Diodes
1
Diode
• Class of non-linear circuits
– having non-linear v-i Characteristics
• Uses
– Generation of :
• DC voltage from the ac power supply
• Different wave (square wave, pulse) form generation
– Protection Circuits
– Digital logic & memory circuits
Creating a Diode
• A diode allows current to flow in one
direction but not the other.
• When you put N-type and P-type silicon
together gives a diode its unique
properties.
Diode
Equivalent circuit in the reverse direction
Equivalent circuit in the forward direction.
Operation
Reverse Bias
• -ve voltage is applied to Anode
• Current through diode = 0 (cut off operation)
• Diode act as open circuit
Forward Bias
+ve voltage applied to Anode
• Current flows through diode
• voltage Drop is zero
(Turned on)
• Diode is short circuit
The two modes of operation of ideal diodes
Forward biased
Reverse biased
Forward Current 10 mA
Reverse Voltage 10 V
Ex 3.2
1
1
iP
1.5v
1.5
iD 
 1.5 A
1
vD  0

vD


vD

1.5v
iD  0
v D  1.5v
Rectifier circuit
Equivalent circuit when vi  0
Input waveform
Output waveform.
Equivalent circuit when vi ≤ 0
Waveform across diode
Exercise 3-3
iD 
10  0
 10mA
1k
t2
1
vD 
 vi dt
t 2  t1 t
1

2

1 
vD    10 sin    0dt 
2  0


1
10
10

 1  1   3.18V
vD  
10 cos  0  
2
2

Battery Charger
24sin  12 V
1
   30 0
2
Conduction Angle    2  120 0
sin  
one  third of
cycle
Figure 3.6 Circuits for Example 3.2.
Diodes are ideal , Find the value of I and V
Example 3.2.
Assumption
Both Diodes are conducting
Assumption
Both Diodes are conducting
V  0,
VB  0
Node A
I D2
10  0

 1mA
10k
Node B
I 5 k  I D1  I D 2 
0   10 
 2mA
5k
From above equation I D1 should be 1mA
It is not possible
Not Possible
Thus assumption of both diode
conducting is wrong
Example 3.2(b).
Assumption # 2
Diodes 1 is not conducting
Diodes 2 is conducting
I D2 
10   10  20

 1.33mA
15
15
VA  10  1.335k   3.3v
VB  1.3310k  10  3.3v
Assumption is correct
VB  VA  3.3 V,
I D1  0, I D2  1.33mA
Figure E3.4
Diodes are ideal , Find the value of I and V
Figure E3.4
Diodes are ideal , Find the value of I and V
I= 2mA
V= 0V
I= 0A
V= 5V
I= 0A
V= -5V
I= 2mA
V= 0V
Figure E3.4 Diodes are ideal , Find the value of I and V
I= 3mA
V= 3V
I= 4mA
V= 1V
Figure P3.2
Diodes are ideal , Find the value of I and V
Figure P3.2 Diodes are ideal , Find the value of I and V
Diode is conducting
I = 0.6 mA
V = -3V
Diode is cut-off
I = 0 mA
V = 3V
Diode is conducting
I = 0.6 mA
V = 3V
Diode is cut-off
I = 0 mA
V = -3V
Problem 3-3
Diodes are ideal , Find the value of I and V
D1 Cut-Off & D2 Conducting
I = 3mA
D1 Cut-Off & D2 Conducting
I = 1mA , V=1 V
Figure P3.4
In ideal diodes circuits, v1
Sketch the waveform of vo
is a 1-kHz, 10V peak sine wave.
v1 is a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
In ideal diodes circuits,
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
Vp+ = 0V
Vp- = - 10V
f = 1 K-Hz
Vo = 0V
Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 10V
Vp- = -10V
f = 1 K-Hz
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
Figure P3.4 In ideal diodes circuits,
Sketch the waveform of vo
v1
s a 1-kHz, 10V peak sine wave.
Figure P3.4
In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 0V
Vp- = -10V
f = 1 K-Hz
V0 = 0V
Vp+ = 10V
Vp- = -5V
f = 1 K-Hz
Figure P3.4 In ideal diodes circuits,
Sketch the waveform of vo
Vp+ = 10V
Vp- = -5V
f = 1 K-Hz
v1
s a 1-kHz, 10V peak sine wave.
Problem 3-4(k)
vi  10V peak @ frequency 1000 H z
vi  10 sin 2000t
For Vi >0 V D1 is cutoff D2 is conducting vo=1V
For Vi < 0 V is conducting D2 is cutoff vo=vi+1V
-9V
Problem 3-4(k)
Figure P3.6
X=A.B
X=A+B
Problem 3-4 (c)
vi  10Vpeak @ frequency1000H z
vi  10 sin 2000t
vo=zero
Problem 3-4(f)
Vi is a 1kHz 10-V peak sine wave.
+ve Half Cycle with 10 V peak
at 1 KHz
Problem 3-4(h)
vi  10Vpeak @ frequency1000H z
vi  10 sin 2000t
vo=zero
Problem 3.5
vi is 10 V peak sine wave and I = 100 mA current source. B is battery
i
of 4.5 V . Sketch and label the B
100 mA
4.5 v
vi  10Vpeak @ frequency1000H z
vi  10 sin 2000t
Solution P3-5
100 mA
vi  10Vpeak @ frequency1000 H z
vi  10 sin 2000t
B  4.5 V
vi  4.5V , D1 conducts D2 cutoff
All curren t flows thru D1, iB  0 A
vi  4.5V , D1 cutoff all current flows thru battery
Conduction angle
10 sin   4.5V    sin 1 (0.45)  26.7 0 ,153.30
Conduction angle    2  126.6 0
Fraction of cycle that i B
126.6
of 100mA flows 
 0.35
360
4.5 v
Problem 3-5
100 mA
4.5 v
Problem 3-5
10
4.5
100 mA
1
1
iBaverage   iB dt  100  0.35T   35mA
T
T
REVERSE POLARITY
PROTECTOR
REVERSE POLARITY
PROTECTOR
• The diode in this circuit protects a radio or
a recorder etc... In the event that the
battery or power source is connected the
wrong way round, the diode does not allow
current to flow.
Problem 3-9
I1
I1
2
2
I3
I3
D1& D2 Conducting
I1=1mA
I3=0.5 mA
I2=0.5 mA
V= 0 V
D1=off, D2=On
I1= I3=0.66 mA
V = -1.7 V
Problem 3-10
D conducting
I=0.225 mA
V=4.5V
D is not conducting
I=0A
V=-2V
Problem 3-16
V
RED GREEN
3V On
0 V Off
-3 V Off
Off
Off
On
D1 conducts
D2 conducts
Quiz No 3 DE 28 EE -A
Sketch vO if vi is 8 sin 
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducting
Solution Quiz No 3
8  4 I1  2 I 2
 2  2 I1  3I 2
2 I 2  2  I 2  1mA
Vo  1 1  2  3V
8V
I1
I2
8
2
I2
 1mA
2
Vo  11  2  3V
vi/2
I
Conduction angle  2  60o
4 sin   2    30
Fraction of Cycle the diode conducts 
  2 1
  33%
2
3
10-10-07
Sketch vO if vi is 10 sin 
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducts
D never conducts
1
+12 V
Vi<5V D2 is cut-off, Vo=5V
Vi>5V D2 is conducts
Vomax
D1
10  5
 5
 7.5V
2
D2
Conduction angle  2  60o
10 sin   5    30
Fraction of Cycle the diode conducts 
  2 1
  33%
2
3
5
22-10-07
Quiz No 3 DE 27 CE -B
Sketch vO if vi is 10 sin 
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducts
D never conducts
1
Vi<5V D2 is cut-off, Vo=Vi
Vi>5V D2 is conducts
Vomax
10  5
 5
 7.5V
2
Conduction angle  2  60o
10 sin   5    30
Fraction of Cycle the diode conducts 
  2 1
  33%
2
3
Problem
•Assume the diodes are ideal,
sketch vo if the input is 10sin
(9)
• Find out the conduction angles
for Diode D1 & D2 (4) and the
fraction of the cycle these diodes
conduct. (2)
 2  vi  1  vo  vi
vi  1
vi  1V  v0 
1  1
4
vipeak  10V  vopeak  4.25V
vi  2V  vo  2V
 2  vi  1  vo  vi
vi  1
vi  1V  v0 
1  1
4
vi  2V  vo  1.25V
vi  2V  vo  2V
v0 
vi  1
1  1V
4
-2V
Two-dimensional representation of the silicon crystal.
14 Electrons
Silicon and Germanium
Silicon Lattice
At room temperature, some of the covalent bonds are broken by
thermal ionization.
Each broken bond gives rise to a free electron and a hole, both of
which become available for current conduction.
Intrinsic Semiconductor
Electrons and holes
Semiconductor Current
The Doping of Semiconductors
.
Valence Electrons
N Type
P Type
p-n Junction
• P Junction
– Concentration of holes is high
– Majority charge carrier are hole
• N Junction
– Concentration of electron is high
– Majority charge carrier are electron
Diffusion Current ID
• Hole diffuse across the junction from the p
side to the n side & similarly electron
• Two current components add together to
form the diffusion current with direction
from p to n side
Drift Current Is
• Diffusion current due to majority carrier
diffusion
• A component due to minority carrier drift
exists across the junction
(a)The pn junction with no applied voltage (open-circuited terminals).
(b) The potential distribution along an axis perpendicular to the junction.
Forward Biased Conduction
•
The polarity of applied voltage which can't produce any current is
called Reverse Bias
•
The polarity of applied voltage which causes charge to flow through
the diode is called Forward Bias.
.
Terminal Characteristics
of
a Junction Diode
The diode i–v relationship
The diode i–v relationship with some scales expanded
and others compressed in order to reveal details.
Terminal Characteristics of a
Junction Diode
• Forward Biased Region v > 0
• Reversed Biased Region v < 0
• Breakdown Region v < -VZK
Forward Biased Region
 nVv

i  Is e
 1




T
• Is Saturation current – Scale Current
– Is is constant at a given temperature
– Is is directly proportional to Cross-Sectional
region of the diode, Is doubles if cross-sectional
area is double
– Is is 10-15 A for small size diode
– Doubles in value for every 10OC rise in
temperature
Forward Biased Region
• Thermal Voltage VT
– VT = kT/q
 nVv


i  Is e
 1




T
• K = Boltzmann’s constant = 1.38 X 10-23 Joules/Kelvin
• T = Absolute Temperature in Kelvin (273 +Temp in Co)
• q = Magnitude of charge = 1.6 X 10-19 Coulombs
– VT @ 20oC is 25.2mV, ~ 25 mV
• n is 1 or 2 depending on the material
and the physical structure of the diode
– n = 1 for Germanium Diode & n=2 for
Silicon
Forward Biased Region
i >> Is
 nVv

i  Is e
 1




T
i  I se
v
nVT
 nVv
ln i  ln I s e


i
v  nVT ln
Is
T

  v ln I
s
 nVT

b Relationship of the current i to the voltage
v holds good over many decades of current
(seven decades, a factor of 107
Forward Biased Region
I1  I s e
v1
nVT
I2  I se
I2
e
I1
v2
nVT
v  v 
2
1
nVT
I2
I2
v2  v1   nVT ln  2.3nVT log
I1
I1
Forward Biased Region
I2
v2  v1   2.3nVT log
I1
• for
I2
 10
I1
v drop changes by
for n = 1
for n = 2
 v  0.5v  cut  in  voltage
v  0.6v
0.8v  0.7v
2.3nVT  60 mV
120 mV
Illustrating the temperature dependence of the diode forward characteristic
At a constant current, the voltage drop decreases by
approximately 2 mV for every 1C increase in
temperature.
Figure E3.9
If V=1V at 20o C, Find V at
400C and 00C
Is
At 20o C Reverse current Is = 1V/1M Ω= 1μ A
Since the reverse leakage current doubles for every 100 C increase,
At 400 C I = 4*1 = 4 μ A
V = 4 μ A * 1MΩ = 4.0 V
At 0 C I = ¼ μ A V = 0.25 V
Forward biased Diode Characteristics
Example 3.3
• A silicon diode displays a forward voltage
of 0.7 V at a current of 1mA. Find Is at n=1
&2
i  I se
 1
2
v
nVT
 I s  ie
3

I s  10 e
3
v
nVT
 0.7
I s  10 e
2510 3
 0.7
 6.9  1016 A
2 2510 3
 8.3  10
10
A
Ex 3.7
Silicon Diode with n=1 has VD=0.7V @
i=1mA. Find voltage drop at i=0.1mA &
10mA
i  I se
 1
v
nVT
 I s  ie
3

I s  10 e
v
nVT
 0.7
2510  3
 6.9  10 15 A
i
10  4
3
For i  0.1mA  V1  VT ln
 25  10 ln
 0.64V
16
Is
6.9  10
i
10  2
3
For i  10mA  V1  VT ln
 25  10 ln
 0.76V
16
Is
6.9  10
Solution P3-18
(a)
At what forward voltage does a diode for which n=2 conduct a
current equal to 1000Is?
(b)
In term if Is what current flows in the same diode when its forward
voltage is 0.7 V
(a)
  2,
i  ISe
Diode current  i  1000IS
v
nVT
 1000 I S  I s e
v
225103
v  0.345V
(b)
v  0.7V
i  ISe
v
nVT
 I se
0.7
0.05
 1.2 106 I S
Problem 3-23
• The circuit shown utilizes three
identical diodes having n=1 and Is=
10 -14 A. Find the value of the
current I required to obtain an output
voltage Vo=2 V. Assume n=1
• If a current of 1mA is drawn away
from the output terminal by a load,
what if the change in the output
voltage. Assume n=1
Solution 3-23
The circuit shown utilizes three identical
diodes having n=1 and Is= 10 -14 A. Find the
value of the current I required to obtain an
output voltage Vo=2 V.
Info available n  1, I S  10 14 A,Vo  2V
The voltage across e ach diode is
I DX  I S e
vDX
ηVT
vo
2
 vDX 
3
3
2
 10
14
e
3
0 .025
 3.81mA
If a current of 1mA is drawn away from the output terminal
by a load, what if the change in the output voltage.
(b)
Load current  1mA, therefore I DY  2.81mA
I DY
e
I DX
(vDY  v DX )
0 .025
e
(vDY  2 / 3 )
0 .025
ΔvoY  vO 2  v01  22.8mV
Problem 3-25
• In the circuit shown,
both diode have n=1,
but D1 has 10 times
the junction area of
D2. What value of V
results?
In the circuit shown, both diode
have n=1, but D1 has 10 times the
junction area of D2. What value
of V results?
Solution 3-25(a)
I D1  I S1e
VD 1
VT
I D 2  I S 2e
VD 2
VT
I S1  10 I S 2
I D1  10 I S 2 e
VD 1
VT
VD 2
VT
I e
I D2
 S 2 VD1  0.1e
I D1
10 I S 2 eVT
V0  VD 2  VD1  VT ln
VD 2 VD 2
VT
..............1
10 I D 2
.................2
I D1
I1  I D2  I D1  I D2  I1  I D1..........3
 I D1  2mA,
I D2  10  2  8mA
V0  VD 2  VD1  0.025  ln
80
 92.2mV
2
solution 2-25 (b)
To obtain a value of 50 mV, what current
I2 id needed.
Vo  50mA, Find I D1, I D 2
I D2  0.01  I D1
I D2
 0.1e
I D1
VD 2 VD 2
VT
I D2

 0.1e 2
0.01  I D 2
I D 2  4.25mA
I D1  (10  4.25)  5.75mA
Problem 3-26
• For the circuit shown,
both diodes are identical,
conducting 10mA at 0.7 V
and 100 mA at 0.8 V.
• Find ‘n’
• Find the value of R for
which V = 80 m V.
Solution 3-26 (a)
Diodes are identical therefore I S , ,VT are same
For Diode 1  VD1  0.7V @ I D1  10mA
For Diode 2  VD2  0.8V @ I D 2  100mA
Find η
I D2
VD 2  VD1  VT ln
I D1
0.8  0.7    0.025  ln
100
10
  1.739
Find R if Vo=80mV
V  VD 2  VD1  VT ln
I D2
I D1
 0.08  1.737  0.025  ln
I D1  1.4mA
R
80
 57.1
1.4
0.01  I D1
I D1
Problem 3.36
Assuming identical diodes for which VD
=0.7V @ ID=1mA. Find R if V0 = 3 V
VDx 
3
 0.75V
4
I DX  I S e
VD 2
I D2 e
 V
I D1
e
VDX
V
T
V
T
D1
e
(VD 2 VD 2 )
T
V
T
I D 2  I D1  e
(VD 2 VD 2 )
.75  0.7  VT ln
R
V
V
T
 1 e
( 0.75  0.7 )
2510 3
 7.389mA
I D2
 I D 2  7.389mA
10 3
10  3
 947
7.389  103
Modeling the Diode
Forward
Characteristics
A simple circuit used to illustrate the analysis of circuits in which
the diode is forward conducting.
VD
ID  ISe
V
T
VDD  VD
ID 
R
Graphical analysis of the circuit using the exponential diode model.
Iterative Analysis using the
Exponential Model
Determined the diode current ID and Diode
voltage VD with VDD =5V and R =1000
ohms. Diode has a current of 1mA @ a VD
of .7 V, and that its voltage drop changes
by 0.1 V for every decade change in
current.
Solution
First
iteration
VD
ID  ISe
VT
VD  0.7V
 4.3mA
I2
V2  V1  2.3VT log
I1
V  2.3VT  0.1V For Every decade change in current
V2  V1  0.1 log
4.3
 0.763V
1.0
Second iteration VD  0.763V
ID  ISe
VD
V
T
 4.237mA
V2  V1  2.3VT ln
I2
I1
4.237
V2  0.763  0.1 log
 0.762V
4.3
Solution
I D  4.237mA,
VD  0.762V
The Piecewise-Linear Model
Approximating the diode forward characteristic with
two straight lines: the piecewise-linear model.
The Piecewise-Linear Model
• Exponential curve is approx into two
straight lines
• Line No 1 with zero slope & Line 2 with a
slope of 1/rd
• The voltage change of less than 50 mV is
observed in case the current change from
0.1 mA to 10 mA.
i 0
v  0V
D
iD 
D
(v D  V D 0 )
rD
v D  VD 0
Piecewise-linear model of the diode forward characteristic and
its equivalent circuit representation.
Piecewise-linear model
The
Constant – Voltage Drop
Model
Constant – Voltage Drop Model
• Forward conducting diode exhibits a
constant voltage drop VD
• The voltage change of less than 50
mV is observed in case the current
change from 0.1 mA to 10 mA.
• Model is used when
– Detailed information about diode
characteristics in not available
Constant-voltage-drop model
The constant-voltage-drop model of the diode forward
characteristics and its equivalent-circuit representation.
The Small – Signal Model
• A small ac signal is superimposed on the
DC components.
• First determined dc Operating Point
• Then small signal operation around the
operating point
– Small portion of the curve is approximated as
almost linear segment of the diode
characteristics.
The Small – Signal Model
Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown are for a diode with n = 2.
The Small – Signal Model
In absence of signal
I D  I s e VD
VT
Once signal is applied
v D (t )  VD  vd (t )
i D (t )  I s e
i D (t )  I s e
i D (t )  I s e
vD
VT
VD
VD
i D (t )  I D e
 vd (t )
VT
V
T
vd (t )
e
vd (t )
V
T
V
T
For very small signal
vd
i D (t )  I D (1 
)
VT
i D (t )  I D  id (t )
vd
 1
VT
The Small – Signal Model
vd
i D (t )  I D (1 
)
ηVT
i D (t )  I D  id (t )
I D vd
id (t ) 
VT
rd 
VT
ID
rd is inversely proportional to ID
Modeling the Diode Forward Characteristic
Table 3.1 (Continued)
Exp 3-6
VDD  10V,vd  1V peak amplitude @ 60Hz
Diode has a current of 1mA @ a VD of .7 V, n  2
Find rd , VD , vd (t )
ID
+
VD
-
+
vd
-
Solution
VDD  VD 10  0.7
ID 

 0.93mA
R
10
VT
2  25
rd 

 53.8
ID
0.93
Small signal
rd
vdpeak  v speak
 5.35mV
R  rd
Input variation of 10% resulted in output voltage
variation of 0.7+5.4mV(0.8%) Voltage regulation
Exercise 3-16
• Design a circuit shown so that Vo=3v
when IL =0 A and Vo changes by 40 mV
per 1mA of diode current.
• (a)
Find the value of R
• (b)
The junction area of each diode
relative to a diode with ).7 V drop at 1mA
current. Assume n=1
Excercise 3-16
rDT
vo 0.04

 3  40
io 10
rDX  40 / 4  10
I DX
Why 4 diodes and not 5? Diodes
will not conduct at 0.6 V
nVT

 2.5mA
rDX
15  3
R
 4.8 K
2.5m
At dc Operating Point VDX  3 / 4  .75V
I D1  1mA, VD1 0.7V
I DX I SX

e
I D1 I S 1
VDX V1
nVT
I SX

 0.34
I S1
The diodes have the junction area 0.34 times the diode
Diode Forward Drop in Voltage
Regulation
• Small signal model is used.
• Voltage remains constant in spite of :
– Changes in load current
– Changes in the dc power supply voltages
• One diode provides constant voltage of
0.7 V and for greater voltages diodes can
be connected in series.
Example 3-7
• A string of three diodes is used to provide
a constant voltage of about 2.1 V. We
want to calculate the percentage change
in this regulated voltage caused by
• (a) a + 10 % change to the power supply
voltage
• (b) Connection of a 1 K ohms load
resistance , Assume n=2
Solution Exp 3-7
P 3-53
• In a particular cct application, ten
“20 mA diodes” ( a 20 mA diode is
a diode that provides a 0.7 V drop
when the current thru it is 20 mA)
connected in parallel operate at a
total current of 0.1 A. For the
diodes closely matched, with n=1,
what current flows in each.
iDx 
0.1
 0.01A
10
What is the corresponding small
signal resistance of each diode and
of the combination?
nVT
rdx 
 2.5
I Dx
2.5
req 
 0.25
10
• If each of the 20 mA diode has a
series resistance of 0.2 ohm
associated with the wire bonds to
the junction.
What is the
equivalent resistance of the 10
parallel connected diodes?
Re q 
1
2.5  0.2  0.27
10
What connection resistance would
single diode need in order to be
totally equivalent?
The diode i–v relationship
Reversed Biased Diode
Leakage current:
•
In the reverse direction there is a small
leakage current up until the reverse
breakdown voltage is reached.
• This leakage is undesirable, obviously the
lower the better.
• Diodes are intended to operate below their
breakdown voltage.
The Reversed Biased Region
 nVv

i  Is e
 1




T
v is negative &  VT (25mV )
i  I S
Current in reserved biased diode circuit is due to leakage
current & increases with increase in reverse voltage
Leakage current is proportional to the junction area &
temperature but doubles for every 10oC rise in
temperature
Breakdown Region
• Once reverse voltage exceeds a threshold value of diode
VZK, this voltage is called breakdown voltage.
VZK
Z – Zener,
K – Knee
• At breakdown knee reverse current increases rapidly
with associated small increase in voltage drop
• Diode breakdown is not destructive if power dissipated
by diode is limited by external circuitry.
• Vertical line for current gives property of voltage
regulation
The diode i–v characteristic with the breakdown
region shown in some detail.
Zener Diode
Zener Diode
• Operation in the Reverse Breakdown
Region
• Very steep i-v curve at breakdown with
almost constant voltage drop region
• Used the designing voltage regulator
• Diode manufactured to operate specifically
in the Breakdown region called Zener or
Breakdown Diode
Zener Diode : Symbol
IZ
- VZ +
Model: Zener
• Manufacturer specify Zener Voltage Vz at a
specified Zener test current Iz, the Max. power
that the device can safely dissipate 0.5 W @ 6.8
v at max 70mA V  I r
z
z z
• rz Dynamic resistance of the Zener and is the
inverse of the slope of the almost linear i-v
curve at operating point Q
• Lower rz, the more constant Zener Voltage
• The most common range of zener voltage is 3.3
volts to 75 volts,
Model for the zener diode.
Model: Zener
Vz  Vzo  rz I z
I z  I zk
Vz  Vzo
Designing of the Zener shunt regulator
+
Supply voltage includes
a large ripple component
Vo
-
Zener regulator
Vo is an output of the zener regulator
that is as constant as possible in spite of
the ripples in the supply voltage VS
and the variations in the load current
Voltage regulator performance can be measured
Line Regulation & Load Regulation
Line Regulation = ΔVo/ΔVs
Load Regulation = ΔVo/ΔIL
Expression of performance : Zener regulator
(Vs -Vo )
(Vo -V zo)

 IL
R
rz
I
+
V
o
-
Vo  Vzo(
•
R
r
)  VS ( z ) - I L(rz ||R)
R  rz
R  rz
Only the first term on right hand side is desirable one
Second and third terms depend upon
Supply Voltage Vs and Load current IL
• Line Regulation =
• Load Regulation =
Vo
rz
/ 
Vs
(rz  R)
ΔVo
 - (rz ||R)
ΔI L
IL
Expression of performance : Zener regulator
I
+
V
o
•
An important consideration for the design is
• To ensure that current through the zener
diode never becomes too low i.e less than
IZK or Izmin
•
Minimum zener current Izmin occurs when
• Supply Voltage Vs is at its minimum VSmin
• Load current IL is at its maximum
ILmax
•
Above design can be made be selecting proper
value of resistor R
R 
(Vs min – VZO - rz I z min )
(I z min  I L max )
IL
where I L max 
VZ
RL
Example 3.8
The circuit with the zener diode replaced with its equivalent circuit model.
Exp 3-8
Example 3-8
V  10v  1v
R  0.5k
V z  6.8v
I z  5mA
rz  20
I zk  0.2mA
I RL  1mA
a) Find No Load
Vo & Vo Line Regulation
Depending upon the manufacturer provide Data
First calculate Vzo
if Vz =6.8 V & Iz=5mA, rZ=20 ohm
Vz  Vzo  rz I z
3
Vzo  Vz  I z rz  6.8  5  20 10  6.7v
Now connecting the Zener diode in the Cct as shown
Calculate actual Iz and resulting Vo
Thus establishing operating Point
Iz 
V  Vzo 10  6.7

 6.35mA
R  rz 500  20
Vo  Vzo  I z rz  6.7  6.35  20 103  6.827V  6.83V
Now carry out Small Signal Analysis
Suppress DC source and calculate resultant change in Vo
Use voltage divider rule
V  rz  1 20
Vo 

 38.5mv
R  rz
520
Line Regulation
Vo 38.5

 3.85mv / v
V
1
b) Find vO if load resistance RL connected
& draws 1mA and load regulation
1mA drawn by load would
decrease by same amount so
Vo  rz I z  20  1mA  20mV

Load Regulation
Vo
 20mV / mA
I z
6.83v
 6.83k
1mA
20  6830
Check
RL || R 
 19.94
6850
exact Calculations
VZ  Vo  VZO  I Z rZ  6.7  5.35  20  6.807V
RL 
IZ 
Vs  VZ
10  6.807

 6.14mA
R  RL || rZ 500  19.94
I Z  6.35  6.14  .21mA  210A
Iz
c) Vo for
I RL
RL  2k
VZ

 3.4mA
RL
I Z  3.4mA
Vo r Z I Z  68mV
• 1) Check
 10
500
2000
Vo 
2000
 10  8v
2500
Zener at Breakdown region
10v
10
0.5k
500
6.63v
Vo
19.8

2k
A

6 .7 v

20
B
A
B
6.7  2000
Voc 
 6.63v
2020
Re q  19.8
d ) RL  500
10v
Vo 
500
10  500
 5v
1000
Zener is not operating
500
@ Vo  V zk
5  6.8v
e) Min value of R for which the diode still
operates in the breakdown region
L
• at Breakdown Region
 10  1v
500
Iz
6 .7 v
I z  I zk  0.2mA
Vz  Vzk  6.7v
Iz
VDD  9v min
RL
0.2mA
9  6.7
 4.6mA
500
I  I zk  I RL
I
I RL  4.6  0.2  4.4mA
RL 
Vzk
6.7
 1.5k

I RL 4.4m
Problem D3.68
Design a 7.5-V zener regulator circuit using a 7.5-V
zener specified at 12mA. The zener has an incremental
resistance rz = 30 Ω and a knee current of 0.5mA. The
regulator operates from a 10-V supply and has a 1.2-kΩ
load.
(a) What is the value of R you have chosen?
(b) What is the regulator output voltage when the supply is
10% high? Is 10% low?
(c) What is the output voltage when both the supply is 10%
high and the load is removed?
(d) What is the smallest possible load resistor that can be
used while the zener operates at a current no lower
than the knee current while the supply is 10% low?
Solution 3-68
rz  30
I Zk  0.5mA
VZ  7.5V
I Z  12mA
7.5  VZO  12  30  10
 VZO  7.14V
I RL
7 .5

 6.25mA
1 .2
3
Design a 7.5-V zener regulator circuit using a
7.5-V zener specified at 12mA. The zener has
an incremental resistance rz = 30 Ω and a knee
current of 0.5mA. The regulator operates from
a 10-V supply and has a 1.2-kΩ load.
(a) What is the value of R you have chosen?
Select
I  10mA
7.5
I RL 
 6.25mA
1.2
So that I Z  3.75mA
Which is  I Zk
R
10  7.5
 250
10
(b) What is the regulator output
voltage when the supply is 10%
high? Is 10% low?
For V   1V
1.2 // 0.03
0.250  (1.2 // .03)
 0.1V
Thus VO  7.4V to  7.6V
VO  1
(c) What is the output voltage when
both the supply is 10% high and
the load is removed?
With V   11V and I L  0
VO  V ZO 
11  VO
0.28
 VO  7.55V
X 0.03
(c) What is the smallest possible load resistor that can
be used while the zener operates at a current no
lower than the knee current while the supply is 10%
low? IZK=0.5mA, VZO=7.14 V
R
3
11V
9  7.155
0.25
 7.38mA
250
VO
2
7.14  0.03 X 0.5
7.155V
0.5mA
1
RL min
7.155

7.38  0.5
 1.04k
RL min
Rectifier Circuit Power Supply
• Power supply must supply dc voltage to be constant in
spite of
– variation is ac line voltage
– Variation in current drawn by load, that is variable load resistance
Rectifier Circuits
• Filter
– Smoothes out pulsating dc but still some time-dependent
components-(ripple) remain in the output
• Voltage Regulation
– Reduces ripples
– Stabilizes magnitude of dc output against variation in
load current
– Regulation by Zener Diode or Voltage regulator I.C
Half Wave Rectifier
Transfer characteristic of
the rectifier circuit
Input and output waveforms, assuming that rD >> R.
Full Wave Rectifier
Input and output waveforms.
Full Wave Rectifier
• Diode in Reverse biased state
Anode @ - Vs
Cathode @ + Vo
• PIV = 2Vs - VDO
Twice as in case of half wave rectifier
Bridge Rectifier
The bridge rectifier: (a) circuit; (b) input and output
waveforms.
Bridge Rectifier
Bridge Rectifier
Bridge Rectifier
D1
• Peak Inverse Voltage
D4
D2
– PIV => consider loop D3, R & D2
– VD3(res) = Vo + VD2
– Vo = Vs – 2VD
– PIV = Vs – 2VD + VD = Vs – VD
– Half of PIV for Full wave Rectifier
D3
Figure 3.28 (a) A simple circuit used to illustrate the
effect of a filter capacitor. (b) Input and output
waveforms assuming an ideal diode.
Peak detector with Load
Figure 3.29 Voltage and current waveforms in the peak
rectifier circuit with CR<<T.
Charge / Discharge Cycle
Peak detector with Load
Vo
iL 
R
iD  iC  iL
dVs
iD  C
 iL
dt
Figure 3.30 Waveforms in the full-wave peak rectifier.
Peak Rectifier : Output Voltage
• When Vr is small
– Vo = Vpeak
– iL is almost constant
– DC components of iL
iL 
VP
R
• Accurate value of output dc voltage
Average Value V  V  1 V
o
P
2
r
Charge / Discharge Cycle
vo  V P e

t
CR
Vo  VP  Vr  VP e
e
T
CR
T
CR
T
 1
CR
T



CR 

 Vr  VP 1  e 


VP
iL 
R
1
Vo  VP  Vr
2


T 

Vr  V P  1  1 

CR 

Vr 
VPT
V
 P
CR
fCR
IL 
IL
Vr 
, provided
fC
VP
R
Vr  V p
Peak Rectifier : Ripple Voltage
• During Discharge cycle
vo  V P e
• At the end of discharge cycle
Vo  VP  Vr  VP e

T
CR
T



CR
 Vr  VP 1  e 


• Since CR >> T
e

T
CR
T
 1
CR

t
CR
Peak Rectifier : Ripple Voltage
T



CR
Vr  VP 1  e 


e

T
CR
T
 1
CR
T 

Vr  V P  1  1 

CR 

Ripple Voltage
VP
Vr 
fCR
VPT
VP
Vr 

CR
fCR
VP
IL 
R
IL
Vr 
, provided
fC
Vr  V p
Peak Rectifier : Conduction Interval
VP cost   VP  Vr
Hence t is small
Coswt   1 
wt 
2!
2
 ...
 t  
  VP  Vr
VP 1 

2


2Vr
 t 
VP
2
When Vr<<Vp, the conduction angle will be small
Deduction
Average Diode Current –During Conduction
iD  iC  iL
iDav  iCav  I L
iCav  iDav  I L
During Charge
Qsupplied  icav t
iCav  i Dav  I L
During Discharge
Qlost  CVr
Qsupplied  Qsupplied  icav t  CV r
Average Diode Current –During Conduction
Qlost  Qsup plied  CVr  icav t
VP T
VP T
Vr 
 CVr 
CR
R
VPT
 iDav  I L t
R
t 
1
2f
2Vr
T

VP
2
T
VP T
 iDav  I L 
R
 2
i Dav
2Vr
VP
2Vr
VP




2VP
2 VP

 I L  i L 1  
Vr
2Vr R

VP
Vr  VP
 i Dav  I L



Deduction
• As waveform of
triangle
is almost right angle r
Vr  VP
iD max  2iDav
Observations
• Diode current flows for short interval and
must replenish the charge lost by the
capacitor. Discharge interval is long &
discharge is through high resistance
rD  RL
• Maximum diode current
CdVi
iD 
 iL
dt
Assuming
that iL is almost constant  I L &
CR  T
iD max

2V p
 iL 1  2
Vr


  2iDav


Example N0 3-9
Consider a peak rectifier fed by a 60 Hz
sinusoidal having a peak value of Vp =
100 V. Let the load resistance R =10 k
Ohms.
(a) Find the value of the capacitance C that
will result in peak to peak ripple of 2 V
(b) Calculate the fraction of the cycle during
which the diode is conduction
(c) Calculate the average and peak value of
the diode current.
Example 3.9
100 Sin2 60t
10k
• Find value of C for Vr=2V (peak to peak)
VP
100
C

 83.3F
Vr fR 2  60 104
• Find fraction of cycles that diode conducts
t 
2Vr
 0.2radian
VP
• => Diode conducts
0.2
 100  3.18%
2
of cycle
Solution Exp 3-9
• Find
iD max & iDav

2VP 

iDav  I L 1  
Vr 

VP
100
IL 

 10mA
R 10000

2  100 
iDav  101  
  324mA
2 

imax  2iDav  648mA
Full wave peak Detector
In full wave rectifier, the capacitor
discharge for almost T/2 time interval. that
mean ripple frequency is twice the input,
V
so
V 
P
r
2 fCR

VP
i Dav  I L 1  
2Vr





VP
imax  2iDav  I L 1  2
2Vr




Applications
• Peak Rectifier – Peak detector is used for
– Detecting the peak of the an input signal for
signal processing systems
– Demodulator for amplitude modulated (AM)
signals
.
Precision Half Wave Rectifier
Super Diode
• Normal Diodes VD= 0.7v are used for
rectifier of input of much larger amplitude
then VD
• For smaller signals detection,
demodulation or rectification Operational
Amplifiers (Op Amp) are used
Wave form Generation Limiting
Clamping
• Limiter Circuit
– Vo is limited between two levels – upper (L+)
and lower (L-) thresholds
Figure 3.33 Applying a sine wave to a limiter can
result in clipping off its two peaks.
Figure 3.34 Soft limiting.
Wave form Generation
Limiting / Clamping
• Double Limiter
– Clips off both negative & positive peaks
• Single Limiter
• Clips off only one side of the input peak
• Application
– Limits the inputs to operation Amplifier to a limit
lower than the breakdown voltage of transistors
of input stage of operational Amplifier
– Half / Full Rectifier for Battery Charger
– Threshold and limiting
Figure 3.35 A variety of basic limiting circuits.
Figure E3.27
Solution Ex 3-27
(a)
(b)
 5  vi  5
 vo  vi
VI  5 Vo D2 Conduct, D1 cut - off
1
 10 


vi  5
vR  
v

5

i

2
10  10 
1
v o  5  v R  v i  2 .5
2
(c)
vi  5 V D1 Conducts & D2 is off
1
 10 
vi  5  vi  5
vR  

2
10  10 
1
1
vo  vi  5  5  vi  2.5
D C Restorer
• The output waveform will have its lower
peak “Clamped” to O V therefore known
as “Clamped Capacitor”
• Output waveform will have a finite average
value & is entirely different and unrelated
to the average value of the input waveform
Application
6
 2v
4
TXR
 4v
0v
4v
DC
Restorers
Figure 3.36 The clamped capacitor or dc restorer with
a square-wave input and no load.
Figure 3.37 The clamped capacitor with a load
resistance R.
Figure 3.38 Voltage doubler: (a) circuit; (b) waveform
of the voltage across D1.
Figure P3.97
Figure P3.98
Figure P3.102
Figure P3.103
Figure P3.105
4
vi
6
VC
D off
0
D on
vo
Diode Off V0  Vi  Vc
The Voltage Doubler
 VP -
C1
D1
VP sin t

 VP -
VP sin t

VD1

2V P

C1 D1  a Clamp circuit
DC Restorer
Special Diode Type
Schottky-Barrier Diode (SBD)
• Shottky-Barrier Diode is formed by
bringing metal into contact with a
moderately doped ‘n’ type semiconductor
material
• Resulting in flow of the conducting current
in one direction from metal anode to the
semiconductor cathode and acts as an
open circuit in the other direction
Schottky-Barrier Diode (SBD)
• Gets two important properties
– SBD switches on-off faster due to current
conducts due to majority carrier b (electrons)
– Forward voltage drop is lower then P-n
junction diode
Varactor
• Variable Capacitor
– Depletion layer acts as junction capacitance
– Depletion layer varies Capacitance
Depletion Region
D
Metallic Plate
Dielectric
– Used for voltage controlled Tuning Circuit
Varactor
• When a reverse voltage is applied to a p-n junction ,
the depletion region, is essentially devoid of
carriers and behaves as the dielectric of a
capacitor.
• The depletion region increases as reverse voltage
across it increases; and since capacitance varies
inversely as dielectric thickness, the junction
capacitance will decrease as the voltage across the p-n
junction increases.
• By varying the reverse voltage across a p-n junction the
junction capacitance can be varied .
Semiconductor diodes
• The tunnel diode, the current through the
device decreases as the voltage is increased
within a certain range; this property, known as
negative resistance, makes it useful as an
amplifier.
• Gunn diodes are negative-resistance diodes
that are the basis of some microwave oscillators.
• Light-sensitive or photosensitive diodes can
be used to measure illumination; the voltage
drop across them depends on the amount of
light that strikes them.
SCR (Thyristor)
• The Silicon Controlled Rectifier (SCR) is simply a
conventional rectifier controlled by a gate signal.
• A gate signal controls the rectifier conduction.
• The rectifier circuit (anode-cathode) has a low forward
resistance and a high reverse resistance.
• It is controlled from an off state (high resistance) to the
on state (low resistance) by a signal applied to the third
terminal, the gate.
• Most SCR applications are in power switching, phase
control, chopper, and inverter circuits.
Photodiode
If reversed biased PN junction is exposed
to incident light – the photons impacting
the junction cause covalent – bond to
break thus give rise to current known as a
photocurrent & is proportional to the
intensity of incident light.
Converts Light energy into a electrical
signals
Photodiode
• Photodiode are manufactured using
Gallium Arsenide (GaAs)
• Photodiodes are important element of
optoelectronics or photonics circuit
(Combination of Electronics & optics) used
for signal processing, storage &
transmission
Photodiode : Applications
• Fiber optics Transmission of telephonic &
TV signals
• Opto-storage are CD ROM computer disks
• Wide bandwidth & low signal attenuation.
• Solar Cell – light energy into Electrical
energy
Light Emitting Diode (LED)
• Inverse of Photodiode
• Converts a forward biased current into light
• GaAs used for manufacturing LEDs
• Used as electronics displays
• Coherent light into a narrow bandwidth laser
diodes
• Fiber Optics & CD ROM
LED
Double heterostructure laser
Optoisolator
• LED
Electrical to light
&
Photodiode
Light to electrical
• Provides complete electrical isolation between electrical
circuits
• Reduces the effects of electrical interference on signal
being fixed within a system
• Reduces risk of shock
• Can be implemented over long distance fiber optics
communication links
Laser Pointer
.
Laser Microphone
End
Problem 3-103
Sketch and label the transfer
Characteristics of the circuit shown over a
+ 10 V range of the input signal. All
diodes are VD =0.7 V @ 1 mA with n=1.
What are the slopes of the characteristic
at the extreme + 10 V levels?
+1 V
Vi
V0
-2 V
-2 V
Problem 3-103
0  Vi  1
Vo  0
Ist Sessional
• Q No 1 (12 Marks) In the circuit shown, input voltage is
a 1kHz, 10 V peak to peak sine wave. The diode is an
ideal diode.
•
(a)
Sketch the waveform resulting at output
terminal vO.
•
(b)
What are its positive and negative peak
values?
Ist Sessional
• Q No 2 (15 Marks) A circuit utilizes three
identical diodes connected in series having n=1
and IS= 10-14 A.
•
(a) Find the value of current required to
obtain an output voltage of 2 V across the three
diodes combined.
•
(b) If a current of 1 mA is drawn away from
the output terminal by a load
•
(i)
What is the change in output voltage?
•
(ii) What is the value of the load?
Ist Sessional
• Q No 3 (13 Marks) For the circuit shown,
sketch the output for the sine wave input of 10
volts peak. Label the positive and negative peak
values assuming that
CR >>T.
Ist Sessional
• Q No 4 (10 Marks) 9.25 V zener diode
exhibits its nominal voltage at a test
current of 28 mA. At this current the
incremental resistance is specified as 7
ohms.
– (a)
Find VZO of the zener model.
– (b)
Find the zener voltage at a current of
10 mA.
Ist Sessional
• Q No 5 (20 Marks) Consider a bridge rectifier
circuit with a filter capacitor C placed across the
load resistor R for the case in which the
transformer secondary delivers a sinusoid of 12
V (rms) having the 60 Hz frequency and
assuming VD = 0.8 V and a load resistance of
100 ohms.
– Find the value of C that results in a ripple voltage no
larger than 1 V peak to peak.
– Find the diode conduction angle.
– Find the load current.
– What is the average load current?
Ist Sessional
• Q No 6 (10 Marks) In a circuit shown, the output
voltage is 2.4 V. Assuming that the diodes are
identical and are having 0.7 V drop at 1mA.
– (a)
– (b)
Find the current following through the resistor R.
What the value of resistor R.
Figure 3.31 The “superdiode” precision half-wave rectifier and its almost-ideal transfer characteristic. Note that when vI > 0
and the diode conducts, the op amp supplies the load current, and the source is conveniently buffered, an added advantage. Not
shown are the op-amp power supplies.
Figure P3.82
Figure P3.91
Figure P3.92
Figure P3.93
Figure P3.105
Figure P3.105
Quiz DE28 EE -B
(10 Marks) 9.25 V zener diode exhibits its
nominal voltage at a test current of 28 mA.
At this current the incremental resistance is
specified as 7 ohms.
–(a)
Find VZO of the zener model.
–(b)
Find the zener voltage at a current of
10 mA.
Quiz DE 28 EE -A
A zener diode whose nominal voltage is
10 V at 10 mA has an incremental
resistance of 50 Ω.
(a) What is the value of VZO of the zener
model?
(b) What voltage do you expect if the diode
current is doubled?