Key educational goals:
Develop the basic principle of operation of a diode. Classify the different
types of diodes and analyze their applications.
Reading/Preparatory activities for class
i)Textbook:
Chapter 10: Sections 10.1-10.7
ii) Power-point file:diodes
1
Questions to guide your reading and to think about ahead of time.
1. In what direction does current normally flow in an ordinary diode?
2. How is an ideal diode different from an actual diode?
3. How is a zener diode different from an ordinary diode?
4. Which type of diodes is used for voltage regulation?
5. Which type of diodes is used for rectification (converting AC to DC)?
6. Which types of diodes are used from clipping and clamping circuits?
7. What circuits elements are used for high voltage dc supply?
Introduction
Chapter 1: Introduction and Chapter 2:Resistive circuits
The main concepts for the module
1. Analyze how an ordinary diode conducts.
2. Evaluate the different models of an ordinary diode.
3. Develop the concept of a load line to determine an operating point in a diode ciruit.
4. Analyze a diode circuit based on assumption/contradiction.
5. Analyze how a zener diode based voltage regulator works.
6. Analyze rectifier circuit operations.
7. Analyze clipping, clamping and voltage multiplier cirucits.
2
Summary
The knowledge gained from this module will be useful for designing voltage regulators, rectifiers,
clamping circuits for protection as may be required to interface transducer outputs to digital
electronics and computers.
For next time
We will next look into the Bipolar Junction Transistors (BJT) whose base-emiiter characteristics
closely follow that of a diode but unlike a diode works like a current controlled current source.
Sample test/exam questions/problems to help you study:
1. Find the currents in diodes D3 and D4 (Fig.1) in the next slide.
2. In Fig.2 (next slide) Vss varies between 15-20V. RL varies between 100-500 ohms. Find the value
of R and its power rating.
3
Fig.1
Fig.2
AC Machines
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Semiconductor Diodes
•
•
•
•
•
•
•
•
•
•
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Diode structure
Diode model approximations
Operating points and load lines
Multiple diode circuits
Rectifier circuits and design
Zener diodes
Voltage regulators using zener diodes
Clipper and clamp circuits using zener diodes
Voltage multipliers
Other types of diodes
Shockley equation
Diodes
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Semiconductor diode
Diodes
6
First approximation of a semiconductor
diode (Ideal-Diode)
iD
iD
R=10 
Vs= 10V
iD = 1A
R=10 
Vs= 10V
Diodes
iD = 0A
7
Second approximation of a semiconductor
diode
iD
Diode on
Diode off
vD
Vf  0.7V
iD
iD
Vs= 10V
R=10 
iD = 0.93 A
R=10 
Vs= 10V
Diodes
iD = 0A
8
Third approximation of a semiconductor
diode
Diode on
iD
vD /iD =Diode body
(bulk) resistance Rf
Diode off
Actual
Characteristics
vD
Vf  0.7V
iD
iD
Vs= 10V
R=10 
iD = 0.62 A (say Rf = 5 )
R=10 
Vs= 10V
Diodes
iD = 0A
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How to Find Out the Exact
Diode Operating point?
Diodes
10
Example of Diode Load Line
If R = 200  , Vss = 3V in the
what is iD and vD?
Operating point
Diodes
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Answer to the Diode Load Line Problem
iD = 9.5 mA, vD = 1.1 V
Diodes
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Multiple Ideal-Diode Circuit Analysis
•Assume either a ‘on’ (short) or ‘off’ (open)state for a diode
•Determine the current direction in an ‘on’ diode.
•Determine the voltage polarity of an ‘off’ diode.
•If the assumption is correct, current flows from anode to
cathode in all ‘on’ diodes; cathode voltage should be positive
with respect to anode in all ‘off’ diodes.
•In a ‘n’ diode circuit 2n such combinations are possible
Diodes
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Example on Diode Circuit
Diodes
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What are the currents in D1,D2,D3,D4?
Assume ideal diodes
Diodes
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Answer and Another Question
iD1= 2 mA, iD2= 0 mA, iD3= 0 mA, iD4= 5 mA
What should be the minimum value of the current source in the
previous figure 10.17(c) in order to make D3 conduct?
Diodes
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Rectifier Circuits
•Rectifier circuits convert AC voltage to DC voltage
•The output DC voltage magnitude is controlled by controlling the input
AC voltage
•Usually classified by two types: Half-wave and Full-wave
•Has many applications such as battery charger, power supply.
Diodes
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Half-Wave Rectifier Circuits
•During the positive half-cycle of the source the diode conducts
•During the negative half-cycle the source diode blocks
•Current through the diode looks similar to the voltage across RL
Diodes
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A simple Battery charger-Example of a
Rectifier
• Can be used to charge a car battery from the alternator
Diodes
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Smoothening the Output Voltage of a
Rectifier-Add a Capacitor across Load
Diodes
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Design of Filter Capacitor
•Assume capacitor takes negligible time to charge
•Assuming constant load current , the charge lost by the capacitor is
Q  ILT (L21)
•The charge lost in the capacitor is reflected in the load voltage as
ripple voltage, Vr and
Vr  Q/C (L22)
•Eliminating Q from (L21) and (L22) one gets
C = ILT / Vr (L23)
where T = 1/f (f= frequency of the input sinusoidal source)
Diodes
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Design of Filter Capacitor(Contd..)
• Load Voltage VL= Vm- Vr /2; where Vm = peak of
the sinusoidal source
Diodes
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Full-wave rectifier
•Normally used with center-tap transformer
•Rectifies both the positive and negative half cycle
•Capacitor size is half that of the half-wave case for the same load
current and ripple voltage
Vout
Vin
Vin
C.T.
Vout
Diodes
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Diode-bridge Full-wave rectifier
•Can be used without a transformer
•Rectifies both the positive and negative half cycle
•Capacitor size is half that of the half-wave case for the same load
current and ripple voltage (why?)
Vin
Vout
Vin
Vout
Diodes
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Comparison of Rectifier Circuits
Half-wave
Full-wave
# of diodes
1
2
4
DC output
voltage(**)
Vin (peak)
Vin (peak) (*)
Vin (peak)
DC diode
current
PIV (***)
IL
0.5IL
0.5IL
2Vin (peak)
2Vin (peak)
Vin (peak)
f
2f
2f
Ripple
Frequency
Full-wave
Bridge
* With 1:2 transformer turns ratio; ** With Large C *** Peak Inverse Voltage (also with Large C)
Diodes
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Example on Rectifier Circuits
Design the capacitor (value and voltage rating) and the
diode (average current and (PIV) peak inverse voltage)
for a rectifier which has to supply 15 VDC and 2 ADC
current. Maximum ripple voltage, 𝑉𝑟 , (peak-to peak) is
1V. The input AC supply frequency is 60 Hz.
i) For a half-wave rectifier
𝐶=
𝐼𝐿
𝑉𝑟
.𝑇 =
𝐼𝐿 1
.
𝑉𝑟 𝑓
=
2 1
.
1 60
= 33,333 𝜇𝐹
𝑉𝑟
2
𝑉𝑚 = 𝑉𝑑𝑐 + = 15 + 0.5 = 15.5 𝑉.
Therefore the capacitor should be 33,333 𝜇𝐹 minimum
and rated for minimum 15.5 V.
The diode should have a minimum PIV of 2 ∗ 15.5𝑉 =
31𝑉 and current rating 2 ADC.
Diodes
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Example on Rectifier Circuits(2)
ii) For a full-wave rectifier
𝐶=
𝐼𝐿 𝑇
𝐼𝐿
𝑉𝑟 2
𝑉𝑟 2𝑓
. =
𝑉𝑚 = 𝑉𝑑𝑐 +
𝑉𝑟
2
.
1
2
= .
1
1 120
= 16,666 𝜇𝐹
= 15 + 0.5 = 15.5 𝑉.
Therefore the capacitor should be 16, 666 𝜇𝐹 minimum
and rated for minimum 15.5 V.
The diode should have a minimum PIV of 2 ∗ 15.5𝑉 =
31𝑉 and current rating 2/2 =1 ADC.
Diodes
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Example on Rectifier Circuits(3)
iii) For a full-wave bridge rectifier
𝐶=
𝐼𝐿 𝑇
𝐼𝐿
𝑉𝑟 2
𝑉𝑟 2𝑓
. =
𝑉𝑚 = 𝑉𝑑𝑐 +
𝑉𝑟
2
.
1
2
= .
1
1 120
= 16,666 𝜇𝐹
= 15 + 0.5 = 15.5 𝑉.
Therefore the capacitor should be 16, 666 𝜇𝐹 minimum
and rated for minimum 15.5 V.
The diode should have a minimum PIV of 15.5𝑉 and
current rating 2/2 =1 ADC.
Diodes
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SPICE simulation of Full-wave
Bridge Rectifier Circuit
Diodes
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Zener Diodes
•Unlike ordinary diodes they are used in the breakdown region
•Zener diodes are used to get reasonably regulated dc voltage when
the input voltage and load resistance vary
•This is achieved by passing controlled current through the zener
diode in the breakdown region
•The nominal breakdown voltage VZ is specified as the zener
voltage (example 4.7 V zener, 8.2 V zener)
• Symbol looks like
Diodes
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Example1 on Zener Diode Voltage
Regulator
VSS = 24V
R = 1.2 k
RL= 6 k;
1.2 k
Find IS and
vL using the
zener
characteristics
Diodes
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Example1 on Zener Diode Voltage
Regulator(contd..)
•Compute the Thevenin equivalent of the previous circuit
with the zener diode as the load
• Thevenin voltage
RL
VT  Vss
 20V;12V
R  RL
RR L
 1;0.6 k
• Thevenin Resistance R T 
R  RL
•We can then write VT +RTiD+vD = 0 and find out
vD,, iD using the zener diode characteristics
•vL = vD and IS = vL /RL + iD
Diodes
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Example1 on Zener Diode Voltage
Regulator(contd..)
Load Line : vD = -VT -RTiD
Diodes
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Answer to Example1 on Zener Diode
Voltage Regulator
vL = 10V
IS = vL /RL + iD = 10/6 +10 mA = 11.67mA
vL = 9.5 V
IS = vL /RL + iD = 9.5/1.2 +5 mA = 12.92mA
Note that load resistance change has caused the output
voltage to change by 0.5V due to non-ideal reverse
characteristics of the zener diode
Diodes
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Example 2 on Zener Diode Voltage
Regulator (ideal-characteristics)
Find v0 for iL= i) 0 ii) 30 iii) 80 mA
Diodes
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Example 2 on Zener Diode Voltage
Regulator
The characteristics shown for the zener diode is the ideal
characteristics- one without any zener bulk resistance. Hence
the zener will be able to provide ideally regulated voltage
unless the load current demand exceeds (15- v0)/100 = 50 mA.
When iL  50 mA the zener absorbs the differential (excess)
current to maintain the load voltage at 10 volts.
When iL  50 mA the zener loses its regulating capacity(since it
cannot generate current!) and the output voltage starts drooping.
Then v0 = 15-100 iL.
Diodes
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Answer to Example 2 on Zener Diode
Voltage Regulator
i) 10V, ii) 10V, iii) 7 V
Another question
What is the maximum possible value iL ? What is the load
resistance under this condition?
Diodes
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Wave-shaping Circuits
•Necessary to shape waveforms
•Many examples can be found in transmitters and receivers in TV or
radar or equipment control and protection circuits
•Uses diodes and zener diodes for clipping and clamping (shaping)
input
Diodes
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A Clipper Circuit (Assume ideal diode)
Diodes
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A Clipper Circuit (Practical
Implementation)
Diodes
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A Clamp (adds dc offset to ac) Circuit
Assumption:
RC (2)/
Diodes
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How Does A Clamp Circuit Work?
•In the positive half cycle C gets charged through D to 10V (peak of
sinewave + 5 V) with the straight plate of C at a higher potential. D
Clips the output to a maximum of -5V.
•In the negative half cycle D is reverse biased. The output can reach
a minimum of –15V (-VC + negative peak of sinewave).
Diodes
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Application of Clamp Circuit –A
Voltage-Doubler
Assumption:
R1C1 or R1C2 (2)/
Diodes
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How does a Voltage-Doubler works
•In the negative half cycle C1 gets charged through D2 to peak of
sinewave (say Vm) with the right plate of C1at a higher potential.
D1is reverse biased since D2 conducts.
•In the positive half cycle D1 is conducting and D2 is reverse biased.
Thus C2 gets charged to the maximum of Vm+ Vmsin(t) or 2Vm
with the top plate of C2 at a higher potential
Diodes
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Voltage-Doubler waveforms
Diodes
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Voltage-Doubler waveforms –different
start point
Diodes
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Other types of Diodes
•LED (Light Emitting Diodes)
•They emit light when forward biased. Typical forward voltage
drop for these diodes are 1.5 - 2.5V for currents between 10 – 20
mA.
•Application areas: Instrument/Equipment panel indicators, Seven
Segment Displays.
•Photo Diodes
•Get forward biased when light falls on them.
•Application areas: Optocouplers for electrical isolation.
•Schottky diodes, Ultrfast recovery diodes
•Application areas:Power Electronics
Diodes
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Shockley Equation
•An equation named after the inventor of Transistor
•Describes approximately the Diode curves like the one shown
below
•Is an important equation to be used in Transistor Analysis
Diodes
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Shockley Equation(2)
  vD  
  1
Shockley equation: i D  IS exp 
  nVT  
kT
VT 
q
Where IS is the saturation current,VT is the thermal voltage,n is the
emission coefficient that takes values between 1 and 2, k is Boltzmann’s
constant ( k = 1.38*10-23 J/K) ,q is the magnitude of electrical charge
on an electron (q = 1.60*10-19 J/K), T is p-n junction temperature in
degree Kelvin.
For example, at T =3000K, IS = 10-14 A, VT  0.026 V
Diodes
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Plus and minuses of Shockley Equation(2)
•Fairly accurate with diode near the forward biased region
•Not a good predictor of IS which happens to be much larger in
magnitude
•Does not predict reverse breakdown
•Usually simpler models for diodes are useful
Diodes
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