chapter 12(analog)
TRANSCRIPT
-
8/9/2019 Chapter 12(Analog)
1/18
Chapter 12 The Frequency Response of
Amplifiers
Section 12.1 The Miller TheoremAmplifiers 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. 1.1!1, we
illustrate a small signal equivalent circuit of a transistor.
Fig. 1.1!1 A small signal equivalent of a transistor
"et 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. 1.1!
. #ote that capacitors e$ist even in low frequency cases. %ut they are not significant
then.
1!1
-
8/9/2019 Chapter 12(Analog)
2/18
Fig. 1.1! 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 &iller 'heorem. "et us consider Fig. 1.1!(.
)a* 'he circuit for &iller 'heorem
1!
-
8/9/2019 Chapter 12(Analog)
3/18
)b* 'he equivalent circuit of the circuit in Fig. 1.1!()a* by removing Z
Fig. 1.1!( +ircuits for $plaining &iller 'heorem
As shown in Fig. 1.1!()a*, there are three nodes, 1 #,# and (# . (# is
grounded. %etween 1# and # , there is an impedance Z . It is further assumed
that kv
v=
1
. -ince we know1
vv
, we may remove Z and have the following
equation
Z
vk
Z
vkv
Z
vvi 11111
*1) === )1.1!1*
We define /Z as follows
*1)/1
k
ZZ
= )1.1!*
Finally, we have
/1
1
1Z
vi = )1.1!(*
-imilarly, we have
*1)*
11)
/
=
=
k
kZ
k
ZZ
)1.1!0*
1!(
-
8/9/2019 Chapter 12(Analog)
4/18
and/
Z
vi = )1.1!*
'he above discussion indicates that we may have an equivalent circuit for the circuit
in Fig. 1.1!()a*, as illustrated in Fig. 1.1!()b*.
'he above discussion is called &iller 'heorem.
For our amplifier, we will have kv
v
gs
ds = . For a capacitor, its corresponding
impedance is GDCjZ
1
= . 'herefore, for an amplifier, we have GDCkjZ *1)1
/1 +=
andGDC
kj
Z
=1
1
1/
. 'his means that we may have two capacitors, namely
GDCkC *1)1 += and GDCk
C
= 11 , shunting the input and output terminals of the
amplifier as shown in Fig. 1.1!0. 'hese two capacitors are often called &iller
capacitors.
Fig. 1.1!0 'he &iller capacitors
From Fig. 1.1!0, we observe the following
)1* For high frequency signals, the capacitors will become short!circuited. 'hus an
amplifier always acts as a low!pass filter.
)* 'he higher the gain, the larger the capacitor 1C . 'his means that the bandwidth
of an amplifier is smaller for a higher gain.
1!0
-
8/9/2019 Chapter 12(Analog)
5/18
In the following, we shall show e$periments to demonstrate the conclusions we
drew in the above.
Experiment 12.1-1 An Amplifier !ith a "o! #ain
In this e$periment, we used the circuit as shown in Fig. 1.1!. 'he circuit
represents a typical low gain amplifier. 'he program is shown in 'able 1.1!1 and the
result is in Fig. 1.1!2. As we can see, the amplifier is indeed a low!pass filter and its
bandwidth is quite broad.
Fig. 1.1! An amplifier with a low gain
'able 1.1!1 3rogram for $periment 1.1!1
$ample 2!1
.protect
.lib /c4mm5(v.l/ ''
.unprotect
.op
.options nomod post
677 1 5 (.(6
8" 1 11 155k
.param W19u
&1 11 ( 5
1!
-
8/9/2019 Chapter 12(Analog)
6/18
:nch "95.(u W9/W1/ m91
:A79/5.;u
-
8/9/2019 Chapter 12(Analog)
7/18
result is shown in Fig. 1.1!@. As can be seen, the bandwidth is significantly reduced.
Fig. 1.1!? An amplifier with a high gain
'able 1.1! 3rogram for $periment 1.1!
$ 2!5
.protect
.lib /c4mm5(v.l/ ''
.unprotect
.op
.options nomod post
677 1 5 6
8m vout vout1 5
8m1 1 11 5
.param W1915u W95u W(9(5u W09(5u
1!?
-
8/9/2019 Chapter 12(Analog)
8/18
&0 ( 11 1
:pch "91u W9/W0/ m91
:A79/5.;u
-
8/9/2019 Chapter 12(Analog)
9/18
Fig. 1.1!@ 'he gain vs frequency for the amplifier in Fig. 1.1!?
Section 12.2 The #ain-&an'!i'th (ro'uct for an Amplifier!ith Fee')ac*
In the previous section, we showed that an amplifier acts a low!pass filter. %esides,
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. 1.!1 shows a schematic diagram of an amplifier with feedback.
Fig. 1.!1 An amplifier with a feedback
From Fig. 1.!1, we have the following
1!;
-
8/9/2019 Chapter 12(Analog)
10/18
-
8/9/2019 Chapter 12(Analog)
11/18
Experiment 12.2-1 The Enlar%in% of the &an'!i'th of an +perational
Amplifier )y ,e%atie Fee')ac*
In this e$periment, we used the circuit in Fig. 1.! as the operational amplifier.
'he open!loop program is in 'able 1.!1 and the bandwidth is shown in Fig. 1.!(.
As can be seen, the bandwidth is quite narrow because of the high gain.
Fig. 1.! 'he operational amplifier for e$periments in -ection 1.
'able 1.!1 3rogram for $periment 1.!1
open loop test
.38>'+'
.>3'I># 3>-'
."I% B+4mm5(v.lB ''
.C#38>'+'
.op
1!11
-
8/9/2019 Chapter 12(Analog)
12/18
677 677D 5 (.(6
6-- 6--D 5 5.56
&1 1 1 677D 677D 3+E W95C "9C
& 1 677D 677D 3+E W95C "9C
&( ( ( 1 677D 3+E W95C "9C
&0 0 ( 677D 3+E W95C "9C
& ( 6% 6--D #+E W9155C "9C
&2 0 6%2 2 6--D #+E W9155C "9C
&? 6i! ? 6--D #+E W9155C "9C
&@ 2 6i: ? 6--D #+E W9155C "9C
&; ? 6%; 6--D 6--D #+E W9155C "9C
&15 6o 0 677D 677D 3+EW915C "9C
&11 6o 6%11 6--D 6--D #+E W95C "9C
6%IA- 6% 5 1.;6
6%IA-2 6%2 5 1.;6
6%IA-; 6%; 5 5.26
6%IA-1 6%11 5 1.?6
6% 6i! 5 1.2v
6in1 11 5 A+ 1
.A+ 7+ 155 1 155555k
6in 6i: 11 1.2v
-
8/9/2019 Chapter 12(Analog)
13/18
Fig. 1.!( 'he bandwidth of the amplifier in Fig. 1.!1
We then incorporate feedback as shown in Fig. 1.!0. 'he program is in 'able
1.! and the new bandwidth is shown in Fig. 1.!. As can be seen, the bandwidth
is significantly enlarged as predicted.
Fig. 1.!0
'able 1.!
open loop test
.38>'+'
1!1(
-
8/9/2019 Chapter 12(Analog)
14/18
.>3'I># 3>-'
."I% B+4mm5(v.lB ''
.C#38>'+'
.op
677 677D 5 (.(6
6-- 6--D 5 5.56
&1 1 1 677D 677D 3+E W95C "9C
& 1 677D 677D 3+E W95C "9C
&( ( ( 1 677D 3+E W95C "9C
&0 0 ( 677D 3+E W95C "9C
& ( 6% 6--D #+E W9155C "9C
&2 0 6%2 2 6--D #+E W9155C "9C
&? 6i! ? 6--D #+E W9155C "9C
&@ 2 6i: ? 6--D #+E W9155C "9C
&; ? 6%; 6--D 6--D #+E W9155C "9C
&15 6o 0 677D 677D 3+EW915C "9C
&11 6o 6%11 6--D 6--D #+E W95C "9C
6%IA- 6% 5 1.;6
6%IA-2 6%2 5 1.;6
6%IA-; 6%; 5 5.26
6%IA-1 6%11 5 1.?6
6% 6i! 5 1.2v
6in1 11 5 A+ 1
.A+ 7+ 155 1 155555k
6in 6i: 11 1.2v
-
8/9/2019 Chapter 12(Analog)
15/18
Fig. 1.!
Experiment 12.2-2 Another Experiment !ith a ifferent Fee')ac* Circuit
In this e$periment, we used the same amplifier circuit as that used in $periment
1.!1. 'he feedback circuit diagram is e$actly as shown in Fig. 1.!1. 'he circuit
is shown in Fig. 1.!2, the program is in 'able 1.!( and the new bandwidth is
shown in Fig. 1.!?. As can be seen, the bandwidth is enlarged.
1!1
-
8/9/2019 Chapter 12(Analog)
16/18
Fig. 1.!2 'he circuit used for $periment 1.!
'able 1.!( 3rogram for $periment 1.!
Eigh =ain Amp
-
8/9/2019 Chapter 12(Analog)
17/18
6-- 6--D 5 56
&1 1 1 677D 677D 3+E W95C "9C
& 1 677D 677D 3+E W95C "9C &( ( ( 1 677D 3+E W95C "9C
&0 0 ( 677D 3+E W95C "9C
& ( 6%? 6--D #+E W9155C
"9C
&2 0 6%@ 2 6--D #+E W9155C
"9C
&? 6i! ? 6--D #+E W9155C
"9C
&@ 2 6i: ? 6--D #+E W9155C
"9C
&; ? 6%; 6--D 6--D #+E W9155C
"9C
&15 vout 0 677D 677D 3+E W915C "9C
&11 vout 6%11 6--D 6--D #+E W95C
"9C
vin: vin 5 A+ 1 sin)1.2 5.55551 1k*
6I#! 6i! 5 1.2
6%IA-? 6%? 5 1.;6
6%IA-@ 6%@ 5 1.;6
6%IA-; 6%; 5 5.26
6%IA-1 6%11 5 1.?6
81 vout @ 15555G
8 @ 5 155G
8( @ 6i: 15G
80 vin 6i: 15G
< transient simulation
-
8/9/2019 Chapter 12(Analog)
18/18
.end
Fig. 1.!? 'he bandwidth produced in $periment 1.!
1!1@