Whites, EE 320
Lecture 34
Page 1 of 9
Lecture 34: MOSFET Common Gate
Amplifier.
We’ll continue our discussion of discrete MOSFET amplifiers
we 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
© 2009 Keith W. Whites
Whites, EE 320
Lecture 34
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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 ro while Rsig appears in
series with 1/gm.
• Input resistance, Rin. Because the gate is grounded, we can see
directly from this small-signal equivalent circuit that
1
Rin =
(4.91),(1)
gm
Actually, this result may not be that readily apparent to you
since while the gate is grounded, the current in the gate is zero
( ig = 0 ).
To verify this result in (1), we can apply a voltage source vx at
the source terminal and calculate the ratio of this voltage to
Whites, EE 320
Lecture 34
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the current directed into the source terminal, which we’ll
define as ix:
vo
i = g m vgs
At the input to this circuit
vx − 0
= ix ⇒ ix = g mvgs
1 / gm
This current ix doesn’t flow through the gate terminal! Instead,
ix flows through the dependent source, then to ground. Indeed,
we see that
ix = − g mvgs = −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
vo = − g m vgs ( RD || RL )
(2)
At the input, we can see that because the gate is grounded
Whites, EE 320
Lecture 34
vi = −vgs
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(3)
Substituting (3) into (2), gives the partial small-signal AC
voltage gain to be
v
Av ≡ o = g m ( RD || RL )
(4.94),(4)
vi
In the case of an open circuit load ( RL → ∞ ), the small-signal
voltage gain becomes
Avo ≡ Av R →∞ = g m RD
(4.95),(5)
L
• Overall small-signal voltage gain, Gv. Using voltage division
at the input to the small-signal equivalent circuit
Rin
vi =
vsig
(6)
Rin + Rsig
Substituting this into
Gv ≡
vo
v v
v
= i o = i Av
vsig vsig vi vsig
N
(7)
= Av
gives the overall small-signal voltage gain of this common
gate amplifier to be
v
Rin
Gv ≡ o =
g m ( RD || RL )
(4.96a),(8)
vsig Rin + Rsig
More specifically, using (1) in this expression
g ( R || R )
Gv = m D L
1 + g m Rsig
(4.96b),(9)
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Lecture 34
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• Overall small-signal current gain, Gi. Using current division
at the output in the small-signal circuit above
− RD
io =
g m vgs
(10)
RD + RL
Because ig = 0 , then at the input we see that
ii = − g mvgs
(11)
Substituting (11) into (10) gives the overall small-signal AC
current gain to be
i
RD
Gi ≡ o =
(12)
ii RD + RL
• Short-circuit small-signal current gain, Ais. The short circuit
small-signal AC current gain can be easily determined from
(12) with RL = 0 as
Ais ≡ Gi R =0 = 1
(13)
L
• Output resistance, Rout. From the small-signal circuit above
with vsig = 0 we find that i = 0 since the gate is grounded.
Consequently,
Rout = RD
(4.97),(14)
Summary
In summary, we find for the CG small-signal amplifier:
o A non-inverting amplifier.
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Lecture 34
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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 RD)
[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 output
resistance, 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 RL = 15 kΩ and Rsig = 50 Ω. What will the
overall voltage gain become for Rsig = 50 Ω? 10 kΩ? 100 kΩ?
The DC analysis results are shown in Fig. E4.30:
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Lecture 34
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(Fig. E4.30)
Using (4.71)
gm =
2 I D 2 ⋅ 0.5 m
=
= 1 mS
VOV 2.5 − 1.5
Based on this DC biasing, the corresponding common gate
amplifier circuit is:
vO
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Lecture 34
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The small-signal equivalent circuit for this amplifier is:
D
g m vgs
G
4.7 M
vo
RD=
15 k
RL=
15 k
ig=0
1/gm
Rout
S
Rsig
+
vsig
-
Rin
The 4.7-MΩ resistor functions to force the gate to ground
potential. But since ig = 0 , it will have no other impact on the
circuit.
• From (1), Rin =
1
1
= −3 = 1 kΩ.
g m 10
• From (14), Rout = RD = 15 kΩ.
• From (5), Avo = g m RD = 10−3 ⋅15 × 103 = 15
V
V
• From (4), Av = g m ( RD || RL ) = 10−3 ⋅ (15k ||15k ) = 7.5
V
V
Whites, EE 320
Lecture 34
g m ( RD || RL )
Av
7.5
=
=
N
1 + g m Rsig ( 4) 1 + g m Rsig 1 + 10−3 ⋅ 50
V
= 7.14
V
RD
15
1 A
=
=
• From (12), Gi =
RD + RL 15 + 15 2 A
• From (9), Gv =
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(15)
• What is the overall voltage gain when:
o Rsig = 1 kΩ? From (15),
Av
7.5
V
Gv =
=
=
3.75
1 + g m Rsig 1 + 10−3 ⋅103
V
7.5
V
o Rsig = 1 0 kΩ? Gv =
=
0.68
1 + 10−3 ⋅10 × 103
V
7.5
V
o Rsig = 1 00 kΩ? Gv =
=
0.074
1 + 10−3 ⋅100 × 103
V
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?