MOSFET EQUIVALENT CIRCUITS
Lesson #4
Section 5.4-6
BME 373 Electronics II –
J.Schesser
20
Small-Signal Equivalent Circuits
• As done for BJTs, we will investigate an equivalent circuit
when the signal variations are small compared to the bias
points
• Some nomenclature:
– The values of the FET parameters at the Q point (i.e., the DC
value) will be denoted by the capital letters with the subscript Q:
IDQ for the Q point drain current.
– The portion of the signal which varies with time will be denoted by
lower case letters and lower case subscripts: id (t) for variable
portion of the drain current
– The total signal will be denoted by lower case letters with upper
case subscripts: iD (t) is the total drain current and
iD (t) = IDQ+ id (t)
vGS (t) =VGSQ + vgs(t)
BME 373 Electronics II –
J.Schesser
21
Small-Signal Equivalent Circuits (Continued)
iD (t)  I DQ  id (t)
vGS (t)  VGSQ  vgs (t)
Recall that
iD (t)  K (vGS (t)  Vto )2
I DQ  id (t)  K (VGSQ  vgs (t)  Vto )2
 K (VGSQ  Vto )2  2K(VGSQ  Vto )vgs (t)  Kv gs (t) 2
But the Q point is also given by
I DQ  K (VGSQ  Vto )2
Therefore,
id (t)  2K(VGSQ  Vto )vgs (t)  Kv gs (t) 2
BME 373 Electronics II –
J.Schesser
22
Small-Signal Equivalent Circuits (Continued)
Assuming the we are dealing with small signals
about the Q point and
vgs (t)  VGSQ  Vto
then
Other forms of g m
id (t)  2K (VGSQ  Vto )vgs (t)
if g m  2 K (VGSQ  Vto )
Let's define, g m , the transconductance
And
as g m  2K (VGSQ  Vto )
I DQ  K (VGSQ  Vto ) 2  (VGSQ  Vto ) 
id (t)  g m vgs (t)
G
+
vgs
S
-
K
Then
also
ig (t)  0
I DQ
gm  2K
id
I DQ
K
 2 KI DQ  2 K I DQ  2 (
W KP
)(
) I DQ
L 2
 (W / L) 2 KP I DQ
D
gmvgs
S
BME 373 Electronics II –
J.Schesser
23
Another More Complex Equivalent Circuit
id
G
D
+
vgs
S
gmvgs
rd
-
S
Up till now we assumed the the FET curves in the saturation region
were flat; however, if there is a slight slope, a small resistance in the
drain should be added to reflect this. So:
Therefore,
id (t )  g m v gs (t ) 
vds (t )
rd
gm 
id
v gs

vds  0
gm 
iD
vGS
iD
vGS
And
v DS VDSQ
Q  point
BME 373 Electronics II –
J.Schesser
1  id
rd vds

v gs  0
1  iD
rd vDS
iD
vDS
vGS VGSQ
Q  point
24
Application of the Small Signal Equivalent Circuit
Common Source Amplifier
VDD 20V
VDD 20V
R1
3Meg
RD
4.7k
out
1uF
in
V
C2
R
C1
100k
0.01uF
RL
10k
M1
Mmodel
+-
R2
1Meg
0
RS
2.7k
CS
100uF
0
0
•
•
0
0
Before we can apply the FET small signal equivalent circuit, we must
review and reduce the circuit elements of this amplifier to those which
are affected by a varying AC signal. In particular,
– The coupling capacitors C1 and C2 are shorted
– The bias resistor RS is shorted by the bypass capacitor CS
– The DC voltage sources are shorted
Note that the DC equivalent circuit is identical to the Self Bias circuit we
previous investigated. Recall IDQ = 0.7833 and VDSQ = 14.203
BME 373 Electronics II –
J.Schesser
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Common Source Amplifier Equivalent Circuit
R
G
+
vin
v+
-
R2
R1
-
io
D
+
vgs
+
gmvgs
rd
S
RD vo
-
RL
S
Assume that saturation region is flat and so rd = ∞ and KP=50mA/V2, Vto=2,
L=10μm, W=400μm, v=100sin(2000πt) mV
g m  2 KP W L I DQ  100e 6 40 0.7833  1.77 mS
100k
v+
-
G
+
vin
1M
3M
-
D
+
vgs
+
1.77mvgs
S
io
BME 373 Electronics II –
J.Schesser
S
4.7k
vo
10k
26
Common Source Amplifier Equivalent Circuit
100k
v+
-
G
+
vin
D
+
750k
vgs
-
+
1.77mvgs
4.7k
vo
S
10k
-
S
vo  io 10k
 (1.77 10 3 v gs
io
4.7
) 10k
4.7  10
 5.66v gs
750
100 sin( 2000t )mV )
850
 500 sin(2000t )mV
 5.66  (
BME 373 Electronics II –
J.Schesser
vo
Av   5.66
vin
vo
Avi   5
v
27
Source Follower
VDD 20 V
•
VDD 20V
R1
2Meg
in
V
+-
R
C1
100k
0.01uF
•
M1
Mmodel
C2
R2
2Meg
RS
out
1uF
RL
1k
•
0
0
0
0
•
•
•
In this circuit, we would like to find
RS to support a Q-point value of the
drain current, IDQ, equal to 10 mA.
Then solve for the gain of the
amplifier using the small signal
equivalent circuit.
Assume KP = 50A/V2, L = 2 m,
W =160 m, Vto = 1 V.
To calculate RS let’s first draw the equivalent DC circuit for
the Q-point.
This becomes the self bias circuit and we have
VGSQ = [R2 / (R1+R2)] x VDD – IDQ * RS
IDQ = K(VGSQ - Vt0 )2
Solving the 2nd equation for VGSQ = 3.236 V and substituting
it into the 1st equation to yield RS = 426.4 
BME 373 Electronics II –
J.Schesser
VDD 20 V
VDD 20V
R1
2Meg
M1
Mmodel
C2
R2
2Meg
0
RS
0
28
Source Follower (Continued)
100k
v+
-
G
+
vin
+
1M
vgs
-
-
+
8.944mvgs
D
•
io
S
v
426.4 o
1k
-
D
Now we apply the small signal equivalent circuit realizing that the drain is
grounded.
6
-3
g m  2 KP W L I DQ  100  10
80 10 10
 8.944mS
(426.4)(1000)
v0  g m vGS  RL'  g m vGS  ( RS  RL )  8.944  10 3  [
]v
RS  RL
(1000  426.4) GS
v0
vin  vGS  v0 
v
g m RL' 0
v0
g m RL'

 .7278
vin 1  g m RL'
BME 373 Electronics II –
J.Schesser
29
Ion Sensing Field Effect Transistor (ISFET)
• Δϕ= RT/F ln (c1/c2)
• R is the gas constant, T the absolute temperature (K) and F the Faraday
constant and ci, are ion concentrations in the solution and oxide.
• Using hydrogen ions can be used to measure pH1 and DNA2
1
2
Bergveld, P. ISFET, Theory and Practice, IEEE SENSOR CONFERENCE TORONTO, OCTOBER 2003
DNA Electronics, http://dnae.co.uk/technology/overview/
BME 373 Electronics II –
J.Schesser
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Homework
• Small Signal Equivalent Circuits
– Problems: 5.28, 5.29, 5.31, 5.32, 5.33
• Common-Source Amplifier
– Problems: 5.36, 5.37, 5.38, 5.40
• Source Follower
– Problems: 5.45, 5.47
BME 373 Electronics II –
J.Schesser
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