Transcript Multi-functional Packaged Antennas for Next

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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 m A  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  ( gmvgs ) RL'
vin  vgs
Av 
v0
  g m RL'
vin
(5.38)
(5.39)
(5.40)
(5.41)
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Common source amplifier contd..
Input resistance
Rin 
Output resistance
vin
 RG  R1 R2
iin
(5.42)
1
1  1
rd
Rd
(5.43)
R0 
Example 5.6 (Modified)
KP  40A / V 2
VDD  20V
Vt 0  1V
 0
 = 0 implies 1/rd = 0 or rd = 
L  15m
W  450m
Find the midband voltage gain, input resistance, and output resistance of the amp.
Thevenin Equivalent
Bias circuit
Circuit
VDD R2 20106
VG 

 4V
R1  R2 5 106
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Common source amplifier contd..
VGSQ 
(2 Rs KVt 0  1)  4 Rs KVG  Rs KVt 0  1
2 Rs K
or
W KP 450 40106
)


 0.6  103
L 2
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2
VGSQ 1  2.55V , VGSQ 2  2.2V , I DQ  1.44m A, VDSQ  12.76V
K (
g m  2 KP
W
L
I DQ  2  40106
450
1.8 103  1.3m S
15
20
1

k; [as, rd  ,  0]
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rd
RL' 
1
1

1 1
1
1
1



3
rd Rd RL 4 10 10 103
AV 
V0
20
  g m RL'  1.9 103  103  5.4
Vin
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4
M
5
1 1
R0  RD rd 
  RD  4k
Rd rd
Rin  R1 R2 
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The Source Follower
Section 5.6
Recall the similarity with
Emitter Follower
To draw the small-signal ac equivalent circuit, we have to first draw the following circuit like remembering that:
 Drain is grounded
 Source is not grounded
 R1  Gate to ground (drain)
 R2  Gate to ground (drain)
 RS  Source to ground (drain)
 gmvgs & rd  drain to source
Thus,
Small-signal ac equivalent circuit
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The Source Follower contd…
How to calculate the output resistance ?
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Current gain:
The current gain Ai = Av Rin/RL , like before.
Fig. 5.33
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Example contd….
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