Common-gate MOSFET circuit
The third amplifier configuration is the
common-gate circuit.
To determine the small-signal voltage and
current gains, and the input and output
impedances, we will use the same smallsignal equivalent circuit for the transistor that
was used previously.
The dc analysis of the common-gate circuit is
the same as that of previous MOSFET
circuits.
THE COMMON-GATE CONFIGURATION
Common-gate
circuit
In the common-gate configuration, the input signal is applied to the
source terminal, output is taken at drain and the gate is at signal ground.
The common-gate configuration shown in Figure is biased with a
constant-current source IQ.
The gate resistor RG prevents the buildup of static charge on the gate
terminal, and the capacitor CG ensures that the gate is at signal
ground.
The coupling capacitor CC1 couples the signal to the source, and
coupling capacitor CC2 couples the output voltage to load resistance
In order to draw the ac
equivalent circuit, we must
go throw the following
steps:
•Capacitors are assumed to
be short circuited.
•DC sources are shorted
and Current source is
opened.
The AC equivalent circuit
for the CG amplifier is shown
in Figure
Small Signal Equivalent Circuit
Figure shows the small signal equivalent circuit of CG amplifier.
Here the MOSFET is replaced by its small signal model.
Here, the small-signal equivalent circuit shown
in figure may appear to be different from those
considered previously.
Input Resistance
Input resistance defined as ratio of input voltage to
the input current.
 Vgs
Ri 
Ii
Applying KCL at node S, we have,
I i   g mVgs   0
 I i  g mVgs  0
I I   g mVgs
 Vgs
1
 Ri 

 g mVgs g m
Voltage gain
Voltage gain is defined as ratio of the output voltage to
the input voltage.,
Applying KVL to output circuit, we get
Vo   g mVgs RD || RL   0
Vo  g mVgs RD || RL   0
 Vo   g mVgs RD || RL 
Applying KVL to input circuit, we get
Vi  I i R si    Vgs   0 Vi  I i R si   Vgs
Vi  I i R si   Vgs  0
But I i   g mVgs
 Vi   g mVgs R si   Vgs
 Vgs 1  g m R si 
Vgs  
Vi
1  g m R si
Sub Vgs value in Vo ,
Vgs  
Vi
1  g m R si

Vo   g mVgs RD || RL 


Vi
RD || RL 
  g m  
 1  g m R si 
Vo 
gm
RD || RL 
1  g m R si
Current Gain
In
many
cases, the
signal input
to
a
commongate circuit
is a current.
Figure shows the small-signal equivalent
common-gate circuit with a Norton equivalent
circuit as the signal source.
We
can
calculate
a
current gain.
The output current Io can be written as
I o  g mVgs 
RD
RD  RL
To find Vgs
At the input we have,
Ii 
 Vgs
RSi
  g mVgs   0
Vgs 1  g m Rs i 
Ii 
Vgs
RSi
 g mVgs   0
Vgs  g mVgs RSi
RSi
 Ii
Rs i
 Ii
I i Rs i
Vgs  
1  g m Rs i
I o  g mVgs 
RD
RD  RL
Sub Vgs value
 
I i Rs i


  gm  

  1  g m Rs i
  RD

 R  R
L
 D
I o  g m Rs
Ai   
I i  1  g m Rs i
 RD

 R  R
L
 D



We may note that if RD >> RL and gmRSi >>1,
then the current gain is essentially unity.
Output Resistance
We can find the output
resistance by setting the
input signal voltage equal to
zero.
From Figure, we see that Vgs
= −gmVgs RSi , which means
that Vgs = 0.
Ro = RD
Consequently, gmVgs = 0.
The
output
resistance,
looking back from the load
resistance, is therefore
For the common-gate MOSFET circuit with the
following parameters: The transistor parameters are
VTN = 1.5 V, kn = 1 mA/V2, and λ = 0. The quiescent
drain current is IDQ = 1 mA. Determine the small
signal voltage gain, current gain, input resistance,
and output resistance.
Given: VTN = 1.5 V, kn = 1 mA/V2, and λ = 0, IDQ = 1 mA, VDD =5
V, Rs = 20k, RG=50K, RD = 4K and RL=8K.
I Q  I DQ  K n VGSQ  VTN 
2
1mA  1mAVGSQ  1.5
RO  RD  4 K
2
V
GSQ  VTN   1mA / 1mA  1
2
VGSQ  VTN 
1
VGSQ  1  VTN ` 1V  1.5V  2.5V
g m  2 K n VGSQ  VYN 
g m  2 (1mA)2.5  1  2mA / V
Ri 
1
1

 500
gm
2mA
Vo 
Vo 
gm
RD || RL 
1  g m R si
2mA
4K || 8K   0.12
1  (2mA)( 20 K )
 g m Rs  RD 


Ai  


 1  g m Rs i  RD  RL 
 (2mA)(20 K )  4 K 

Ai  
  0.32
 1  (2mA)(20 K )  4 K  8K 
Solution:
The quiescent gate-to-source voltage is determined from
I Q  I DQ  K n VGSQ  VTN 
2
V
 VTN  
2
GSQ
I DQ
VGSQ 

I DQ
Kn
g m  2 K n VGSQ  VYN 
Kn
 VGSQ  VTN 
The small-signal transconductance is
I DQ
Kn
 VTN
1mA
 1V  2V
1mA
 22  1  2 mA/ V
ro 
1
1

    0
 I DQ
0
The output current as
 g m Rs
I o I i 
 1  g m Rs i
 RD

 R  R
L
 D



I o  g m Rs  RD 



I i  1  g m Rs i  RD  RL 
The output voltage is, Vo = I o R L
 g m Rs
 I i 
 1  g m Rs i
 RL RD

 R  R
L
 D



 2mA50 K    10 K  4 K 

 0.1sin t



1  2mA50 K  10 K  4 K 
 0.283 sin t V
Figure given below is the ac equivalent circuit of a common-gate amplifier. The
transistor parameters are VTN = 0.4 V, kn = 100μA/V2, and λ = 0. The quiescent drain
current is IDQ = 0.25 mA. Determine the transistor W/L ratio and the value of RD such
that the small-signal voltage gain is Av = Vo/Vi = 20 and the input resistance is Ri =
500Ω.
gm  2 Kn I D
2
1 ' W 
K n  I D
2 L
Here , K n'  100 A / V 2  .0.1 mA / V 2
W 
 0.0125   1
L
1
0.1 W 0.25  1
2
L

W

L

1
0.0125

1
9
0.111
W
 9 2  81  80
L