Chapter 2

winetediousElectronics - Devices

Oct 7, 2013 (3 years and 10 months ago)

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2
-
1

Chapter 2 MOSFET Amplifiers

Section 2.1

The Basic Principles of
Common Source
NMOS

Amplifiers

In Fig. 2.1
-
1, we show an NMOS circuit with its I
-
V curves and
a
load line.


Fig. 2.1
-
1

An NMOS circuit without any AC



In Fig. 2.1
-
2, an input

signal

is added to
. We shall show later

will be amplified. Since the source is signal grounded, w
e shall call this kind of
amplifiers common source amplifiers.

It must be noted tha
t the signal is a small
signal and is also an alternating current (AC) signal. It will become clear that this
kind of amplifiers can hardly amplify large signals.



2
-
2


Fig 2.1
-
2

An NMOS
circuit

with

an

AC signal



Consider Fig.
2.1
-
3. As shown in Fig. 2.1
-
3,
the input signal induces an AC
voltage between gate and source. Let us denote this AC voltage by
.
Throughout this lecture notes, we shall use small letters to denote AC parameters.
Obviously,
.
In Fig. 2.1
-
2, we can see that





Since the amplitude of

is larger than that of
, this circuit functions as an
amplifier.

It is important to note the DC voltage

still exists. In some sense,
we may say that
(AC) rides on
(DC). But, so far as amplification is
concerned, we are only interested in the AC output voltage signal
.


2
-
3


Fig. 2.1
-
3

The amplification of



Suppose a smaller load

is used, as shown in Fig. 2.1
-
4. This amplifier will
not function very well now.


Fig. 2.
1
-
4

The behavior of an amplifier with a small load



It will be shown later that a large load is usually desirable and needed. But,
this
may drive the transistor out of saturation, as shown in Fig. 2.1
-
5. Once the transistor
is out of saturation, the ou
tput signal will be distorted and the circuit is no longer an

2
-
4

amplifier any more.


Fig. 2.1
-
5

A

large

driving the transistor out of saturation



On the other hand, if

is not pro
per, the output signal will also be distorted,
as shown in Fig.2.1
-
6.



Fig. 2.1
-
6

An

improper

causing the
distortion

of





2
-
5

Section 2.2

The DC Analysis of An NMOS Common
Source Amplifier


In

this section, we shall show that the performance of an amplifier can be easily
understood by having a DC analysis of the circuit. That is, we
temporarily

ignore
the existence of the AC input signal and concentrate our mind on the DC parameters.



Consi
der Fig. 2.2
-
1.



Fig. 2.2
-
1

An amplifier circuit



We shall first see how the value of the load

affects the performance of the
amplifier. Fig. 2.2
-
2 shows three possible cases of
. As we showed before, if
the load

is small, the change of

will induce only a small change of
,
which is also

in our case. On the other hand, for a large
, even a small
change of

will induce a large change of
.


2
-
6


Fig. 2.2
-
2

I
-
V

curves and different load lines for the circuit in Fig. 2.2
-
1



Let us assume that
th
e load

is small. Fig. 2.2
-
3 shows the relationship
between

and
. In this case,

falls slowly as

rises. We added
the AC small signals on
to the diagram. It can be seen that

is quite small.
Thus the amplifier does not function well as an amplifier
.



Fig. 2.2
-
3


vs

for small



Fig. 2.2
-
4 shows the situation where

is large. It can be seen that the sharp
drop of

will induce a large output

and thus a large amplification.


2
-
7


Fig. 2.2
-
4


v
s

for large



It should also be seen from Fig. 2.2
-
5 that in this case, there is a very narrow
region for the proper biasing voltage
. In other

words, we must be very careful
in selecting the biasing voltage. We shall see later in many experiments that a slight
deviation from a proper

may cause great trouble for the amplifier.

Thus we
may say that a DC analysis of an amp
lifier is always important as it gives us
information about how to set the gate bias voltage.



Fig. 2.2
-
5

The proper selection of the gate biasing voltage


2
-
8

Section 2.3

The Creation of the AC Current
i
out

in the
Transistor


In the above se
ction, we presented the DC analysis of an amplifier. This
explains why the circuit works as an amplifier. But, it does not give us any idea
about the gain of the amplifier. In other words, it does not let us know how much the
input signal voltage is amp
lified. To find the gain, an AC analysis is needed.


An amplifier behaves as an amplifier because an AC current is created because
of the input voltage
, which is also

In the following, we shall first expla
in
how this AC current, namely
, is produced. We the
n

proceed
to
show

how this

affects
.



We first point out that the transistor must be in the saturation region. This can
be easily s
een in Fig. 2.3
-
1. If the transistor is in the triode region, the output
voltage would be highly distorted.

It should be noted that a transistor is out of
saturation if it has a very large load.



Fig. 2.3
-
1

The case where t
he transistor is out of saturation



2
-
9


Once the transistor is in the saturatio
n region, we may use Equation (1.2
-
3)

which we display again as follows:





(2.3
-
1)


In the above equation, the third
term
is
very small

because

is very small

and
t
he first term is a DC term. Thus
so far as AC signal is concerned,
we have





(2.3
-
2)


We define





(2.3
-
3)


Eq
uivalently, we have




(2.3
-
4)



Equation (2.3
-
4) indicates that the input voltage

creates an AC current

linearly because

is a
constant

once

is fixed
.

It should be noted that there
is still a DC current
. We may say that
(AC) rides on this
(DC)

as
shown in Fig. 2.3
-
2
.



Fig. 2.3
-
2

riding on


2
-
10

Section 2.4

The Determination of
v
out


In the above section, we showed that the input voltage

creates an AC current
.
This current of

course would in turn create an AC output voltage
. We shall
explain its mechanism in this section.




Let us redraw again the amplifier circuit as shown in Fig. 2.4
-
1.



Fig. 2.4
-
1

An amplifier circ
uit


We first show that the

will affect the output voltage because of the
existence of
.

This can be seen by examining Fig. 2.4
-
2 which illustrates how

affects
.

Note the following:


(1)

The input signal

changes
. That is,

induces an AC voltage
.

(2)

This voltage

in turn induces a change of
current

. That is,
the

2
-
11

AC voltage

induces an AC output current labeled as
.

(3)

As can be seen from Fig. 2.4
-
2,

finally changes
. That is, it
induces an A
C output voltage, labeled as
.


Fig. 2.4
-
2

The inducing of

by


As can be from Fig. 2.4
-
2, when

rises,

falls

and when

falls,

rises. Thus,
the polarity of

is opposite to the polarity of
. From Fig. 2.4
-
1,
we have





(2.4
-
1)

This shows that there is an AC output voltage caused by the existence of
.
However, we shall not let the AC output voltage simply be equal to
-

at this
moment because there is another factor which

will affect
, as we will explain
below.



We usually assume that in the saturation region, the I
-
V curve of the transistor is
rather flat. In reality, it is not so flat, as illustrated in Fig. 2.4
-
3.



2
-
12


Fig. 2.4
-
3

The meaning of


Traditionally, we are familiar with the fact that the change of voltage will cause a
change of current. In this case, it is opposite.
If there is a small
signal AC
current
, a sm
all signal AC voltage

will be induced.

In other words, the
AC current

will induce an AC voltage
. Thus, we may
define

the output
impedance

as





(2.4
-
2)


to illustrate the relationship between

and
. It must be noted that this

exists because of the I
-
V curves, not because o
f the load
.
If the slope of the I
-
V
curve is very flat,

will be quite large; otherwise, it will be quite small.

A
large

indicates that a small

will in
duce

a l
arge
.



From the above discussion, we know that

will be affected by
,

and
. In the
following
, we shall explain how

is determined.



Let us take a look at the amplifier circuit in Fig. 2.4
-
1. There are DC
voltages

in the circuits. They are

and
. If they are not proper, the amplifier will
not work.
On the

other hand,
if they are proper, since


is an AC voltage,
it

will
not be affected by the existence of
these
DC
voltages
. We therefore first
short
-
circuit all of the DC
voltages
, namely

and
.
One mu
st note that one
terminal of

is connected to

and another
terminal

is connected to ground, as
shown in Fig. 2.4
-
4(a). Once we short
-
circuit
,

will be connected t
o
ground and t
he amplifier circuit becomes that shown in Fig. 2.4
-
4
(b)
.

In Fig 2.4
-
4(b),
there is also an

connected between D and S(ground).



2
-
13


(a) An amplifier circuit with DC voltage

(b)The amplifier c
ircuit with DC
voltage
s
short
-
circuited


Fig. 2.4
-
4

The short
-
circuiting of DC voltage
s

in an amplifier circuit



As we discussed before,

a

small signal input voltage

induces a small
signal current
. The r
elationship between

and

can be found in Equation
(2.3
-
4).
As
shown in Equation (2.3
-
4),

and

are of the same polarity.
This
is illustrated in Fig. 2.4
-
5. Note

that there is no connection between Node G and
Node S as there is no current flowing into the gate. That is, there is a
voltage

between
G and S, but no current from G.





Fig. 2.4
-
5

Small signal equivalent circuit of a tran
sistor


Because of the short
-
circuiting of DC
voltages
,

is across D and S

as shown
in Fig. 2.4.4(b)
. So is
.


The entire small signal equivalent circuit of the
amplifier is shown in Fig.2.4
-
6
(b)
.



2
-
14


(a) An amplifier circuit

(b) The small signal equivalent circuit

(c) The small signal
equivalent

circuit further
s
i
mpl
if
i
ed


Fig. 2.4
-
6

The small signal equivalent circuit of an amplifier circuit



From Fig. 2.4
-
6
(c)
, we
can see that t
he current will flow through the
parallel

connection of

and
, which is expressed as
. The value of Z can
be found as follows:














(2.4
-
3)


Note that the polarity of

is opposite to that of
, as expected by
examining Fig. 2.4
-
2. The meaning of the opposition of polarity is illustra
ted in Fig.
2.4
-
7.


2
-
15


Fig. 2.4
-
7

The meaning of polarity of AC signals



Let us note that

is usually much larger than

because in reality, the I
-
V
curves of a transistor are rather
flat and a flat I
-
V curve produces a large
.
On the

other hand, as we pointed
out
before,

cannot be too large because a large

will drive the transistor into the non
-
saturation region.

When a large resistor is in
parallel with another resistor, it can be ignored. Thus,

is usually ignored and we
have





(2.4
-
4)


We denote the gain of the amplifier b
y
. Then





(2.4
-
5)



That a large

will produce a better amplifier can also be seen by the DC
analysis. The reader should go back to Section 2.2
again. Take a look at the
figures from Fig. 2.2
-
2 to Fig. 2.2
-
5. As seen in these figures, we can see that a large

will produce a sharp input
-
output relationship and thus a high gain.

Of course,
it must be under the condition tha
t the load will not drive the transistor out of
saturation.



In the above sections, we only talked about NMOS amplifiers. The same
discussion can be used to explain how a PMOS amplifier works. A typical PMOS
amplifier is shown in Fig. 2.4
-
8.


2
-
16


Fig. 2.4
-
8 A PMOS amplifier

Section 2.5

Experiments of the D
e
sign of NMOS
Transistors


To design an NMOS amplifier, we must pay attention to the operating point.
Fig. 2.5
-
1 shows how the operating
point
is determined. It is essentially determi
ned
by selecting an appropriate

and an appropriate
.



Fig. 2.5
-
1

The determination of operating points



In the following, we shall show a set of experiments to demonstrate how
d
ifferent parameters, including

and
,

will affect the performance of the
amplifier.

Experiment 2.5
-
1

An Appropriate Operating Point


2
-
17


The purpose of this experiment is to demonstrate that an appropriate operati
ng
point can be
determined
. The circuit is shown in Fig. 2.5
-
2, the SPICE program is in
Table 2.5
-
1 and the operating point is shown Fig. 2.5
-
3. The input and output
signals are shown in Fig. 2.5
-
4.

The gain of this amplifier was found to be 20.



Fig. 2.5
-
2

The amplifier circuit with

for Experiment 2.5
-
1


Table 2.5
-
1

The program for Experiment 2.5
-
1

EX2
-
1

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VDD

1

0

3.3v

R1 1 2 100k

V2 2 0 0v


.param

W1=5u

M1

2

3

0

0


+nch L=0.35u

W='W1' m=1


+AD='0.95u*W1' PD='2*(0.95u+W1)'


2
-
18

+AS='0.95u*W1' PS='2*(0.95u+W1)'


*Fig2.5
-
3*

VGS_1

3

0

0.65v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)


*Fig2.5
-
4*

*VGS_2

3

4

0.65v

*Vin 4 0 sin(0v 0.01v 10Meg)

*.tran 0.1ns 600ns


.end



Fig. 2.5
-
3

The operating points of the circuit in Fig. 2.5
-
2


Load Line

I
DS

I
-
V Curve

V
DS


2
-
19


Fig. 2.5
-
4

The gain of the ampli
fier in Experiment 2.5
-
1

Experiment 2
.5
-
2 The Operating Point with the Same
V
GS
, but a Smaller Load


In this experiment, we used a smaller load as shown in Fig.
2.5
-
5. The program
is in Table 2.5
-
2, the operating point is shown in Fig. 2.5
-
6 and the inpu
t and output
voltages are shown in Fig. 2.5
-
7.

As can be seen in Fig. 2.5
-
7, the gain is now
smaller.


Fig. 2.5
-
5

The amplifier circuit with

for Experiment 2.5
-
2

v
out

v
in


2
-
20


Table 2.5
-
2

Program for Experiment 2.
5
-
2

EX2
-
2

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VDD

1

0

3.3v

R1 1 2 10k

V2 2 0 0v


.param

W1=5u

M1

2

3

0

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)'


*Fig2.5
-
6
*

VGS_1

3

0

0.65v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)


*Fig2.5
-
7
*

*VGS_2

3

4

0.65v

*Vin 4 0 sin(0v 0.01v 10Meg)

*.tran 0.1ns 600ns


.end



2
-
21


Fig. 2.5
-
6

The operating

point of the amplifier of Experiment 2.5
-
2



Fig. 2.5
-
7

The gain of the amplifier in Experiment 2.5
-
2


v
out

v
in

I
DS

Load Line

V
DS

I
-
V Curve



2
-
22

Experiment 2
.5
-
3 An Amplifier with a Higher
V
GS


If

is too high, the transistor may be outside

of saturation and the output
signal will be distorted. The purpose of this experiment is to demonstrate this point.
We raised

from 0.65V to 0.72V. As can be seen,
the transistor is indeed out of
saturation and
the output signal
is

indeed distorted. The testing is shown below.


Fig.2.5
-
8

The amplifier circuit with

for Experiment 2.5
-
3


Table 2.5
-
3

Program for Experiment 2.5
-
3

EX2
-
3

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VDD

1

0

3.3v

R1 1 2 100k

V2 2 0 0v


.param

W1=5u

M1

2

3

0

0


+nch L=0.35u

W='W1' m=1



2
-
23

+AD='0.95u*W1' PD='2*(0.95u+W1)'

+AS='0.95u*W1' PS='2*(0.95u+W1)'


*Fig2.5
-
9
*

VGS_1

3

0

0.
72v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)


*Fig2.5
-
10
*

*VGS_2

3

4

0.72v

*Vin 4 0 sin(0v 0.01v 10Meg)

*.tran 0.1ns 600ns


.end



Fig. 2.5
-
9

The operating points of the amplifier of Experiment 2.5
-
3


I
DS

I
-
V Curve

V
DS

Load Line


2
-
24


Fig. 2.5
-
10

The
distortion

of output signals of the amplifier in Experiment 2.5
-
3


Experiment 2.5
-
4 An Amplifier with a Too Low Bias Voltage

V
GS


In this experiment, we used a low bias voltage. That is, we let

be 0.6V. The
program is in Table 2.5
-
4 and the IV
-
curve is in Fig. 2.5
-
11 and the result is in Fig.
2.5
-
12.
As can be seen, there is no amplification.


Table 2.5
-
4 Program for Experiment 2.5
-
4

EX2
-
4

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TT

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VDD

1

0

3.3v

R1 1 2 100k

V2 2 0 0v



2
-
25

.param

W1=5u

M1

2

3

0

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)'


*Fig2.5
-
11*

VGS_1

3

0

0.6v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)


*Fig2.5
-
1
2
*

*
VGS_2

3

4

0.6v

*
Vin 4 0 sin(0v 0.01v 10Meg)

*
.tran 0.1ns 600ns


.end



Fig. 2.5
-
11

The IV
-
curve of Experiment 2.5
-
4

I
-
V Cruve

Load Line


2
-
26


Fig. 2.5
-
12

The result of Experiment 2.5
-
4


Experiment 2.5
-
5

An Amplifier with a Too L
arge Load



In this experiment, we use a very large load, 2000K. The program is in Table
2.5
-
5, the IV
-
curve is in Fig. 2.5
-
13 and the result is in Fig. 2.5
-
14.

Again, we can
see that
the transistor is out of saturation and
t
here is no amplication.



Table 2.5
-
5 The program with a very large load

EX2
-
5

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VDD

1

0

3.3v

R1 1 2 2000k

V2 2 0 0v


.param

W1=5u

M1

2

3

0

0


+nch L=0.35u

W='W1' m=1


+AD='0.95u*W1' PD='2*(0.95u+W1)'

v
in

v
out


2
-
27

+AS='0.95u*W1' PS='2*(0.95u+W1)'


*Fig2.5
-
1
3
*

VGS
_1

3

0

0.6
5
v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)


*Fig2.5
-
1
4
*

*
VGS_2

3

4

0.65v

*
Vin 4 0 sin(0v 0.01v 10Meg)

*
.tran 0.1ns 6
00ns


.end



Fig. 2.5
-
13

The IV
-
curve for Experiment 2.5
-
5

I
-
V Cruve

Load Line


2
-
28


Fig. 2.5
-
14

The result of Experiment 2.5
-
5


Experiment 2.5
-
6 The Input Output Curve of a PMOS Amplifier



In this experiment, we test a PMOS amplifier whose circuit is shown in Fig.

2.5
-
15.


Fig. 2.5
-
15 The circuit of a PMOS amplifier for Experiment 2.6
-
6.


Table 2.5
-
6 The program for Experiment 2.5
-
6

CWLu20110310HW1

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.unprotect

.op

v
in

v
out


2
-
29

.options nomod post


VDD

1

0

3.3v


M1

2

3

1_1

1_1

PCH

W=5u

L=0.35u



VG

3

0

0v

RL

2

0

100k

Rdm1

1

1_1

0


.DC VG 0 3.3v 0.1v

.end



Fig. 2.5
-
16 T
h
e input output curve of the circuit in Fig. 2.5
-
15


Experiment 2.5
-
7 The A
mplification of
the PMOS Amplifier in Fig. 2
.5
-
15




In this experiment, we shall see whether the PMOS amplifier in Fig. 2.5
-
15
works. The program is in Table 2.5
-
7 and the result is in Fig. 2.5
-
17. It can be seen
Vout

V
GS


2
-
30

that the amplifier works

while the gain is around 9.


Table 2.5
-
7 The program for E
xperiment 2.5
-
7

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VDD

1

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3.3v


M1

2

3

1_1

1_1

PCH

W=5u

L=0.35u



RL

2

0

100k

Rdm1

1

1_1

0


VG

3

4

2.4v

Vin 4 0 sin(0v 0.1v 10k)


.tran 0.1us 1000us

.end














Fig. 2.5
-
17 The result of Experiment 2.5
-
7


Experiment 2.5
-
8 The Input
-
Output Curves for Different Values of Lo
ads

Vout

Vin


2
-
31



In this experiment, we obtained input
-
output curves for different loads. The
program is in Table 2.5
-
8 and the result is i
n Fig. 2.5
-
18.


Table 2.5
-
8

The program for Experiment 2.5
-
8

CWLu20110310HW1

.protect

.lib


\
浭〳㔵0⹬




⹵湰.潴散t

⹯.

⹯灴.潮猠湯o潤⁰潳o


⹰.牡洠m‽‱〰


噄V

1

0

㌮㍶




2

3

ㅟ1

ㅟ1

PCH

W=㕵

L=〮0㕵





3

0





2

0

x

R摭d

1

ㅟ1

0


⹄䌠噇‰″
⸳瘠〮ㅶ⁳睥e瀠p‱〰欠㄰〰欠㌰に


⹥湤








2
-
32


Fig. 2.5
-
18 The input
-
output curves for different loads of a PMOS amplifier


Experiment 2.5
-
9 An Improved PMOS Amplifier.



Based upon the result obtained in Experiment 2.5
-
8, we redesigned the PMOS
a
mplifier by using the bias voltage to be 2.6V and the load to be 1000K. The
program is in Table 2.5
-
9 and the result is in Fig. 2.5
-
19. The gain was increased to
13.


Table 2.5
-
9 The program of an improved PMOS amplifier

CWLu20110317HW2

.protect

.lib 'c
:
\
mm0355v.l' TT

.unprotect

.op

.options nomod post


VDD

1

0

3.3v


M1

2

3

1_1

1_1

PCH

W=5u

L=0.35u


V
GS

V
out

R
L
=100K

RL=1000K


2
-
33


RL

2

0

1000k

Rdm1

1

1_1

0


VG

3

4

2.6v

Vin 4 0 sin(0v 0.1v 10k)


.tran 0.1us 1000us

.end





Fig. 2.5
-
19 The result of the improved P
MOS amplifier


Experiment 2.5
-
10

An Improved
N
MOS Amplifier

by
Changing of the
W
idth
of the
T
ransistor

In this experiment, we used an NMOS with
smaller
width=2u
as shown in Fig.
2.5
-
20. The program is in Table 2.5
-
10, the operating point is shown in Fig.

2.5
-
21
and the input and output voltages are shown in Fig. 2.5
-
22.


v
in

v
out


2
-
34


Fig. 2.5
-
20


Table 2.5
-
10 The program of
Experiment 2.5
-
10

EX2
-
1

.protect

.lib 'D:
\
model
\
tsmc
\
MIXED035
\
mm0355v.l' TT

.unprotect

.op

.options nomod post


VD
D

1

0

3.3v

R1 1 2 100k

*V2 2 0 0v


.param

W1=2u

M1

2

3

0

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)'


*Fig2.5
-
21
*

*VGS_1

3

0

0.65v

*.DC V2 0 3.3v 0.1v

*.PROBE I(R1) I(M1)


*Fig2.5
-
22
*


2
-
35

VGS_2

3

4

0.65v

Vin 4 0 sin(0v 0.01v 10Meg)

.tran 0.1ns 600ns


.end



Fig. 2.5
-
21



Fig. 2.5
-
22



2
-
36

Section 2.6

Some
Guidelines

for Designing MOSFET
Amplifiers

By now, the reader may have some feeling about the designing of an NMOS, or
PM
OS, amplifier. The important thing is to determine an appropriate operating point.
Suppose that we start with the situation as shown Fig. 2.6
-
1.



Fig. 2.6
-
1

The determination of
the
operating point of NMOS


We may either in
crease

which will result in the situation shown in Fig.
2.6
-
2(a) or increase

which will result in the situation in Fig. 2.6
-
2
(b).



Fig. 2.6
-
2

The adjusting
of
parameters of an a
mplifier to obtain appropriate
operating

points


We

must note that an increase of

will result in a larger
. It is not very
desirable

to have a large

as this will increase the power con
sumption.



In general, we would like to have a small

which corresponds to a low
.

2
-
37

Note that

has to be higher than
. Usually,

is not s
ignificantly higher
than
. If

is around 0.55V, it is appropriate to set

to be 0.65V.


`

Let us consider Fig. 2.6
-
3. In this case, assume that we have a rather

large

.
This will not allow

to be large
because

a large

will drive the transistor into
the non
-
saturation region. Thus we must use a smaller load if

is large, which is
not
good
.



Fig. 2.6
-
3

The
problem

of using a large


The reader is also encouraged to take a look at Fig. 2.2
-
5 to see why

cannot
be too large.




There is one comprehensive way to tackle the d
esign of

NMOS and PMOS
amplifiers: to perform an input/output analysis. This allows us to select an
appropriate load resistor and an appropriate bias voltage.



In any case,

this kind of amplifiers suffer from

one drawback. We like

to
be large. Yet,

cannot be too large because a large

will often drive the
transistor out of saturation. In the next chapter, we shall introduce CMOS circuits
which will avoid this problem



We should n
ote that by changing the width of a transistor, we will also change
the performance of an
amplifier
.



In conclusion, the following parameters must be appropriately set: the load

of the amplifier
,

the width of the transistor and the

gate bias voltage.



2
-
38

Section 2.7
Exercise 2


1.

Explain why the following is an amplifier

in Fig. 2.7
-
1
.



Fig. 2.7
-
1 The figure for Problem 1 in Exercise 2


2.

In the above circuit, explain why you cannot use a very small load?

3.

In the above circuit, explain why you cannot use a very large load?

4.

Consider the

vs

curve.

(a)

Do you want a curve which is rather sharp or a curve which is rather flat?
Why?

(b)

How do you achieve a sharp

vs

curve?

5.

Derive the following formula:
.

6.

Explain why we need to short circuit all constant voltage sources in the small
signal
equivalent

circuit of an amplifier.

7.

Consider
F
ig
. 2.7
-
2
.



2
-
39


Fig. 2.7
-
2

The figures for Problem 7 of Exercise 2


(a)

Explain the meaning of
.

(b)

Explain why

can be usually ignored.

(c)

Explain how this equivalent circuit is obtained.

(d)

Derive the
following

formula:
.

(e)

Explain the meaning of the negative sign.

8.

In general, what is the value of
?