electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
Chapter 17
Transistors
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Objectives:
• List some common transistor applications
• Describe the basic structure of Bipolar
Junction Transistors (BJT’s)
• Explain the difference between npn and pnp
transistors
• Explain transistor biasing
• Discuss transistor current and voltage
relationships
• Define voltage and current gain
characteristics of BJT’s
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Transistor Applications
• Transistors can be used to amplify
small signals
• Transistors can function as an
electronic switch
• Transistors can form the basis for
oscillator circuits
• Transistors are the building blocks of
digital logic & microprocessors
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
Bipolar junction transistors (BJTs)
The BJT is a transistor with three regions and two pn
junctions. The regions are named the emitter, the base, and
the collector and each is connected to a lead.
There are two types of
BJTs – npn and pnp.
B (Base)
C (Collector)
n
p
n
Base-Collector
junction
B
Base-Emitter
junction
E (Emitter)
Electronics Fundamentals 8th edition
Floyd/Buchla
Separating the regions
are two junctions. C
p
n
p
E
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
BJT biasing
Transistors have both AC and DC characteristics. For
normal DC operation, the base-emitter junction is forwardbiased and the base collector junction is reverse-biased.
For the pnp
npn transistor, this
condition requires that the base
is more negative
positive than
thanthe
theemitter
and the and
emitter
collector
the collector
is more is more
positive than
negative
thanthe
thebase.
base.
BC reversebiased
B
+
+
BE forwardbiased
C
+
pnp
npn
E
+
Why do we need to “bias” the transistor?
Because it is a non-linear device, and we want it to operate
In the “sweet spot” where it is most linear.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
BJT currents
A small base current (IB) is able to control a larger collector
current (IC). Some important current relationships for a BJT
are:
IC
IE  IC  IB
IB
I C  β DC I B
b is usually between 20 and 300,
depending on the particular transistor.
Electronics Fundamentals 8th edition
Floyd/Buchla
I
I
IE
I
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Voltage-divider DC bias
1. Because the base current is small, the simple voltage divider
approximation is useful for calculating the base voltage.
 R2 
VB  
VCC
R

R
 1
2 
2. After calculating VB, you can find VE
by subtracting 0.7 V for VBE.
3. Next, calculate IE by applying
Ohm’s law to RE:
V
IE  E
RE
4. Then apply the approximation, I C  I E
R1
RC
VB
R2
+Vcc
VC
VE
RE
5. Finally, you can find the collector
voltage from VC  VCC  I C RC
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
Voltage-divider bias
Calculate VB, VE, and VC for the circuit.
 R2 
6.8 k


VB  
V

 CC 
15 V = 3.02 V
 27 k + 6.8 k 
 R1  R2 
VE = VB  0.7 V = 2.32 V
IE 
VE 2.32 V

 2.32 mA
RE 1.0 k
I C  I E  2.32 mA
+15 V
R1
27 k
RC
2.2 k
2N3904
R2
6.8 k
RE
1.0 k
VC  VCC  IC RC  15 V   2.32 mA  2.2 k  9.90 V
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
Collector characteristic curves
The collector characteristic curves are a family of
curves that show how collector current varies with
the collector-emitter
IC
voltage for a given IB.
IB6
The curves are divided into
three regions:
The
breakdown
region
The
saturation
region
The
active
region
is the
isbias
after“sweet
thewhen
active
occurs
theregion
basespot”
for
and
is isoperation.
characterized
emitter
and the baselinear
Thisby
is
rapid
increase
collector
collector
junctions
are
the
region
forinoperation
current.
Operation
in this
both
forward
biased.
as
a class-A
amplifier.
region may destroy the
transistor.
Electronics Fundamentals 8th edition
Floyd/Buchla
IB5
IB4
IB3
IB2
IB1
IB = 0
0
VCE
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Chapter 7
Summary
CE amplifier
In a common-emitter amplifier, the input signal is applied
to the base and the output is taken from the collector. The
signal is larger but inverted at the output.
VCC
R1
RC
Output coupling
capacitor
C2
RE
C3 Bypass
capacitor
C1
Input coupling
capacitor
Electronics Fundamentals 8th edition
Floyd/Buchla
R2
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Voltage gain of a CE amplifier
Calculate the voltage gain of the CE amplifier. The dc
conditions were calculated earlier; VC was 9.9volts
VCC = +15 V
C1
AV = Vout/Vin = RC/RE
= 2.2k/1k = - 2.2
RC
C2
2.2 k
2N3904
2.2 mF
Sometimes the gain will be shown with
a negative sign to indicate phase
inversion.
Electronics Fundamentals 8th edition
Floyd/Buchla
R1
27 k
R2
6.8 k
1.0 mF
RE
1.0 k
C3
100 mF
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
Voltage gain of a CE amplifier with bypass capacitor
Adding a bypass capacitor maintains the DC bias, but
increases the AC gain because RE2 is bypassed for AC signals.
Calculate the voltage gain of the CE amplifier. The dc conditions
were calculated earlier;
VC was +9.9volts
VCC = +15 V
AV = Vout/Vin = RC/RE1
C1
R1
27 k
2N3904
= 2.2k/.2k = - 11
2.2 mF
R2
What is the largest input voltage
6.8 k
allowed before clipping occurs?
Max Vpeak out is Vcc-Vc = 5.1Vp or 10.2Vpp
Since Av = -11, the max input is 10.2/11 = .93Vpp
Electronics Fundamentals 8th edition
Floyd/Buchla
RC
C2
2.2 k
1.0 mF
RRE =200
1.0 k
E1
C3
100 mF
C=100uf
RE2=800
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
CC amplifier
In a common-collector amplifier, the input signal is applied
to the base and the output is taken from the emitter. There is
no voltage gain, but there is potential power gain due to
increased current capacity through the transistor.
The output voltage is nearly
the same as the input; there is
no phase reversal as in the
CE amplifier.
The input resistance is larger
than in the equivalent CE
amplifier because the emitter
resistor is not bypassed.
Electronics Fundamentals 8th edition
Floyd/Buchla
VCC
R1
C1
R2
RE
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Class B amplifiers
The class B amplifier is more efficient than the class A amplifier
because it only conducts output current to the load when an input
is applied. It is widely used in power amplifiers.
VCC
The amplifier shown uses
complementary transistors – one is
R1
npn and the other is pnp – in order
C1
Q1
to conduct on both half cycles of
the input.
D1
C3
The bias method shown avoids
cross-over distortion by
bringing the transistors just
above cutoff using diodes,
making this a Class AB amp.
Electronics Fundamentals 8th edition
Floyd/Buchla
C2
D2
Q2
RL
R2
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
The BJT as a switch
BJTs are used in switching applications when it is
necessary to provide current drive to a load.
VCC
In switching applications, the transistor
is either in cutoff or in saturation.
In cutoff, the input voltage is too
small to forward-bias the
transistor. The output (collector) IIN = 0
voltage will be equal to VCC.
When IIN is sufficient to saturate
the transistor, the transistor acts
like a closed switch. The output is
near 0 V.
Electronics Fundamentals 8th edition
Floyd/Buchla
VCC
RC
RC
VOUT
0 CC
V
=V
IIN > IC(sat)/bDC
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
The FET
The field-effect transistor (FET) is a voltage controlled
device where gate voltage controls drain current. There
are two types of FETs – the JFET and the MOSFET.
JFETs have a conductive channel
with a source and drain connection
on the ends. Channel current is
controlled by the gate voltage.
G (Gate)
The gate is always operated with
reverse bias on the pn junction formed
between the gate and the channel. As
the reverse bias is increased, the
channel current decreases.
Electronics Fundamentals 8th edition
Floyd/Buchla
D
D (Drain)
p
n
p
p
S (Source)
n-channel JFET
G
n
n
S
p-channel JFET
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
The FET
The MOSFET differs from the JFET in that it has an
insulated gate instead of a pn junction between the gate
and channel.
Like JFETs, MOSFETs have a conductive channel with the
source and drain connections on it.
D (Drain)
Channel current is
controlled by the gate
voltage. The required gate
voltage depends on the type
of MOSFET.
Electronics Fundamentals 8th edition
Floyd/Buchla
p
n
G (Gate)
Channel
D
p
G
n
Substrate
S (Source)
S
n-channel
MOSFET
p-channel
MOSFET
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
The FET
In addition to the channel designation, MOSFETs are
subdivided into two types – depletion mode (D-mode)
or enhancement mode (E-mode).
The D-MOSFET has a physical channel which can be enhanced or
depleted with bias. For this reason, the D-MOSFET can be operated
with either negative bias or positive bias.
The E-mode MOSFET has no
physical channel. It can only be
operated with positive bias (Emode). Positive bias induces a
channel and enables conduction as
shown here with a p-channel device.
Electronics Fundamentals 8th edition
Floyd/Buchla
D
D
n
G
p
n
G
p
n
n
S
S
induced channel
E-MOSFET
E-MOSFET with bias
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
FET amplifiers
FET amplifiers are voltage controlled and generally do
not have as much gain or linearity as BJT amplifiers.
The major advantage of FETs is high input resistance.
The input resistance of a FET
amplifier depends on the bias
resistors. For the CS amplifier
shown, Rin(tot) = RG because
the gate-source resistance is a
reverse biased pn junction. RG
is made higher than the bias
resistors in a BJT amplifier
because of the negligible
input current to the FET.
Electronics Fundamentals 8th edition
Floyd/Buchla
+VDD
RD
C2
C1
RG
RS
C3
Common-source amplifier
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Conductance
An important parameter for FETs is the conductance.
(roughly analogous to the Beta in the BJT.) Recall that
conductance is the reciprocal of resistance, so from Ohm’s
I
law:
gm  d
with units expressed in mhos, or Siemens
Vgs
Sometimes the prefix “trans” is added, transconductance, to indicate the
current and voltage are not measured in the same circuit (gate and drain).
For the common source amplifier, the drain current multiplied by
the drain resistor is the output voltage. The voltage gain (ratio of
output voltage to input voltage) can then be developed as:
Av 
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout I d Rd

 g m Rd
Vin
Vgs
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Summary
FET amplifiers
If voltage gain is not required, the common-drain (CD)
amplifier is a simple high input resistance amplifier.
Although the voltage gain is
less than 1, the power gain is
high because of the high
input resistance. The circuit
shown has the advantage of
only two resistors (RS
represents the load). The
output voltage is slightly
smaller and in phase with the
input, which is similar to the
CC circuit for a BJT.
Electronics Fundamentals 8th edition
Floyd/Buchla
+VDD
C
RG
RS
Common-drain amplifier
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Feedback oscillators
An oscillator is a circuit that generates a repetitive
waveform on its output. A feedback oscillator uses
positive feedback from the output to sustain oscillations.
Conditions for oscillations are
1.
The phase shift around the
loop must be 0o.
2.
The closed loop gain must
be 1 (unity gain).
Electronics Fundamentals 8th edition
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Av
Acl = AvB = 1
Feedback
B
circuit
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Feedback oscillators
The Colpitts and Hartley oscillators are examples of a
feedback oscillator. The amplifier sections are nearly
identical, but the LC feedback network is different.
To obtain the required signal
reinforcement (positive
feedback), the amplifier
inverts the signal (180o) and
the feedback network shifts
the phase another 180o.
An additional capacitor is in
the amplifier to block dc.
Hartley
Colpitts oscillator
oscillator
feedback network
network.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Feedback oscillators
The Pierce oscillator is another example of a feedback
oscillator that is commonly used with a crystal as the
resonating feedback element.
This circuit uses a FET or
MOSFET for high input
resistance. The CS amplifier
circuit provides 180o phase
shift and the crystal shifts the
phase another 180o. The
crystal could be replaced with
an LC resonant circuit, but
with less accuracy.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Chapter 7
Summary
Example
The crystal is replaced with a parallel LC tank circuit.
L = 82uH, and C = 33pF. What is the resonant frequency?
fr 
1
2 LC
= 3 mHz
Electronics Fundamentals 8th edition
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C
L
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Chapter 7
Selected Key Terms
Bipolar junction A transistor with three doped semiconductor
transistor (BJT) regions separated by two pn junctions.
Class A An amplifier that conducts for the entire input
amplifier cycle and produces an output signal that is a
replica of the input signal in terms of its
waveshape. Even with no input, there is bias
current flowing in the output, so it is relatively
inefficient.
The state of a transistor in which the output
Saturation current is maximum and further increases of
the input variable have no effect on the output.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Selected Key Terms
Cutoff The non-conducting state of a transistor.
Amplification The process of producing a larger voltage,
current or power using a smaller input signal
as a pattern.
Common- A BJT amplifier configuration in which the
emitter (CE) emitter is the common terminal.
An amplifier that conducts for half the input
Class B
cycle. It is an efficient amp because no output
amplifier
current flows when there is no input signal.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Selected Key Terms
Junction field- A type of FET that operates with a reverseeffect transistor biased junction to control current in a channel.
(JFET)
MOSFET Metal-oxide semiconductor field-effect
transistor.
Depletion mode The condition in a FET when an applied
signal causes the channel to be depleted of
majority carriers (normally on).
Enhancement The condition in a FET when an applied
mode signal causes the channel to have an
abundance of majority carriers (normally off).
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Selected Key Terms
Common-Source An FET amplifier configuration in which the
source is the common terminal. Similar to
the Common Emitter circuit for the BJT.
Oscillator A circuit that produces a repetitive waveform
on its output with only a dc supply voltage as
an input.
Feedback The process of returning a portion of a
circuit’s output signal to the input. Negative
feedback provides stability. Positive
feedback creates oscillations.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
2. In a common-emitter amplifier, the output ac signal will
normally
a. have greater voltage than the input.
b. have greater power than the input.
c. be inverted.
d. all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
3. In a common-collector amplifier, the output ac signal
will normally
a. have greater voltage than the input.
b. have greater power than the input.
c. be inverted.
d. all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
4. The type of amplifier shown is a
a. common-collector.
b. common-emitter.
VCC
c. common-drain.
d. none of the above.
C1
R1
R2
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RE
© 2010 Pearson Education, Upper Saddle
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Chapter 7
Quiz
5. A major advantage of FET amplifiers over BJT amplifiers
is that generally they have
a. higher gain.
b. greater linearity.
c. higher input resistance.
d. all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
6. A type of field effect transistor that can operate in either
depletion or enhancement mode is an
a. D-MOSFET.
b. MOSFET.
c. JFET.
d. none of the above.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
7. For an FET, transconductance is the ratio of
a. drain voltage to drain current.
b. gate-source voltage to drain current.
c. gate-source current to drain voltage.
d. drain current to gate-source voltage.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 7
Quiz
8. A transistor circuit shown is a
a. D-MOSFET with voltage-divider bias.
+VDD
b. E-MOSFET with voltage-divider bias.
c. D-MOSFETwith self-bias.
d. E-MOSFET with self bias.
R1
RD
R2
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Chapter 7
Quiz
9. Colpitts and Hartley oscillators both have
a. positive feedback.
b. amplification.
c. a closed loop gain of 1.
d. all of the above.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Chapter 7
Quiz
10. If you were troubleshooting the circuit shown here,
you would expect the gate voltage to be
a. more positive than the drain voltage.
+VDD
b. more positive than the source voltage.
c. equal to zero volts.
R1
RD
d. equal to +VDD
R2
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Chapter 7
Quiz
Answers:
6. a
Electronics Fundamentals 8th edition
Floyd/Buchla
2. d
7. d
3. b
8. b
4. a
9. d
5. c
10. b
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