Department of Engineering Science and Physics
College of Staten Island
Introduction to Solid State Electronics
Semiconductors: These are the materials, which do not have free electrons to
support the flow of electrical current through them at room temperature.
However, valence electron may become free electrons if sufficient energy is
induced into the material, for example, by heating the material.
Solid State Devices: These are nonlinear devices (diodes, transistors, etc.),
which are made by creating “junction(s)” of two types of semiconductor materials:
(1) p-type materials and (2) n-type materials. Both p-type and n-type materials
are created by adding “impurities” into two “pure” semiconductor elements: (1)
Silicon and (2) Germanium. Here we shall discuss how p-type or n-type materials
are created using Silicon, which undergoes through a process known as
“doping.” The process of creating p-type or n-type materials using Germanium is
essentially similar.
Silicon-based p-type or n-type materials:
Silicon atom has 14 protons and 14 electrons. Figure 1 shows the atomic
structure of a Silicon atom. The electrons in the outer ring are called valence
electrons. In pure silicon the atoms are held tightly together through what is
known as “covalent bond,” in which the four valence electrons are shared with
four neighboring atoms to form covalent bonds. Unlike conductors pure
semiconductors do not have free electrons at room temperature to support the
flow of electric current for any useful purpose. Figure 2 shows the crystal
structure of pure Silicon.
Figure 1: Atomic structure of the Silicon atom.
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Department of Engineering Science and Physics
College of Staten Island
Figure 2: Crystal Structure of pure Silicon.
Doping:
It is the process of mixing an impurity with a pure semiconductor material. It is
important to carefully choose impurity materials because impurities’ atoms should
be able to “fit” into the crystal structure of the semiconductor material without
disturbing the lattice. Two kinds of impurities are used:
1. The impurities with five valence electrons in the outer ring are called
“donors.” The semiconductor material doped with a donor impurity is
known as n-type semiconductor material. Phosphorous, Arsenic, and
Antimony are examples of donor impurities.
2. The impurities with three valence electrons in the outer ring are called
“accepters.” The semiconductor material doped with an accepter impurity
is known as p-type semiconductor material.
N-type Material:
A Phosphorous atom has a total of 15 electrons, which leaves five electrons in
the outer ring. When Phosphorous is added to Silicon as impurity one
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Department of Engineering Science and Physics
College of Staten Island
Phosphorous atom makes covalent bonds with four Silicon atoms using four out
of five of the electrons in the outer ring. This leaves the fifth phosphorous
electron not being attached to any particular atom and thus can move around as
a free electron. These free electrons support the flow of electrical current through
the doped semiconductor material. The number of free electrons depends on the
amount of impurity added. If a large amount of Phosphorous is mixed the
semiconductor is known as “heavily doped.” Heavily doped n-type Silicon has
more free electrons than a “lightly doped” n-type Silicon, in which a relatively
small amount of impurity (phosphorous) is added. Figure 3 shows the crystal
structure of n-type Silicon semiconductor.
Figure 3: Crystal Structure of the n-type Silicon semiconductor.
P-type Material:
A Boron atom has a total of 5 electrons, which leaves three electrons in the outer
ring. When Boron is added to Silicon as impurity one Boron atom makes covalent
bonds with three Silicon atoms using the three electrons in the outer ring. This
leaves the crystal structure with one missing covalent bond between the Boron
and a Silicon atom. This missing covalent bond is represented as a “hole” in the
crystal structure of the p-type material. The term hole implies that there is an
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Department of Engineering Science and Physics
College of Staten Island
Figure 4: Crystal Structure of the p-type Silicon semiconductor.
Figure 4: Crystal Structure of the p-type Silicon semiconductor after an electron
fills a hole leaving behind a hole.
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Department of Engineering Science and Physics
College of Staten Island
opportunity for an electron from the neighboring covalent bond to jump in and
complete the missing covalent bond (filling the hole). Figure 4 shows the crystal
structure of the p-type Silicon semiconductor. Note that when an electron jumps
in and fills a hole to complete the missing covalent bond, it leaves a hole at its
original location (the now broken covalent bond). Figure 5 shows the crystal
structure of the p-type Silicon semiconductor after an electron jumps in to fill the
hole (leaving a hole at its place). This phenomenon continues throughout the ptype semiconductor giving the perception of “free holes” analogous to free
electrons in n-type semiconductor. Figure 6 depicts a representation of the p-type
semiconductor with free holes and of the n-type semiconductor with free
electrons. It is important to note that even though the p-type semiconductor has
free holes and the n-type semiconductor has free electrons, both semiconductors
are electrically neutral.
Figure 6: Representation of the p-type and n-type semiconductors.
P-N Junction:
The p-type or n-type semiconductors are of little use by themselves. However, an
interesting phenomenon takes place when a p-type semiconductor is placed next
to an n-type semiconductor, thus forming a “junction.” At the junction some of the
free electrons from the n-type semiconductor cross the junction and fill the holes.
This movement of electrons across the junction leaves the region in n-type
semiconductor adjacent to the junction with excess positive charge. The
electrons filling the holes in the region adjacent to the junction on the p-type
semiconductor create excess negative charge. Figure 7 shows the dynamics of a
p-n junction. This oppositely charged region on both sides of the junction, known
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Department of Engineering Science and Physics
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as the “depletion region,” creates a potential barrier, which is 0.7 volts for Silicon
based semiconductor and 0.2 for Germanium based semiconductor.
Figure 8: The p-n junction.
The p-n junction forms a simple solid-state device known as the diode. Let’s now
examine the behavior of the p-n junction when a voltage source is connected
across the device. If we connect the positive and negative terminals of a power
supply to the p-type and n-type semiconductors, respectively, the junction is said
to be forward-biased. Figure 9 shows the dynamics of the junction under this
condition. When the potential applied across the p-n junction is sufficiently high
Figure 9: A forward-biased p-n junction.
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Department of Engineering Science and Physics
College of Staten Island
(greater than 0.7 volts for Silicon based semiconductor and greater than 0.2 volts
for the germanium based semiconductor), the free electrons from the n-type
material will be able to jump across the depletion region and go towards the
positive terminal of the power supply. This causes the ID to flow through the p-n
junction, from positive terminal to the negative terminal of the power supply.
On the other hand, if we connect the positive and negative terminals of a power
supply to the n-type and p-type semiconductors, respectively, the junction is said
to be reverse-biased. Figure 10 shows the dynamics of the junction under this
condition. When the potential applied across the p-n junction with this polarity,
the free electrons in the n-type semiconductor are attracted towards the positive
voltage and move away from depletion region. The holes on the p-type
semiconductor are attracted to the negative terminal of the power supply. This, in
effect, widens the depletion region and essentially makes it impossible to have a
current flow from positive terminal to the negative terminal of the power supply. In
this way, a p-n junction (diode) essentially acts like a valve, in which air can be
pumped in one direction but cannot come out (flow) in the other direction.
Figure 10: A reverse-biased p-n junction.
The behavior of the p-n junction under forward -and reverse-biased conditions is
very useful in several applications. Here, we’ll consider one such application
known as rectification. Rectification is a process of converting an AC signal to a
DC signal. Figure 11 shows a simple circuit of a rectifier. This particular rectifier
is called half-wave rectifier as it only produces the positive half cycle of the
input signal at the output of the rectifier. Note that during the positive half cycle,
the diode is forward biased and the current ID flows through the diode. During
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Department of Engineering Science and Physics
College of Staten Island
the negative half cycle, however, the diode becomes reverse-biased and
prevents the current flow through the diode.
Figure 11: Half-wave rectifier.
The average output voltage can be found as follows:
The input voltage is a sine wave with a period T. The output is also a sine wave
with a period T and a peak voltage of Vout max . However, the negative half of the
sine wave is missing. Therefore, the average voltage at the output can be
computed as:
Vout _ av
Vout _ max
1T
= ∫Vout _ max sin ωtdt =
T 0
T
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T/2
∫
0
sin
Vout _ max
2Πt
dt =
= 0.318Vout _ max
T
Π
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Department of Engineering Science and Physics
College of Staten Island
Applications of Diode
Diode operating characteristics:
As mentioned earlier, a silicon diode has potential barrier of 0.7 volts at the p-n junction.
We’ll refer to this potential barrier as VO N. The diode will conduct electrical current if it
is forward-biased and the applied voltage is greater than VON. If the diode is reversebiased, it will not conduct electrical current until the voltage is greater than what is know
as breakdown voltage (referred to as peak reverse voltage on manufacturers’ data
sheets). There is, however, a small amount of “leakage current” present while the diode is
in reverse biased mode. At the breakdown voltage, a large amount of current flows
through the device, which may permanently damage the device. The breakdown voltage
for 1N4001 diode is 50 volts peak. Figure 1 shows the diode operating characteristics for
the silicon-based devices. Note that the operating characteristics in the reverse-biased
region are not drawn to the scale.
Figure 1: Operating characteristics for a typical diode (Silicon).
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Department of Engineering Science and Physics
College of Staten Island
Rectification:
The behavior of the p-n junction under forward -and reverse-biased conditions is
very useful in several applications. Here, we’ll consider one such application
known as rectification. Rectification is a process of converting an AC signal to a
DC signal. Figure 2 shows a simple circuit of a rectifier. This particular rectifier is
called half-wave rectifier as it only produces the positive half cycle of the input
signal at the output of the rectifier. Note that during the positive half cycle, the
diode is forward biased and the current ID flows through the diode. During the
negative half cycle, however, the diode becomes reverse-biased and prevents
the current flow through the diode.
Figure 11: Half-wave rectifier.
The average output voltage can be found as follows:
The input voltage is a sine wave with a period T. The output is also a sine wave
with a period T and a peak voltage of Vout max . However, the negative half of the
sine wave is missing. Therefore, the average voltage at the output can be
computed as:
Vout _ av =
Vout _ max
1T
Vout _ max sin ω tdt =
∫
T 0
T
T /2
∫
0
sin
Vout _ max
2Πt
dt =
= 0. 318Vout _ max
T
Π
Note that
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Department of Engineering Science and Physics
College of Staten Island
2Π
T
Half-wave rectification is not very efficient because the power from
transferred during only the positive half cycle. Another process,
rectification, transfers the power from input to the output during
negative half cycles. Figures 3 and 4 show one possible way
rectification using a “bridge rectifier.”
ω = 2Π f =
input to the output is
known as full-wave
both the positive and
to perform full-wave
Figure 3: Bridge rectifier. The current flow is shown for the positive half cycle.
Note that in Fig. 3, the diodes A and C are forward-biased and the diodes B and
D (with dotted circle around them) are reverse-biased during the positive half
cycle. Therefore, the current from the AC source passes through the diode A,
leaves the bridge at the terminal marked “+” and then passes through the load
resistor RL. The current then enters the bridge at the terminal marked “–,“ passes
through the diode C and returns to the AC source. The output voltage at this
point is the same as the input voltage during the positive half cycle.
Figure 4 shows the current flow during the negative half cycle of the input
voltage. During the negative half cycle the diodes B and D are forward-biased
and the diodes A and C (with dotted circle around them) are reverse-biased.
Therefore, the current from the AC source passes through the diode D, leaves
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Department of Engineering Science and Physics
College of Staten Island
the bridge at the terminal marked “+” and then passes through the load resistor
RL. The current then enters the bridge at the terminal marked “–,“ passes through
the diode B and returns to the AC source. The output voltage still has the same
polarity and, therefore, the same as the input voltage during the positive half
cycle. That is, the negative half cycle has, in effect, revered at the output. In this
way, we get power transferred from the input to the output of the circuit for both
the positive and negative half cycles; however, the output voltage remains
positive during both half cycles. Therefore, the average output voltage for the fullwave rectifier is twice of the average output voltage for the half-wave rectifier.
Figure 4: Bridge rectifier. The current flow is shown for the negative half cycle.
Note that the output voltage after rectification is unidirectional (DC). However, the
amplitude of the output voltage is not constant. This phenomenon is known as
the ripple effect. In the pulsating output voltage has the frequency equal to the
input AC voltage in the case of half-wave rectification and the frequency twice
that of the input AC voltage in the case of full-wave rectification. The difference
between the maximum and the minimum output voltage of the unidirectional
pulsating output voltage is known as ripple. In many applications requiring DC
voltage it is highly desirable to have a relatively constant unidirectional voltage;
that is, the output voltage should have minimum ripple. The ripple can be
reduced by filtering the output of the rectifier.
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Department of Engineering Science and Physics
College of Staten Island
Filtering:
Figure 5 shows a half-wave rectifier with a filter. During the positive cycle the
capacitor is charged to Vout_max . As the out decreases from Vout_max , the capacitor
starts discharging through the load resistor RL. If the value of the capacitor is
properly chosen, it can be made to discharge at a rate slower than the change in
the output voltage. The idea is to discharge the capacitor such that it is not fully
discharged before the next positive cycle starts. At the next positive cycle the
capacitor is charged again to V out_max .
Figure 5: Half-wave rectifier with filter.
Figure 6 shows the full-wave rectifier with filter. Note that output voltage is
present during both the positive and negative half cycles of the input voltage.
Therefore, the capacitor starts charging again after a relatively short period of
time once it begins discharging when compared to half-wave rectifier.
If RL and an acceptable value of ripple are known, the value of the capacitor can
be found using the following approximate formula.
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Department of Engineering Science and Physics
College of Staten Island
Zener Diode
Zener diode is a special diode that is fabricated in such a way as to make the
breakdown characteristics very steep. Furthermore, the device is made to work in
the reverse breakdown region. Figure 7 shows the operating characteristics of
Zener diode. Note that when the reverse biased voltage reaches VZ there is a
sharp increase in the reverse biased current (IZ). It is important to limit the current
through Zener diode to IZmax to avoid permanent damage to the device.
Figure 7: Operating characteristics of the Zener diode.
Zener diode is used to build voltage regulators. A voltage regulator provides a
constant voltage to the load under varying source voltage or varying load
condition. Figure 8 shows a simple circuit for a voltage regulator using Zener
diode. The current through Zener diode IZ can be found as follows:
IR = IZ + IL
also, I R =
VS − VZ
− IL .
Ri
Note that,
IZ =
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VS − VZ
, therefore,
Ri
(2)
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Department of Engineering Science and Physics
College of Staten Island
VS − VZ VS −VZ
=
.
(3)
IR
IZ + I L
Let’s now examine the conditions under which the current through Zener diode is
minimum or maximum. It can be seen from Eq. 2 that IZ is a minimum when VS is
a minimum and IL is maximum. That is,
Ri =
I Z min =
VS min − VZ
− I L max .
Ri
(4)
Also, IZ is a maximum when V S is a maximum and IL is minimum. That is,
I Z max =
VS max − VZ
− I L min .
Ri
(5)
Note that the IZmin given by Eq. 3 must be greater than the minimum required
current in the breakdown region specified by the manufacturer of the Zener
diode. Similarly, ZI max given by Eq. 4 must be less than the maximum allowed
current in the breakdown region specified by the manufacturer of the Zener
diode. As a rule of thumb, IZmin should be not less than 10% of IZmax to assure that
the diode operates in the constant breakdown voltage region. That is,
IZmin = 0.1 IZmax.
(6)
From Eqs. 4, 5, and 6 IZmax is given by
I Z max =
I L min (VZ − VS min ) + I L max (VS max − VZ )
.
VS min − 0. 9VZ − 0.1VS max
(7)
Figure 8: Zener voltage regulator.
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Department of Engineering Science and Physics
College of Staten Island
Design Example 1:
Design a full wave rectifier for the following conditions:
1. The minimum output voltage should be at least 90% of the maximum
output voltage.
2. The load resistance may vary from 1.5 KO to 2.0 KO.
Assume ideal diodes are used. The peak input voltage is 100V.
Solution:
We’ll use the full wave bridge rectifier. Since we are assuming that ideal diodes
are used, the maximum output voltage VOmax is the same as input peak voltage.
That is, VOmax = 100V. The minimum output voltage should be atleast 0.9x VOmax.
Therefore, VOmin = 0.9 x 100 = 90V. The value of the capacitor can be found using
the following Eq.
C=
VO max
∆Vfp RL
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Department of Engineering Science and Physics
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Where,
? V = V Omax - VOmin = 100 – 90 = 10V.
VOmax = 100V
fP = 120 (Note that fP = 120 for full wave rectifier and fP = 60 for the half wave
rectifier).
RL = 1.5 KO (Note that we must choose smallest value of RL if we are given a
range of values for RL).
C=
100
= 55.6 x10 −6 = 55.6 µF
3
10x120 x1.5 x10
Design Example 2:
Design a 10V Zener regulator for the following conditions:
1. The load current IL ranges from 100 mA to 200 mA.
2. The source voltage ranges from 14 V to 20 V.
Solution:
Form condition 1 we have:
ILmin = 100 mA and ILmax = 200 mA
Form condition 2 we have:
VSmin = 14 V and V Smax = 20 V
We can find IZmax using Eq. 7. That is,
I Z max =
I L min (VZ − VS min ) + I L max (VS max − VZ )
VS min − 0. 9VZ − 0.1VS max
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Department of Engineering Science and Physics
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I Z max =
100 x10 − 3 (10 − 14) + 200 x10 − 3 (20 − 10)
= 0.53 A
14 − 9 − 2
We’ll use Eq. 5 to calculate Ri (given below).
I Z max =
Ri =
VS max − VZ
− I L min . This equation can be rearranged as:
Ri
VS max − VZ
20 − 10
=
= 15. 9Ω
I Z max + I L min 0.53 + 0.1
Note that ILmin = 100 mA = 100x10-3 A = 0.1 A.
Now we’ll calculate power ratings for the resister and Zener diode. Power can be
calculated as:
Power = V x I = I2 x R = V2 / R.
Power rating for Zener diode:
the voltage across the Zener diode is VZ = 10 V. The maximum current through
the Zener diode is IZmax = 0.53A. Therefore, the power rating for the Zener diode
is:
P = 10 X 0.53 = 5.3 Watts
Power rating for Resistor R i:
The current through the resistor Ri is given by
IR = IZmax + ILmin = 0.53 + 0.1 = 0.63 A.
The maximum voltage across Ri is given by
V = VSmax – VZ = 20 –10 = 10 V
Therefore, the power rating of Ri is:
10 X 0.63 = 6.3 Watts.
This completes the design of the voltage regulator.
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