Refer Figure 18.12 in the textbook.
R4
13k
in figure 18.12 is
. It is required to maintain the output voltage
R4
R4
zero for the change in
value. Determine new
value of the circuit.
The value of
VGS 3
Use the formula and calculate the value of
2I D3
VGS 3 VTN
K n3
VTN
I D3
Substitute 1 V for
VGS 3 1V
, 0.8 mA for
2 0.8 10
4 10
and
mA
V2
K n3
for
in the equation.
1.632 V
Q3
Apply Ohms law and write the equation source to gate voltage for the transistor
VGS 3 I D1 R3
R3
Rewrite the equation to find
V
R3 GS 3
I D1
VGS 3
I D1
Substitute 1.632 V for
1.632
R3
0.5mA
and 0.5 mA for
3.264 k
VGS 4
Use the formula and calculate the value of
2I D4
VGS 4 VTN
Kn
VTN
VGS 4 1V
2 4 10
10 10
10
I D4
Substitute 1 V for
, 4 mA for
3
and
mA
V2
Kn
for
in the equation.
1.894 V
R4
Apply KVL at the node
VR 4 VO VGS 4 VSS 0
R4
. Let the voltage across
VR 4
be
R4
Apply Ohms law to the resistor
I D 3 R4 VO VGS 4 VSS 0
and rearrange the equation.
I D 3 R4 VO VGS 4 VSS
R4
VO VGS 4 VSS
I D3
VGS 4
VO
VSS
I D3
Substitute 1.894 V for
, 0 for
, 10 V for
and 0.8 mA for
VSS 5 5
Here, the net voltage of
is
which is equal to 10 V.
0 1.894 10
R4
0.8 103
14.86 k
15 k
15 k
R4
Hence, the value of new
is
gm4
Use the formula and calculate the transconductance
gm4 2Kn I D 4
10
I D4
Substitute 4 mA for
and
g m 4 2 10 103 4 103
8.94 mS
in the equation.
mA
V2
Kn
for
Q4
for transistor
Avt 4
Write the equation for
g m 4 R1 R2
Avt 4
1 g m 4 R1 R2
and substitute the values.
8.94 mS 10 k 10 k
1 8.94 mS 10 k 10 k
0.994
f P1
Use the equation and find
f P1
1
2 R1 P R2 CGD 2 CGS 2 1 Avgs 2
g m 4
R1 , R2
10 k
Substitute
for
Avgs 2
in the equation.
gm 4
CGD 2
, 8.94 mS for
, 1 pF for
CGS 2
, 5 pF for
and 0.5 for
f P1
1
2 10 k P 10 k
1pF 5 pF 1 0.5
8.94 mS
9.04 MHz
9.04 MHz
Hence, the poles of the amplifier-1 is
gm2
Use the formula and calculate the value of
ID2
mA for
.
gm2 2Kn I D 2
10
ID2
Substitute 0.5 mA for
gm2 2Kn I D 2
and
mA
V2
. Here, consider the standard current 0.5
Kn
for
2 10 103 0.5 10 3
3.16 mS
M2
g m1
At the source of
gm 2
, the resistances are parallel to each other. So,
is parallel to
.
CGS 1 CGS 2
fP2
Also, the capacitance is
. Consider this and write the equation to find
from the textbook.
1 gm2
fP2
2 CGS 2
gm2
Substitute 3.16 mS for
1 3.16 mS
fP2
2 5 pF
CGS 2
, and 5 pF for
101MHz
101MHz
Hence, the poles of the amplifier-2 is
gm3
Use the formula and calculate the value of
gm3 2 K p I D3
4
I D3
Substitute 0.8 mA for
and
g m 3 2 4 103 0.8 103
mA
V2
Kp
for
2.53mS
fP4
Write the equation to find
from the textbook.
g m3
1
2 CGS 3 CGD 4 CGS 4 1 Avgs 4
fP4
gm2
CGS 3
Substitute 2.53 mS for
, 5 pF for
Avgs 4
.
1
2.53mS
fP4
2 5 pF 1pF 5 pF 1 0.994
66.8 MHz
CGS 4
, 5 pF for
CGD 4
, 1 pF for
, and 0.994 for
66.8MHz
Hence, the poles of the amplifier-4 is
f P5
Write the equation to find pole frequency estimate
textbook.
1
f P5
1
2
P R1 R2 CGS 4
gm4
gm4
Substitute 8.94 mS for
M5
at the source of
from the
CGS 4 10 k
R1 , and R2
, 5 pF for
,
for
.
f P5
1
P 10 k 10 k 5 pF
8.94 mS
286 MHz
2
286 MHz
Hence, the poles of the amplifier-5 is
m
Write the formula to find the phase margin
.
f
fT
fT
1
m 180 90 tan 1 T tan 1 T tan 1
tan
f P 2
f P1
fP4
f P 5
fT
It is required to find the value of
when the
70
equation to
.
f
f
70 180 90 tan 1 T tan 1 T tan 1
f P 2
f P1
f P1
phase margin is
. So, equate the
fT
fT
1
tan
fP4
f P 5
fP2
fP4
Substitute 9.04 MHz for
, 101 MHz for
, 66.8 MHz for
.
1
fT
fT
1
180 90 tan
tan
9.04 MHz
101MHz
70
1
fT
fT
tan 1
tan
66.8
MHz
286
MHz
70
f P5
, and 286 MHz for
1
fT
fT
tan
9.04 MHz
101MHz
1
fT
fT
tan 1
tan
66.8 MHz
286 MHz
1
tan
70 90
1
fT
fT
tan
9.04 MHz
101MHz
1
fT
fT
tan 1
tan
66.8
MHz
286
MHz
1
tan
20
fT
10.005 MHz
Solve the equation and the value of
is
10.005 MHz
fT
Hence, the value of
is
.
fB
To know the value of dominant pole
equation and calculate T.
g R g R
4
T Avt 4 m1 3 m3
2
2
, the attenuation value is required. Write the
Avt 4
g m1
gm3
Substitute 0.994 for
, 3.16 mS for
, 2.53 mS for
R
11.9 k
4
for
in the equation.
3.16 mS 3.27 k 2.53mS 11.9 k
T 0.994
2
2
R3
3.27 k
for
, and
0.994 5.167 15.05
77.4
fB
Calculate the value of dominant pole
f
fB T
T
fT
Substitute 10.005 MHz for
10.005 MHz
fB
77.4
129 kHz
, and 77.4 for T.
Calculate the value of compensation capacitance at the dominant pole frequency. Write
CC
the equation and find
.
1
fB
2 R3 CGD CC 1 Avt 3
Avt 3
Calculate the voltage gain
Avt 3 g m 3 R4
2.53 103 11.9 103
30.1
f B 3.27 k
R3
CGD , CGD 4
CGS
Substitute 129 kHz for
,
for , 1 pF for
, 5 pF for
, and 30.1
Avt 3
for
.
1
129kHz
2 3.27 k 1pF CC 1+30.1
1pF CC
1
2 3.27 k 129kHz 1+30.1
1pF CC 12.132 1012
CC 11.13pF
11.13pF
Hence, the value of required miller capacitance is
