Download Tutorial 3Answer 1. Zener diode has two types of reverse

Tutorial 3Answer
1. Zener diode has two types of reverse breakdown. State and explain these reverse
breakdown.
Avalanche breakdown is a high-field effect that occurs when the electrostatic field
strength associated with the p-n junction is strong enough to pull electrons out of the
valence band within the depletion region.
Zener breakdown is a type of reverse breakdown that occurs at relatively low reverse
voltages. The n-type and p-type materials of a zener diode are heavily doped, resulting
in a very narrow depletion region. Therefore, the electric field existing within this region
is intense enough to pull electrons from their valence bands and create current at a low
reverse voltage (VR).
2. An 7.5 V zener diode (7.5 V at 30 0C) has a positive temperature coefficient of 0.05
%/0C. What is the zener voltage at 60 0C?
The change in zener voltage is
ΔVZ = VZ × TC × ΔT
= (7.5V)(0.05 %/oC)(60 oC – 30 oC)
= (7.5 V)(0.0005/oC)(30 oC) = 112.5mV
Notice that 0.05%/oC was converted to 0.0005/oC. The zener voltage at 60 oC is
VZ + ΔVZ = 7.5V + 112.5 mV = 7.61V
3. Determine the minimum and the maximum input voltages that can be regulated by the
zener diode in Figure below.
From the data sheet in Figure, the following information for the IN4733 is obtained:
VZ = 5.1 V at IZT = 49 mA, IZK = 1 mA, and
ZZ = 7 Ω at IZT.
VOUT ≈ 5.1V – ΔVZ = 5.1 V – (IZT – IZK)ZZ
= 5.1 V – (48 mA)(7 Ω) = 5.1 V – 0.336 V
= 4.76 V
VIN(min) = IZKR + VOUT
= (1 mA)(100 Ω) + 4.76 V = 4.86 V
PD (max)
1W
I ZM 

 196mA
VZ
5.1V
VOUT ≈ 5.1V – ΔVZ = 5.1 V + (IZM – IZT)ZZ
= 5.1 V + (147 mA)(7 Ω) = 5.1 V + 1.03 V
= 6.13 V
VIN(max) = IZMR + VOUT
= (196 mA)(100 Ω) + 6.13 V = 25.7 V
4. Determine the minimum and the maximum load currents for which the zener diode in
Figure will maintain regulation. What is the minimum RL that can be used? VZ = 12V,
IZK = 1 mA, and IZM = 50 mA. Assume ZZ = 0 Ω and VZ remains a constant 12 V
over the range of current values, for simplicity
When IL = 0 A (RL = ∞), IZ is maximum
V  VZ
I Z (max)  I T  IN
R
24V  12V

 25.5mA
470
Since IZ(max) is less than IZM, 0 A is an acceptable minimum value for IL because the
zener can handle all of the 25.5 mA. IL(min) = 0 A
The maximum value of IL occurs when IZ is minimum (IZ = IZK),
IL(max) = IT – IZK = 25.5 mA – 1mA = 24.5 mA
The minimum value of RL is
RL(min)=VZ/IL(max) = 12 V/24.5 mA = 490 Ω
5. Determine the output voltage for each zener limiting circuit in Figure below.
5.4V
0
-7,5V
6. Briefly explain the basic operation of LED.
The basic operation of LED is as illustrated in Fig. 3-14:
“When the device is forward-biased, electrons cross the p-n junction from the n-type
material and recombine with holes in the p-type material. These free electrons are in the
conduction band and at a higher energy than the holes in the valence band.
When recombination takes place, the recombining electrons release energy in the form
photons.
A large exposed surface area on one layer of the semiconductive material permits the
photons to be emitted as visible light.”
This process is called electroluminescence.
Various impurities are added during the doping process to establish the wavelength of the
emitted light. The wavelength determines the color of visible light.