the gate is zero

( 0g

i = ).

To verify this result in (1), we can apply a voltage source v x at

the source terminal and calculate the ratio of this voltage to

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the current directed into the source terminal, which we’ll

define as i x:

m gsi g v=

ov

At the input to this circuit

0

1/ x

x

m

vi

g

−= ⇒

x m gsi g v=

This current i x doesn’t flow through the gate terminal! Instead,i x flows through the dependent source, then to ground. Indeed,

we see that

x m gsi g v i= − = −

Tricky! In any event, the input resistance in (1) has been

verified.

• Partial small-signal voltage gains, Av and Avo. At the output

side of the small-signal circuit

( )||o m gs D Lv g v R R= − (2)

At the input, we can see that because the gate is grounded

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i gsv v= − (3)

Substituting (3) into (2), gives the partial small-signal AC

voltage gain to be

( )||ov m D Li

v A g R R

v≡ = (4.94),(4)

In the case of an open circuit load ( L

R → ∞), the small-signal

voltage gain becomes

Lvo v m D R

A A g R→∞

≡ = (4.95),(5)

• Overall small-signal voltage gain, Gv. Using voltage division

at the input to the small-signal equivalent circuit

insig

in sig

i

Rv v

R R=

+ (6)

Substituting this into

sig sig sig

v

o i o iv v

i

A

v v v vG Av v v v

=

≡ = = (7)

gives the overall small-signal voltage gain of this common

gate amplifier to be

( )insig in sig

||ov m D L

v RG g R R

v R R≡ =

+ (4.96a),(8)

More specifically, using (1) in this expression

( )

sig

||

1m D L

v

m

g R RG

g R=

+ (4.96b),(9)

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

at the output in the small-signal circuit above

D

o m gs D L

Ri g v

R R

−=

+

(10)

Because 0g

i = , then at the input we see that

i m gsi g v= − (11)

Substituting (11) into (10) gives the overall small-signal AC

current gain to be

o Di

i D L

i RG

i R R

≡ =

+

(12)

• Short-circuit small-signal current gain, Ais. The short circuit

small-signal AC current gain can be easily determined from

(12) with 0 L

R = as

01

Lis i R

A G=

≡ = (13)

• Output resistance, Rout. From the small-signal circuit above

with sig 0v = we find that 0i = since the gate is grounded.

Consequently,

out D R R= (4.97),(14)

Summary

In summary, we find for the CG small-signal amplifier:

o A non-inverting amplifier.

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o Moderate input resistance [see (1)].

o Moderately large small-signal voltage gain [see (9)], but

smaller than CS amplifier.

o

Small-signal current gain less than one [see (12)].o Potentially large output resistance (dependent on R D)

[see (14)].

Similar to the BJT CB amplifier we discussed in Lecture 20, the

CG amplifier finds use as a current buffer amplifier . It has the

relatively small input resistance, relatively large outputresistance, and Gi less than (and potentially near) one

characteristics of such amplifiers. (Does this amplifier provide

any power gain for a signal?)

Example N34.1 (based on text exercise 4.34). Use the circuit of

Fig. E4.30 to design a common gate amplifier. Find Rin, Rout, Avo,

Av, Gv, and Gi for 15 L R = k Ω and sig 50 R = Ω. What will the

overall voltage gain become for sig 50 R = Ω? 10 k Ω? 100 k Ω?

The DC analysis results are shown in Fig. E4.30:

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(Fig. E4.30)

Using (4.71) 2 2 0.5 m1

2.5 1.5 D

m

OV

I g

V

⋅= = =

− mS

Based on this DC biasing, the corresponding common gate

amplifier circuit is:

Ov

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The small-signal equivalent circuit for this amplifier is:

S

m gsg v R D=15 k

D

G

ov

vsig

ig=0

+

-

1/gm

R L=15 k

4.7 M

Rsig

Rin

Rout

The 4.7-MΩ resistor functions to force the gate to ground

potential. But since 0g

i = , it will have no other impact on the

circuit.

•

From (1), in 31 1 110m

Rg

−= = = k Ω.

• From (14), out 15 D R R= = k Ω.

• From (5), 3 3V

10 15 10 15

V

vo m D A g R

−= = ⋅ × =

• From (4), ( ) ( )3

V|| 10 15k ||15k 7.5

Vv m D L A g R R

−= = ⋅ =

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• From (9),( )

( ) 34sig

|| 7.5

1 1 1 10 50m D L v

v

m m sig

g R R AG

g R g R −

= = =+ + + ⋅

(15)

V

7.14 V=

• From (12),

15 1 A

15 15 2 A D

i

D L

RG

R R= = =

+ +

• What is the overall voltage gain when:

o

sig 1 R = k Ω? From (15),

3 3sig

7.5 V3.75

1 1 10 10 Vv

v

m

AG

g R −

= = =+ + ⋅

o sig 1 R = 0 k Ω? 3 37.5 V

0.681 10 10 10 V

vG

−= =

+ ⋅ ×

o sig 1 R = 00 k Ω? 3 37.5 V

0.074

1 10 100 10 V

vG −= =+ ⋅ ×

We see from these calculations that the overall voltage gain

decreases substantially as Rsig increases. Can you explain what

is physically happening to cause this to occur?

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