Electronic II (ECE235b)
MOSFET Amplifiers
Anestis Dounavis
The University of Western Ontario
Faculty of Engineering Science
MOSFET Amplifiers
5.7 Single stage BJT Amplifiers
There are basically 3 configurations for implementing single-stage MOSFET amplifiers:
1. Common source (CS) amplifier,
2. Common gate (CG) amplifier,
3. Common drain (CD) amplifier.
All 3 configurations use the same structure with the same biasing configurations:
5.7.1 The Basic Structure
Figure 4.42 shows the basic circuit to implement the various configurations of the MOS amplifier. The
various DC currents and voltages are also shown.
Figure 4.42 Basic structure of the circuit used to realize single-stage discrete-circuit MOS amplifier
configurations.
Exercise 4.30
Consider the circuit of Figure 4.42. VDD=VSS=10V, I=0.5mA, RG=4.7M, RD=15k, Vt=1.5V, and
K=1mA/V2. Find VOV, VGS, VG, VS and VD. Calculate gm and ro, assuming VA=75V.
Figure E4.30
4.7.3 The Common-Source (CS) Amplifier
The Common-Source configuration is shown in Figure 4.43.
Figure 4.43 (a) Common-source amplifier based on the circuit of Fig. 4.42. (b) Equivalent circuit of
the amplifier for small-signal analysis. (c) Small-signal analysis performed directly on the amplifier
circuit with the MOSFET model implicitly utilized.
Input resistance:
Rin 
vi
 RG
ii
Relationship between vsig and vi:
v i  vsig
Rin
RG
 vsig
Rin  R sig
RG  Rsig
Output resistance:
Rout  RD || ro
Output voltage:
v 0   g m v gs (r0 || R D || R L )
Open circuit voltage gain:
Voltage gain:
Overall voltage gain:
Av 
Avo 
where v gs  vi
vo
  g m ( r0 || R D )
vi RL  
vo
  g m ( r0 || RD || RL )
vi
Gv 
vo
RG

g m ( r0 || RD || RL )
vsig
RG  R sig
From our analysis the CS amplifier has a very high input resistance, a moderately high voltage gain and
a relatively high output resistance.
4.7.4 The Common-Source (CS) Amplifier with a Source Resistance
The Common-Source configuration with a source resistance is shown in Figure 4.44.
Figure 4.44 (a) Common-source amplifier with a resistance RS in the source lead. (b) Small-signal
equivalent circuit with ro neglected.
Input resistance:
Rin 
vi
 RG
ii
Relationship between vsig and vi:
v i  vsig
Relationship between vgs andvi:
v gs  vi


1/ g m
vi

1 / g m  RS 1  g m R S
RS is used to control the magnitude of the signal vgs and thus ensure that vgs does not become
too large and cause unacceptable nonlinear distortion.
In addition, Rs improves the frequency bandwidth.
vi
id  i 
Output voltage:
v 0  i D ( RD || RL )  
1 / g m  RS
Open circuit voltage gain:
Avo 
Voltage gain:
Av 
Overall voltage gain:

gm vi
1  g m RS
The current iD:

Rin
RG
 vsig
Rin  R sig
RG  Rsig
g m ( R D || R L )
vi
1  g m RS
vo
g R
 m D
vi RL  
1  g m RS
vo
g ( R || RL )
 m D
vi
1 g m RS
Gv 
vo
RG
g m ( RD || RL )

vsig
RG  Rsig 1  g m RS
The voltage gain is reduced by a factor of 1  g m RS .
Output resistance:
Rout  R D
To summarize:
 The inclusion of Rs results in a gain reduction by a factor of (1+gmRs).
 Rs increases the DC bias stability since the Rs absorbs part of the voltage away from the 1/gm
resistor
 Rs improves the frequency bandwidth.
Problem 4.77
For the CS amplifier shown in Figure 4.77
a) Vt=1V and K=2mA/V2, find VGS, ID, and VD.
b) Find gm and ro if VA=100V.
c) Find Rin, Rout, vgs/vsig, vo/vgs and vo/vsig.
Remove the capacitor at the source, ignoring ro and repeat the analysis.
Figure P4.77
4.7.5 Common-Gate (CG) Amplifier
The Common-Gate amplifier or grounded gate amplifier is shown in Figure 4.45. The ro resistor is not
included to simplify the analysis.
Figure 4.45 (a) A common-gate amplifier based on the circuit of Fig. 4.42. (b) A small-signal
equivalent circuit of the amplifier in (a). (c) The common-gate amplifier fed with a current-signal input.
Input resistance:
Rin 
vi
1

ii
gm
Relationship between vsig and vi:
v i  vsig
Relationship between vgs andvi:
v gs  vi
Rin
1
 vsig
Rin  R sig
1  g m Rsig
1/ g m
vi

1 / g m  RS 1  g m R S
vi
 gm vi
Rin
The current ii:
i i 
The current id:
i d  i  ii  vi g m
Output voltage:
v 0  i d ( R D || R L )  g m ( R D || RL )vi
Open circuit voltage gain:
Avo 
Voltage gain:
Av 
Overall voltage gain:
Output resistance:
vo
 g m RD
vi RL  
vo
 g m ( RD || RL )
vi
Gv 
vo
Rin
g ( R || RL )

g m ( RD || RL )  m D
vsig Rin  Rsig
1  g m Rsig
Rout  R D
To summarize:
 Unlike the CS amplifier, which is inverting, the CG amplifier is noninverting.
 CS has a very high input resistance: the input resistance of the CG amplifier is low.
 While the Av values of both CS and CG amplifier are nearly identical, the overall voltage gain
of the CG amplifier is smaller due to its low input resistance.
 CG amplifier can be used as a unity-gain current amplifier.
 The high frequency performance of CG amplifier is superior to CS amplifier.
4.7.6 Common-Drain (CD) Amplifier or Source-Follower Amplifier
The Common-Drain amplifier is shown in Figure 4.45.
Figure 4.46 (a) A common-drain or source-follower amplifier. (b) Small-signal equivalent-circuit
model. (c) Small-signal analysis performed directly on the circuit. (d) Circuit for determining the output
resistance Rout of the source follower.
Input resistance:
Rin 
vi
 RG
ii
Relationship between vsig and vi:
Output voltage:
Avo 
Voltage gain:
Av 
Output resistance:
Rin
RG
 vsig
Rin  R sig
Rsig  RG
v 0  i d ( R D || R L )  g m ( R D || RL )vi
Open circuit voltage gain:
Overall voltage gain:
v i  vsig
vo
r0

vi RL   r0  1 / g m
vo
RL || r0

vi R L || r0  1 / g m
Gv 
vo
RG
RL || r0

vsig RG  Rsig RL || r0  1 / g m
Rout  (1 / g m ) || r0
Summary:
 The common drain has a high impedance
 A relatively low output resistance
 The voltage gain is less than one but close to unity.
Problem 4.87
For the circuit shown let Vt=1V, K=0.8mA/V2, and VA=40V
1. Find Rs, RD, and RG so that ID=0.1mA, the largest possible value for RD is used while a maximum
signal swing at the drain of plus or minus 2.5V is possible and the input resistance at the gate is
10MW.
2. Find the values of ro and gm.
3. If terminal X is grounded and terminal Z is connected to a current source delivering a signal
current of 10A having a resistance of 100k, find the voltage signal that can be measured at Y.
Find Rin and Rout. For simplicity, neglect the effect of ro.
4. If terminal Y is grounded, find the voltage gain from X to Z with Z open-circuited. Find the
output resistance and input resistance of the source follower?
Figure P4.87
JFET Example
The JFET characteristics can be described by the same equations used for MOSFETs.
 The n-channel transistor is turned off (cutoff region) when vGS <= Vp. When 0 >= vGS > Vt,
and vDS is positive the transistor is on.
 The transistor is said to be in the triode region when vDS< vGS – Vp.
 The transistor is said to be in the saturation region (pinch-off) when vDS>= vGS – Vp
The equations relating iD and the applied voltages are
1 2 

i D  K ( vGS  VP )v DS  v DS
(Triode region)

2

1
i D  K (v GS  VP ) 2
(Saturation region)
2
1
i D  K ( vGS  VP ) 2 (1  v DS )
(Saturation region [channel length modulation])
2
where K  2 I DSS / VP2 . The current IDSS is the drain current when vGS=0V. Note that for a n-channel
JFET Vp is negative, whereas for a n-channel MOSFET Vp is positive.
Example 1
Determine the following for the circuit shown: VGS, ID, VDS, VS, VG, VD.
IDSS=8mA
Vp=-6V
Example 2
For the self-biasing configuration of example 1, input impedance, output impedance, and voltage gain
Vi/Vo.