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The sinusoid signal will be distorted if the ac input signal is large. However, the distortion can be neglected if the
signal is small.
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Graphical solutions
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Fig. 5.18
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Small signal equivalent circuits
Assuming operation in the saturation region
Where
For small signal case
is negligible
gm = transconductance
Where,
The gate current for FET is negligible
Small signal equivalent circuit
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Dependence of gm on Q-point and device parameters
We know that
and
VGSQ  Vto 
IDQ
K
But from (5.3)
Therefore
μn- surface mobility of electron
Cox- capacitance of gate per unit area
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More complex equivalent circuits
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At higher frequencies small capacitance have to be added between device terminals
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Also in the saturation region iD versus VDS is considered to be constant. This is not actually
the case. The drain current, iD increases slightly as VDS increases. In order to take care of that
we must add a drain resistance rd in the small signal model.
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Example
Determine the values of gm and rd the MOSFET characteristics shown below.
From equation 5.34, we have,
obtain iD = 6.7 mA at VDS = 4 V and iD = 8 mA
at VDS = 14 V.
Thus, the reciprocal of rd is calculated as:
8  6.7  mA  0.13 103 S
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i
 D 
14  4V
rd vDS
Thus, rd = 7.7 KΩ
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Common source amplifier
C1, C2 – Coupling capacitors  short circuit for AC
signals and open circuit for DC bias calculation
CS – bypass capacitor  small impedance for AC
Voltage Gain
RL' 
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1  1  1
rd
Rd
RL
v0  ( g mvgs ) RL'
vin  v gs
Av 
v0
  g m RL'
vin
(5.38)
(5.39)
(5.40)
(5.41)
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