ELECTRICAL SCIENCE-2
(15B11EC211)
UNIT-5
Lecture-1
1
Topics to be Discussed
• Introduction
• Diode Representations
• The P-N Junction Under No Bias, Reverse Bias, and Forward Bias
• V-I Characteristic of Ideal Diode, and Practical Diode
• Si Diode Vs Ge Diode
• Effect of Temperature
• Example
• Junction Capacitances
• Transition Capacitance
• Diffusion Capacitance
• References
2
Introduction [1, 2]
• The semiconductor diode is basically a p-n junction diode.
• A p-n junction diode is two-terminal or two-electrode semiconductor device, which
allows the electric current in only one direction while blocks the electric current in
opposite or reverse direction.
• If the diode is forward biased, it allows the electric current flow. On the other hand,
if the diode is reverse biased, it blocks the electric current flow.
• A p-n junction is formed at the boundary between a p-type and n-type
semiconductor created in a single crystal of semiconductor by doping.
• Methods:
• Ion implantation
• Diffusion of dopants
• Epitaxy (growing a layer of crystal doped with one type of dopant on top of a layer of crystal
doped with another type of dopant).
3
Diode Representations
4
The p-n Junction Under No Bias [1, 2]
• In a p-n junction, without an external applied voltage, an equilibrium condition is
reached in which a potential difference is formed across the junction. This potential
difference is called built-in potential. It is also called barrier potential or contact
potential or diffusion potential.
5
The p-n Junction Under No Bias [1, 2]
• The regions nearby the p-n interfaces lose their neutrality and become charged, forming
the space charge region or depletion layer or transition region.
• Therefore, under no-bias conditions, no external current exists as Diffusion Current (Idif)
= Drift Current (Idr).
6
The p-n Junction Under Reverse Bias [1, 2]
• The small amount of current that exists under reverse bias condition is called reverse saturation current (Is).
• It also causes widening of the depletion layer as the carriers move away from the junction, leaving more
donor and acceptor ions depleted.
• When the potential formed by the widened depletion layer equals the applied voltage, the current will cease
except for the small thermal current (due to minority carriers).
7
The p-n Junction Under Reverse Bias [1, 2]
•
•
•
•
Net barrier potential, VB increases.
Diffusion current, Idif due to majority carriers reduces to almost zero.
Drift current, Idr slightly increases.
The net current, IR remains constant till breakdown, hence called saturation current, Is.
8
The p-n Junction Under Forward Bias[1, 2]
• Positive biases lower the bands in p-region. As a result, electrons in an n-region and
holes in a p-region will have smaller barriers to overcome and diffuse to the other side.
This leads to a shrinking depletion region and increased conductivity.
• At the junction the electrons and holes combine so that a continuous current can be
maintained.
9
The p-n Junction Under Forward Bias [1, 2]
•
•
•
•
•
Applied voltage opposes the contact potential.
Net barrier potential, VB is reduced.
Diffusion current, Idif increases.
Drift current, Idr slightly decreases.
The forward biasing current is as follows: I F
= I dif − I dr
10
V-I Characteristic of Ideal Diode [1, 2]
• The V-I characteristic of the ideal diode is highly non-linear; although it consists of two straight line
segments, these are at 90° to one another. A non-linear curve that consists of straight line segments is said to
be piece-wise linear.
11
V-I Characteristics of Practical Diode [1, 2]
• The V-I characteristics or voltage-current characteristics of the practical diode is shown in the below
figures.
• The horizontal line in the below figures represents the amount of voltage applied across the p-n junction
diode whereas the vertical line represents the amount of current flows in the p-n junction diode.
12
• The forward voltage at which the current through the junction starts increasing rapidly, is called
the knee voltage or cut-in voltage. It is generally 0.7V for a silicon diode and 0.2V or 0.3V for a
germanium diode.
• The reverse voltage at which p-n junction of a diode breaks down with sudden rise in reverse
current is known as break down voltage.
13
• Diode Equation:
where,
I D = I S (e
qVD
 kT
− 1)
IS = Reverse saturation current
q = Charge of electron (1.6×10-19C)
VD = Applied forward bias voltage across diode in Volts
k = Boltzmann’s constant (1.38×10-23J/K)
η = Non-ideality or emission coefficient or semiconductor constant (typically 1 for Ge and 2 for Si )
T = Junction temperature in Kelvins (Note: If the temperature is given in Celsius (C) then it can be
converted to Kelvin (K) by the help of following relation, C+273 = K)
• Thermal Voltage: VT = kT/q = 26mV at room temperature (300K)
• So Diode Equation can also be represented as follows:
I D = I S (e
VD
VT
− 1)
14
• As per biasing conditions, the working of diode can be explained in following three
cases.
• Case 1:
No bias: When VD = 0, ID = 0
• Case 2:
Forward bias: VD is positive
VD
VT
ID  ISe
• Case 3:
Reverse bias: VD is negative, diode current quickly saturates.
I D = −I S
15
Example [2]
Find the applied voltage on a forward biased diode if the current is 1mA and reverse saturation
current is 10-10A. Temperature is 25°C and take ideality factor as 1.5.
Solution:
16
Silicon and Germanium Semiconductor Diodes [1]
• For designing the diodes, silicon is more preferred over germanium.
• The p-n junction diodes made from silicon semiconductors work at high temperature than the germanium
semiconductor diodes.
• Forward bias voltage for silicon semiconductor diode is approximately 0.7 volts whereas for germanium
semiconductor diode is approximately 0.3 volts.
• Silicon semiconductor diodes do not allow the electric current flow, if the voltage applied on the silicon diode
is less than 0.7 volts.
• Germanium semiconductor diodes do not allow the electric current flow, if the voltage applied on the
germanium diode is less than 0.3 volts.
• Most importantly, the cost of silicon semiconductors is low when compared with the germanium
semiconductors.
17
Si Diode Vs Ge Diode
Diode Characteristics
Vt is cut-in, knee, offset, turn-on, or threshold voltage.
Is is reverse saturation current.
Vt ≡ 0.7V (for Si diodes)
Vt ≡ 0.3V (for Ge diodes)
18
Effect of Temperature [2]
❑ Effect of temperature:
• Reverse saturation current, IS = IS0 2(∆T/10) = IS0 2(T2 – T1)/10.
• For every 10 degree rise in temperature, IS doubles.
• The width of the depletion layer decreases with rise in temperature and hence barrier
potential also decreases with rise in temperature.
• Barrier potential decreases by 2 mV for each Celsius degree rise in temperature.
• Therefore, temperature and applied reverse bias are very important factors in designs
which changes the reverse saturation current.
19
Effect of Temperature [2]
Diode Characteristics with Variations in Temperature
20
Example [2]
The threshold voltage of a silicon diode is 0.7V at 25°C. What will be the value of threshold voltage if
the junction temperature rises to 100°C.
Solution:
21
Example [2]
The reverse saturation current of a diode at 25°C is 1.5 × 10-9A. What will be reverse current at
temperature 30°C?
Solution:
22
Junction Capacitances [2, 3]
❑ Junction Capacitance:
• There are two kinds of junction capacitances as follows:
1. Transition or Depletion Capacitance, CT : This is predominant in reverse bias. It
decreases with applied voltage.
2. Storage or Diffusion Capacitance, CD: This is predominant in forward bias. It is
normally of the order of several hundred pF.
23
Transition Capacitance [2, 3]
❑ Transition Capacitance (CT):
d
d
+
C=
Q =
Q−
dV
dV
A
CT =
d
where, ε = Permittivity of the semiconductor, A = Area of plates or p-type and n-type regions, d = Width of
depletion region, Q = Charge across the capacitance
• In reverse biased region, the transition or depletion region capacitance (or charge storage
capacitance) exists.
• It depends upon area of the junction, the width of depletion region ( the width of
depletion region depends on reversed biased voltage) and dielectric constant of the
material.
• CT decreases with increase in the reverse biased voltage.
• Varicap/Varactor diode (variable capacitance diode) make use of variation in CT .
24
Diffusion Capacitance [2, 3]
❑ Diffusion capacitance:
dQ
CD =
where,
dV
C = Diffusion capacitance
D
dQ = Change in number of minority carriers stored outside the depletion region
dV = Change in voltage applied across diode
• Diffusion capacitance is directly proportional to the electric current or applied voltage. If large electric
current flows through the diode, a large amount of charge is accumulated near the depletion layer. As a
result, large diffusion capacitance occurs.
• In the similar way, if small electric current flows through the diode, only a small amount of charge is
accumulated near the depletion layer. As a result, small diffusion capacitance occurs.
• When the width of depletion region decreases, the diffusion capacitance increases. The diffusion
capacitance value will be in the range of nano farads (nF) to micro farads (μF).
• In case of forward bias region, diffusion capacitance is prominent and transition capacitance may neglected.
25
References
[1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and Circuit Theory’,
PHI, 7e, 2001.
[2] D.C. Kulshreshtha, ‘Electronic Devices and Circuits’, New Age, 2e, 2006.
[3] A. S. Sedra,, and K. C. Smith, ‘Microelectronic Circuits’. Oxford Oxford
University Press, 7th Edition, 2012.
26
ELECTRICAL SCIENCE-2
(15B11EC211)
UNIT-5
Lecture-2
27
Topics to be Discussed
•
•
•
•
•
•
•
•
•
•
•
Diode Resistances
Static or DC Resistance
Dynamic or AC Resistance
Reverse Resistance
Diode Equivalent Circuits
Circuit Model of a Diode
Simplified Circuit Model
Other Types of Diodes
Advantages of p-n Junction Diode
Possible Applications of Diodes
References
28
Diode Resistances [1, 2]
1. Static or DC Resistance (High value): It is the resistance offered by the
diode to a dc current.
2. Dynamic or AC Resistance (Low value): It is the resistance offered by
the diode to ac current.
3. Reverse Resistance (Very high value): It is the resistance offered by the
diode in reverse bias condition.
29
Diode Resistances [2]
1. Static or DC Resistance of diode:
2. Dynamic or AC Resistance of diode:
3. Reverse Resistance of diode:
4. Average Resistance of diode:
(defined between limits of operation)
30
Static or DC Resistance [1, 2]
• When forward biased voltage is applied to a diode that is connected to a DC circuit, a
DC or direct current flows through the diode.
• Direct current or electric current is nothing but the flow of charge carriers (free
electrons or holes) through a conductor. In DC circuit, the charge carriers flow
steadily in single direction or forward direction.
• The resistance offered by a p-n junction diode when it is connected to a DC circuit is
called static resistance.
• Static resistance is also defined as the ratio of DC voltage applied across diode to the
DC current or direct current flowing through the diode.
• The resistance offered by the p-n junction diode under forward biased condition is
denoted as RF.
31
Static or DC Resistance [2]
VD OB OB
RF =
=
=
= cot 
I D OA BP
32
Example [2]
Determine the DC resistance of a silicon diode, whose V-I characteristics is
given in Fig. at (a) ID = 2 mA, (b) ID = 20mA, and (c) VD = -10 V.
33
Solution:
34
Dynamic or AC Resistance [2]
• The dynamic resistance is the resistance offered by the p-n junction diode when AC
voltage is applied.
• When forward biased voltage is applied to a diode that is connected to AC circuit, an
AC or alternating current flows though the diode.
• In AC circuit, charge carriers or electric current does not flow in single direction. It
flows in both forward and reverse direction.
• Dynamic resistance is also defined as the ratio of change in voltage to the change in
current. It is denoted as rf.
35
Dynamic or AC Resistance [2]
Vd
rf =
= cot β
I d
36
Example [2]
The V-I characteristics of a silicon diode in forward bias is given in Fig. Calculate the
DC and AC resistances of the diode when the diode current is 20 mA.
37
Solution:
38
Dynamic Resistance of a Diode [2]
Estimate the value of AC resistance that would be offered by a semiconductor diode at (a)
ID = 10mA, (b) ID = 20mA.
Solution:
VT 25(mV )
rf =

I D I D (mA)
39
Reverse Resistance [1, 2]
• Reverse resistance is the resistance offered by the p-n junction diode when it is
reverse biased.
• When reverse biased voltage is applied to the p-n junction diode, the width of
depletion region increases. This depletion region acts as barrier to the electric current.
• Hence, a large amount of electric current is blocked by the depletion region. Thus,
reverse biased diode offers large resistance to the electric current.
• The resistance offered by the reverse biased p-n junction diode is very large
compared to the forward biased diode. The reverse resistance is in the range of mega
ohms (MΩ).
40
Diode Equivalent Circuits [1]
• An equivalent circuit is a combination of elements properly chosen to best represent
the actual terminal characteristics of a device, system, or such in a particular
operating region.
• Piece-wise Linear Equivalent Circuit:
41
Diode Equivalent Circuits [1]
Simplified Equivalent Circuit:
• For most applications, the resistance, rav is sufficiently small to be ignored in
comparison to the other elements of the network.
• The removal of rav from the equivalent circuit is the same as implying that the
characteristic of the diode appears as shown in Figure.
Fig. Simplified equivalent circuit for silicon semiconductor diode
42
Diode Equivalent Circuits [1]
Ideal Equivalent Circuit :
• Now that rav has been removed from the equivalent circuit let us take it a step further
and establish that a 0.7V level can often be ignored in comparison to the applied
voltage level.
• In this case, the equivalent circuit will be reduced to that of an ideal diode as shown in
following Figure.
43
Circuit Model of a Diode [2]
44
Circuit Model of a Diode [2]
45
Simplified Circuit Model [2]
46
Example [2]
For the circuit shown in given Figure, find the current through the diode ‘D’ assuming
the diode to be (a) ideal, (b) real with piecewise linear circuit model having threshold
voltage Vt = 0.3 V, and average resistance rav = rf = 25 Ω.
47
Solution:
(a) Assuming diode ‘D’ to be ideal: the current through the diode ‘D’ is given as
V AA 10
ID =
=
= 222mA
R1
45
(b) Considering the diode to be real with piecewise linear circuit model
VTH = open circuit voltage across AB
VTH = VAA
R2
5
= 10 
= 1V
R1 + R2
45 + 5
48
RTH =
R1 R2
45  5
=
= 4.5
R1 + R2 45 + 5
VTH − Vt 1.0 − 0.3 0.7
ID =
=
=
= 23.7mA
RTH + rf 4.5 + 25 29.5
49
Example [2]
Determine the current ‘I’ in the circuit shown in Fig. Assume the diodes to be of silicon
and forward resistance of diodes to be zero.
50
Solution:
51
Example [2]
Calculate the current through 48 Ω resistor in the circuit shown in Fig. Assume the
diodes to be of silicon and forward resistance of each diode is 1 Ω.
52
Solution:
53
Other Types of Diodes [2]
The various types of diodes are as follows:
• p-n junction diode
• Zener diode
• Tunnel diode
• Schottky diode
• Varactor diode
• Diac
• Triac
• Silicon Controlled Rectifier (SCR) or Thyristor
• Light Emitting Diode
• Photodiode
• Laser diode
54
Other Types of Diodes [2]
(Thyristor)
55
Advantages of p-n Junction Diode
• A p-n junction diode is the simplest form of all the semiconductor devices. Moreover,
diodes plays a major role in many electronic devices.
• A p-n junction diode can be used to convert the alternating current (AC) to the direct
current (DC). These diodes are used in power supply devices.
• If the diode is forward biased, it allows the current flow. On the other hand, if it is
reverse biased, it blocks the current flow.
• In other words, the p-n junction diode becomes on when it is forward biased whereas the
p-n junction diode becomes off when it is reversed biased (i.e. it acts as switch).
• Thus, the p-n junction diode is used as electronic switch in digital logic circuits.
56
Possible Applications of Diodes
• Diodes are used in rectification.
• Diodes are used in clamping circuits for DC restoration.
• Diodes are used in clipping circuits for wave shaping.
• Diodes are used in voltage multipliers.
• Diodes are used as switch in digital logic circuits used in computers.
• Diodes are used in demodulation circuits.
• Laser diodes are used in optical communications.
• Light Emitting Diodes (LEDs) are used in digital displays.
• Diodes are used in voltage regulators.
57
References
[1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and Circuit Theory’,
PHI, 7e, 2001.
[2] D.C. Kulshreshtha, ‘Electronic Devices and Circuits’, New Age, 2e, 2006.
58
ELECTRICAL SCIENCE-2
(15B11EC211)
UNIT-5
Lecture-3
59
Topics to be Discussed
•
•
•
•
•
•
•
•
•
•
•
Rectifier Circuits
Half Wave Rectifier
Parameters of the Half Wave Rectifier
Efficiency of Half Wave Rectifier
Ripple Factor
Peak Inverse Voltage
Peak Factor
Form Factor
Effect of Using a Silicon diode with VT = 0.7V
Examples
References
60
Rectifier Circuits [1, 2]
• A rectifier is a circuit which converts the alternating current (AC) input power into
a direct current (DC) output power.
• One of the most important applications of diodes is in the design of rectifier circuits.
• A diode rectifier forms an essential building block of the dc power supplies required to
power electronic equipment.
• A block diagram of such a power supply is shown below.
61
Rectifier Circuits [1, 2]
• Two Types of rectifications are possible.
– Half Wave Rectifier
– Full Wave Rectifier
62
Half Wave Rectifier [1, 2]
• The process of removing one half of the input signal to establish a dc level is called half wave
rectification.
• In Half wave rectification, the rectifier conducts current during positive half cycle of input ac signal
only. Negative half cycle is suppressed.
63
Half Wave Rectifier [1, 2]
Operation:
• Case 1: During “positive” half cycle (0 to T/2), the diode is forward biased.
64
Half Wave Rectifier [1, 2]
• Case 2: During “negative” half cycle (T/2 to T), the diode is reverse biased.
Therefore, NO current flows through the diode or circuit.
65
Parameters of the Half Wave Rectifier [2]
• Let vi = Vm sinωt be the input voltage to the rectifier, where Vm is the peak input
voltage.
• Let rf is the resistance in the forward direction i.e. in the ‘ON’ state, and RR (= ∞)
in the reverse direction i.e. in the ‘OFF’ state.
 Im
i=
0
for 0  t  
for   t  2
where,
Vm
Im =
rf + RL
66
DC output current or Average Current (Idc):
2
I dc
(
)

2


Im
1

Area
1
1 

I m − cos t 0 =
= 0.318 I m
=
=
idt =
I m sin tdt + 0dt =
 2

Base 2
2 
0

0




or I dc
If RL>>rf, then
I dc
Vm
=
 r f + RL
(
Im =
)
Vm
rf + RL
Vm
=
RL
DC output voltage or Average Voltage (Vdc):
Vdc = I dc  RL =
If RL>>rf, then
Vdc =
Vm
Im

 RL
=
Vm
Vm
 RL =
rf
 r f + RL

 1 +
 RL
(
)





67
RMS Current (Irms):
I rms =
1
2
2

i 2 d t =
0

2


1
2
2
 I m sin tdt + 0dt 
2 


0

I m2
=
2
RMS Voltage (Vrms):
If RL>>rf, then



0

(1 − cos 2t ) dt = I m
2
Vrms
Im
= I rms  RL =
 RL
2
Vrms
Vm
=
2
2
=
Vm
Vm
 RL =
rf
2 r f + RL

21 +
 RL
(
)




Im =
Vm
rf + RL
68
Efficiency of Half Wave Rectifier [2]
• The ratio of DC output power to the applied input AC power is known as rectifier
efficiency.
D𝐶output power 𝑃𝑑𝑐
Rectifierefficiency,𝜂 =
=
Total power
𝑃𝑎𝑐
2
 Im 
   RL
 
2
Pdc
I dc
 RL
RL
4
= 2
=
= 2
2
Pac I rms  RL + r f
 RL + r f
 Im 
   RL + r f
 2 
0.406
=
rf 

1 +

 R 
L 

(
)
(
)
(
)
• The maximum efficiency of a half wave rectifier is equal to 40.6% (i.e. ηmax = 40.6%).
69
Ripple Factor [2]
• Ripple Factor is the ratio of rms value of ac component present in the
rectified output to the average value/dc value of rectified output. It is a
dimensionless quantity and denoted by ‘r’.
2
2
I 2rms − I dc
Vrms
− Vdc2
Ripple factor (r ) =
=
2
I dc
Vdc2
Ripple factor of Half wave rectifier:
2
 Im 
 
2

Ripple factor (r ) =
− 1 = 1.21
2
 Im 
 
 
70
Peak Inverse Voltage [1, 2]
• Peak inverse voltage (PIV) or peak reverse voltage (PRV) is the maximum voltage
that the diode can withstand without breakdown during reverse bias condition.
VPIV  Vm
71
Peak Factor [2]
• Peak Factor is defined as the ratio of maximum/peak value to the
RMS value of an alternating quantity.
Vm
Peak value I m
Peak Factor =
=
=
rms value I rms Vrms
• For the half wave rectifier :
Im
Im
Peak Factor =
=
=2
I rms I m
2
72
Form Factor [2]
• The ratio of the RMS value to the average value of an alternating
quantity (current or voltage) is called Form Factor.
I rms Vrms
rms value
Form Factor =
=
=
average value I avg Vavg
• For the half wave rectifier :
I rms
Form Factor =
I avg
Im

2
=
= = 1.57
Im 2

73
Effect of Using a Silicon Diode with VT = 0.7V [1]
• For the forward-bias region in the HWR:
74
Half Wave Rectifier
Properties
Ideal diode
Diode with VT
Im/π
Im/π
Vm/(rf+RL)
(Vm-VT)/(rf+RL)
0.318Vm
0.318(Vm-VT)
Maximum efficiency
0.406/(1+(rf/RL))
0.406/(1+(rf/RL))
Peak Inverse Voltage
Vm
Vm
D.C. Current (Idc)
Maximum Current (Im)
D. C. voltage (Vdc)
75
Example [1]
(a) Sketch the output vo and determine the dc level of the output for the network of
given figure.
(b) Repeat part (a) if the ideal diode is replaced by a silicon diode.
76
Solution:
(a) In this situation the diode will conduct during the negative part of the input as
shown in Fig., and vo will appear as shown in the figure.
For the full period, the dc level is
Vdc = −0.318Vm = −0.318  20 = −6.36V
• The negative sign indicates that the polarity of the output is opposite to the
defined
polarity.
77
(b) Using a silicon diode, the circuit and the output vo will appear as shown in the figure
below.
Vdc = −0.318(Vm − VT ) = −0.318  (20 − 0.7 ) = −6.14V
78
Example [2]
A diode having internal resistance rf = 20Ω is used for half-wave rectification. If the applied
voltage v = 50sinωt and load resistance RL= 800 Ω, find: (i) Im, Idc, Irms (ii) AC power input
and DC power output (iii) DC output voltage (iv) Efficiency of rectification.
Solution: v = 50sinωt
Maximum voltage, Vm = 50V
(i)
79
Example [2]
A half-wave rectifier is used to supply 50V DC to a resistive load of 800 Ω. The diode
has a resistance of 25 Ω. Calculate AC voltage required (Vm).
Solution: Output DC voltage, Vdc = 50V
Diode resistance, rf = 25 Ω
Load resistance, RL= 800Ω
Let Vm be the maximum value of AC voltage required
80
Example [2]
An AC supply of 230 V is applied to a half-wave rectifier circuit through a transformer of
turn ratio 10 : 1. Find (i) the output DC voltage and (ii) the peak inverse voltage. Assume
the diode to be ideal.
81
Solution: RMS primary voltage = 230V
Maximum primary voltage (Vpm)
Vpm = 2  r.m.s primary voltage = 2  230 = 325.3
Vsm = V pm 
N2
1
= 325.3  = 32.53
N1
10
(i) output DC voltage (Vdc):
Vdc = I dc  RL =
Im

 RL =
Vsm

=
32.53

= 10.36V
(ii) Peak inverse voltage: During the negative half-cycle of a.c. supply, the diode is
reverse biased and hence conducts no current. Therefore, the maximum secondary
voltage appears across the diode.
VPIV = Vsm = 32.53V
82
References
[1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and Circuit
Theory’, PHI, 7e, 2001.
[2] D.C. Kulshreshtha, ‘Electronic Devices and Circuits’, New Age,
2e, 2006.
83
ELECTRICAL SCIENCE-2
(15B11EC211)
UNIT-5
Lecture-4
84
Topics to be Discussed
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Full Wave Rectifier
Center Tapped Full Wave Rectifier
Parameters of the Full Wave Rectifier
Efficiency of Full Wave Rectifier
Ripple Factor
Peak Inverse Voltage
Peak Factor and Form Factor
Effect of Using a Silicon diode with VT = 0.7V
Examples
Full Wave Bridge Rectifier
Peak Inverse Voltage
Effect of Using a Silicon diode with VT = 0.7V
Examples
Advantages and Disadvantages of Full Wave Bridge Rectifier
References
85
Full Wave Rectifier [1, 2]
• The full wave rectifier is a very useful circuit for AC to DC conversion and it is designed using the
diodes.
• Unlike the half wave rectifier, the full wave rectifier gives output during both half cycles.
• The output of the Full wave rectifier is pulsating DC.
• The ripple in the output waveform can be reduced using the filter.
• There are two types of full wave rectifier circuits as follows:
– Center tapped full wave rectifier
– Full wave bridge rectifier
86
Center Tapped Full Wave Rectifier [1]
• A popular full wave rectifier appears in following Figure with only two diodes but
requiring a center-tapped (CT) transformer to establish the input signal across
each section of the secondary of the transformer.
87
Center Tapped Full Wave Rectifier [1]
Operation:
• Case 1: During the positive portion of ‘vi’ applied to the primary of the transformer,
the network will appear as shown in Figure.
88
Center Tapped Full Wave Rectifier [1]
• Case: 2 During the negative portion of the input, the network appears as shown in Figure.
• Reversing the roles of the diodes but maintaining the same polarity for the voltage across the load
resistor, R.
• The net effect is the same output as that appearing in positive half cycle with the same DC levels.
89
Parameters of the Full Wave Rectifier [1, 2]
• Let vi = Vm sinωt be the input voltage to the rectifier, where Vm is the peak input
voltage.
• Let rf is the resistance in the forward direction i.e. in the ‘ON’ state, and RR (= ∞) in
the reverse direction i.e. in the ‘OFF’ state.
 Im
i1 = 
0
for 0  t  
for   t  2
 0
i2 = 
I m
for 0  t  
for   t  2
where,
Vm
Im =
rf + RL
90
Average Current (Idc):
I dc =
1
2
2

i1dt +
0
1
2
2
 i dt
2
0

 1
  I m sin tdt + 0  +

 2
0

I m I m 2I m
=
+
=
= 0.636I m
1
=
2



2Vm
or I dc =
 (rf + RL )
If RL>>rf, then I = 2Vm
dc
RL
Average Voltage(Vdc):
Vdc = I dc  RL =
If RL>>rf, then
Vdc =
2Vm

 2

 0 +  I m sin tdt 





Im =
Im

 RL
=
Vm
rf + RL
2Vm
2Vm
 RL =
rf 
 (rf + RL )

 1 + 
 RL 
91
RMS Current (Irms): Since current is of the same form in the two halves.
I rms




1
2
2
2
=
i d t =
  I m sin tdt 

 0
 0

1
=
I m2



(1 + cos 2t ) dt =
0
RMS Voltage (Vrms):
Vrms
2
Im
2
Vm
Im
= I rms  RL =
 RL= 2(r + R )  RL =
f
L
2
If RL>>rf, then
Vrms
Vm
=
2
Vm
rf 


2 1 +
 RL 
92
Efficiency of Full Wave Rectifier [1, 2]
• The ratio of DC power output to the applied input AC power is known as rectifier
efficiency.
DC power output Pdc
Rectifier efficiency,  =
=
Input AC power Pac
2
 2I m 

  RL
  
2
Pdc
I dc
 RL
RL
8
= 2
=
= 2
2
Pac I rms  RL + r f
 RL + r f
 Im 

  RL + r f
 2
0.812
=
rf 

1 +

 R 
L 

(
)
(
)
(
)
• The efficiency of a center tapped full wave rectifier is equal to 81.2% (i.e. ηmax =
81.2%)
93
Ripple Factor [1, 2]
• Ripple Factor is the ratio of RMS value of ac component present in the rectified output to
the average value of rectified output. It is a dimensionless quantity and denoted by ‘r’.
2
2
I 2rms − I dc
Vrms
− Vdc2
Ripple factor (r ) =
=
2
I dc
Vdc2
Ripple factor of full wave rectifier:
2
 Im 


2

Ripple factor (r ) =
− 1 = 0.48
2
 2I m 


  
94
Peak Inverse Voltage [1, 2]
• Peak inverse voltage (PIV) or peak reverse voltage (PRV) is the maximum voltage that
the diode can withstand without breakdown during reverse bias condition.
• The required PIV rating for the Center tapped full wave rectifier can be determined as
shown below.
• Applying Kirchhoff”s voltage law,
VPIV = Vm + vo = Vm + Vm
VPIV  2Vm
95
Peak Factor and Form Factor [1, 2]
• For the full wave rectifier, the expressions for peak factor
and form factor are as follows:
Peak Factor:Peak
Factor =
Im
I r .m. s
Im
=
= 2 = 1.414
Im
2
Im
Form Factor =
Form Factor:
I r .m.s

2
=
=
= 1.11
2I m 2 2
I avg

96
•
Effect of Using a Silicon Diode with VT = 0.7V
[1]
For the forward-bias region in the FWR:
• The applied signal must now be at least VT = 0.7V before the diode can turn ‘ON’.
• When conducting, the difference between vo and vi is a fixed level of VT = 0.7V and vo =vi -VT.
Vdc  0.636 (Vm − VT )
97
Properties
Center Tapped Full wave rectifier
Ideal diode
D.C. Current (Idc)
Maximum Current
(Im)
D. C. voltage (Vdc)
2Im/π
2Im/π
Vm/(rf+RL)
(Vm-VT)/(rf+RL)
0.636Vm
0.636(Vm-VT)
Maximum efficiency 0.812/(1+(rf/RL))
Peak Inverse Voltage
Diode with VT
2Vm
0.812/(1+(rf/RL))
2Vm-VT
98
Example [2]
Problem:
A full-wave rectifier uses two diodes, the internal resistance of each diode may be assumed constant at
20Ω. The transformer rms secondary voltage from center tap to each end of secondary is 50V and load
resistance is 980Ω. Find: (i) the average load current (ii) the r.m.s. value of load current.
Solution: rf = 20 Ω, RL = 980 Ω
99
Example [2]
• In the center-tap circuit shown in Figure, the diodes are assumed to be ideal
i.e. having zero internal resistance. Find: (i) DC output voltage (ii) Peak inverse
voltage (iii) Rectification efficiency.
100
Solution:
RMS primary voltage = 230V
RMS secondary voltage = 230×(1/5) = 46V
Max. voltage across secondary = 46 √2 = 65V
Max voltage across half secondary winding (Vm) = 65/2 = 32.5 V
101
Full Wave Bridge Rectifier [1, 2]
• Another type of circuit that produces the same output waveform as the full wave rectifier circuit, is the Full
Wave Bridge Rectifier.
• The main advantage of this bridge circuit is that It does not require a special center tapped
transformer, thereby reducing its size and cost.
102
Full Wave Bridge Rectifier [1, 2]
• The four diodes labelled D1 to D4 are arranged in “series pairs” with only two diodes
conducting current during each half cycle.
103
Full Wave Bridge Rectifier
Operation:
• Case 1: During the positive half cycle of the supply, diodes D1 and D2 conduct in series while
diodes D3 and D4 are reverse biased.
104
Full Wave Bridge Rectifier
• Case 2: During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but
diodes D1 and D2 switch “OFF” as they are now reverse biased.
Note: The current flowing through the load is the same direction as before in the positive half cycle.
105
Peak Inverse Voltage [1,2]
• The required PIV rating for the full-wave bridge rectifier can be determined as
shown below.
• Applying Kirchhoff’s voltage law,
VPIV = vo = Vm
V PIV  Vm
106
•
Effect of Using a Silicon Diode with VT = 0.7V
[1]
For the forward-bias region in the FWR:
• The applied signal must now be at least VT = 0.7V before the diode can turn ‘ON’.
• When conducting, the difference between vo and vi is a fixed level of VT = 0.7V and vo =vi -2VT.
Vdc  0.636 (Vm − 2VT )
107
Properties
Full Wave Bridge Rectifier
Ideal diode
D.C. Current (Idc)
Diode with VT
2Im/π
2Im/π
Vm/(2rf+RL)
(Vm-2VT)/(2rf+RL)
0.636Vm
0.636(Vm-2VT)
Maximum efficiency
0.812/(1+(2rf/RL))
0.812/(1+(2rf/RL))
Peak Inverse Voltage
Vm
Vm-VT
Maximum Current (Im)
D. C. voltage (Vdc)
108
Example [2]
In the bridge type circuit shown in Figure, the diodes are assumed to be ideal. Find : (i)
d.c. output voltage (ii) peak inverse voltage (iii) rectification efficiency. Assume primary
to secondary turns to be 4:1.
109
Solution:
RMS primary voltage = 230V
RMS. secondary voltage = 230×(1/4) = 57.5V
Max. Voltage across secondary = 57.5 √2 = 81.3 V
110
Example [2]
The bridge rectifier shown in Figure, uses silicon diodes. Find (i) DC output voltage (ii)
DC output current. Use simplified model for the diodes
111
Solution:
AC voltage across transformer secondary is 12V r.m.s.
Peak secondary voltage (Vm) = 12√2=16.97 V
112
Advantages and Disadvantages of Full Wave
Bridge Rectifier
Advantages:
• Need for center tap transformer is eliminated.
• PIV is one half of that of center tap circuit.
• Output is twice than that of center tap circuit.
Disadvantages:
• Requires 4 diodes.
113
References
[1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and Circuit
Theory’, PHI, 7e, 2001.
[2] D.C. Kulshreshtha, ‘Electronic Devices and Circuits’, New Age,
2e, 2006.
114
Electrical Science-2 (15B11EC211)
Unit-5
Clipper Circuits
Lecture-5
115
Outline
• Content
–
–
–
–
–
–
–
–
–
Introduction to clippers
Classifications
Series clipper
Parallel clipper
Biased clipper
Examples
Combination clipper
Summary of clippers
Reference
116
Introduction of clipper
•
•
Clippers are networks that employ diodes to “clip” away a portion of an input
signal without distorting the remaining part of the applied waveform.
A circuit which use to remove the peak of the wave is called clipper.
Ref 1
117
Why clipper used?
•
•
The purpose of clippers is to protect a circuit against too high or too low voltages.
A simplest clipper circuit →
Vin
Ref 1
118
Classification of clippers
• According to non- linear devices used
– Diode clippers and Transistor clippers
• According to circuit configuration
– Series clippers and Parallel clippers
• According to biasing
– Biased clippers and Unbiased clippers.
• According to level of clipping
– Positive clippers, Negative clippers and combination clippers
Ref 1
119
Series and Parallel clippers
If ,Vi < Vd, diode OFF, V0=0
D
R.B
Vi
R
V0
Vi
R
V0
Series configuration
Ref 1
120
Series…
If, Vi > Vd, diode ON, V0=Vi - Vd
Vd
+
Vi
F.B
R
V0
Where, Vd=0 V (for ideal diode)
Vd=0.7 V (practical diode or
silicon diode)
Input and output waveform
Red line represent Practical diode
and Blue line for Ideal diode
121
Parallel
If ,Vi > Vd, positive half cycle
diode ON, V0=Vd
R
Vi
D
V0
R
Parallel configuration
Vi
Vd
F.B
+
-
V0
Where, Vd=0 V (for ideal diode)
Vd=0.7 V (practical diode or
silicon diode)
Ref 1
122
Parallel...
If ,Vi < Vd, for negative half cycle
diode OFF V0= Vi
R
Vi
D
R.B
V0
Input and output waveform
Red line represent Practical diode
and Blue line for Ideal diode
Ref 1
123
Biased positive clipper
•
•
•
When the input signal voltage is positive.
If Vi < VR, diode is reversed biased and does not conduct. Therefore, Vo = Vi
If Vi > VR, diode is forward biased and thus, Vo= VR+Vd
Ref 1
124
Biased negative clipper
• When the input signal voltage is negative.
• If Vi < VR, diode is reversed biased and does not conduct. Therefore, Vo =
Vi
• If Vi < VR, diode is forward biased and thus, Vo= VR-Vd
Vi
V0
VR
Ref 1
125
Biased series clippers
D
D
+V -
+V -
R
R
Vi
Fig. 1
V0
R.B
R
Vi<VR
i
R
V0=o
Fig. 2
Ref 1
126
F.B
The output voltage
V0=Vi-VR-Vd
Vi
+V R
Vi>VR
Vd
R
KVL
V0
Vo
Vm –VR -Vd
Vm
Diode on
VR
T/2
T
T/2
T
Diode off
Vi=VR diode change the state
127
Example1
•
Determine the output waveform for the sinusoidal input of given Fig.1 consider
ideal diode.
Fig. 1
Solution:
For ideal diode Vd=0
Ref 1
128
•
•
•
we find that the transition from one state to the other will occur when,
𝑉𝑖 + 5 = 0
𝑉𝑖 = −5
Fig .2
129
Using Fig. 3 , we find that for conditions when the diode is on and the diode current is
established the output voltage will be the following, as determined using Kirchhoff’s
voltage law:
𝑉0 = 𝑉𝑖 + 5
Fig. 3
130
Practice problem:
Find the output voltage for the network examined in above example if the
applied signal is the square wave as,
Ans.
Ref 1
131
Example 2
•
Determine the output waveform of given network
Ref 1
132
•
The transition level of the input voltage can be found from below Fig. by
substituting the short-circuit equivalent and remembering the diode current is 0 mA
at the instant of transition. The result is a change in state when Vi = 4 V.
Ref 1
133
Practice problem:
Repeat above using a silicon diode with Vd = 0.7 V.
Ans
Ref 1
134
Combination (positive and negative) clipper
•
•
When a portion of both positive and negative of each half cycle of the input
voltage is to be clipped (or removed), combination clipper is employed.
The circuit for such a clipper is shown in below fig.
Ref 1
135
•
•
Combination clipper…
During the positive half cycle, the diode D1 is forward biased by the input supply voltage Vi and reverse biased
by the reference voltage V1. On the other hand, the diode D2 is reverse biased by both input supply voltage
Vi and reference voltage V2
– if Vi<V1, V0=Vi , D1 and D2 off
– If Vi>V1, Vo=V1, D1 on and D2 off
During the negative half cycle, the diode D1 is reverse biased by both input supply voltage Vi and reference
voltage V1. On the other hand, the diode D2 is forward biased by the input supply voltage Vi and reverse biased
by the reference voltage V2.
– if Vi<V2, V0= -Vi , D1 and D2 off
– If Vi>V2, Vo=V1, D1 off and D2 on
Ref 1
136
Summary of clippers
Ref 1
137
138
References
• [1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and
Circuit Theory’, PHI, 7e, 2001.
139
Electrical Science-2 (15B11EC211)
Unit-5
Clamper Circuit
Lecture-6
140
Outline
• Content
– Introduction to clampers
– Classifications
– Positive clampers
– Negative clampers
– Biased clampers
– Examples
– Summary of clampers
– Reference
141
Introduction of clampers
•
A clamper is a network
constructed of a diode, a
resistor, and a capacitor
that shifts a waveform to a
different dc level without
changing the appearance
of the applied signal.
Ref 1
142
Classifications
• Negative clampers
• Positive clampers
• Biased clampers
Ref 1
143
Negative clamper
• The Negative Clamping
circuit consists of a
diode connected in
parallel with the load.
• This type of clamping
circuit shifts the input
waveform in a negative
direction, as a result the
waveform lies below a
DC reference voltage.
144
• During the positive half cycle (interval 0 to T/2)
• 𝑉𝑜 = 0 V for this time interval.
• During this same interval of time, the time constant determined by τ= RC
is very small
145
•
•
•
•
During the negative half cycle (interval T/2 to T)
Applying Kirchhoff’s voltage law around the input loop results in,
−𝑉 − 𝑉 − 𝑉0 = 0
𝑉0 = −2𝑉
146
Positive clampers
• This type of clamping circuit shifts the input waveform in a positive
direction, as a result the waveform lies above a DC reference voltage.
147
• During the positive half wave cycle → diode goes off
• 𝑉0 = 2𝑉
• During the negative half wave cycle →
148
Biased clamper
• The additional dc supply connected within diode as shown in figure.
• For the positive half cycle
• If ,𝑉𝑖 < 𝑉1, diode is OFF, 𝑉0 = 𝑉𝑖 − 𝑉𝑐
• If, 𝑉𝑖 > 𝑉1, diode is ON, 𝑉0 = 𝑉1 and the capacitor charge up to voltage
𝑉𝑐 = 𝑉𝑖 − 𝑉1
• In this case the resistor R is not shorted out by the diode.
149
• For the negative half cycle,
• 𝑉0 = −𝑉𝑖 − 𝑉𝑐 ≡ −𝑉𝑖 − 𝑉𝑖 + 𝑉1 = −2𝑉𝑖 + 𝑉1
150
Example 1
• Determine the output voltage of given network
151
Solution
•
period t1-t2 of the input signal since
the diode is in its short-circuit state. For
this interval the network will appear as
shown in Fig.
152
•
For the period t2 to t3 the network will appear as shown in Fig.
153
•
•
The resulting output appears in Fig. with the input signal.
You can see the output swing of 30 V matches the input swing.
input
output
154
Practice problem
• Repeat Example 1 using a silicon diode with Vd 0.7 V.
155
Summary of clampers
156
References
• [1] R. Boylestad and L. Nashelsky, ‘Electronic Devices and
Circuit Theory’, PHI, 7e, 2001.
157
ELECTRICAL SCIENCE-II
(15B11EC211)
Zener Diode and Applications
UNIT 5
Lecture 7 & 8
158
Content
•
•
•
•
•
•
•
Introduction to Zener Diode
Zener Diode I-V Characteristic
Temperature Coefficient of Zener Diode
Examples
Zener Diode Application as Regulator
Example
References
159
Introduction to Zener Diode [1, 2]
Zener Diodes are basically the same as the standard PN junction diodes but they
are specially designed to have a low and specified Reverse Breakdown
Voltage which takes advantage of any reverse voltage applied to it.
➢
The symbol of Zener diode is shown in Fig. 1(a).
➢
The standard PN junction diode and the Zener diode
are presented side by side in Fig. 1 to ensure that the
direction of conduction of each is clearly understood
together with the required polarity of the applied
voltage.
(a)
(b)
Fig. 1 (a) Zener diode (b) standard PN
junction diode
160
Zener Diode I-V characteristic [1, 2]
• The Zener diode, when biased in the forward
direction, that is Anode positive with respect to
its Cathode, behaves just like a normal PN
junction diode passing the rated current.
• However, unlike a conventional diode which
blocks any flow of current through itself when
reverse biased, that is the Cathode becomes
more positive than the Anode, as soon as the
reverse voltage reaches a pre-determined value,
the Zener diode begins to conduct in the reverse
direction.
161
Zener Diode I-V characteristic [1, 2]
• This is because when the reverse voltage applied across the Zener diode exceeds the rated
voltage of the device a process called Avalanche Breakdown occurs in the semiconductor
depletion layer and a current starts to flow through the diode to limit this increase in voltage.
• The Zener Diode is used in its “reverse bias” or reverse breakdown mode, i.e. the diodes anode
connected to the negative supply. From the I-V characteristics curve, we can see that the Zener
diode has a region in its reverse bias characteristics of almost a constant negative voltage
regardless of the value of the current flowing through the diode.
• This voltage remains almost constant even with large changes in current providing the Zener
diodes current remains between the breakdown current IZ(min) and its maximum current
rating IZ(max).
• The location of the Zener region can be controlled by varying the doping levels. An increase in
doping that produces an increase in the number of added impurities, will decrease
the Zener potential.
162
Zener Diode I-V characteristic [1, 2]
• This ability of the Zener diode to control itself can be used to great effect to regulate or
stabilize a voltage source against supply or load variations.
• The function of a voltage regulator is to provide a constant output voltage to a load
connected in parallel with it in spite of the ripples in the supply voltage or variations in the
load current.
163
Temperature Coefficient of Zener Diode [1]
• The Zener potential of a Zener diode is very sensitive to the temperature of operation.
The temperature coefficient can be used to find the change in Zener potential due to a
change in temperature using the following equation:
VZ VZ
TC =
 100% / C
T1 − T0
where, T1 is the new temperature level
T0 is room temperature in an enclosed cabinet (25°C)
TC is the temperature coefficient
VZ is the nominal Zener potential at 25°C.
 VZ is the change in Zener potential
164
Example-1 [1]
Example- Analyze the 10V Zener diode with typical temperature coefficient of
+0.072%/°C, if the temperature is increased to 100°C (the boiling point of water).
Solution
TC VZ
VZ =
(T1 − T0 )
100%
VZ =
(0.072)(10)
(100C − 25C )
100%
VZ = 0.54V
The resulting Zener potential is now
VZ = VZ + VZ =10.54V
165
Practice Example-2 [1]
Example- At what temperature will the 10-V Zener diode, with typical temperature
coefficient of +0.072%/°C, have a nominal voltage of 10.75 V?
Solution
VZ = 10.75 − 10 = 0.75V
VZ (100)
T1 − T0 =
TCVZ
0.75(100)
T1 − T0 =
= 104.17C
0.072(10)
T1 = 25C + 104.17C
T1 = 129.17C
166
Practice Example-3 [1]
Example- Determine the temperature coefficient of a 5-V Zener diode (rated 25°C
value) if the nominal voltage drops to 4.8 V at a temperature of 100 °C
Solution
VZ = 5 − 4.8 = 0.2V
VZ VZ
TC =
 100% / C
T1 − T0
0 .2 5
TC =
 100% / C
100 − 25
TC = 0.053% / C
167
Zener Diode Applications [1, 2]
• The most common application of the Zener diode is
to use Zener diode as a regulator.
• The basic configuration appears as shown in Fig. 2.
• The following three conditions surrounding the
analysis of the basic Zener regulator are
considered.
Fig. 2. Basic Zener regulator
1. Vi and RL Fixed
2. Fixed Vi, Variable RL
3. Fixed RL, Variable Vi
168
1. Vi and RL Fixed [1, 2]
1. Determine the state of the Zener diode by removing it from the network and
calculating the voltage across the resulting open circuit.
RLVi
V = VL =
R + RL
• If V ≥ VZ, the Zener diode is on, and the appropriate equivalent
model can be substituted.
• If V ≤ VZ, the diode is off, and the open-circuit equivalence is
substituted.
Fig. 3 Determining the state of the
Zener diode
169
Cont…
2. Substitute the appropriate equivalent circuit and solve for the desired unknowns.
For the network of Fig. 2, the “on” state will result in the equivalent network of Fig. 4. Since
voltages across parallel elements must be the same, we find that
VL = VZ
By applying KCL, we find that
IZ = IR – IL
where
V
V − VL
V
IL = L
and I R = R = i
RL
R
R
The power dissipated by the Zener diode is
determined by
Fig. 4 Substituting the Zener
equivalent for the “on” situation
PZ = VZIZ
that must be less than the PZM specified for the device.
170
Example-4 [1]
a.
For the Zener diode network of Fig. 5 , determine
VL, VR, IZ, and PZ.
b. Repeat part (a) with RL = 3 k.
Solution (a)
In order to determine the state of the Zener diode, redraw
the network as shown in Fig. 6.
V=
Fig. 5
RLVi
1.2 (16)
=
= 8.73V
R + RL 1 + 1.2
Since V = 8.73 V is less than VZ =10 V, the diode is in the
“off” state. Substituting the open-circuit equivalent results
in the same network as in Fig. 6 , where we find that
VL = V = 8.73 V
VR = Vi – VL = 16 – 8.73 = 7.27V
IZ = 0, PZ = VZIZ = 0
Fig. 6
171
Cont…
Solution (b)
Redraw the network as shown in Fig. 6, the value of V is now
given as
V=
RLVi
3 (16)
=
= 12V
R + RL 1 + 3
Fig. 7
Since V = 12 V is greater than VZ =10 V, the diode is in the “on” state. Substituting the
appropriate equivalent circuit results in the network shown in Fig. 7, where we find that
VL = VZ = 10 V
VR = Vi – VL = 16 – 10 = 6V
IL =
VL 10
=
= 3.33mA,
RL 3k
IR =
VR 6
= = 6mA
R 1k
IZ = IR – IL = 6 – 3.33 = 2.67 mA, PZ = VZIZ = 10(2.67) = 26.7mW
172
Example-5 [1]
EXAMPLE The network of Fig. 8 is designed to limit the
voltage to 20 V during the positive portion of the applied
voltage and to 0 V for a negative excursion of the applied
voltage. Check its operation and plot the waveform of the
voltage across the system for the applied signal. Assume the
system has a very high input resistance so it will not affect the
behavior of the network.
Fig. 8
Solution
Once the voltage across the Zener diode reaches 20 V the
Zener diode will turn on
Fig. 9
173
Cont…
For the negative region of the applied signal the silicon diode is reverse
biased and presents an open circuit
Fig. 10
The voltage across the system will therefore appear as shown in Fig. 11.
Fig. 11
174
2. Fixed Vi and Variable RL [1, 2]
In this case, the applied dc voltage is fixed and the load resistor is variable. To determine the
minimum load resistance of Fig. 1 that will turn the Zener diode on, simply calculate the value of
RL that will result in a load voltage VL = VZ. That is,
RLVi
VZ = VL =
R + RL
Solving for RL, we have:
RL min
RVZ
=
Vi − VZ
(1)
(2)
Any load resistance value greater than the RL obtained from (2) will ensure that the Zener diode
is in the “on” state and the diode can be replaced by its VZ source equivalent. The condition
defined by (2) establishes the minimum RL, but in turn specifies the maximum IL as
I L max
VL
VZ
=
=
RL RL min
(3)
175
Cont…
Once the diode is in the “on” state, the voltage across R remains fixed at
(4)
VR = Vi – VZ
and IR remains fixed at
IR =
VR
R
(5)
(6)
The Zener current,
IZ = IR – IL
resulting in a minimum IZ when IL is a maximum and a maximum IZ when IL is a minimum
value, since IR is constant.
Since IZ is limited to IZM as provided on the data sheet, it does affect the range of RL and
therefore IL. Substituting IZM for IZ establishes the minimum IL as:
and the maximum load resistance as:
ILmin = IR – IZM
(7)
VZ
=
I L min
(8)
RL max
176
Example
a.
For the network of Fig. 2 , determine the range of RL
and
IL
that
will
result
in
VRL
being maintained at 10 V.
b. Determine the maximum wattage rating of the diode.
Solution (a)
Determine the value of RL that will turn on the Zener diode
RL min
Fig. 2
RVZ
(1k )(10)
=
=
= 250 ohm
Vi − VZ 50 − 10
The voltage across the resistor R is
VR = Vi – VZ = 50 – 10 = 40 V
and the magnitude of IR is given as
IR =
VR 40
=
= 40 mA
R 1k
177
Cont…
The minimum level of IL is
ILmin = IR – IZM = 40 – 32 = 8 mA
The maximum value of RL
RL max =
VZ
10
=
= 1.25 k ohm
I L min 8 mA
Fig. 2
The results reveal that for the network of Fig. 2 with a fixed Vi, the output voltage will
remain fixed at 10 V for a range of load resistance that extends from 250 Ω to 1.25 kΩ.
Solution (b)
Pmax = VZIZM = 10(32 mA) = 320 mW
178
3. Fixed RL and Variable Vi [1, 2]
For fixed values of RL in Figure, the voltage Vi must be sufficiently large to turn the Zener
diode on. The minimum turn-on voltage Vi = Vimin is determined by
RLVi
VZ = VL =
R + RL
Vi min
( RL + R )VZ
=
RL
(9)
(10)
The maximum value of Vi is limited by the maximum Zener current IZM. Since
IZM = IR – IL,
(11)
IRmax = IZM + IL
(12)
179
Cont…
Since IL is fixed at VZ/RL and IZM is the maximum value of IZ, the maximum Vi is defined
by
Vimax = VRmax + VZ
(13)
Vimax = IRmaxR + VZ
(14)
180
Example
Example- Determine the range of values of Vi that will
maintain the Zener diode of Fig. 3 in the “on” state.
Solution
Determine the minimum value of Vi
Vi min
Fig. 3
( RL + R )VZ (1200 + 220)20
=
=
= 23.67V
RL
1200
The fixed load current IL is then determined by
IL =
VL VZ
20
=
=
=16.67 mA
RL RL 1200
181
Cont…
The maximum value of IR is then determined by (12),
IRmax = IZM + IL = 60 + 16.67 = 76.67 mA
To determine the maximum value of Vi , apply (14),
Fig. 3
Vimax = IRmaxR + VZ
Vimax = (76.67 mA)(220) + 20
Vimax = 36.87 V
The results reveal that for the network of Fig. 3 with a fixed RL, the output voltage will
remain fixed at 20 V for a range of input voltage that extends from 23.67 V to 36.87 V.
182
Practice Example
Example- An unloaded zener voltage regulator has an input source voltage of 20 V, a
series resistance of 330 Ω, and a zener voltage of 12 V. How much current is flowing
through the zener diode ?
Solution
Vi = 20 V, R = 330 Ω, VZ = 12
Since load is disconnected, IL = 0.
Therefore,
Vi − VZ 20 − 12
IZ = IR =
=
= 24.2 mA
R
330
183
Practice Example
Example- An unloaded zener voltage regulator has an input source voltage of 20 V, a
series resistance of 330 Ω, and a zener voltage of 12 V. If the series resistor has a
tolerance of ±10 %, what is the maximum zener current ??
Solution
Vi = 20 V, R = 330 Ω, VZ = 12
Minimum value of series resistor, Rmin = 330 – 0.1×330 = 297 Ω. Thus,
maximum value of zener current is
Vi − VZ 20 − 12
IZ = IR =
=
= 26.9 mA
Rmin
297
184
Practice Example
a.
Design the network of Fig. 4 to maintain VL at 12
V for load variation (IL) from 0 mA to 200 mA.
That is, determine RS and VZ.
b. Determine PZ max for the Zener diode of part (a).
Solution (a)
Given, VL = 12V, ILmin = 0, ILmax = 200 mA.
Fig. 4
The Zener diode must be on to maintain VL at 12V and VZ must also be equal to VL. Therefore,
VL = VZ = 12V
The minimum value of load resistance to make the Zener diode on is given by
RL min
RSVZ
RS (12)
=
=
= 3RS
Vi − VZ 16 − 12
185
Cont…
The load current is maximum when load resistance is minimum and given by
I L max =
VZ
,
RL min

200 mA =
12
,
3RS
RS = 20Ω
Once the Zener diode is in the “on” state, the voltage across RS remains fixed at
VRS = Vi – VZ = 16 – 12 = 4V
The current through the RS is also fixed and given as
VRS
4
IR =
=
= 0 .2 A
RS 20
186
Cont…
The current through the Zener diode is maximum (IZM) when the load current is minimum and
given as
IZM = IR – ILmin = 0.2 – 0 = 0.2A
Solution (b)
The maximum power rating of Zener diode is given as
PZM = IZMVZ = 0.2(12) = 2.4 W
187
References
1. Robert L. Boylestad, Louis Nashelsky, “Electronic Devices and Circuit Theory”, 11thed, Prentice Hall of India,
2014.
2. A. S. Sedra,, and K. C. Smith, ‘Microelectronic Circuits’. Oxford Oxford University Press, 7th Edition,
2012.
188