Transcript Document

electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
chapter 17
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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
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River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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
River, NJ 07458. All Rights Reserved.
Chapter 17
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
I C  α DC I E
I C  β DC I B
IB
I
I
IE
I
Electronics Fundamentals 8th edition
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© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
Voltage-divider bias
Because the base current is small, the approximation
 R2 
VB  
VCC is useful for calculating the
R

R
 1
2 
After calculating VB, you can find VE
by subtracting 0.7 V for VBE.
R1
Next, calculate IE by applying Ohm’s
law to RE:
VE
IE 
RE
base voltage.
RC
VB
R2
VC
VE
RE
Then apply the approximation I C  I E
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 17
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 17
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
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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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 =
3.0 k
Q-point
2
0
0
4
VL = 7.2 V
Electronics Fundamentals 8th edition
Floyd/Buchla
8
12
V (V)
IL = 2.4 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
Load lines
For the transistor, assume the base current is
established at 10 mA 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 mA base current.
IB = 25 mA
IC (mA)
2.0 k
6
+12 V
IB = 20 mA
IB = 15 mA
4
IB = 10 mA
2
Bias circuit
0
IB = 5.0 mA
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.
Chapter 17
Summary
Signal 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 mA and 15 mA
due to the input signal. What is the
change in the collector current?
IC (mA)
6
4
2
IB = 25 mA
IB = 20 mA
IB = 15 mA
IB = 10 mA
IB = 5.0 mA
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.
Chapter 17
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
River, NJ 07458. All Rights Reserved.
Chapter 17
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
Av 
Vout RC

Vin
re
2.2 k

 204
10.8 
C1
R1
27 k
RC
C2
2.2 k
2N3904
2.2 mF
R2
6.8 k
1.0 mF
RE
1.0 k
C3
100 mF
Sometimes the gain will be shown with a
negative sign to indicate phase inversion.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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 (bac+1), which is
approximately equal to bac.
Using this approximation,
Rin(tot) = R1||R2||bacre.
Electronics Fundamentals 8th edition
Floyd/Buchla
VCC = +15 V
C1
R1
27 k
RC
C2
2.2 k
2N3904
2.2 mF
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 17
Summary
Input resistance of a CE amplifier
Calculate the input resistance of the CE amplifier. The
transistor is a 2N3904 with an average bac of 200. The value of re
was found previously to be 10.8 . Thus, bacre = 2.16 k.
VCC = +15 V
Rin(tot) = R1||R2||bacre
= 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 bac, which can vary.
Electronics Fundamentals 8th edition
Floyd/Buchla
C1
R1
27 k
RC
C2
2.2 k
2N3904
2.2 mF
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 17
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 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
VCC
R1
C1
R2
RE
© 2010 Pearson Education, Upper Saddle
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Chapter 17
Summary
CC amplifier
Calculate re and Rin(tot) for the CC amplifier. Use b = 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
VE 7.76 V

 7.76 mA
RE 1.0 k
25 mV
25 mV
re 

 3.2 
IE
7.76 mA
VCC = +15 V
IE 
R1
22 k
C1
2N3904
10 mF
Rin(tot) = R1||R2||bac(re+ RE)
= 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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
Class B amplifiers
The class B amplifier is more efficient than the class A
amplifier. It is widely used in power amplifiers.
The amplifier shown uses
complementary transistors – one is
an npn and the other is a pnp.
The bias method shown avoids
cross-over distortion by
bringing the transistors just
above cutoff using diodes.
VCC
C1
R1
Q1
D1
C2
D2
C3
Q2
RL
R2
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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
River, NJ 07458. All Rights Reserved.
Chapter 17
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
River, NJ 07458. All Rights Reserved.
Chapter 17
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 17
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 (D-mode) or positive bias (E-mode).
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 17
Summary
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
VS
+VS
RS
n-channel
RG
RS
p-channel
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
n-channel D-MOSFET
with zero bias
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
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
The bias voltage is normally
set to make the gate more
positive than the source by an
amount exceeding VGS(th).
DD
RD
R1
RG
RD
R2
Drain-feedback bias
Electronics Fundamentals 8th edition
Floyd/Buchla
Voltage-divider bias
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
Transconductance
An important parameter for FETs is the transconductance.
Recall that conductance is the reciprocal of resistance, so
from Ohm’s law:
gm 
Id
Vgs
The prefix “trans” is added 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 17
Summary
FET amplifiers
If voltage gain is no required, the common-drain
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 is smaller and in
phase with the input.
+VDD
C
RG
RS
Common-drain amplifier
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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
Floyd/Buchla
Av
Acl = AvB = 1
Feedback
B
circuit
© 2010 Pearson Education, Upper Saddle
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Chapter 17
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
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
Summary
Troubleshooting
Assume a troubleshooter measures dc parameters for a circuit that has
no signal output. For the amplifier shown, which of these dc readings
would indicate a problem? If wrong, what problem is indicated?
a) VS = 1.5 V Reading is too high;
+VDD= 12 V
there may be a drain-source short.
b) VD = +12 V Reading = VDD,
indicating no drain current.
c) VG = 0 V Reading is expected.
RD
3.9 k C2
C1
0.1 mF
RG
10 M
Electronics Fundamentals 8th edition
Floyd/Buchla
RS
560 
C3
100 mF
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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.
Saturation The state of a transistor in which the output
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 17
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.
Chapter 17
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.
Chapter 17
Selected Key Terms
Common-source An FET amplifier configuration in which the
source is the common terminal.
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 in such a
way as to create certain specified operating
conditions.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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.
Chapter 17
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 17
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.
Chapter 17
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
Electronics Fundamentals 8th edition
Floyd/Buchla
RE
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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 17
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.
Chapter 17
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
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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.
Chapter 17
Quiz
9. A Colpitts or Hartley oscillator both have
a. positive feedback.
b. amplification.
c. a closed loop gain of 1.
d. all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 17
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.
Chapter 17
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. c
6. a
2. d
7. d
3. b
8. b
4. a
9. d
5. c
10. b
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.