3 0 2
CHAPTER 4
)
TA 3 LE 4.3
M O S FIELD-EFFECT TRANSISTORS
(MOSFETs)
Characteristic Parameters of Amplifiers
4.7
SINGLE-STAGE MOS
AMPLIFIERS
Equivalent Circuits
B
Circuit
A:
A'
WV — o -
+
©
v
sis
Definitions
g
-o
sig
Input resistance with no load:
ff
"out
R, =
=
—
o-
-AAA,
Output resistance:
^
©
•
<;
•
Input resistance:
K
"in =
V
>
-
h
Open-circuit voltage gain:
E>
A' .
sig
V,
Rr
Voltage gain:
v^ = 0
©
Open-circuit overall voltage gain:
G
Short-circuit current gain:
A
is
-
wv
-AAA,—c-
R,
v„
=
^ vo —
Overall voltage gain:
~
Relationships
Current gain:
v«
0
R
h
v
Avo
R,+R
Short-circuit transconductance:
G
m
= L
Vi
n
A
R,=0
=
C
7?
vo
sig
L
0
Rr
G„ = G„
o
R,
R + R
R<
.
R; + 7?.;
G
A
v c
+R
ia
Output resistance of amplifier proper:
5. W h e n evaluating the gain A from t h e open-circuit value A , R is t h e output resis­
t a n c e to u s e . This is b e c a u s e A is b a s e d o n feeding t h e amplifier with an ideal voltage
signal v . T h i s should b e evident from E q u i v a l e n t Circuit A in T a b l e 4 . 3 . O n the other
hand, if w e are evaluating the overall voltage gain G from its open-circuit value G ,
the output resistance to u s e is R . This is b e c a u s e G is b a s e d o n feeding the a m p l i ­
fier w i t h v , w h i c h h a s an internal resistance 7 \ . This should b e evident from
E q u i v a l e n t Circuit C in T a b l e 4 . 3 .
v
v0
0
v
t
v
out
sig
v0
v
sig
6. W e u r g e t h e r e a d e r to carefully e x a m i n e and reflect o n t h e definitions and the six
relationships p r e s e n t e d in T a b l e 4 . 3 . E x a m p l e 4.11 should h e l p in this regard.
303
304
CHAPTER 4
MOS FIELD-EFFECT TRANSISTORS
4.7
(MOSFETs)
SINGLE-STAGE MOS AMPLIFIERS
resulting in
iv
= 2.86 k Q
out
The value of R can be determined from
in
A transistor amplifier is fed with a signal source having an open-circuit voltage v of 10 m V and
an internal resistance i ? of 100 k Q . The voltage v at the amplifier input and the output voltage
v are measured both without and with a load resistance R = 10 k Q connected to the amplifier
output. The measured results are as follows:
sig
sig
v,
R
t
0
v
ûg
L
v,(mV)
| Without Ä
70 !
m
10
R
+
in
100
which yields
J With R connected
= 400 k Q
R.
L
n
Find all the amplifier parameters.
+ ^SIG
Thus,
v.. (mV)
t
R
The short-circuit transconductance G can be found.as follows:
m
Solution
=1.4—
3 = 7 mA/V
G„ = ^¡2
i?
m
0
First, we use the data obtained for R = °° to determine
L
and the current gain A can be determined as follows:
90
Ko = j
t
and
= io v / v
A
VQ/RL
=
'
90
G„„ —
10
9 V/V
^>Rm
=
VR
v, R
in
L
= a"R = 8.7510x — = 350 A / A
Now, since
L
Finally, w e determine the short-circuit current gain A
is
G„„
Table 4.3,
R, + R, G
9 =
x 10
R,- + 100 '
as follows. F r o m Equivalent Circuit A in
the short-circuit output current is
However, to determine v w e need to know the value of R
t
which gives
in
obtained with R = 0. Toward that
L
end, note that from Equivalent Circuit C, the output short-circuit current can be found as
Ri = 900 k Q
iosc ~
Next, we use the data obtained when R = 10 k Q is connected to the amplifier output to determine
G V ig/R x
v0
s
ou
L
Now, equating the two expressions for i
and substituting for G
osc
8.75 V / V
vo
G„
and
70
= ^ = 7 V/V
G
v
The values of A and A
v
vo
4
B
-
v
Rr
A
.
and for v from
can be used to determine R as follows:
A
R:
-
n
Vi
=
D I
v
^r~\
A
I N A . =Tr~
0
results in
t
IN J
r +r„
l
10
S.75 = 10W + R
-o
SIG
si
^'m\R
=o £
0
:81.8kQ
which gives
R=
0
Similarly, we use the values of G and G
v
vo
1.43 k Q
W e now can use
to determine R
om
Rr
R
from
^vo^i ^INJFI
^'OSC
=Q^^o
to obtain
R
L + out
10
7 = 910 + Ä„
A
= W
=
1 0
x
8 1
. g / i .43
=
572 A / A
by
306
$ËjS CHAPTER
4
4.7
M O S FIELD-EFFECT TRANSISTORS (MOSFETs)
SINGLE-STAGE MOS AMPLIFIERS
DD
A
4.31 (a) If in the amplifier of Example 4 . 1 1 , W , is doubled, find the values for A'.. ? . and R .
for R doubled (but ^ unchanged; i.e., 100 k Q ) . (c) R e p e a t for both 7? and R doubled.
si;
L
(b) Repeat
mt
s i g
sig
L
Arts, (a) 400 k Q , 5.83 V / V , 4.03 k Q ; (b) 538 k Q , 7.87 V / V , 2.86 k Q ; (c) 538 k Q , 6.8 V / V , 4.03 k Q
p
c
cl
«sig
4.7.3 The Common-Source (CS) Amplifier
-AAA,
T h e c o m m o n - s o u r c e (CS) or g r o u n d e d - s o u r c e configuration is the m o s t widely u s e d of all
M O S F E T amplifier circuits. A c o m m o n - s o u r c e amplifier realized using the circuit of
Fig. 4.42 is s h o w n in Fig. 4.43(a). O b s e r v e that to establish a signal g r o u n d , or an ac g r o u n d
as it is s o m e t i m e s called, at the source, w e h a v e connected a large capacitor, C , b e t w e e n the
source and ground. This capacitor, usually in the pF range, is required to provide a very small
i m p e d a n c e (ideally, zero i m p e d a n c e ; i.e., in effect, a short circuit) at all signal frequencies of
interest. In this w a y , the signal current passes through C to g r o u n d and thus bypasses the out­
put resistance of current source I (and any other circuit c o m p o n e n t that m i g h t b e connected to
the M O S F E T source); hence, C is called a b y p a s s capacitor. Obviously, the l o w e r the sig­
nal frequency, the less effective the b y p a s s capacitor b e c o m e s . This issue will b e studied in
Section 4.9. F o r our purposes h e r e w e shall a s s u m e that C is acting as a perfect short circuit
and thus is establishing a zero signal voltage at t h e M O S F E T source.
o—
1 '© I
s
s
(a)
s
i,
s
In order n o t to disturb the dc bias current and voltages, t h e signal to be amplified, s h o w n
as v o l t a g e source v with an internal resistance R , is c o n n e c t e d to t h e gate t h r o u g h a large
capacitor C . Capacitor C , k n o w n as a c o u p l i n g c a p a c i t o r , is required to act as a perfect
short circuit at all signal frequencies of interest w h i l e b l o c k i n g d c . H e r e again, w e n o t e that
as the signal frequency is l o w e r e d , the i m p e d a n c e of C (i.e., l/j(0C )
will increase and
its effectiveness as a coupling capacitor will b e c o r r e s p o n d i n g l y r e d u c e d . T h i s p r o b l e m too
will b e c o n s i d e r e d in Section 4.9 w h e n t h e d e p e n d e n c e of t h e amplifier o p e r a t i o n o n fre­
q u e n c y is studied. F o r our p u r p o s e s h e r e w e shall a s s u m e C is acting as a perfect short
circuit as far as t h e signal is c o n c e r n e d . B e f o r e l e a v i n g C , w e s h o u l d p o i n t out that in sit­
uations w h e r e t h e signal source c a n p r o v i d e an a p p r o p r i a t e dc p a t h to g r o u n d , t h e gate c a n
b e c o n n e c t e d directly to the signal s o u r c e a n d b o t h R a n d C c a n b e d i s p e n s e d with.
$ig
sia
C 1
cl
cl
-AAA,
K= 0
ch
+
©
v
'
D
-o-
G
-o—
- O V„
'Rn
>R .
r
cl
cl
(b)
C 1
G
cl
T h e v o l t a g e signal resulting at the drain is c o u p l e d to t h e l o a d resistance R via another
coupling capacitor C . W e shall a s s u m e that C acts as a perfect short circuit at all signal
frequencies of interest and thus that t h e o u t p u t v o l t a g e v = v . N o t e that R can b e either an
actual load resistor, to w h i c h the amplifier is r e q u i r e d to p r o v i d e its output v o l t a g e signal, or
it can b e t h e i n p u t resistance of another amplifier stage in cases w h e r e m o r e than o n e stage
of amplification is needed. ( W e will study m u l t i s t a g e amplifiers in C h a p t e r 7.)
L
C2
C2
a
d
L
T o d e t e r m i n e the t e r m i n a l characteristics of t h e C S a m p l i f i e r — t h a t is, its i n p u t resis­
t a n c e , v o l t a g e gain, a n d o u t p u t r e s i s t a n c e — w e r e p l a c e t h e M O S F E T with its small-signal
m o d e l . T h e resulting circuit is s h o w n in F i g . 4 . 4 3 ( b ) . A t the o u t s e t w e o b s e r v e that this
amplifier is unilateral. T h e r e f o r e R d o e s n o t d e p e n d o n R , a n d t h u s R ^ R . A l s o , R
will not d e p e n d o n R , and thus R
= R . A n a l y s i s of this circuit is straightforward and
p r o c e e d s in a step-by-step m a n n e r , from t h e signal source to the amplifier load. A t t h e input
iB
sig
oat
L
in
t
oat
0
i
g
= 0
%=
R*.
-AAAr
h
-g ^(.r \\R \\R )
m
0
D
L
' Rj
©
(c)
(4.78)
FIGURE 4 . 4 3 (a) Common-source amplifier based on the circuit of Fig. 4.42. (b) Equivalent circuit of the
amplifier for small-signal analysis, (c) Small-signal analysis performed directly on the amplifier circuit with
the MOSFET model implicitly utilized.
J
3 0 7