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1
FETFREQUENCY RESPONSE
LOW FREQUENCY
2
LOW FREQUENCY – COMMON SOURCE
R Si C 1
R D C 2
+V DD
v i R G
R S
C 3
R L
v o
3
vi gmVgs
S
DG
Vgs
+
-
RS
RDRLRG
RSi
C3
C1C2
Input RC circuit Output RC
circuit
Bypass RC circuit
Low-frequency equivalent circuit
4
The cutoff frequencies defined by the input , output and bypass circuits can be obtained by the following formulas.
112
1
CRf
Cc where
RC1=RSi+RG
222
1
CRf
Cc where
RC2=RD+RL
332
1
CRf
Cc where
RC3=RS||1/gm
Input RC circuit
Output RC circuit
Bypass RC circuit
5
EXAMPLE
Determine the lower cutoff frequency for the FET amplifier. Given K = 0.4mA/V2, VTN= 1V, = 0
10K
v i
R Si C 1
0.01 mFR G
1 MWR S
1 kW
C 2
0.5 mF
R D
4.7 kW
R L
2.2 kW
v o
+V DD
20 V
C 3
2 mF
6
HzMKCR
fC
c 8.15)01.0)(110(2
1
2
1
11
m
HzmKCR
fC
c 73.238)2)(211(2
1
2
1
33
m
HzKKCR
fC
c 13.46)5.0)(2.27.4(2
1
2
1
22
m
Input RC circuit
Output RC circuit
Bypass RC circuit
7
Since fc in bypass RC circuit, is the largest of the three cutoff frequencies and is separated from the next highest frequency by more than two octaves, the dominant pole approximation is applicable. Hence the low cutoff frequency fH for the amplifier is
fH = fc = 238.73Hz
8
LOW FREQUENCY – COMMON GATE
Analyse for mid-band gain and lower cut-off frequency
V 1TNV 2mA/V 3NK 0 pF 15gsC pF 4gdC
9
.
.
R D
R L
R i C C 1
C C 2
R S
V DD = +5 V
V SS = -5 V
v o
v i
Q
2 kW
10 kW
5 kW
4 kW
1 mF
2 mF
10
R S
R D
V DD = +5 V
V SS = -5 V
5 kW
10 kW
I G = 0
I D
I D
V S
V D
DC ANALYSIS
V 349.1GSV
mA/V 093.2mg
11
AC ANALYSIS – MID-BAND
. .
.
R i
2 kW
R S
R D R Lg mv gs
v i
v o
G D
S
10 kW
5 kW 4 kW
+v s
-
+
v gs
-i i
i s
RRR DL //'
12
sgs vv
'Lgsmo Rvgv
'Lsm Rvg
'Lms
o Rgv
v
Ssiii RiRiv
gsmis vgii
smi
si
S
s vgR
vv
R
v
smi
s
S
s
i
i vgR
v
R
v
R
v
13
iSmSi
S
i
s
RRgRR
R
v
v
i
s
s
o
i
ov v
v
v
v
v
vA
iSmSi
SLm RRgRR
RRg
'
14
210093.2102
105//4093.2
vA
V/V 864.0
15
AC ANALYSIS – LOW FREQUENCY
. .
.
R i
2 kW
R S
v i
S
10 kW
C C 1
C C 2
R D
5 kW 4 kWR L
v o
1 mF
2 mF
g mv gs
DG
+
v gs
-
R in
16
W
4561010093.21
10
1 43
4
Sm
Sin Rg
RR
Effective resistance across CC1 is;
W k 456.2456.021 iniC RRR
rad/s 40710110456.2
1163
111
CCL CR
Hz 8.642
407
21
1
LLf
17
rad/s 56102109
1163
222
CCL CR
Effective resistance across CC2 is;
W k 9542 LDC RRR
Hz 8.82
56
22
2
LLf
Since the two frequencies are separated by more than two octaves, the dominant pole approximation can be applied. The lower cut-off frequency is the higher of the two frequencies i.e. 64.8 Hz
18
LOW FREQUENCY – COMMON DRAIN
.
.
R 1
R 2
R i
R S R L
C 1
C 2
V DD = 10 V
v o
v i 1 kW
17.1 MW
50 MW
10 kW8 kW
10 µF
1 µF
mA 12.3DSSI
V 25.1tV
pF 4.3gdC pF 6.3gsC
0
Analyse for AM & fL
19
R 1
R 2
R S
V DD = 10 V
17.1 MW
50 MW
8 kW
I D
I D
V G
I G = 0
V S
DC ANALYSIS
V 548.0GSV
mA/V 8.2mg
20
.
.
.
+
v gs
_
g mv gsR G
R i
v i
R L '
G D
Sv o
+
v g
_
1 kW
AC ANALYSIS – MID-BAND
21 // RRRG
50//1.17
W M 74.12
LSL RRR //'
10//8W k 44.4
21
ogsg vvv oggs vvv
'' LogmLgsmo RvvgRvgv
'1
'
Lm
Lm
g
o
Rg
Rg
v
v
iisG
Gg vv
RR
Rv
1i
g
v
v
22
'1
'
Lm
Lm
i
g
g
o
i
oM Rg
Rg
v
v
v
v
v
vA
Substituting values;
V/V 93.0
10//812.31
10//812.3
1
Om
OmM Rg
RgA
23
AC ANALYSIS – LOW FREQUENCY
.
.
.
g mv gsR G
R i
R S R L
C 1 G D
S
v i
v o
+
v g
_
+
v gs
_ C 2
24
The effective resistance across C1 is;
WWW M 74.12M1.17//50k 11 GsC RRR
mHz 2.110101074.122
1
2
166
11
CR
fCs
L
25
To find the effective resistance across C2, we apply a voltage source vx as follows;
..
R i C 1
R G
G D
+
v gs
_
v x
i x
R S
g mv gs
i S
S
26
x
xi i
vR
S
xS R
vi
xgs vv
At node S;
xgsmS ivgi or xxmS
x ivgR
v
xS
Smx v
R
Rgi
1
Sm
S
x
xi Rg
R
i
vR
1
27
The output circuit is simplified to;
R i R L
C 2
The effective resistance across C2 is;
iLC RRR 2
Sm
SL Rg
RR
1
W
k 342.1080000028.01
800010000
28
222 2
1
CRf
CL
Hz 4.151010342.102
163
By dominant pole approximation, the lower cut-off frequency is;
Hz 4.152 LL ff
29
• The cut-off frequency is also known as 3-dB frequency or half-power frequency or corner frequency.
• The lower cut-off frequency is determined by the components (R and C) external to the transistor.
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