electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
Lesson 2: Transistors and Applications
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Introduction
A transistor is a semiconductor device that controls current between
two terminals based on the current or voltage at a third terminal.
It is used for amplification or switching of electrical signals.
The basic structure of the bipolar junction transistor, BJT, determines
its operating characteristics.
DC bias is important to the operation of transistors in terms of setting
up proper currents and voltages in a transistor circuit.
Two important parameters are αDC and βDC
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
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Lesson 2
Bipolar junction transistors (BJTs)
FIGURE 17–2 Transistor symbols.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
BJT biasing
For normal 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.
Electronics Fundamentals 8th edition
Floyd/Buchla
BC reversebiased
+
pnp
npn
+
+
+
BE forwardbiased
© 2010 Pearson Education, Upper Saddle
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Lesson 2
BJT currents
IE  IC  IB
FIGURE 17–4 Transistor currents.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
I C  αDC I E
Where αDC (dc alpha) = IC/ IE
I C  βDC I B
IB
I
I
IE
I
Where βDC (dc beta) = IC/ IB
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Voltage-divider bias
Because the base current is small, the approximation
 R2 
is useful for calculating the base voltage.
VB  
 VCC
 R1  R2 
After calculating VB, you can find
VE by subtracting 0.7 V for VBE.
VE = VB - VBE
Next, calculate IE by applying Ohm’s
law to RE:
V
IE  E
RE
Then apply the approximation I C  I E
R1
RC
VB
R2
VC
VE
RE
Finally, you can find the collector voltage
from VC  VCC  I C RC
VCE  VC  VE
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
V
2.32 V
IE  E 
 2.32 mA
RE 1.0 k
I C  I E  2.32 mA
+15 V
R1
RC
27 k
2.2 k
2N3904
R2
RE
6.8 k
1.0 k
VC  VCC  I C 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.
Lesson 2
Voltage-divider bias
Determine VB, VE, VC, VCE, IB, IE, and IC in the given
Figure, The 2N3904 is a general-purpose transistor with a
typical βDC = 200.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
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
active
region
region
is after
saturation
region
the
is
after
saturation
the active
region.
region
occurs
when
the
baseThis
and
isisisthe
characterized
region
for by
emitter
and
the baseoperation
rapid
increase
of class-A
in collector
collector
junctions
are
operation.
current.
Operation
in this
both forward
biased.
region may destroy the
transistor.
Electronics Fundamentals 8th edition
Floyd/Buchla
IB5
IB4
IB3
IB2
IB1
IB = 0
0
VCE
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Collector characteristic curves
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Collector characteristic curves
Draw the family of collector characteristics curves for the
circuit in the given figure for IB = 5 µA to 25 µA in 5 µA
increments. Assume βDC = 100
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Collector characteristic curves
Draw the family of collector characteristics curves for the
circuit in the given figure for IB = 5 µA to 25 µA in 5 µA
increments. Assume βDC = 100
IB
IC
5 µA
0.5 mA
10 µA
1.0 mA
15 µA
1.5 mA
20 µA
2.0 mA
25 µA
2.5 mA
Electronics Fundamentals 8th edition
Floyd/Buchla
I C   DC I B
I C  100  5A  0.5mA
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Lesson 2
Load lines
A load line is an IV curve that represents the response of
a circuit that is external to a specified load.
For example, the load line for the
Thevenin circuit can be found by
calculating the two end points: the
current with a shorted load, and the
output voltage with no load.
2.0 k
+12 V
Electronics Fundamentals 8th edition
Floyd/Buchla
INL
= 6.0
0 mA
mA
SL =
VNL
= 012VV
SL =
I (mA)
6
Load line
4
2
0
0
4
8
12
V (V)
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Load lines
The IV response for any load will intersect the load line
and enables you to read the load current and load
voltage directly from the graph.
I (mA)
Read the load current
and load voltage from the graph if
a 3.0 k resistor is the load.
2.0 k
+12 V
IV curve for
3.0 k resistor
6
4
RL =
Q-point
2
3.0 k
0
0
4
8
VL = 7.2 V
Electronics Fundamentals 8th edition
Floyd/Buchla
12
V (V)
IL = 2.4 mA
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Load lines
The load line concept can be extended to a transistor
circuit. For example, if the transistor is connected as a
load, the transistor characteristic
I (mA)
curve and the base current
establish the Q-point.
6
2.0 k
4
+12 V
2
0
Electronics Fundamentals 8th edition
Floyd/Buchla
0
4
8
12
V (V)
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Load lines
Load lines can illustrate the operating conditions for a
transistor circuit. Assume the IV curves are as shown:
If you add a transistor load to the last
circuit, the base current will establish
the Q-point. Assume the base current
is represented by the blue line.
2.0 k
+12 V
I (mA)
6
4
2
0
Electronics Fundamentals 8th edition
Floyd/Buchla
For this base current,
the Q-point is:
0
4
8
12
V (V)
The load voltage (VCE) and current
(IC) can be read from the graph.
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Load lines
For the transistor, assume the base current is
established at 10 A by the bias circuit. Show the Qpoint and read the value of VCE and IC. The Q-point is the
intersection of the
load line with the
10 A base current.
IB = 25 A
IC (mA)
2.0 k
6
+12 V
IB = 20 A
IB = 15 A
4
IB = 10 A
2
Bias circuit
0
IB = 5.0 A
0
4
8
12
VCE (V)
VCE = 7.0 V; IC = 2.4 mA
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Signal (ac) operation
When a signal is applied to a transistor circuit, the output can have a
larger amplitude because the small base current controls a larger collector
current.
This increase is called amplification
The ratio of the ac collector current (Ic) to the ac base current (Ib) is
designated by βac (the ac beta) of hfe
Ic
 ac 
Ib
FIGURE 17–13 An amplifier with voltage-divider bias with capacitively
coupled input signal. Vin and Vout are with respect to ground.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Signal (ac) operation
When a signal is applied to a transistor circuit, the
output can have a larger amplitude because the small
base current controls a larger collector current.
For the load line and
characteristic curves from the last
example (Q-point shown) assume IB
varies between 5.0 A and 15 A
due to the input signal. What is the
change in the collector current?
IC (mA)
6
4
2
IB = 25 A
IB = 20 A
IB = 15 A
IB = 10 A
IB = 5.0 A
The operation along
0
VCE (V)
0
4
8
12
the load line is shown in red.
Reading the collector current, IC varies from 1.2 mA to 3.8 mA.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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.
VC
C
R1
RC
Output coupling
capacitor
C2
C1
Input coupling
capacitor
Electronics Fundamentals 8th edition
Floyd/Buchla
R2
RE
C3 Bypass
capacitor
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Lesson 2
Summary
CE amplifier
The bypass capacitor increases voltage gain. It shorts the signal around
the emitter resistor, RE, in order to increase the voltage gain. To
understand why let us consider the amplifier without the bypass
capacitor as explained the preceding equations.
VC
C
R1
RC
Output coupling
capacitor
C2
C1
Input coupling
capacitor
Electronics Fundamentals 8th edition
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R2
RE
C3 Bypass
capacitor
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Lesson 2
Formulas
Lowercase italic subscript indicate signal (ac) voltages and signal
(alternating currents)
Av (Voltage gain) = Vout / Vin
Vout = IcRC
The signal voltage at the base is approximately equal to
Vb ≡ Vin ≡ Ie (re + RE)
where re is the internal emitter resistance of the transistor.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Formulas without the bypass capacitor
Av now can be expressed as
Av = Vout / Vin = IcRC / Ie (re + RE)
Since Ic ≡ Ie, the currents cancel and the gain is the ratio of the resistance.
Av = RC / (re + RE)
If RE is much greater than re, then
Av ≡ RC / RE
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Formulas with the bypass capacitor
If the bypass capacitor is connected across RE, it effectively shorts the
signal to ground leaving only re in the emitter. Thus the voltage gain of
the CE amplifier with the bypass capacitor shorting RE is:
Av = RC / re
The transistor parameter re is important because it determines the
voltage gain of a CE amplifier in conjunction with RC. A formula for
estimating re is given without derivation in the following equation:
re≡ 25mV / IE
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Summary
Voltage gain of a CE amplifier
Calculate the voltage gain of the CE amplifier. The dc
conditions were calculated earlier; IE was found to be 2.32 mA.
re 
VCC = +15 V
25 mV
25 mV

 10.8 
IE
2.32 mA
V
R
Av  out  C
Vin
re
2.2 k

 204
10.8 
RC
R1
C1
27 k
2.2 k
2N3904
2.2 F
R2
6.8 k
C2
1.0 F
RE
1.0 k
C3
100 F
Sometimes the gain will be shown with a
negative sign to indicate phase inversion.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Lesson 2
Phase Inversion
The output voltage at the collector is 180 degrees out of phase with
the input voltage at the base. Therefore, the CE amplifier is
characterized by a phase inversion between the input and output. The
inversion is sometimes indicated by a negative voltage gain.
AC Input Resistance
Rin ≡ βac Ib re / Ib
The Ib terms cancel, leaving
Rin = Vb / Ib
Rin ≡ βac re
Total Input Resistance of a CE amplifier:
Vb = Iere
Ie ≡ βac Ib
Electronics Fundamentals 8th edition
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Rin(tot) = R1║R2║Rin
Rin(tot) = R1║R2║βac re
RC has no effect because of the reverse-biased,
base-collector junction.
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Current Gain:
The signal current gain of a CE amplifier is
Ai= Ic / Is
Where Is is the source current and is calculated by Vin /
Rin(tot)
Power Gain:
The power gain of a CE amplifier is the product of the voltage gain
and the current gain.
Ap≡ Av Ai
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Decibel (dB) Measurement
The decibel (dB) is a logarithmic measurement of the ratio of one
voltage to another or one power to another, which can be used to
express the input-to-output relationship .
VOLTAGE RATIO: dB = 20 log (Vout / Vin)
POWER RATIO: dB = 10 log (Pout / Pin)
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Lesson 2
Input resistance of a CE amplifier
The input resistance of a CE amplifier is an ac resistance
that includes the bias resistors and the resistance of the
emitter circuit as seen by the base.
Because IB << IE, the emitter
resistance appears to be
much larger when viewed
from the base circuit. The
factor is (ac+1), which is
approximately equal to ac.
Using this approximation,
Rin(tot) = R1||R2||acre.
Electronics Fundamentals 8th edition
Floyd/Buchla
VCC = +15 V
RC
R1
C1
27 k
2.2 k
2N3904
2.2 F
R2
6.8 k
C2
1.0 F
RE
1.0 k
C3
100 F
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Input resistance of a CE amplifier
FIGURE 17–20 Total input resistance.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
Input resistance of a CE amplifier
Calculate the input resistance of the CE amplifier. The
transistor is a 2N3904 with an average ac of 200. The value of re
was found previously to be 10.8 . Thus, acre = 2.16 k.
VCC = +15 V
Rin(tot) = R1||R2||acre
= 27 k||6.8 k||2.16 k
= 1.55 k
Notice that the input resistance of
this configuration is dependent on
the value of ac, which can vary.
Electronics Fundamentals 8th edition
Floyd/Buchla
RC
R1
C1
27 k
2.2 k
2N3904
2.2 F
R2
6.8 k
C2
1.0 F
RE
1.0 k
C3
100 F
© 2010 Pearson Education, Upper Saddle
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Lesson 2
CE amplifier
Determine the voltage gain, current gain, and power
gain for the CE amplifier given. DC  ac = 100. Also, express the
voltage and power gains in decibels.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
CC amplifier (emitter-follower)
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 power gain.
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
VC
C
R1
C1
R2
RE
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Lesson 2
The voltage gain of a CC amplifier is approximately 1, but
the current gain is always greater than 1.
Voltage Gain
Av (Voltage Gain)
Av = Vout / Vin
Vout = IeRE
Vin= Ie (re + RE)
Electronics Fundamentals 8th edition
Floyd/Buchla
Av = IeRE / Ie (re + RE)
The gain expression simplifies to:
Av = RE / (re + RE)
It is important to notice here that the
gain is always less than 1. Because re is
normally much less than the RE, then a
good approximation is Av = 1
© 2010 Pearson Education, Upper Saddle
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Lesson 2
AC Input Resistance
The emitter-follower is characterized by a high input resistance, which
makes it a very useful circuit. Because of the very high input
resistance , the emitter follower can be used as a buffer to minimize
loading effects when one circuit is driving another.
Rin = Vb / Ib
Vb = Ie (re + RE)
Ie ≡ βac Ib
Rin ≡ βac Ib (re + RE) / Ib
The Ib terms cancel, leaving
Rin ≡ βac (re + RE)
If RE is at least ten times larger than re , then the
input resistance at the base is:
Rin ≡ βac RE
Total Input Resistance of a CC amplifier:
Rin(tot) = R1║R2║Rin
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Lesson 2
Current Gain:
The signal current gain for the emitter-follower is
Ai= Ie / Is
Where Is is the signal current and is calculated by Vin / Rin(tot)
Since Ie = Vout / RE and Is = Vin / Rin(tot) then Ai can also be expressed as
assuming Vout / Vin = 1:
Ai= (Vout / RE )/ (Vin / Rin(tot) ) = Rin(tot) / RE
Power Gain:
The power gain of is the product of the voltage gain and the current gain. For
the emitter-follower, the power gain is approximately equal to the current gain
because the voltage gain is approximately equal to 1.
Ap≡ Ai
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Lesson 2
CC amplifier
Calculate re and Rin(tot) for the CC amplifier. Use  = 200.
 R2 
27 k


VB  
V

 CC 
15 V = 8.26 V
R

R
22
k

+
27
k



 1
2 
VE  VB  0.7 V = 7.76 V
V = +15 V
CC
VE 7.76 V

 7.76 mA
RE 1.0 k
25 mV
25 mV
re 

 3.2 
IE
7.76 mA
IE 
Rin(tot) = R1||R2||ac(re+ RE)
R1
C1
22 k
2N3904
10 F
= 22 k||27 k|| 200 (1.0 k) = 9.15 k
R2
27 k
RE
1.0 k
Because re is small compared to RE, it
has almost no affect on Rin(tot).
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
CC amplifier
Determine the input resistance of the emitter-follower in the given
figure. Also find the voltage gain, current gain and power gain.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
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)/DC
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Lesson 2
The BJT as a switch
FIGURE 17–30 Ideal switching action of a transistor.
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
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Lesson 2
The BJT as a switch
FIGURE 17–30 Ideal switching action of a
transistor.
Conditions in Cutoff A transistor is in cutoff when the base-emitter junction is not forward biased.
VCE( cutoff )  VCC
Conditions in Saturation When the base-emitter junction is forward-biased and there is enough base
current to produce a maximum collector current, the transistor is saturated.
The minimum value of base current to produce saturation is:
I B (min) 
I C ( sat)
 DC
Electronics Fundamentals 8th edition
Floyd/Buchla
VBE )  0.7V
VRB  VIN  0.7V
I C ( sat) 
RB (max) 
VCC
RC
VRB
I B (min)
© 2010 Pearson Education, Upper Saddle
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Lesson 2
The BJT as a switch
(a) For the transistor switching circuit in the given figure, what is
VCE when VIN = 0 V?
(b) What minimum value of IB will saturate this transistor if the
βDC is 200?
(c) Calculate the maximum value of RB when VIN = 5 V.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
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Lesson 2
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)
p
n
p
p
The gate is always operated with
reverse bias on the pn junction formed
between the gate and the channel. As
S (Source)
the reverse bias is increased, the
n-channel JFET
channel current decreases.
Electronics Fundamentals 8th edition
Floyd/Buchla
D
D (Drain)
G
n n
S
p-channel JFET
© 2010 Pearson Education, Upper Saddle
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Lesson 2
The FET
FIGURE 17–33 Effects of VGG on channel width and drain current
(VGG = VGS).
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET
FIGURE 17–34 JFET schematic symbols.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET: MOSFET
The MOSFET (Metal Oxide Semiconductor FET)
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
p-channel
MOSFET
MOSFET
© 2010 Pearson Education, Upper Saddle
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Lesson 2
The FET: MOSFET
FIGURE 17–35 Representation of the basic structure of D-MOSFETs.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET: MOSFET
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 (D-mode) or
positive bias (E-mode).
FIGURE 17–36 Operation of n-channel D-MOSFET.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET: MOSFET
FIGURE 17–37 D-MOSFET schematic symbols.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET: MOSFET
The E-mode MOSFET has no physical channel. It can only be operated
with positive bias (E-mode). Positive bias induces a channel and enables
conduction as shown here with a p-channel device.
G
D
D
n
n
p
G
p
n
n
S
S
induced channel
E-MOSFET
E-MOSFET with bias
FIGURE 17–38 E-MOSFET construction and
operation.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
The FET: MOSFET
FIGURE 17–39 E-MOSFET schematic symbols.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
JFET biasing
JFETs are depletion mode devices – they must be
operated such that the gate-source junction is reverse
biased.
+VDD
VDD
The simplest way to bias a JFET is
to use a small resistor is in series
with the source and a high value
resistor from the gate to ground. VG = 0 V
The voltage drop across the source
resistor essentially reverse biases
RG
the gate-source junction.
Because of the reverse-biased
junction, there is almost no
current in RG. Thus, VG = 0 V.
Electronics Fundamentals 8th edition
Floyd/Buchla
RD
RD
VG = 0 V
V S
+VS
RS
n-channel
RG
RS
p-channel
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
JFET biasing
For the n-channel JFET, the gate-tosource voltage is VGS  VG  VS  0V  I D RS
VGS   I D RS
For the p-channel JFET, the gate-tosource voltage is VGS   I D RS
The drain voltage with respect to ground
is V  V  I R
D
DD
D
D
Since VS=IDRS, the drain-to-source
voltage is V  V  I R  R 
DS
DD
Electronics Fundamentals 8th edition
Floyd/Buchla
D
D
FIGURE 17–40 Self-biased JFETs (IS = ID in
all FETs).
S
© 2010 Pearson Education, Upper Saddle
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Lesson 2
JFET biasing
Find VDS and VGS in the JFET circuit below, given that ID =
5mA
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
D-MOSFET biasing
D-MOSFETs can be operated in either depletion mode
or in enhancement-mode. For this reason, they can be
+VDD
biased with various bias circuits.
The simplest bias method for a DMOSFET is called zero bias. In this
method, the source is connected directly
to ground and the gate is connected to
ground through a high value resistor.
RD
VG = 0 V
RG
Only n-channel D-MOSFETs
are available, so this is the
only type shown.
n-channel D-MOSFET
with zero bias
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
D-MOSFET biasing
FIGURE 17–42 A zero-biased D-MOSFET.
Recall that depletion/enhancement
MOSFETs can be operated with either
positive or negative values of VGS.
A simple bias method, called zero bias, is
to set VGS = 0 V so that an ac signal at the
gate varies the gate-to-source voltage
above and below this bias point.
Since VGS = 0 V, ID = IDSS as indicated.
IDSS is defined as the drain current when
VGS = 0 V.
The drain-to-source voltage is expressed
as V  V  I R
DS
DD
DSS
D
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
D-MOSFET biasing
Determine the drain-to-source voltage in the circuit of the
given figure. The MOSFET data sheet give VGS(off) = -8 V and
IDSS = 12 mA
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
E-MOSFET biasing
E-MOSFETs can use bias circuits similar to BJTs but
larger value resistors are normally selected because of
the very high input resistance.
+VDD
+V
DD
The bias voltage is normally
set to make the gate more
positive than the source by an
amount exceeding VGS(th).
RD
R1
RG
RD
R2
Drain-feedback bias
Electronics Fundamentals 8th edition
Floyd/Buchla
Voltage-divider bias
© 2010 Pearson Education, Upper Saddle
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Lesson 2
E-MOSFET biasing
FIGURE 17–44 E-MOSFET biasing arrangements.
Recall that enhancement-only MOSFETs
must have a VGS greater than the
threshold value VGS(th) .
In either bias arrangement, the purpose is
to make the gate voltage more positive
than the source by an amount exceeding
VGS(th) .
In the drain-feedback circuit, there is
negligible gate current and, therefore, no
voltage drop across RG . As a result,
VGS = VDS.
 R2 
Equation for the voltage-divider bias is
VDD
VGS  
given by
R

R
2 
 1
Electronics Fundamentals 8th edition
Floyd/Buchla
VDS  VDD  I D RD
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
E-MOSFET biasing
Determine the amount of drain current in the given figure. The
MOSFET has a VGS(th) of 3 V.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Selected Key Terms
Bipolar A transistor with three doped semiconductor
junction regions separated by two pn junctions.
transistor
(BJT) An amplifier that conducts for the entire input
cycle and produces an output signal that is a
Class A replica of the input signal in terms of its
amplifier waveshape.
The state of a transistor in which the output
current is maximum and further increases of
Saturation 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.
Lesson 2
Selected Key Terms
Cutoff The non-conducting state of a transistor.
Q-point The dc operating (bias) point of an amplifier.
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.
Class B An amplifier that conducts for half the input
amplifier cycle.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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 the channel is
depleted of majority carriers.
Enhancement The condition in a FET when the channel has
mode an abundance of majority carriers.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
1. The Thevenin circuit shown has a load line that crosses
the y-axis at
a. +10 V.
b. +5 V.
5.0 k
+10 V
c. 2 mA.
d. the origin.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
1. The Thevenin circuit shown has a load line that crosses
the y-axis at
a. +10 V.
b. +5 V.
5.0 k
+10 V
c. 2 mA.
d. the origin.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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.
Lesson 2
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.
Lesson 2
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. have all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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. have all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
4. The type of amplifier shown is a
a. common-collector.
b. common-emitter.
VC
C
c. common-drain.
d. none of the above.
C1
R1
R2
Electronics Fundamentals 8th edition
Floyd/Buchla
RE
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
4. The type of amplifier shown is a
a. common-collector.
b. common-emitter.
VC
C
c. common-drain.
d. none of the above.
C1
R1
R2
Electronics Fundamentals 8th edition
Floyd/Buchla
RE
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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.
Lesson 2
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.
Lesson 2
Quiz
6. A type of field effect transistor that can operate in either
depletion or enhancement mode is an
a. D-MOSFET.
b. E-MOSFET.
c. JFET.
d. none of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
6. A type of field effect transistor that can operate in either
depletion or enhancement mode is an
a. D-MOSFET.
b. E-MOSFET.
c. JFET.
d. none of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
Quiz
8. A transistor circuit shown is a
a. D-MOSFET with voltage-divider bias.
b. E-MOSFET with voltage-divider bias.
+VDD
c. D-MOSFETwith self-bias.
d. E-MOSFET with self bias.
R1
RD
R2
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lesson 2
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
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.