Chapter 12 The Frequency Response of
Amplifiers
Section 12.1
The Miller Theorem
Amplifiers were introduced before. In the previous chapters, we only talked about
the gain of an amplifier. We did not talk about the frequency of the input signal. In
fact, we deliberately ignored frequencies to simplify the discussion. In Fig. 12.1-1,
we illustrate a small signal equivalent circuit of a transistor.
Rg
G
D
vi
gmVgs
RL
S
Fig. 12.1-1
A small signal equivalent of a transistor
Let us point out that the above equivalent circuit is for low frequencies only.
As the frequency of the input signal gets higher, capacitors appear as shown in Fig.
12.1-2. Note that capacitors exist even in low frequency cases. But they are not
significant then.
12-1
CGD
Rg
G
D
CGS
vi
RL
gmVgs
S
Fig. 12.1-2
A small signal equivalent circuit with capacitors considered
With a capacitor introduced between the gate and drain terminals, it will be
helpful for us to learn the Miller Theorem. Let us consider Fig. 12.1-3.
i1
N1
i2
N2
Z
Z1
v1
v2
N3
(a) The circuit for Miller Theorem
12-2
Z2
Z2'
Z 1'
Z 2' 
Z
Z1 ' 
1 k
kZ
k 1
(b) The equivalent circuit of the circuit in Fig. 12.1-3(a) by removing Z
Fig. 12.1-3 Circuits for Explaining Miller Theorem
As shown in Fig. 12.1-3(a), there are three nodes, N 1 , N 2 and N 3 . N 3 is
grounded. Between N 1 and N 2 , there is an impedance Z .
that v 2
v1
v
 k . Since we know 2
v1
It is further assumed
, we may remove Z and have the following
equation:
i1 
v 2  v1 kv1  v1 (k  1)v1


Z
Z
Z
(12.1-1)
We define Z ' as follows:
Z1 ' 
Z
(1  k )
(12.1-2)
Finally, we have:
i1 
v1
Z1 '
(12.1-3)
Similarly, we have
12-3
Z2 '
and
i2 
Z
1
(1  )
k

kZ
(k  1)
(12.1-4)
v2
Z2 '
(12.1-5)
The above discussion indicates that we may have an equivalent circuit for the circuit
in Fig. 12.1-3(a), as illustrated in Fig. 12.1-3(b).
The above discussion is called Miller Theorem.
For our amplifier, we will have
impedance
Z1 ' 
is
Z
1
j (1  k )C GD
1
jCGD
and Z 2 ' 
.
vds
 k . For a capacitor, its corresponding
v gs
Therefore,
for
an
amplifier,
we
have
1
. This means that we may have two
 1
j 1  CGD
 k
 1
capacitors, namely C1  (1  k )CGD and C 2  1  C GD , shunting the input and
 k
output terminals of the amplifier as shown in Fig. 12.1-4. These two capacitors are
often called Miller capacitors.
A(s)
C1=(1+k)CGD
C2=(1-1/k)CGD
Fig. 12.1-4
The Miller capacitors
From Fig. 12.1-4, we observe the following:
(1) For high frequency signals, the capacitors will become short-circuited. Thus an
amplifier always acts as a low-pass filter.
12-4
(2) The higher the gain, the larger the capacitor C1 .
This means that the
bandwidth of an amplifier is smaller for a higher gain.
In the following, we shall show experiments to demonstrate the conclusions we
drew in the above.
Experiment 12.1-1: An Amplifier with a Low Gain
In this experiment, we used the circuit as shown in Fig. 12.1-5. The circuit
represents a typical low gain amplifier. The program is shown in Table 12.1-1 and
the result is in Fig. 12.1-6. As we can see, the amplifier is indeed a low-pass filter
and its bandwidth is quite broad.
VDD=3.3V
RL=100k
D
G
5u/0.35u
S
VG=1V
vout
vin
Fig. 12.1-5 An amplifier with a low gain
Table 12.1-1
•
•
Example 6-1
.protect
•
•
•
•
•
•
•
•
.lib 'c:\mm0355v.l' TT
.unprotect
.op
.options nomod post
VDD 1
RL
1
0
11
Program for Experiment 12.1-1
3.3V
100k
12-5
•
.param W1=5u
•
•
•
•
•
•
•
•
•
•
M1
11 2
3
0
+nch L=0.35u W='W1' m=1
+AD='0.95u*W1' PD='2*(0.95u+W1)'
+AS='0.95u*W1' PS='2*(0.95u+W1)'
•
•
.END
VG
Vin1
.AC
2
0
3
0
DEC
0.65v
AC 1
100 1
10000000k
.PLOT AC VDB(11)
Fig. 12.1-6
The gain vs frequency for the amplifier in Fig. 12.1-5
Experiment 12.1-2: An Amplifier with a Higher Gain
In this experiment, we used the circuit shown in Fig. 12.1-7. The gain of this
12-6
amplifier is much higher as explained before. The program is in Table 12.1-2 and
the result is shown in Fig. 12.1-8.
reduced.
As can be seen, the bandwidth is significantly
VDD=3.3V
VDD=3.3V
M1
50/2
1
M3
50/
2
3
M2
50/2
2
M4
50/
2
M10
150/2
4
VBIAS5 = 1.9V
5
V-
M5
100/
2
M7
100/
2
M6
100/2
VBIAS6 = 1.9V
vo
6
M8
100/2
7
vout
V+
VBIAS11 = 1.75V
M11
50/2
M9
100/2
VBIAS9 =0.6V
VSS=0V
Fig. 12.1-7
VSS=0V
An amplifier with a high gain
Table 12.1-2 Program for Experiment 12.1-2
• Ex 6-20
• .protect
• .lib 'c:\mm0355v.l' TT
• .unprotect
• .op
• .options nomod post
• VDD 1
0
5V
• Rm2 vout vout_1 0
• Rm1 1
1_1 0
12-7
• .param W1=10u W2=20u W3=30u W4=30u
•
•
•
•
M4
3
2
1_1 1
+pch L=1u W='W4' m=1
+AD='0.95u*W4' PD='2*(0.95u+W4)'
+AS='0.95u*W4' PS='2*(0.95u+W4)'
•
•
•
•
M3
vout 4
3
1
+pch L=0.5u
W='W3' m=1
+AD='0.95u*W3' PD='2*(0.95u+W3)'
+AS='0.95u*W3' PS='2*(0.95u+W3)'
• M2
vout_1
• +nch L=0.5u
6
7
0
W='W2' m=1
• +AD='0.95u*W2' PD='2*(0.95u+W2)'
• +AS='0.95u*W2' PS='2*(0.95u+W2)'
• M1
7
8
0
0
• +nch L=1u W='W1' m=1
• +AD='0.95u*W1' PD='2*(0.95u+W1)'
• +AS='0.95u*W1' PS='2*(0.95u+W1)'
•
•
•
•
•
•
•
•
Vin
9
0
AC 1
.AC DEC
100 1
VG1 8
9
0.817V
VG2 6
0
1.8V
VG3 4
0
3V
VG4 2
0
4V
*.tf v(vout) vin
*.tran 0.1us 600us
• .plot
• .end
1000000k
VDB(vout_1)
12-8
Fig. 12.1-8
Section 12.2
The gain vs frequency for the amplifier in Fig. 12.1-7
The Gain-Bandwidth Product for an
Amplifier with Feedback
In the previous section, we showed that an amplifier acts a low-pass filter. Besides,
the bandwidth decreases as the gain increases. In this section, we shall show that the
gain-bandwidth product for an amplifier with feedback is a constant.
Fig. 12.2-1 shows a schematic diagram of an amplifier with feedback.
vin
+
A(s)
k
Fig. 12.2-1
An amplifier with a feedback
12-9
From Fig. 12.2-1, we have the following:
A( S )(vin  kvout )  vout
(12.2-1)
vout (1  A( S )k )  A( S )vin
(12.2-2)
Thus, we have
G( S ) 
vout
A( S )

vin 1  A( S )k
(12.2-3)
It must be understood that this is a negative feedback and will reduce the gain.
Based upon the discussion presented in Section 12.1, we understand that an
amplifier can be roughly viewed as a low-pass filter. For simplicity, we assume that
the transfer function of our amplifier is a first-order one. That is, we have
A( S ) 
a 0
S  0
(12.2-4)
Substituting (12.2-4) into (12.2-3), we have:
a 0
S  0
a 0
a 0
G(S ) 


ka 0
S   0  k 0 S  (1  k ) 0
1
S  0
(12.2-5)
From Equation (12.2-4), by setting S to be 0, we obtain the open-loop gain to be a.
a
From Equation (12.2-5), the close-loop gain for low frequency is
. Thus, we
(1  k )
know that the gain is reduced from a to
a
. Again, from Equation (12.2-4), we
(1  k )
obtain the bandwidth for the open-loop case to be  0 . From Equation (12.2-5), we
observe that the bandwidth is increased from  0 to (1  k )0 . Thus for the
 a 
(1  k ) 0   a 0 which is
close-loop case, the gain-bandwidth product is 
 (1  k ) 
a constant.
12-10
In the following, we shall show some experiments related to the frequency
response of amplifiers.
Experiment 12.2-1: The Enlarging of the Bandwidth of an Operational
Amplifier by Negative Feedback
In this experiment, we used the circuit in Fig. 12.2-2 as the operational amplifier.
The open-loop program is in Table 12.2-1 and the bandwidth is shown in Fig. 12.2-3.
As can be seen, the bandwidth is quite narrow because of the high gain.
VDD=3.3V
VDD=3.3V
M1
50/2
1
M3
50/
2
3
M2
50/2
2
M4
50/
2
M10
150/2
4
VBIAS5 = 1.9V
5
v-
M5
100/
2
M7
100/
2
M6
100/2
VBIAS6 = 1.9V
vo
6
M8
100/2
7
VBIAS11 = 1.75V
VSS=0V
VSS=0V
The operational amplifier for experiments in Section 12.2
Table 12.2-1
•
•
•
•
•
M11
50/2
M9
100/2
VBIAS9 =0.6V
Fig. 12.2-2
vout
v+
Program for Experiment 12.2-1
open loop test
.PROTECT
.OPTION POST
.LIB "C:\mm0355v.l" TT
.UNPROTECT
12-11
• .op
• VDD VDD!
• VSS VSS!
0
0
• M1
PCH
1
1
W=50U
• M2
PCH
2
1
W=50U
• M3
PCH
3
3
W=50U
• M4
PCH
4
3
W=50U
3.3V
0.025V
VDD!
VDD!
VDD!
VDD!
L=2U
L=2U
1
VDD!
2
VDD!
L=2U
L=2U
• M5
3
VB5
5
NCH W=100U
L=2U
VSS!
• M6
4
VB6
6
NCH W=100U
L=2U
VSS!
• M7
5
L=2U
Vi-
7
VSS! NCH
W=100U
• M8
6
L=2U
Vi+
7
VSS! NCH
W=100U
• M9
7
L=2U
VB9
VSS!
• M10 Vo
L=2U
4
VDD!
• M11 Vo
L=2U
VB11
• VBIAS5
• VBIAS6
•
•
•
•
•
•
VB5 0
VB6 0
VSS!
1.9V
1.9V
VBIAS9
VB9 0
0.6V
VBIAS1 VB11
0
1.75V
VB
Vi- 0
1.65v
Vin1 11 0
AC 1
.AC DEC
100 1
100000k
Vin2 Vi+ 11 1.65v
• *Ri
12
0
3.3k
12-12
VSS!
NCH
W=100U
VDD! PCH W=150U
VSS! NCH
W=50U
• *Rf
12
Vo 220k
• .PLOT AC VDB(Vo)
• .END
Fig. 12.2-3
The bandwidth of the amplifier in Fig. 12.2-1
We then incorporate feedback as shown in Fig. 12.2-4. The program is in Table
12.2-2 and the new bandwidth is shown in Fig. 12.2-5. As can be seen, the
bandwidth is significantly enlarged as predicted.
Rf
Ri
+
Vin
12-13
Vout
Fig. 12.2-4
Table 12.2-2
•
•
•
•
•
•
open loop test
.PROTECT
.OPTION POST
.LIB "C:\mm0355v.l" TT
.UNPROTECT
.op
• VDD VDD!
• VSS VSS!
0
0
• M1
PCH
1
1
W=50U
• M2
PCH
2
1
W=50U
• M3
PCH
3
3
W=50U
• M4
PCH
4
3
W=50U
3.3V
0.025V
VDD!
VDD!
VDD!
VDD!
L=2U
L=2U
1
VDD!
2
VDD!
L=2U
L=2U
• M5
3
VB5
5
NCH W=100U
L=2U
VSS!
• M6
4
VB6
6
NCH W=100U
L=2U
VSS!
• M7
5
L=2U
Vi-
7
VSS! NCH
W=100U
• M8
6
L=2U
Vi+
7
VSS! NCH
W=100U
• M9
7
L=2U
VB9
VSS!
• M10 Vo
L=2U
4
VDD!
• M11 Vo
L=2U
VB11
• VBIAS5
• VBIAS6
• VBIAS9
VB5 0
VB6 0
VB9 0
VSS!
1.9V
1.9V
0.6V
12-14
VSS!
NCH
W=100U
VDD! PCH W=150U
VSS! NCH
W=50U
• VBIAS1
•
•
•
•
VB
Vin1
.AC
Vin2
• *Ri
• *Rf
VB11
Vi- 0
11 0
DEC
Vi+ 11
12
12
0
1.65v
AC 1
100 1
1.65v
1.75V
100000k
0
3.3k
Vo 220k
• .PLOT AC VDB(Vo)
• .END
Fig. 12.2-5
Experiment 12.2-2: Another Experiment with a Different Feedback Circuit
In this experiment, we used the same amplifier circuit as that used in Experiment
12.2-1. The feedback circuit diagram is exactly as shown in Fig. 12.2-1. The
circuit is shown in Fig. 12.2-6, the program is in Table 12.2-3 and the new bandwidth
12-15
is shown in Fig. 12.2-7.
As can be seen, the bandwidth is enlarged.
VDD=3.3V
VDD=3.3V
M1
50/2
1
M3
50/
2
3
M2
50/2
2
M4
50/
2
M10
150/
2
4
VBIAS5 = 1.9V
v-
M5
100/
2
5
M7
100/
2
M6
100/
2
M8
100/
2
VBIAS6 =
vout
1.9V
6
vo
v+
M11
50/2
7
1.65V
R4
M9
100/2
VBIAS9 =0.6V
VBIAS11 = 1.75V
R1
AC
VSS=0V
VSS=0V
1.65V
R3
Fig. 12.2-6
The circuit used for Experiment 12.2-2
Table 12.2-3
Program for Experiment 12.2-2
• High Gain Amp
• *********************************
•
•
•
•
•
.PROTECT
.OPTION POST
.lib "C:\model\tsmc\MIXED035\mm0355v.l" TT
.UNPROTECT
.op
• VDD VDD!
0
3.3V
12-16
R2
• VSS
•
•
•
•
•
VSS!
M1
1
M2
2
M3
3
M4
4
M5
3
L=2U
0
0V
1
VDD!
1
VDD!
3
1
3
2
VB7 5
VDD!
VDD!
VDD!
VDD!
VSS!
• M6
4
L=2U
VB8 6
VSS!
NCH
W=100U
• M7
5
L=2U
Vi- 7
VSS!
NCH
W=100U
• M8
6
L=2U
Vi+ 7
VSS!
NCH
W=100U
• M9
7
L=2U
VB9 VSS!
• M10 vout 4
VDD!
• M11 vout VB11 VSS!
L=2U
•
•
•
•
•
•
vin+ vin
VIN- ViVBIAS7
VBIAS8
VBIAS9
VBIAS1
• R1
• R2
• R3
0
AC
0
1.65
VB7 0
1.9V
VB8 0
1.9V
VB9 0
0.6V
VB11
PCH
PCH
PCH
PCH
VSS!
VDD!
PCH
VSS!
0
vout 8
10000K
8
0
100K
8
Vi+ 10K
.probe v(vin) v(vout)
.tf V(vout) vin+
.AC DEC 10 100
.PLOT AC VDB(vout)
.end
NCH
W=100U
W=150U
NCH
L=2U
W=50U
1 sin(1.65 0.00001 1k)
• R4
vin Vi+ 10K
• * transient simulation ***
•
•
•
•
•
W=50U
L=2U
W=50U
L=2U
W=50U
L=2U
W=50U
L=2U
NCH
W=100U
1G
12-17
1.75V
Fig. 12.2-7
The bandwidth produced in Experiment 12.2-2
12-18