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Electronic Devices
Ninth Edition
Floyd
Chapter 9
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
In a CS amplifier, the input
signal is applied to the gate
and the output signal is taken
from the drain. The amplifier
has higher input resistance
and lower gain than the
equivalent CE amplifier.
+VDD
RD
C3
Vout
C1
RL
Vin
RG
RS
C2
The voltage gain is given by the equation Av = gmRd.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
Recall that conductance is the reciprocal of resistance and admittance
is the reciprocal of impedance. Data sheets typically specify the
forward transfer admittance, yfs rather than transconductance, gm. The
definition of yfs is
I
y fs 
DYNAMIC CHARACTERISTICS
Forward Transfer Admittance
(VDS = 15 Vdc, VGS = 0)
D
VG
Symbol
2N5457
2N5458
|Yfs|
Min
Typ
1000
1500
3000
4000
Max
5000
5500
Unit
m mhos
An alternate gain expression for a CS amplifier is Av = yfsRd.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
You can estimate what the transfer
characteristic looks like from values on
the specification sheet, but keep in mind
that large variations are common with
JFETs. For example, the range of
specified values for a 2N5458 is shown.
OFF CHARACTERISTICS
Gate-Source Cutoff Voltage
(VDS = 15 Vdc, iD = 10 nAdc)
Electronic Devices, 9th edition
Thomas L. Floyd
9
2
– VGS (V)
Symbol
2N5457
2N5458
ON CHARACTERISTICS
Zero Gate-Source Drain Current
(VDS = 15 Vdc, VGS = 0)
ID (mA)
2N5457
2N5458
V GS(off)
Min
-0.5
-1.0
–7
–1
Typ
-
Max
-6.0
-7.0
Symbol
Min
Typ
Max
I DSS
1.0
2.0
3.0
6.0
5.0
9.0
0
Unit
Vdc
Unit
mAdc
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
To analyze the CS amplifier. you need to start with dc values. It is useful
to estimate ID based on typical values; specific circuits will vary from
this estimate.
VDD
+12 V
For a typical 2N5458, what is
the drain current?
RD
2.7 kW
Vout
C1
2N5458
From the specification sheet, the
typical IDSS = 6.0 mA and VGS(off)
= -4 V. These values can be
plotted along with the load line
to obtain a graphical solution.
Electronic Devices, 9th edition
Thomas L. Floyd
0.1 mF
Vin
100 mV
RG
10 MW
RS
470 W
C2
10 mF
See the following slide…
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
(continued)
A graphical solution is illustrated.
On the transconductance curve,
plot the load line for the source
resistor.
Load line for 470 W resistor
ID (mA)
6
Then read the current and
voltage at the Q-point.
ID = 2.8 mA and
VGS = -1.3 V
Electronic Devices, 9th edition
Thomas L. Floyd
Q
– VGS (V)
–4
-1.3 V
2.8 mA
0
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
(continued)

I D RS 

Alternatively, you can obtain ID using Equation 9-2: I D  I DSS 1 V
GS(off) 

2
The solution to this quadratic equation is simplified using a calculator
that can handle quadratic equations.
ID=IDSS (1–(–ID RS/VG...
After entering the equation, enter the
ID= .0027494671581759
IDSS= .006
known values, but leave ID open.
RS= 470
For the typical values for the 2N5458,
enter absolute
VGSOFF= 4.0
value
(IDSS = 6 mA and VGS(off) = -4 V) with
bound=(–1E 99,1E 99)
a source resistance of 470 W, we find
GRAPH RANGE ZOOM TRACE SOLVE
2.75 mA.
press F5
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
Assume IDSS is 6.0 mA, VGS(off) is -4 V, and VGS = -1.3 V as
found previously. What is the expected gain?
Output is
inverted
gm0 
VDD
+12 V
2  6.0 mA 
2 I DSS

 3.0 mS
4
V
VGS(off)

V 
g m  g m 0 1 - GS 
 V

GS(off) 

 -1.3 V 
 3.0 mS 1 
 -4.0 V 
2.02 mS
RD
2.7 kW
Vout
C1
2N5458
0.1 mF
Vin
100 mV
RG
10 MW
RS
470 W
C2
10 mF
Av = gmRd = (2.02 mS)(2.7 kW) = 5.45
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Source Amplifier
The gain is reduced when a load is connected to the amplifier because
the total ac drain resistance (Rd) is reduced.
VDD
+12 V
How does the addition of the
10 kW load affect the gain?
RD
2.7 kW
Rd 
RD RL
RD  RL
 2.7 kW 10 kW 

2.7 kW  10 kW
 2.13 kW
Vout
C1
2N5458
0.1 mF
Vin
100 mV
RG
10 MW
RS
470 W
C2
10 mF
RL
10 kW
Av = gmRd = (2.02 mS)(2.13 kW) = 4.29
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The D-MOSFET
In operation, the D-MOSFET has the unique property in that it can be
operated with zero bias, allowing the signal to swing above and below
ground. This means that it can operate in either D-mode or E-mode.
En
ha
nc
em
ent
ID
+VDD
Q
RD C2
Vout
tio
le
p
e
D
C1
RL
Vin
–VGS
n
Id
0
+VGS
RG
Vgs
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The E-MOSFET
The E-MOSFET is a normally off device. The n-channel device is biased
on by making the gate positive with respect to the source. A voltagedivider biased E-MOSFET amplifier is shown.
ID
Enhancement
+VDD
RD
R1
C3
Vout
C1
Id
RL
Vin
Q
IDQ
R2
RS
C2
0
VGS
VGS(th)
Vgs
VGSQ
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The E-MOSFET
The E-MOSFET amplifier in
Example 9-8 is illustrated in
Multisim using a 2N7000 MOSFET.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Common-Drain Amplifier
In a CD amplifier, the input
signal is applied to the gate
and the output signal is taken
from the source. There is no
drain resistor, because it is
common to the input and
output signals.
+VDD
C1
Vin
C2
RG
RS
Vout
RL
g m Rs
The voltage gain is given by the equation Av 
1  g m Rs
The voltage gain is always < 1, but the power gain is not.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Cascode Amplifier
The cascode connection is
a combination of CS and
CG amplifiers. This forms
a good high-frequency
amplifier. The input and
output signals at 10 MHz
are shown for this circuit
on the following slide…
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Cascode Amplifier
The input signal for the
cascode amplifier is
shown in red; the output
is blue. What is the gain?
The peak of the input is
24.7 mV.
The peak of the output
is 2.33 V.
AV = 94.3
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Class-D Amplifier
MOSFETs are useful as class-D amplifiers, which are very efficient
because they operate as switching amplifiers. They use pulse width
modulation, a process in which the input signal is converted to a series
of pulses. The pulse width varies proportionally to the amplitude of the
input signal.
Pulse-width modulation is easy to set up in Multisim. The
following slide shows the circuit. A sine wave is compared to a
faster triangle wave of the about the same amplitude using a
comparator (a 741 op-amp can be used at low frequencies).
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Class-D Amplifier
A circuit that you can use
in lab or in Multisim to
observe pulse width
modulation in action. The
scope display is shown on
the following slide…
Electronic Devices, 9th edition
Thomas L. Floyd
Op-amp set
up as a
comparator
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Class-D Amplifier
The signal is the yellow sine wave and is compared
repeatedly to the triangle (cyan). The result of the
comparison is the output (magenta).
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Class-D Amplifier
The modulated signal is amplified
by class-B complementary
MOSFET transistors. The output
is filtered by a low-pass filter to
recover the original signal and
remove the higher modulation
frequency.
PWM is also useful in control
applications such as motor
controllers. MOSFETs are widely
used in these applications because
of fast switching time and low onstate resistance.
Electronic Devices, 9th edition
Thomas L. Floyd
+VDD
Q1
Modulated
input
Low-pass
filter
RL
Q2
–VDD
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summary
The Analog Switch
MOSFETs are also used as analog switches to connect or disconnect an
analog signal. Analog switches are available in IC form – for example
the CD4066 is a quad analog switch that used parallel n- and p-channel
MOSFETs. The configuration shown allows signals to be passed in
either direction.
Advantages of MOSFETs are
that they have relatively low
on-state resistance and they
can be used at high
frequencies, such as found in
video applications.
Electronic Devices, 9th edition
Thomas L. Floyd
IN/OUT
OUT/IN
Control
Simplified internal construction of
a bidirectional IC analog switch.
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Selected Key Terms
Common-source A FET amplifier configuration in which the
source is the (ac) grounded terminal.
Common-drain A FET amplifier configuration in which the
drain is the (ac) grounded terminal.
Source-follower The common-drain amplifier.
Class-D A nonlinear amplifier in which the transistors
amplifier are operated as switches.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Selected Key Terms
Pulse-width A process in which a signal is converted to a
modulation series of pulses with widths that vary
proportionally to the signal amplitude.
Analog switch A device that switches an analog signal on
and off.
CMOS Complementary MOS.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
1. Compared to a common-emitter amplifier, a commonsource amplifier generally will have
a. higher gain and higher input resistance
b. higher gain and lower input resistance
c. lower gain and higher input resistance
d. lower gain and lower input resistance
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
2. The abbreviation yfs means
a. forward transfer admittance
b. forward on-state resistance
c. reverse transfer susceptance
d. reverse on-state conductance
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
3. The plot shown is a graphical solution for a self-biased
FET amplifier. The red line represents the
a. gate resistor
ID (mA)
b. source resistor
c. drain resistor
6
d. none of the above
– VGS (V)
Electronic Devices, 9th edition
Thomas L. Floyd
–4
0
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
4. The resistance represented by the red line is
a. 150 W
b. 240 W
ID (mA)
c. 470 W
d. 666 W
6
– VGS (V)
Electronic Devices, 9th edition
Thomas L. Floyd
–4
0
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
g m Rs
5. The gain equation Av 
is used to calculate the
1  g m Rs
gain of
a. a CS amplifier
b. a CD amplifier
c. a CG amplifier
d. any of the above
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
6. A FET that can be biased with zero bias is a
a. an n-channel JFET
b. a D-MOSFET
c. an E-MOSFET
d. all of the above
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
7. The cascode amplifier shown
uses
a. A CS and a CD stage
b. Two CS stages
c. Two CD stages
d. none of the above
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
8. The principle circuit used in creating a pulse width
modulator is a
a. peak detector
b. clipper
c. comparator
d. low-pass filter
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
9. The circuit is an amplifier for a pulse width modulated
signal. The load has the demodulated signal. The yellow
box represents a
+V
DD
a. peak detector
Q1
b. clipper
c. comparator
d. low-pass filter
Modulated
input
RL
Q2
–VDD
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
10. When the control signal is active, the output of an
analog switch should look like
a. the input signal
b. a square wave
c. a modulated pulse
d. a dc level
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Quiz
Answers:
Electronic Devices, 9th edition
Thomas L. Floyd
1. c
6. b
2. a
7. d
3. b
8. c
4. d
9. d
5. b
10. a
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.