Download Solution Tutorial 1

Solutions – Tutorial Set 1 EMT 111/4 – Electronic Devices
1. What is the definition of semiconductor devices?
Electronic component made by semiconductor materials and can be used in a lot
of applications because it’s compact, suitable in large electromagnet spectrum
(from dc to UV frequency), its ability to conduct current and voltage in large
value and can be integrated in IC.
Important elements of electronic system.
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Explain with your own words an atom, electron, valence electron, free electron,
ionization and shell. What is the maximum number of electrons that can exist in
the third shell?
An atom
- the smallest particle of an element that retains the characteristics of that
element.
Electrons
- The basic particles of negative charge. Electrons move around the atomic
nuclei.
Valence electrons
- Electron in the valence shell and contribute to chemical reactions and
bonding within the structure of a material and determine its electrical
properties.
Free electrons
- The escape valence electron from valence shell. This phenomena happens
when we gives the external energy to the materials.
Ionization
- The process of losing a valence electron. This phenomena happens when
valence electron have a sufficient amount of energy and then escape from
the outer shell and the atom’s influence.
Shell
- Orbits that have been grouped into energy bands.
N e = 2n 2 = 2(3) 2 = 18
The maximum number of electrons in 3rd shell = 18 electrons.
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3. Discuss the different in terms of energy band between insulator, semiconductor and
conductor. Draw the figures to explain your answer.
Insulator – very wide energy gap. Valence electrons do not jump into the conduction
band except under breakdown conditions, i.e extremely high voltage, thus makes
insulator doesn’t conduct electrical current under normal operation.
Semiconductor – a much narrow energy gap compare to insulator. This gap permits some
valence electrons to jump into conduction band and become free-electrons. This gap also
makes semiconductor suitable for devices applications for a various temperature.
Conductor – the valence band and conduction band are overlap. It is because a large
number of free electrons in conduction band without any external energy. This overlap
makes conductor are not suitable for devices applications for high temperature
conditions.
4. Why does semiconductor have fewer free electrons than a conductor?
The valence electrons of semiconductors are more tightly bound to the atoms than
those of conductors.
5. How the covalent bond formed?
By sharing of valence electrons with neighboring atoms.
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6. Describe the properties of n and p-type semiconductors in terms of bonding diagram
and energy diagram.
In terms of bonding diagram:
p-type
n-type
* to increase the number of conduction
band electrons in intrinsic s/c (Si).
* pentavalent (5 e valence) atom are added
(As, P, Bi Sb).
* 4 atom’s valence electrons are used to
form the covalent bonds and leaving one
extra electron – becomes a conduction
electrons (do not leaves a hole in valence
band) and can be carefully controlled by
number of impurity atoms.
- called donor atom.
- electrons are the majority carriers
- holes are the minority carriers.
*to increase the number of holes in
intrinsic s/c (Si).
* trivalent (3 e valence) atom are added
(B, In, Ga).
* 3 atom’s valence electrons are used to
form the covalent bonds and create one
holes when each trivalent atom is added.
* number of holes can be carefully
controlled by number of impurity atoms.
- called acceptor atom.
- holes are the majority carriers.
- electrons are the minority
carriers.
Energy diagram:
Doping Process
- an additional energy level is created in the energy band structure.
n-type:
- an additional energy level (donor energy level) is created near the conduction band.
p-type:
- an additional energy level (acceptor energy level) is created near the valence band.
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7. Discuss the bias of a diode.
Forward Bias
- conditions that allows current through the pn junction.
- The negative terminal of VBIAS is connected to n-region, while positive
terminal connected to p-region.
- VBIAS must be greater than the barrier potential.
- The negative side of the bias-voltage source ‘pushes’ free electrons
toward the pn junction.
- Provides electron current and hole current.
- As more electrons flow into the depletion region, the number of positive
ions is reduced, and as more holes effectively flow into depletion region
on the other side of the pn junction, the number of negative ions reduced.
- Cause depletion region narrow.
- In terms of barrier potential, free electrons provided enough energy to
climb the energy hill and cross the depletion region.
- Efree e = VBARRIER.
- Additional small voltage drop occurs across p and n region due to internal
resistance – dynamic resistance (doped semiconductive materials).
Reverse Bias
- conditions that essentially prevents current through the diode.
- Positive terminal of VBIAS is connected to n region while the negative
terminal is connected to p region of diode.
- Because unlike charges attract, the positive side of the bias-voltage source
‘pulls’ the free-electrons away the pn junction – additional positive ions
are created.
- In the p-region, electrons from negative side of voltage source enter as
valence electrons and move from hole to hole toward the depletion region
and create additional negative ions.
- Cause the depletion region widens.
- Reverse current exist after the transitions current dies out- cause by the
minority carriers in the n and p regions – produced thermally by electronhole pairs.
- VREVERSE-BIAS = VBREAKDOWN , reverse current drastically increase and will
damage the diode.
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8. Analyze the voltage-current (V-I) characteristic curve of a diode. And what happens
to the barrier potential when the temperature increases?
FORWARD-BIASED
- Forward current IF increases very little (mA) upward along the vertical
axis until the forward voltage VF across pn junction reaches
approximately 0.7V at the knee of the curve.
- After this point (0.7V), IF increases rapidly but VF remains at
approximately 0.7V.
- Normal operation for forward-biased diode is above knee of the curve.
- The resistance of the forward-biased is not constant over the entire curve
– called dynamic or ac resistance, r’d.
REVERSE-BIASED
- Reverse voltage VR increases to the left along the horizontal axis, and the
reverse current IR increases downward along the vertical axis.
- Very little reverse current (µA or nA) until VR reaches approximately VBR
at the knee of the curve.
- After this point (VBR), IR increases rapidly resulting overheating and
possible diode damage but VR remains at approximately VBR.
The barrier potential decreases as temperature increases.
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9. Discuss the operation of diodes and explain the three diode models.
Discuss basic operation of diode
- Diode is a single pn junction device with conductive contacts and wire
leads connected to each region. Part of diode is n-type and other part is ptype semiconductor, respectively.
- Diode is in forward-biased when voltage source is connected with the
positive terminal at anode and negative terminal at cathode, respectively.
- IF is from anode to cathode.
- Diode in reverse-biased when the voltage source positive terminal at
cathode and negative terminal at anode, respectively.
Ideal Model
- is a simple switch, with forward-biased for closed (on) switch and reversebiased in open (off) switch as shown in Figure 1 below.
Fig. 1 (Ideal Diode Model)
-
Barrier potential, forward dynamic resistance and reverse current are
neglected.
VF = 0 because barrier potential and dynamic resistance are neglected.
The forward current determined by Ohm’s law, IF = VBIAS / RLIMIT.
IR = 0, since reverse current is neglected.
The reverse voltage equals the bias voltage, VR = VBIAS
Use the ideal model to troubleshoot or trying to figure out the operation of
a circuit and are not concerned with more exact values of voltage or
current.
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Practical Model
- Adds the barrier potential in series to the ideal switch model. When the
diode is forward-biased, it is equivalent to closed (on) switch.
- When diode in reverse, the switch is open (off) and just as in the ideal
model. Fig. 2 below shows the practical model of diode.
Fig. 2 (Practical Diode Model)
-
-
Forward-dynamic resistance and reverse current are neglected.
The diode assumed to have a voltage across it (right-shift from Fig. 2(c))
when forward-biased (VF = 0.7V (silicon) and VF = 0.3V (Ge)).
The barrier potential does not affect reverse-bias, so it is not a factor.
The forward current determine by Kirchhoff’s voltage law and
− VF
V
I F = BIAS
RLIMIT
The diode assumed to have zero reverse current, IR = 0 and VR = VBIAS.
Complete Model
- complete model of a diode consists of the barrier potential, the small r’d,
and large internal reverse resistance r’R.
- Reverse current taken into account because it provides a path for the
reverse current which is included in this diode model.
- Fig. 3 shows the complete diode model.
Fig. 3 (Complete Diode Model)
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-
-
When diode is forward-biased, it acts as a closed (on) switch in series with
the barier potential voltage and the small r’d.
When diode in reverse, it acts as open switch (off) in parallel with the
large r’R.
Barrier potential does not affect reverse bias, so it is not a factor.
The diode assumed to have a voltage across it when forward-biased
(right-shift), since barrier potential and r’d are included. This voltage, VF,
consists of the barrier potential voltage plus the small voltage drop across
the dynamic resistance.
Curve slopes because the voltage drop due to dynamic resistance
increases as the current increases.
The reverse current is indicated by the portion of the curve to the left of
the origin.
For the complete model of diode:
V F = 0.7V + I F r ' d and I F = (VBIAS − 0.7V ) /( RLIMIT + r ' d )
10. Assume that the diode in Fig. 1-36(a) (Flyod, pg. 32: Example 1-1) fails open.
What is the voltage across the diode and the voltage across the limiting resistor?
Diode is defective, then VD = 10V, thus no current, VLIMIT = 0V
11. Define the following term:
(i)
Laser = Light Amplification by Stimulated Emission of Radiation
(ii)
LED = Light Emitting Diode
(iii) Varaktor = A variable Capacitance Diode
12. Sketch the symbols of the following diodes:
(i)
Varaktor
:
(ii)
PIN
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Part B : Calculation
01.(a) Determine the forward voltage and forward current for the germanium.
Diode, if given Rlimit = 0.5 kΩ and Vbias = 20 V for each of the diode models. Also
find the voltage across the limiting resistor in each case. Assume rd’ = 100 Ω at
the determined value of forward current.
Answer :
a.
Ideal model :
VF = 0 V
IF =
Vbias
20V
=
= 40mA
Rlim it 0.5 × 10 3 Ω
Practical model :
V F = 0.3 V
IF =
Vbias − VF (20 − 0.3)V
=
= 39.4 mA
Rlim it
0.5 × 10 3 Ω
V RLIMIT = I F RLIMIT = (39.4 × 10 −3 A) × (0.5 × 10 3 Ω) = 19.7 V
Complete model :
V − VF
(20 − 0.3)V
I F = bias
=
= 32.83 mA
'
Rlim it + rd (0.5 × 10 3 Ω) + 100Ω
V F = 0.3 + I F rd' = 0.3 + 32.83 10 −3 A × 100Ω = 3.583 V
V RLIMIT = I F RLIMIT = 32.83 × 10 −3 A × 0.5 × 10 3 Ω = 16.415 V
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(b) Determine the reverse voltage and reverse current for the silicon.
Diode, if given Rlimit = 2 kΩ and Vbias = 8 V for each of the diode models. Also
find the voltage across the limiting resistor in each case. Assume IR = 2 µA at
the determined value of forward current.
b.
Ideal model :
IR = 0 A
V R = Vbias = 8 V
V RLIMIT = 0 V
Practical model :
IR = 0 A
V R = Vbias = 8 V
V RLIMIT = 0 V
Complete model :
I R = 2 µA
V R = I R RRLIMIT = 2 × 10 −6 A × 2 × 10 3 Ω = 4 mV
V R = Vbias − VRLIMIT = 8V − 0.004V = 7.996 V
02.
Determine the peak value of the output voltage for half–wave rectifier (use
silicon diode), if given the input voltage is 400 V and the turn ratio is 0.8.
Answer :
V p ( pri ) = V p (in) = 400 V
V p (sec ) = n V p (in) = 0.8 × 400 V = 320 V
V p (out ) = V p (sec) − 0.7V = 320V − 0.7V = 319.3 V
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03.
Derive the average value of the half-wave rectifier output voltage (Vavg) as given
in equation :
Vp
Vavg =
π
where : Vp is the peak value of the voltage
Answer :
The average value of a half-wave rectifier sine wave is the area under the
curve divided by the period (2π). The equation for a sine wave is:
V = V p sin θ
V AVG =
π
Vp
area
1
=
V p sin θ dθ =
− cos θ
∫
2π
2π 0
2π
(
[(− cos
2π
V AVG =
Vp
V AVG =
Vp
π ) − (− cos 0)] =
Vp
2π
)]
π
0
[− (−1) − (−1)] =
Vp
2π
[2] =
π
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04.
Based on the filtered bridge rectifier in Figure 1, answer the following questions
(All diodes are 1N4001).
(i). Determine the peak primary voltage, the peak secondary voltage, and the
unfiltered full-wave rectified peak voltage.
Answer :
Vp (pri) = √2 Vrms = √2 x 100V = 141.42 V
Vp (sec) = n Vp (pri)
n = Nsec/Npri = 1/10 = 0.1
Vp (sec) = n Vp (pri) = 0.1 x 141.42V = 14.142 V
Vp (out) = V (rect) = Vp (sec) – 1.4V = 14.142 V– 1.4 V = 12.742 V
(ii). Determine the peak to peak voltage ripple Vr(pp) at the output and the value
of the dc output voltage,.VDC
Answer :
The frequency of a full-wave rectified voltage become is 120 Hz (from 60 Hz
(frequency input) x 2 = 120 Hz).
1
1
Vr ( pp) =
V p (rect ) =
12.742V = 2.94 V
f RL C
120 Hz × 1.2 × 10 3 Ω × 30 × 10 −6 F
(
)
(
)
⎛
⎞
1
1
⎞
⎟⎟V p (rect ) = ⎛⎜1 −
V DC = ⎜⎜1 −
⎟ 12.742V = 11.27 V
3
−6
2
f
R
C
2
120
Hz
1
.
2
10
30
10
F
×
×
×
Ω
×
×
⎝
⎠
L
⎝
⎠
(iii). Calculate the ripple percentage?
Answer :
V ( pp ) 2.94V
r= r
=
= 0.26
VDC
11.27V
The percentage ripple is 26 %
(iv). How to reduce the ripple percentage in Figure 1?
(i)
increase the value of the filter capacitor
(ii)
increase the load resistance
Figure 1
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05.
Figure 2
A loaded zener regulator is shown in Figure 2. Given VZ = 8 V at IZT = 0.05 A,
IZK = 0.003 A, ZZ = 20 Ω and IZM = 0.01 A. Determine the maximum permissible load
currents.
Answer :
VZ (min) = VZ − ∆I Z Z Z
= 8V − (0.05 A − 0.003 A)(20Ω)
= 7.06 V
V R = Vin − VZ (min) = 25V − 7.06V = 17.94 V
VR 17.94V
=
= 179.4 mA
R
100
I L (max) = I T − I ZK = 0.1794 − 0.01 = 0.1694 A = 169.4 mA
IT =
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