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Chapter 9 Instructor Notes
After a brief introduction to semiconductors materials in Section 9.1, Section 9.2 introduces the pn
junction and the semiconductor diode equation and i-v curves. These sections could be bypassed in favor
of a more intuitive presentation, provided in Section 9.3 with an explanation of the basic circuit behavior of
the diode, beginning with large-signal models. This explanation is supplemented by the sidebar
Make the
Connection: Hydraulic Check Valves (pp. 466-467), which continues the electric-hydraulic analogy
introduced in earlier chapters, and provides an intuitive explanation of the operation of the semiconductor
diodes
The box Focus on Methodology: Determining the Conduction State of a Diode (p. 467) helps the
student understand how diode state of conduction can be determined. The large-signal circuit model
material is probably sufficient for the purposes of most introductory courses.
5nstructors who are interested in introducing the subject of small-signal models, in preparation for
a the study of transistor small-signal amplifiers, may find the rest of the section, covering small-signal
diode models (pp. 474-483), of interest. The concept of operating point is introduced, with a review of the
load-line equation and the definition of the quiescent operating point of a device. The box Focus on
Methodology: Determining the Operating Point of a Diode (p. 475) summarizes this material. Section 3.
can be recalled (or introduced for the first time) at this point. The solution methods used in Examples 9.5,
9.6, 9.7 emphasize the use of simple circuit models for the diode, together with the use of Thèvenin
equivalent circuits; this method is quite general, and will reinforce the importance (and understanding) of
the concept of equivalent circuits in analyzing more advanced circuits.
The use of Device Data Sheets is continued in this chapter with a summary data sheet on a general
purpose diode. The box Focus on Methodology: Using device data sheets (pp. 476-477) is designed to
familiarize the student with the basic contents and function of device data sheets. The instructor may wish
to expand on this introductory presentation by asking students to locate data sheets in the accompanying
CD-ROM or on the web, and to identify specific devices and their parameters for use in homework
problems.
Section 9.4 discusses various diode rectifier circuits; Section 9.5 introduces DC power supplies;
Zener diode circuits and voltage regulation. Section 9.6 analyzes various signal processing circuits, and
introduces two application examples in the boxes Focus on Measurements: Peak Detector Circuit for
Capacitive Displacement Transducer (pp. 499-501), which is tied to two earlier boxeson the capacitive
displacement transducer (pp. 147-148 and pp. 175-177), and Focus on Measurements: Diode Thermometer
(pp. 502-503). The latter example can be tied to a laboratory experiment
Finally, Section 9.7 introduces
photodiodes and solar cells and includes the box Focus on Measurements: Opto-Isolators (p. 506)
The homework problems present a graded variety of problems, mostly related to the 17 examples
and application examples presented in the text.
1
With many thanks to Bill Ribbens, who first suggested this idea to me some 20 years ago.
2
G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3
rd
Edition, Kendall-Hunt, 1998.
Learning Objectives
- Understand the basic principles underlying the physics of semiconductor devices in
general, and of the pn junction in particular. Become familiar with the diode equation
and i-v characteristic. Sections 1, 2.
- Use various circuit models of the semiconductor diode in simple circuits. These are
divided into two classes: large signal model, useful to study rectifier circuits, and small
signal models, useful in signal processing applications. Section 2.
- Study practical full-wave rectifier circuits and learn to analyze and determine the
practical specifications of a rectifier using large-signal diode models. Section 3.
- Understand the basic operation of Zener diodes as voltage references, and use simple
circuit models to analyze elementary voltage regulators. Section 4.
- Use the diode models presented in Section 2 to analyze the operation of various
practical diode circuits in signal processing applications. Section 5.
- Understand the basic principle of operation of photodiodes, including solar cells,
photosensors and light-emitting diodes. Section 6.
Section 9.1: Electrical Conduction in Semiconductor Devices
Section 9.2: The pn Junction and the Semiconductor Diode
Problem 9.
Solution:
Known quantities:
The Ionized Acceptor Density for a doped Silicon:
3
17
m
−
a a
N N , = 0
d
N
Find:
a) If this material is an N or P type extrinsic semiconductor.
b) Which are the majority and which the minority charge carriers.
c) The density of majority and minority carriers.
Analysis:
a) Each acceptor dopant atom introduces an additional positive charge carrier and a negative atomic ion.
The ion is NOT a charge carrier. The density of positive carriers [holes] increases because of the
doping so the material is extrinsic P type Silicon.
b) The majority carriers are the positive carriers or valence band holes; the minority carriers are the
negative carriers or conduction band free electrons.
c) + 0 − − = 0
−
po po a
CNE : p n N ,
po
io
po
p
n
CPE: n
2
2
−
a
po
io
po
N
p
n
p , ( ) ( ) 0
2 2
−
po po a io
p p N n
Using the quadratic equation:
3
16 16 17
12
32
2
17 17
12
2
2
m
p N N n.
po a a io
− −
where the negative answer is physically impossible.
Now use the CPE to obtain the minority carrier density:
3
15
17
32
2
m
p
n
n
po
io
po
Note that because of the doping, the hole density is now about 100 times the electron density. The
thermally produced carriers present in the intrinsic Silicon before doping has a small effect on the carrier
densities in the extrinsic Silicon. At higher doping levels, the effect becomes negligible.
As temperature increases, the densities of the thermally produced carriers increase and their effect on the
final carrier densities increase. At very high temperatures [about 175 C for Silicon] the thermally produced
carriers primarily determine the final carrier densities and the doping has a negligible effect [ie, the
semiconductor behaves as an intrinsic material]. This is why semiconductors cannot operate in high
temperature environments.
Analysis:
At a temperature of absolute zero, ALL valence electrons in a semiconductor are contained in a covalent
bond and there are NO charge carriers.
The internal or thermal energy of a solid material is caused by the vibration of the atoms and electrons
about their equilibrium position. As the temperature of the material increases, its thermal vibrational energy
increases. Some electrons will gain sufficient energy to escape the covalent bond in the valence band, and
"jump" past the energy gap into the conduction band. As a consequence, TWO charge carriers are
generated. The conduction or free electron in the conduction band is a negative charge carrier. The
vacancy in the valence band covalent bond or "hole" is a positive charge carrier.
A conduction band electron may also give up energy and recombine with a valence band hole.
The generation and recombination rates both increase with temperature. At any particular temperature,
they are equal and produce equal equilibrium densities of electrons and holes. The equilibrium carrier
densities increase with temperature. For Silicon at T = 300 K [approximately room temperature]:
3
16
3
10
m
carriers
cm
carriers
n p..
io io
[A number of carriers is a dimensionless quantity and may be omitted from the units.]
Almost all semiconductors devices are "doped" to achieve DIFFERENT densities of positive and negative
carriers. A "P-type" semiconductor has a higher density of positive carriers and an "N-type"
semiconductor has a higher density of negative carriers. However, at high temperatures the density of
thermally produced carriers becomes very large and significantly reduces or nullifies the effects of the
doping, ie, the positive and negative carrier densities become nearly equal. For this reason, semiconductor
devices cannot be used in high temperature applications. The limit in temperature depends on the
semiconductor material.
Problem 9.
Solution:
Find:
Describe the properties of donor and acceptor dopant atoms and how they affect the densities of charge
carriers in a semiconductor material.
Analysis:
An "intrinsic" semiconductor material is undoped. When dopant atoms are added the material becomes an
"extrinsic" semiconductor. Doping results in the replacement of an intrinsic atom with a dopant atom. As
few as one out of every million intrinsic atoms may be replaced.
A "donor" dopant atom has 5 valence electrons. Only 4 are required to complete the bonding structure in
the semiconductor material. The 5th electron requires very little energy to escape to the conduction band
and become a negative charge carrier. This leaves behind a donor atom with one missing electron or a
negative atomic ion. The ion is immobile and cannot move through the material; therefore, IT IS NOT A
CHARGE CARRIER.
Each donor contributes an additional negative carrier to the material. The increased density of negative
carriers results in an increased recombination rate which reduces the density of positive carriers. [The
PRODUCT of the two densities remains constant.] Materials doped with donor atoms are N type extrinsic
semiconductors. The majority carriers are conduction band electrons, the minority carriers are valence
band holes.
An "acceptor" dopant atom has 3 valence electrons; however, 4 are required to complete the bonding
structure in the semiconductor material. The "missing" 4th electron causes a vacancy or hole in the
bonding structure. Another valence electron may move to and occupy this hole thus eliminating it and
generating another hole and a negative atomic ion. The ion is immobile and cannot move through the
material; therefore, IT IS NOT A CHARGE CARRIER.
Each acceptor contributes an additional positive carrier to the material. The increased density of positive
carriers results in an increased recombination rate which reduces the density of negative carriers. [The
PRODUCT of the two densities remains constant.] Materials doped with acceptor atoms are P type
extrinsic semiconductors. The majority carriers are valance band holes, the minority carriers are
conduction band electrons.
Problem 9.
Solution:
Find:
Physically describe the behavior of the charge carriers and ionized dopant atoms in the vicinity of a
semiconductor PN junction that causes the potential [energy] barrier that tends to prevent charge carriers
from crossing the junction.
Analysis:
Semiconductor atoms are not shown in the two figures.
The circles represent ionized dopant atoms and the
uncircled plus and minus signs represent charge carriers.
The dotted line in the first figure represents how far the
depletion/space charge region extends into the P and N
regions.
Near the junction, the negative carriers in the N material
recombine with the positive carriers in the P material.
This forms a small region on either side of the junction
which is depleted of charge carriers; however, the ionized
dopant atoms are immobile and remain. Therefore, in the
N material the region has a net positive charge and in the
P material the region has a net negative charge. This
region is called either a "depletion" region [depleted of
carriers] or a "space charge" region [due to the dopant
ions].
If a conduction band free electron [a majority carrier] in the N material tries to cross into the P material, it
encounters and is repelled by the net negative charge [due to the ionized acceptor atoms] in the depleted
part of the P material.
If a valence band hole [a majority carrier] in the P material tries to
cross into the the N material, it encounters and is repelled by the net
positive charge [due to the ionized donor atoms] in the depleted part
of the N material.
The repulsion of the carriers is characterized as a Coulombic
"potential [energy] barrier". With no voltage applied across the
junction, Ohm's law requires the current to be zero. Actually, very,
very small equal but opposite currents do flow across the junction
but the net curent is zero.
The barrier can be decreased by applying a "forward bias" voltage
across the junction. This allows more carriers to cross the junction
and when this voltage is greater than a certain value [0.7 V for
Silicon] a significant current [milliamps] flows.
The barrier can be increased by applying a "reverse bias" voltage
across the junction. This increases the barrier and fewer majority
carriers have sufficient energy to cross the junction, ie, the current
essentially ceases. However, there is a VERY, VERY small
"reverse saturation current" [in the femtoamps range] due to the minority carriers. Since the minority
carriers are thermally produced, this current is dependent on temperature.
Figure F10SA.
Figure F10SB.
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.7 using the offset diode model.
Find:
A plot of v L
versus v
S
.
Analysis:
The circuit can be represented as shown in the following figure.
For
γ
v < V
1
, the diode is reverse biased, and = 0
L
v.
In term of
S
v , we have
1
R
R
v v V
S
L S γ
For
γ
v ≥ V
1
, i.e.
1
R
R
v V
S
S γ
the diode is forward biased, and
S L
S S
S
L
S L
S L
S
R R R
R R
V
R
v
v v V
R R R
R
V
R
v
v
1
1
1
1
1
γ
γ
γ
.
Where
−
S L
S S
S
R R R
R
R
R
V V
1 , tan
1
1
1
α
γ
.
v
S
v
L
0
α
V
S
11
Problem 9.
Solution:
Known quantities:
For the circuit of Figure P9.7: = 6 V
S
v , and the resistances = = = 1 kΩ
1 S L
R R R.
Find:
Determine
D
i and
D
v graphically.
Assumptions:
Use the diode characteristic of the 1N461A.
Analysis:
Replace the diode with an open circuit, and = 3 V
OC
D
v.
Replace the diode with a short circuit, and 2 mA
SC
D
i.
These are the end points of the load line.
The load line is superimposed on the diode characteristic in the figure.
From the intersection of the load line and the diode characteristic,
we see that ≈ 1. 5
D
i mA and ≈ 0. 75
D
v V.
Problem 9.
Solution:
Known quantities:
The current I = 1 mA, that make the diode to be above the knee of its i − v characteristic.
Find:
a) The value of R to establish a 5 mA current in the circuit.
b) With the value of R established in the preceding part, what is the minimum value to which the voltage
E could be reduced and still maintain diode current above the knee.
Assumptions:
V = 0. 7 V
γ
.
Analysis:
a)
3
−
R
b)
086 07 156 V
3
E...
E.
I
min
min
−
Load line
Assumptions:
The drop across each forward biased diode is 0.7 V.
Analysis:
a)
2
D and
4
D are forward biased;
1
D and
3
D are reverse biased. v 5 0_._ 7 4_._ 3 V
out
b)
1
D and
2
D are reverse biased;
3
D is forward biased. v 10 0_._ 7 9_._ 3 V
out
c)
1
D is reverse biased;
2
D is forward biased. v 5 0_._ 7 4_._ 3 V
out
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.15; v ( t ) sin ( , t )
S
= 10 2000 π.
Find:
The output waveform and the voltage transfer characteristic.
Assumptions:
The diode is ideal.
Analysis:
For v S
< 8 V, v
o
= 4 V. For v
S
≥ 8 V, v
o
= v
S
/2.
The voltage transfer characteristic is
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.15: v ( t ) 10 sin( 2 , 000 t ).
S
= π
Find:
The output waveform and the voltage transfer characteristic.
Assumptions:
The diode has an offset = 0. 6 V.
γ
V
Analysis:
For v S
< 6_._ 8 V, v
o
= 4 - V
γ
= 3.4 V. For v
S
≥ 6.8 V, v
o
= v
S
/2.
The voltage transfer characteristic is
v
o
0
8 v
S
v
o
0
8
16
t
v
(V)
5
0.5 1 1.5 2
0
4
(ms)
o
Problem 9.
Solution:
Known quantities:
Same as Problem 9.15 but with v ( t ) ( t )
S
= 1. 5 sin 2 , 000 π , the battery equal to 1 V and the resistors of
1kΩ.
Find:
The output waveform and the voltage transfer characteristic.
Assumptions:
The diode has an offset = 0. 6 V
γ
V , and = 200 Ω.
D
r
Analysis:
For v S
< 2 (1- V
γ
) = 0.8 V,
S
D
D
S
o
v
r
r
V
v
v
γ
For v S
≥ 0.8 V, v
o
= v
S
/2.
The voltage transfer characteristic is
v S
v
o
0 1.6 v
S
Device:
T
D
V
V
D
I I e
0
0
I
I
V V ln
D
D T
= , Circuit, KVL: − + + = 0
S D D
V I R V ,
R
V V
I
S D
D
A simultaneous solution (the lower left Q point) of the device and circuit characteristics is required. To do
this iteratively, initially assume a value for the diode voltage, say 0.7 V for a Silicon device. Then:
- Using the initial or new diode voltage and the circuit characteristic, determine a new diode current.
- Using this new diode current and the device characteristic, determine a new diode voltage.
ITERATE or REPEAT until convergence is obtained.
Voltage New Current New Voltage
700 mV 0.5 ma 377.2 mV
377.2 mV 0.5461 ma 379.5 mV
379.5 mV 0.5458 ma 379.5 mV <<< Convergence
Convergence is obtained after only 3 iterations and:
DQ
I
,
V 379_._ 5 mV
DQ
This is a much more accurate solution than that of Part a). The two solutions differ significantly because
the reverse saturation current given is atypically large.
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.18; the diode is fabricated from Silicon and
0
T
D
V
V
D
I I e. At
T = 300 K, 2 0310 A
15
0
−
I =. , = ≈ 26 mV
q
kT
V
T
, v 5_._ 3 V 7 cos ( t ) mV
S
= + ω ,
s
rad
ω = 377 , R = 4_._ 6 k.
Find:
Determine, using superposition and the offset voltage model for the diode, the DC or Q-point current
through the diode.
Analysis:
Suppress the AC component of the source voltage. Construct the DC equivalent circuit using the threshold
(or offset voltage model). The Q point is at the intersection of the device (diode) characteristic and the
circuit characteristic given by the KVL below. Here, the device characteristic is approximated by the
threshold voltage model giving the approximate Q point at the upper right.
The DC source voltage will tend to make the diode conduct. Assume the diode is on. Then:
V V 0. 7 V
D Don
−
,
S DQ D
KVL V I R V
,
1_._ 0 mA
R
V V
I
S Don
DQ
−
The current is positive so the assumption above that the diode is on is valid.
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.18; the diode is fabricated from Silicon and
0
T
D
V
V
D
I I e. At
T = 300 K, 250 10 A
12
0
−
I = , = ≈ 26 mV
q
kT
V
T
, v 4_._ 2 V 110 cos ( t ) mV
S
= + ω ,
s
rad
ω = 377 , R = 7 k. The DC operating point is: I 0_._ 546 mA
DQ
= , V 379_._ 5 mV
DQ
Find:
The equivalent small-signal AC resistance of the diode at room temperature at the Q point given.
Analysis:
0
I
V
I
V
I e
V
dV
dI
dI
dV
r
DQ
T
Q
D
T
Q
V
V
T
Q
D
Q D
D
D
D
T
D
Figure F22SB.
= = 0. 74 ⋅ 0. 021 = 15 mV
D DD
P v i
c. For R L
=100 Ω, we have
5 0. 757 4. 24 V
0. 0424 A
0. 757 V
L
D L
D
v
i i
v
Similarly, for R L
= 500 Ω, we have
5 0. 717 4. 283 V
0. 04283 A
0. 717 V
L
D L
D
v
i i
v
Problem 9.
Solution:
Known quantities:
A diode with the i-v characteristic in Figure 9.32, connected in series to a 2 V source and a load resistance
of 200
Ω .
Find:
a. The load current and voltage
b. The power dissipated by the diode.
c. The load current and voltage if the load is changed to 100 Ω and 300 Ω.
Analysis:
a. The operating point can be determined by using the load-line analysis
The load line is
D
D
L
D
D
v
v
R
v
i 0. 01 0. 005
By drawing this line on the top of Figure 9.32, the following operating point is obtained
= ≈ 6 mA; ≈ 0. 73 V⇒ = 2 − = 2 − 0. 73 = 1. 27 V
D L D L D
i i v v v
b. = = 0. 73 ⋅ 0. 006 = 4. 38 mV
D DD
P v i
c. For R L
=100 Ω, we have
2 0. 825 1. 175 V
- 75 mA
0. 825 V
L
D L
D
v
i i
v
0 0.2 0.4 0.6 0.8 1
0
Diode i−v curve
Diode voltage, V
Diode current, A
R
L
=100 Ω
R
L
=500 Ω
d. For R L
=300 Ω, we have
2 0. 7 1. 3 V
- 3 mA
0. 7 V
L
D L
D
v
i i
v
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.18; the diode is fabricated from Silicon and
0
T
D
V
V
D
I I e
. At
T = 300 K, 250 10 A
12
0
−
I = , = ≈ 26 mV
q
kT
V
T
, v 4_._ 2 V 110 cos ( t ) mV
S
= + ω ,
s
rad
ω = 377 , R = 7 k. The DC operating point is: I 0_._ 548 mA
DQ
= , V 0. 365 V
DQ
r 47_._ 45
d
Find:
Determine, using superposition the AC voltage across the diode and the AC current through it.
Analysis:
Suppress the DC component of the source voltage. Replace the diode with its AC equivalent resistance;
then:
d
d
d S
R r
r
V V
7406 0 V
V
d
, v ( t ) 740_._ 6 cos ( t ) V
d
= ω
1561 0 A
r
V
I
d
d
d
, i 15_._ 61 cos ( t ) A
d
= ω.
The total solution is then:
0 54810 1561 ( ) 10 A
− 3 − 6
i =. +. cos t
d
ω
( ) 0 365 7406 ( ) 10 V
− 6
v t =. +. cos t
d
ω
Section 9.4: Rectifier Circuits
Problem 9.
Solution:
Known quantities:
The circuit of Figure P9.26. The input voltage is sinusoidal with an amplitude of 5 V.
Find:
The average value of the output voltage.
Assumptions:
V = 0. 7 V
γ
.
Analysis:
The capacitor will charge to 5 V− 0_._ 7 V= 4_._ 3 Vand, therefore, the input sine wave will be shifted up
4.3 V to produce the output. As a result, after the cycle (the capacitor builds up its stored charge during
the third quarter cycle), the average value of the output will be 4.3 V.
Problem 9.
Solution:
Known quantities:
The rectifier circuit of Figure P9.27; v ( t ) = Asin ( 2 π 100 t ) V. The conduction must begin during each
positive half-cycle at an angle no greater than 5 °.
Find:
The minimum peak value
A
that the AC source must produce.
Assumptions:
V = 0. 7 V
γ
.
Analysis:
( )
( )
. V
sin
A sin. A
min min
°
°
Problem 9.
Solution:
Known quantities:
A half-wave rectifier is to provide an average voltage of 50 V at its output.
Find:
a) Draw a schematic diagram of the circuit.
b) Sketch the output voltage waveshape.
c) Determine the peak value of the input voltage.
d) Sketch the input voltage waveshape.
e) The rms voltage at the input.
Analysis:
a)
b)
c)
v 0_._ 318 v 50 v 157_._ 2 V
ave peak peak
d)
e)
1112 V
V
rms
in
Problem 9.
Solution:
Known quantities:
A rectifier circuit similar to that of Figure 9.25. Load Resistance 100
Ω , AC source voltage 30 V (rms).
Find:
The peak and average current in the load.
Assumptions:
Ideal diode.
Analysis:
The peak voltage is
0. 424 A
= 30 2 V⇒ = = =
R
V
V I
peak
peak peak
R
R R
The average current in the load is
