Neamen D. Microelectronics: Circuit Analysis and Design
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258 Part 1 Semiconductor Devices and Basic Applications
*Test Your Understanding
TYU 4.12 The transistor parameters of the circuit in Figure 4.49 are
V
TN1
=
V
TN2
= 0.6
V,
K
n1
= 1.5
mA/V
2
,
K
n2
= 2
mA/V
2
, and
λ
1
= λ
2
= 0
. (a) Find
I
DQ1
,
I
DQ2
,
V
DSQ1
, and
V
DSQ2
. (b) Determine the small-signal voltage gain. (c) Find the
output resistance
R
o
. (Ans. (a)
I
DQ1
= 0.3845
mA,
I
DQ2
= 0.349
mA,
V
DSQ1
=
2.31
V,
V
DSQ2
= 7.21
V; (b)
A
v
=−20.3
; (c)
R
o
= 402 )
4.9 BASIC JFET AMPLIFIERS
Objective: • Develop the small-signal model of JFET devices and
analyze basic JFET amplifiers.
Like MOSFETs, JFETs can be used to amplify small time-varying signals. Initially,
we will develop the small-signal model and equivalent circuit of the JFET. We will
then use the model in the analysis of JFET amplifiers.
Small-Signal Equivalent Circuit
Figure 4.53 shows a JFET circuit with a time-varying signal applied to the gate. The
instantaneous gate-to-source voltage is
v
GS
= V
GS
+v
i
= V
GS
+v
gs
(4.53)
where v
gs
is the small-signal gate-to-source voltage. Assuming the transistor is biased
in the saturation region, the instantaneous drain current is
i
D
= I
DSS
1 −
v
GS
V
P
2
(4.54)
where
I
DSS
is the saturation current and V
P
is the pinchoff voltage. Substituting
Equation (4.53) into (4.54), we obtain
i
D
= I
DSS
1 −
V
GS
V
P
−
v
gs
V
P
2
(4.55)
If we expand the squared term, we have
i
D
= I
DSS
1 −
V
GS
V
P
2
−2I
DSS
1 −
V
GS
V
P
v
gs
V
P
+ I
DSS
v
gs
V
P
2
(4.56)
The first term in Equation (4.56) is the dc or quiescent drain current I
DQ
, the second
term is the time-varying drain current component, which is linearly related to the signal
voltage v
gs
, and the third term is proportional to the square of the signal voltage. As in
the case of the MOSFET, the third term produces a nonlinear distortion in the output
current. To minimize this distortion, we will usually impose the following condition:
v
gs
V
P
2
1 −
V
GS
V
P
(4.57)
4.9.1
V
DD
i
D
i
G
+
–
v
i
+
–
V
GS
R
D
v
DS
+
–
v
GS
+
–
v
O
Figure 4.53 JFET common-
source circuit with time-
varying signal source in
series with gate dc source
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Chapter 4 Basic FET Amplifiers 259
Equation (4.57) represents the small-signal condition that must be satisfied for JFET
amplifiers to be linear.
Neglecting the term
v
2
gs
in Equation (4.56), we can write
i
D
= I
DQ
+i
d
(4.58)
where the time-varying signal current is
i
d
=+
2I
DSS
(−V
P
)
1 −
V
GS
V
P
v
gs
(4.59)
The constant relating the small-signal drain current and small-signal gate-to-source
voltage is the transconductance g
m
. We can write
i
d
= g
m
v
gs
(4.60)
where
g
m
=+
2I
DSS
(−V
P
)
1 −
V
GS
V
P
(4.61)
Since V
P
is negative for n-channel JFETs, the transconductance is positive. A rela-
tionship that applies to both n-channel and p-channel JFETs is
g
m
=
2I
DSS
|
V
P
|
1 −
V
GS
V
P
(4.62)
We can also obtain the transconductance from
g
m
=
∂i
D
∂v
GS
v
GS
=V
GSQ
(4.63)
Since the transconductance is directly proportional to the saturation current
I
DSS
, the transconductance is also a function of the width-to-length ratio of the
transistor.
Since we are looking into a reverse-biased pn junction, we assume that the input
gate current i
g
is zero, which means that the small-signal input resistance is infinite.
Equation (4.54) can be expanded to take into account the finite output resistance of a
JFET biased in the saturation region. The equation becomes
i
D
= I
DSS
1 −
v
GS
V
P
2
(1 + λv
DS
)
(4.64)
The small-signal output resistance is
r
o
=
∂i
D
∂v
DS
−1
v
GS
=const.
(4.65)
Using Equation (4.64), we obtain
r
o
=
λI
DSS
1 −
V
GS
V
P
2
−1
(4.66(a))
or
r
o
∼
=
[λI
DQ
]
−1
=
1
λI
DQ
(4.66(b))
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260 Part 1 Semiconductor Devices and Basic Applications
The small-signal equivalent circuit of the n-channel JFET, shown in Figure 4.54,
is exactly the same as that of the n-channel MOSFET. The small-signal equivalent
circuit of the p-channel JFET is also the same as that of the p-channel MOSFET.
However, the polarity of the controlling gate-to-source voltage and the direction of
the dependent current source are reversed from those of the n-channel device.
Small-Signal Analysis
Since the small-signal equivalent circuit of the JFET is the same as that of the MOS-
FET, the small-signal analyses of the two types of circuits are identical. For illustra-
tion purposes, we will analyze two JFET circuits.
EXAMPLE 4.18
Objective: Determine the small-signal voltage gain of a JFET amplifier.
Consider the circuit shown in Figure 4.55 with transistor parameters
I
DSS
=
12 mA,
V
P
=−4
V, and
λ = 0.008 V
−1
. Determine the small-signal voltage gain
A
v
= v
o
/v
i
.
Solution: The dc quiescent gate-to-source voltage is determined from
V
GSQ
=
R
2
R
1
+ R
2
V
DD
− I
DQ
R
S
where
I
DQ
= I
DSS
1 −
V
GSQ
V
P
2
Combining these two equations produces
V
GSQ
=
180
180 + 420
(20) − (12)(2.7)
1 −
V
GSQ
(−4)
2
which reduces to
2.025V
2
GSQ
+17.25V
GSQ
+26.4 = 0
4.9.2
V
gs
g
m
V
gs
r
o
–
+
GD
S
Figure 4.54 Small-signal
equivalent circuit of
n-channel JFET
v
i
C
C1
C
S
v
o
R
L
= 4 kΩ
R
2
= 180 kΩ
R
S
= 2.7 kΩ
R
1
= 420 kΩ
R
D
= 2.7 kΩ
C
C 2
V
DD
= 20 V
+
–
Figure 4.55 Common-source JFET circuit with source resistor and source bypass capacitor
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Chapter 4 Basic FET Amplifiers 261
The appropriate solution is
V
GSQ
=−2.0V
The quiescent drain current is
I
DQ
= I
DSS
1 −
V
GSQ
V
P
2
= (12)
1 −
(−2.0)
(−4)
2
= 3.00 mA
The small-signal parameters are then
g
m
=
2I
DSS
(−V
P
)
1 −
V
GS
V
P
=
2(12)
(4)
1 −
(−2.0)
(−4)
= 3.00 mA/V
and
r
o
=
1
λI
DQ
=
1
(0.008)(3.00)
= 41.7k
The small-signal equivalent circuit is shown in Figure 4.56.
Since
V
gs
= V
i
, the small-signal voltage gain is
A
v
=
V
o
V
i
=−g
m
(r
o
R
D
R
L
)
or
A
v
=−(3.0)(41.72.74) =−4.66
V
o
g
m
V
gs
V
gs
–
+
V
i
r
o
R
1
⎪⎪ R
2
+
–
R
D
R
L
Figure 4.56 Small-signal equivalent circuit of common-source JFET, assuming bypass
capacitor acts as a short circuit
Comment: The voltage gain of JFET amplifiers is the same order of magnitude as
that of MOSFET amplifiers.
EXERCISE PROBLEM
Ex 4.18: For the JFET amplifier shown in Figure 4.55, the transistor parame-
ters are:
I
DSS
= 4
mA,
V
P
=−3
V, and
λ = 0.005 V
−1
. Let
R
L
= 4
k,
R
S
=
2.7 k, and
R
1
+ R
2
= 500
k. Redesign the circuit such that
I
DQ
=
1.2 mA and
V
DSQ
= 12
V. Calculate the small-signal voltage gain. (Ans.
R
D
=
3.97 k,
R
1
= 453
k,
R
2
= 47
k,
A
v
=−2.87
)
DESIGN EXAMPLE 4.19
Objective: Design a JFET source-follower circuit with a specified small-signal volt-
age gain.
For the source-follower circuit shown in Figure 4.57, the transistor parameters
are:
I
DSS
= 12
mA,
V
P
=−4
V, and
λ = 0.01 V
−1
. Determine R
S
and
I
DQ
such that
the small-signal voltage gain is at least
A
v
= v
o
/v
i
= 0.90
.
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262 Part 1 Semiconductor Devices and Basic Applications
Solution: The small-signal equivalent circuit is shown in Figure 4.58. The output
voltage is
V
o
= g
m
V
gs
(R
S
R
L
r
o
)
Also
V
i
= V
gs
+ V
o
or
V
gs
= V
i
− V
o
Therefore, the output voltage is
V
o
= g
m
(V
i
− V
o
)(R
S
R
L
r
o
)
The small-signal voltage gain becomes
A
v
=
V
o
V
i
=
g
m
(R
S
R
L
r
o
)
1 + g
m
(R
S
R
L
r
o
)
As a first approximation, assume r
o
is sufficiently large for the effect of r
o
to be
neglected.
The transconductance is
g
m
=
2I
DSS
(−V
P
)
1 −
V
GS
V
P
=
2(12)
4
1 −
V
GS
(−4)
If we pick a nominal transconductance value of
g
m
= 2
mA/V, then
V
GS
=−2.67
V
and the quiescent drain current is
I
DQ
= I
DSS
1 −
V
GS
V
P
2
= (12)
1 −
(−2.67)
(−4)
2
= 1.335 mA
The value of R
S
is then determined from
R
S
=
−V
GS
−(−10)
I
DQ
=
2.67 + 10
1.335
= 9.49 k
Also, the value of r
o
is
r
o
=
1
λI
DQ
=
1
(0.01)(1.335)
= 74.9k
v
i
C
C1
v
o
R
L
= 10 kΩ
R
G
= 50 kΩ
+10 V
–10 V
R
S
C
C2
+
–
Figure 4.57 JFET source-follower circuit
V
gs
g
mVgs
r
o
–
+
V
o
R
S
R
L
V
i
R
G
+
–
Figure 4.58 Small-signal equivalent
circuit of JFET source-follower circuit
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Chapter 4 Basic FET Amplifiers 263
The small-signal voltage gain, including the effect of r
o
,is
A
v
=
g
m
(R
S
R
L
r
o
)
1 + g
m
(R
S
R
L
r
o
)
=
(2)(9.491074.9)
1 + (2)(9.491074.9)
= 0.902
Comment: This particular design meets the design criteria, but the solution is not
unique.
EXERCISE PROBLEM
Ex 4.19: Reconsider the source-follower circuit shown in Figure 4.57 with tran-
sistor parameters
I
DSS
= 8
mA,
V
P
=−3.5
V, and
λ = 0.01 V
−1
. (a) Design the
circuit such that
I
DQ
= 2
mA. (b) Calculate the small-signal voltage gain if R
L
approaches infinity. (c) Determine the value of R
L
at which the small-signal
gain is reduced by 20 percent from its value for (b). (Ans. (a)
R
S
= 5.88
k,
(b)
A
v
= 0.923
,
R
L
= 1.61
k)
In Example 4.19, we chose a value of transconductance and continued through the
design. A more detailed examination shows that both g
m
and R
S
depend upon the drain
current I
DQ
in such a way that the product g
m
R
S
is approximately a constant. This means
the small-signal voltage gain is insensitive to the initial value of the transconductance.
Test Your Understanding
TYU 4.13 Reconsider the JFET amplifier shown in Figure 4.55 with transistor para-
meters given in Example 4.l8. Determine the small-signal voltage gain if a 20 k
resistor is in series with the signal source v
i
. (Ans.
A
v
=−3.98
)
*TYU 4.14 For the circuit shown in Figure 4.59, the transistor parameters are:
I
DSS
= 6
mA,
|V
P
|=2
V, and
λ = 0
. (a) Calculate the quiescent drain current and
drain-to-source voltage of each transistor. (b) Determine the overall small-signal
voltage gain
A
v
= v
o
/v
i
. (Ans. (a)
I
DQ1
= 1
mA,
V
SDQ1
= 12
V,
I
DQ2
= 1.27
mA,
V
DSQ2
= 14.9
V; (b)
A
v
=−2.05
)
v
i
C
S
Q
1
v
o
R
L
= 2 kΩ
R
2
= 430 kΩ
R
1
= 70 kΩ
R
S1
= 4 kΩ
R
S2
= 4 kΩR
D1
= 4 kΩ
C
C2
C
C1
V
DD
= 20 V
Q
2
+
–
Figure 4.59 Figure for Exercise TYU 4.14
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264 Part 1 Semiconductor Devices and Basic Applications
4.10 DESIGN APPLICATION:
A TWO-STAGE AMPLIFIER
Objective: • Design a two-stage MOSFET circuit to amplify the
output of a sensor.
Specifications: Assume the resistance R
2
in the voltage divider circuit in Figure 4.60
varies linearly as a function of temperature, pressure, or some other variable. The
output of the amplifier is to be zero volts when
δ = 0
.
Design Approach: The amplifier configuration to be designed is shown in Fig-
ure 4.60. A resistor R
1
will be chosen such that the voltage divider between R
1
and R
2
will produce a dc voltage v
I
that is negative. A negative gate voltage to M
1
then means
that the resistance R
S1
does not need to be so large.
Choices: Assume NMOS and PMOS transistors are available with parameters
V
TN
= 1
V,
V
TP
=−1
V,
K
n
= K
p
= 2
mA/V
2
,and
λ
n
= λ
p
∼
=
0
.
Solution (voltage divider analysis): The voltage v
I
can be written as
v
I
=
R(1 +δ)
R(1 +δ) +3R
(10) − 5 =
(1 + δ)(10)
4 + δ
−5
or
v
I
=
(1 + δ)(10) − 5(4 +δ)
4 + δ
=
−10 + 5δ
4 + δ
Assuming that
δ 4
, we then have
v
I
=−2.5 + 1.25δ
R
D1
R
S1
R
S2
R
D2
v
I
v
o
M
2
M
1
R
1
= 3 R
R
2
= R (1 + d )
V
+
= 5 V
V
+
= 5 V
V
–
= –5 V
V
–
= –5 V
Figure 4.60 Two-stage MOSFET amplifier for design application
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Chapter 4 Basic FET Amplifiers 265
Solution (DC Design): We will choose
I
D1
= 0.5
mA and
I
D2
= 1
mA. The gate-
to-source voltages are determined to be:
0.5 = 2(V
GS1
−1)
2
⇒ V
GS1
= 1.5V
and
1 = 2(V
SG2
−1)
2
⇒ V
SG2
= 1.707 V
We find
V
S1
= V
I
− V
GS1
=−2.5 − 1.5 =−4V
. The resistor R
S1
is then
R
S1
=
V
S1
− V
−
I
D1
=
−4 − (−5)
0.5
= 2k
Letting
V
D1
= 1.5
V, we find the resistor
R
D1
to be
R
D1
=
V
+
− V
D1
I
D1
=
5 − 1.5
0.5
= 7k
We have
V
S2
= V
D1
+ V
SG2
= 1.5 +1.707 = 3.207 V
. Then
R
S2
=
V
+
− V
S2
I
D2
=
5 − 3.207
1
= 1.79 k
For
V
O
= 0
, we find
R
D2
=
V
O
− V
−
I
D2
=
0 − (−5)
1
= 5k
Solution (ac Analysis): The small-signal equivalent circuit is shown in Figure 4.61.
We find
V
2
=−g
m1
V
gs1
R
D1
and
V
gs1
= V
i
/(1 + g
m1
R
S1
)
. We also find
V
o
=
g
m2
V
sg2
R
D2
and
V
sg2
=−V
2
/(1 + g
m2
R
S2
)
. Combining terms, we find
V
o
=
g
m1
g
m2
R
D1
R
D2
(1 + g
m1
R
S1
)(1 + g
m2
R
S2
)
V
i
The ac input signal is
V
i
= 1.25 δ
, so we have
V
o
=
(1.25)g
m1
g
m2
R
D1
R
D2
(1 + g
m1
R
S1
)(1 + g
m2
R
S2
)
δ
+
–
–
–
+
+
V
gs1
V
sg2
R
S1
R
S2
R
D2
R
D1
V
2
V
o
V
i
g
m1
V
gs1
g
m2
V
sg2
Figure 4.61 Small-signal equivalent circuit of two-stage MOSFET amplifier for design
application
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266 Part 1 Semiconductor Devices and Basic Applications
We find that
g
m1
= 2
K
n
I
D1
= 2
(2)(0.5) = 2mA/V
and
g
m2
= 2
K
p
I
D2
= 2
(2)(1) = 2.828 mA/V
We then find
V
o
=
(1.25)(2)(2.828)(7)(5)
[1 + (2)(2)][1 +(2.828)(1.79)]
δ
or
V
o
= 8.16δ
Comment: Since the low-frequency input impedance to the gate of the NMOS is
essentially infinite, there is no loading effect on the voltage divider circuit.
Design Pointer: As mentioned previously, by choosing the value of R
1
to be larger
than R
2
, the dc voltage to the gate of M
1
is negative. A negative gate voltage implies
that the required value of R
S1
is reduced and can still establish the required current.
Since the drain voltage at M
1
is positive, then by using a PMOS transistor in the
second stage, the source resistor value of R
S2
is also reduced. Smaller source
resistances generate larger voltage gains.
4.11 SUMMARY
• The application of MOSFET transistors in linear amplifiers was emphasized in
this chapter. The basic process by which a transistor circuit can amplify a small
time-varying input signal was discussed.
• A small-signal equivalent circuit for the MOSFET transistor, which is used in
the analysis and design of linear amplifiers, was developed.
• The three basic amplifier configurations were considered: the common-source,
source-follower, and common-gate. These three circuits form the basic building
blocks for more complex integrated circuits.
• The common-source circuit amplifies a time-varying voltage.
• The small-signal voltage gain of a source-follower circuit is approximately
unity, but has a low output resistance.
• The common-gate circuit amplifies a time-varying voltage, and has a low input
resistance and a large output resistance.
• Analysis of n-channel circuits with enhancement-load or depletion-load devices
was performed. In addition, the analysis of CMOS circuits was carried out.
These circuits are examples of all MOSFET circuits, which are developed
throughout the remainder of the text.
• The small-signal equivalent circuit of a JFET was developed and used in the
analysis of several configurations of JFET amplifiers.
• As an application, MOS transistors were incorporated in the design of a two-
stage amplifier.
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Chapter 4 Basic FET Amplifiers 267
CHECKPOINT
After studying this chapter, the reader should have the ability to:
✓ Explain graphically the amplification process in a simple MOSFET amplifier
circuit.
✓ Describe the small-signal equivalent circuit of the MOSFET and to determine
the values of the small-signal parameters.
✓ Apply the small-signal equivalent circuit to various MOSFET amplifier circuits
to obtain the time-varying circuit characteristics.
✓ Characterize the small-signal voltage gain and output resistance of the common-
source, source-follower, and common-gate amplifiers.
✓ Describe the operation of an NMOS amplifier with either an enhancement load,
a depletion load, or a PMOS load.
✓ Apply the MOSFET small-signal equivalent circuit in the analysis of multistage
amplifier circuits.
✓ Describe the operation and analyze basic JFET amplifier circuits.
REVIEW QUESTIONS
1. Discuss, using the concept of a load line, how a simple common-source circuit
can amplify a time-varying signal.
2. How does the transistor width-to-length ratio affect the small-signal voltage gain
of a common-source amplifier?
3. Discuss the physical meaning of the small-signal circuit parameter r
o
.
4. How does the body effect change the small-signal equivalent circuit of the
MOSFET?
5. Sketch a simple common-source amplifier circuit and discuss the general ac
circuit characteristics (voltage gain and output resistance).
6. Discuss the general conditions under which a common-source amplifier would
be used.
7. Why, in general, is the magnitude of the voltage gain of a common-source
amplifier relatively small?
8. What are the changes in dc and ac characteristics of a common-source amplifier
when a source resistor and a source bypass capacitor are incorporated in the
design?
9. Sketch a simple source-follower amplifier circuit and discuss the general ac
circuit characteristics (voltage gain and output resistance).
10. Sketch a simple common-gate amplifier circuit and discuss the general ac circuit
characteristics (voltage gain and output resistance).
11. Discuss the general conditions under which a source-follower or a common-gate
amplifier would be used.
12. Compare the ac circuit characteristics of the common-source, source-follower,
and common-gate circuits.
13. State the advantage of using transistors in place of resistors in MOSFET inte-
grated circuits.
14. State at least two reasons why a multistage amplifier circuit would be required in
a design compared to using a single-stage circuit.
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