The presentation describes diode application; series / parallel diode configuration, half wave and full wave rectifier, clipper, clamper, zener diode as a shunt regulator, voltage multiplier

1. Diode Applications
REC 101: Basic Electronics Unit 1
PN junction diode: Introduction of Semiconductor Materials Semiconductor Diode: Depletion
layer, V-I characteristics, ideal and practical, diode resistance, capacitance, Diode Equivalent
Circuits, Transition and Diffusion Capacitance, Zener Diodes breakdown mechanism (Zener
and avalanche) Diode Application: Series , Parallel and Series, Parallel Diode Configuration,
Half and Full Wave rectification, Clippers, Clampers, Zener diode as shunt regulator, Voltage-
Multiplier Circuits Special Purpose two terminal Devices :Light-Emitting Diodes, Varactor
(Varicap) Diodes, Tunnel Diodes, Liquid-Crystal Displays.
9/11/2017 1
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad

2. Diode Application: Series, Parallel and
Series, Parallel Configuration
• The forward resistance of diode is so small
compared to the other series elements of network
that it can be ignored (if not specified)
• In general, a diode is in the ON state if the current
established by the sources is such that, its direction
matches to the arrow of diode symbol, and VD  VK
otherwise it is in OFF state
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
2
Diode VK (V)
Ideal 0
Ge 0.3
Si 0.7
GaAs 1.2
Steps to find the state of diode
1. Replace the diode with equivalent circuit.
2. mentally replace diode with a resistance
and check direction of current through it.
3. If current direction matches to diode arrow
then diode is ON otherwise it is OFF
Practical
VK
ideal ideal
VK
Rf
OR
Diode Equivalent circuit
VK knee voltage
Rf diode forward resistance

3. Diode Application: Series , Parallel and
Series, Parallel Diode Configuration,
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
3
ideal
0.7 V
10 V 100 
Si
10 V 100 
1.Replace diode with
equivalent circuit
ideal
0.7 V
10 V 100 
2.Replace ideal diode with
fictitious resistance and
find the current direction
3. As current matches with
diode direction, it is ON.
Short-circuit the diode
and solve the circuit
mA93A093.0
100
7.010


I
0.7 V
10 V 100 
I

4. Diode Application: rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
4
• One of the common application of diode is ‘rectifier’ which
converts AC signal into DC.
• There are many rectifier circuits
– Half Wave rectifier
– Full wave rectifier; Bridge type & Centre tapped
• Practical rectifier consists of; Transformer, Rectifier, Filter &
Regulator stage
• Rectifier output for one cycle (period) of input is analyzed
AC
Input
+
DC
output
-
Transformer
Stage
Rectifier
Stage
Regulator
stage
Filter
Stage

5. Diode Application: Half Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
5
t
Vi(t)
T
Vm
-Vm
+
Vo(t)=Vi(t)
-
+
Vi(t)
-
Rt
Vi(t)
T
Vm
-Vm
Vo
T
Vm
t
Positive Half cycle
(0  t< T/2)
Vi(t) >0 and diode is ON
So Vo(t) = Vi(t)
+ -
+
V0(t)
-
+
Vi(t)
-
R
Half wave rectifier circuit 
T
tSinVtV mi


2
where,)( 
The process of removing one-
half the input ac signal to
establish a dc level is called
halfwave rectification
V0(t)
T
Vm
t
+
Vi(t)
-
Rt
Vi(t)
T
Vm
-Vm
Negative Half cycle
(T/2  t< T)
Vi(t) <0 and diode is OFF
So Vo(t) = 0
+
Vo(t)=0
-






TtTfor
TtfortV
tV i
2/0
2/0)(
)(periodofoutputcombinedTherefore 0

6. Diode Application: Half Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
6
t
Vi(t)
T
Vm
-Vm
V0(t)
T
Vm
t
2T
2T
Vdc= 0.318Vm
 
  






mm
T
m
dc
T
m
TT
i
T
dc
V
Cos
T
Cos
V
tCos
T
V
V
dttSinV
T
dtdttV
T
dttV
T
V






















 
)0(
22
1
.0)(
1
)(
1
2
0
2
0
2
0
2
00
0
VDC: is periodic average of output wave
mdc VV 318.0 





TtTfor
TtfortV
tV i
2/0
2/0)(
)(0
 
       
2
0
4
2sin
42
2cos1
2
.0)(
1
)(
1
2
2
0
2
2
0
22
0
2
2
0
2
22
0
2
0
2
0
2
mm
T
m
T
m
T
m
dc
T
m
TT
i
T
orms
VV
t
T
V
t
T
V
dtt
T
V
V
dttSin
T
V
dtdttV
T
dttV
T
V

















Vrms: square root of periodic average of squared output
mrms VV 5.0

7. Diode Application: Half Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
7
Peak inverse voltage (PIV or PRV) :
Applying KVL, in loop of reverse bias diode. PIV  Vm
-
Vm
+
R
- PIV +
V0(t)=0 V
Ripple factor: ratio of Vac to Vdc
  21.1157.11
2
12
factorformiskwhere
11factorRipple
2
2
2
f
2
222
222
































m
m
f
dc
rms
dc
dcrms
dc
ac
dcacrms
V
V
k
V
V
V
VV
V
V
VVV

8. Diode Application:
Bridge type Full Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
8
t
Vi(t)
T
Vm
-Vm
t
Vi(t)
T
Vm
-Vm
Vo(t)
T
Vm
t
+
Vi(t)
-
R
- Vo(t) +
Positive Half cycle
(0  t< T/2)
Vi(t)>0 so
D2 and D3 is ON
D1 and D4 is ON
So Vo(t) = Vi(t)
+
Vi(t)
-
R
- Vo(t) +
D1 D2
D3 D4
The process of utilizing full
period of ac signal to establish
a dc level is called fullwave
rectification
A bridge with 4 diodes as
shown in figure is most
familiar full wave rectifier.

9. Diode Application:
Bridge type Full Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
9
t
Vi(t)
T
Vm
-Vm
V0(t)
T
Vm
t
t
Vi(t)
T
Vm
-Vm
Vo(t)
T
Vm
t
+
Vi(t)
-
R
- Vo(t) +
Negative Half cycle
(T/2  t< T)
Vi(t)<0 and
D2 and D3 is ON
D1 and D4 is ON
So Vo(t) = - Vi(t)
 

 m
T
m
T
i
T
dc
V
dttSinV
T
dttV
T
dttV
T
V
22
)(2
1
)(
1 2
0
2
00
0 








 
VDC: As the area above the axis for one full cycle
is now twice of that for a half-wave system, dc
level has also been doubled
mdc VV 636.0






TtTfortV
TtfortV
tV
i
i
2/)(
2/0)(
)(periodofoutputcombinedTherefore 0
Vdc= 0.636Vm

10. Diode Application:
Bridge type Full Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
10
 
22
2
)(
2
)(
1 22
0
2
22
0
2
0
2 mm
T
m
T
i
T
orms
VV
dttSin
T
V
dttV
T
dttV
T
V   
Vrms:
mrms VV 707.0
Peak inverse voltage (PIV or PRV)
Applying KVL, in loop containing
input and reverse bias diode.
PIV  Vm
+
Vm
-
R
- Vo(t) +
-
Vm
+
R
- Vo(t) +
Ripple factor: ratio of Vac to Vdc
  48.01109.11
22
1
2
21
2
2
2
2



























m
m
dc
rms
V
V
V
V

11. Diode Application:
Centre tapped Full Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
11
t
Vi(t)
T
Vm
-Vm
t
Vi(t)
T
Vm
-Vm
Vo(t)
T
Vm
t
R
1:2
+
Vi(t)
- - Vo(t) +
+
Vi(t)
-
+
Vi(t)
-
Positive Half cycle
(0  t< T/2)
Vi(t) >0 so
D1 is ON & D2 is OFF
So V0(t) = Vi(t)
+
Vi(t)
-
R
- Vo(t) +
+
Vi(t)
-
+
Vi(t)
-
1:2
D1
D2
Another type of fullwave
rectifier is Centre tapped
It uses 2 diodes with a centre
tapped transformer.
t
Vi(t)
T
Vm
-Vm
Vo(t)
T
Vm
t
R
1:2
+
Vi(t)
- - Vo(t) +
+
Vi(t)
-
+
Vi(t)
-
Negative Half cycle
(T/2  t< T)
Vi(t) <0 so
D1 is OFF & D2 is ON
So V0 (t) = -Vi(t)

12. Diode Application:
Centre tapped Full Wave rectifier
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
12
VDC: equal to that of bridge type full wave rectifier
mdc VV 636.0






TtTfortV
TtfortV
tV
i
i
2/)(
2/0)(
)(periodofoutputcombinedTherefore 0
t
Vi(t)
T
Vm
-Vm
Vo(t)
T
Vm
t
Vdc= 0.636VmVrms: equal to that of bridge type full wave rectifier
mrms VV 707.0
Peak inverse voltage (PIV or PRV)
Applying KVL, in loop containing input
and reverse bias diode.
PIV  2Vm
R
1:2
+
Vm
- - Vm +
+
Vm
-
+
Vm
-- PIV +
Ripple factor: ratio of Vac to Vdc
48.0

13. Diode Application: Clippers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
13
Clippers are networks that employ diodes to “clip” away a
portion of an input signal without distorting the remaining
part of the applied waveform.
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
Vm
-Vm
If Vi(t) 0, Diode ON, Vo(t)= Vi(t)
If Vi(t) <0, Diode OFF, Vo(t) =0
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
-Vm
If Vi(t) >0, Diode OFF, Vo(t)= 0
If Vi(t) 0, Diode ON, Vo(t) =Vi(t)
Negative Series Clipper
Positive Series Clipper

14. Diode Application: Clippers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
14
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
Vm-VVV V
-V
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
-(Vm+V)
If Vi(t) -V0, Diode ON, Vo(t)= Vi(t) -V
If Vi(t) -V<0, Diode OFF, Vo(t) =0
If Vi(t) -V>0, Diode OFF, Vo(t)= 0
If Vi(t) -V  0, Diode ON, Vo(t) =Vi(t) -V
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R
t
Vo(t)
Vm+V
V
V
V
-V
t
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
-(Vm-V)
If Vi(t) +V0, Diode ON, Vo(t)= Vi(t) +V
If Vi(t) +V<0, Diode OFF, Vo(t) =0
If Vi(t)+V>0, Diode OFF, Vo(t)= 0
If Vi(t) +V  0, Diode ON, Vo(t) =Vi(t) +V
T
Negative biased Series Clipper Positive biased Series Clipper

15. Diode Application: Clippers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
15
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
Vm
If Vi(t) >0, Diode OFF, Vo(t)= Vi(t)
If Vi(t)  0, Diode ON, Vo(t) =0
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-Vm
If Vi(t)  0, Diode ON, Vo(t)= 0
If Vi(t) <0, Diode OFF, Vo(t) =Vi(t)
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-Vm
If Vi(t)-V 0, Diode ON, Vo(t)= V
If Vi(t)-V <0, Diode OFF, Vo(t) = Vi(t)
V
V V
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
Vm
If Vi(t)-V >0, Diode OFF, Vo(t)= Vi(t)
If Vi(t)-V 0, Diode ON, Vo(t) = V
V
V V
Negative parallel Clipper Positive parallel Clipper
Negative biased parallel Clipper Positive biased parallel Clipper

16. Diode Application: Clippers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
16
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-Vm
If Vi(t)+V  0, Diode ON, Vo(t)= -V
If Vi(t)+V <0, Diode OFF, Vo(t) = Vi(t)
VV -V
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
Vm
If Vi(t)+V >0, Diode OFF, Vo(t)= Vi(t)
If Vi(t)+V 0, Diode ON, Vo(t) = -V
V-V
Negative biased parallel Clipper Positive biased parallel Clipper
-V
+
V0(t)
-
+
Vi(t)
-
R
t
Vi(t)
T
Vm
-Vm
If Vi(t)- V1 0, Diode D1 ON, & Vi(t)+ V2 >0, Diode D2 OFF, Vo(t)= V1
If Vi(t)- V1 <0, Diode D1 OFF, & Vi(t)+ V2 >0, Diode D2 OFF, Vo(t)= Vi(t)
If Vi(t)- V1 <0, Diode D1 OFF, & Vi(t)+ V2 0, Diode D2 ON, Vo(t)= -V2
V1
V1
t
Vo(t)
T
V1
V2
-V2
-V2
D1 D2

17. Diode Application: Clampers
Clamping networks have capacitor connected in series with a
diode and resistance in parallel.
Steps to solve clamper circuit
1. Start analysis by examining the response of the portion of
the input signal that will forward bias the diode
2. determine maximum voltage with polarity on capacitor
(assumed that capacitor is charged instantaneously)
3. Assume that during the period when the diode is in OFF state
the capacitor holds on to its established voltage level
4. Now determine the output
5. Check that total swing of the output matches to the input
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
17

18. Diode Application: Clampers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
18
Step 1. In +ve half cycle, Diode is ON and maximum voltage on
charged capacitor voltage Vm with polarities shown
V1
+
V0(t)
-
+
Vi(t)
-
R
C
+
V0(t)
-
+
Vi(t)
-
R
C
+ Vm -
t
Vi(t)
T
Vm
-Vm
t
Vi(t)
T
Vm
-Vm

19. Diode Application: Clampers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
19
2. Assuming capacitor has charged upto to Vm and holds in when
diode is in OFF state
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
Vm
-Vm
Now as Vi(t)-Vm0, diode is ON and Vo(t)=0
+
V0(t)
-
+
Vi(t)
-
R
C
+ Vm -
+
V0(t)
-
+
Vi(t)
-
R
C
+ Vm -
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-2Vm
Now as Vi(t)-Vm<0, diode is OFF and Vo(t)= Vi(t)-Vm =-2Vm
Positive part of cycle
(0  t< T/2)
Vi(t) =Vm so
Diode is ON
So V0(t) = 0
Negative part of cycle
(T/2  t< 0)
Vi(t) =-Vm so
Diode is OFF
So V0(t) = Vi(t)-Vm= -2Vm

20. Diode Application: Clampers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
20
The output is drawn
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-2Vm
-2Vm
-2Vm

21. Diode Application: Clampers
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
21
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
-2Vm
+
V0(t)
-
+
Vi(t)
-
R
C
Vm
+ -
Vm
+
V0(t)
-
+
Vi(t)
-
R
C
t
Vi(t)
T
Vm
-Vm
t
Vo(t)
T
2Vm
- m +
t
Vi(t)
T
Vm
-Vm
Vm+V +
V0(t)
-
+
Vi(t)
-
R
C
V
- +
t
Vo(t)
T
2Vm+V
V
t
Vi(t)
T
Vm
-Vm
V
+
V0(t)
-
+
Vi(t)
-
R
C
(Vm+V)
+ -
t
Vo(t)
T
-V
-2Vm-V
t
Vi(t)
T
Vm
-Vm
Vm-V +
V0(t)
-
+
Vi(t)
-
R
C
V
- +
t
Vo(t)
T
2Vm-V
-V
t
Vi(t)
T
Vm
-Vm
V
+
V0(t)
-
+
Vi(t)
-
R
C
(Vm-V)
+ -
t
Vo(t)
T
V
-2Vm+V

22. Diode Application: Zener diode as shunt regulator
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
22
The use of the Zener diode as a regulator is very common.
Steps to analyse Zener diode networks
1. Determine the state of the Zener diode by removing it from the
network and calculating the voltage across the resulting open circuit
2. Substitute the appropriate equivalent circuit and solve for the
desired unknowns + VD -
Forward bias VD  0
+ VK -
+ VD -
+ VD - - VZ +
Reverse bias VZ <VD < 0
Breakdown VD = VZ

23. 9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
23
+
V0
-
R
+
VZ
-
IZ
RL
Vi
+
V0
-
R
+
V
-
RL
Vi
L
iL
RR
VR
V

If V  VZ Then Zener is ON (breakdown region)
If V < VZ Then Zener is OFF
If diode is ON, then circuit shall be
+
V0
-
R
RL
Vi VZ
IR
IZ IL
ZZZ
L
Z
i
L
ZZi
Z
LRZ
ZO
IVP
RR
V
R
V
R
V
R
VV
I
III
VVV












11
Diode Application: Zener diode as shunt regulator

24. 9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
24
L
iL
RR
VR
V


If V  VZ Then Zener is ON (breakdown region)
If V < VZ Then Zener is OFF
If diode is ON, then circuit shall be
sheetdataperascurrentdiodemaxiswhere
soconstant,isas
breakdowninONbetodiodeforresistanceLoadMin
thus
min
minmin
max
min
min
ZMZMRL
R
L
Z
L
L
L
L
Zi
Z
L
L
iL
ZO
IIII
I
R
V
R
V
I
R
VV
RV
R
RR
VR
VVV






+
V0
-
R
+
VZ
-
IZ
RL
Vi
+
V0
-
R
+
V
-
RL
Vi
+
V0
-
R
RL
Vi VZ
IR
IZ IL
Diode Application: Zener diode as shunt regulator

25. Diode Application: Voltage-Multiplier Circuits
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
25
• Voltage multipliers use clamping action to increase peak rectified
voltages without increasing the transformer’s voltage rating.
• Multiplication factors of two, three, and four are common.
• Voltage multipliers are used in high-voltage, low-current
• applications such as cathode-ray tubes (CRTs) and particle
accelerators

26. Diode Application: Voltage-Multiplier Circuits
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
26
Half wave Voltage doubler
t
Vi(t)
T
Vm
-Vm
D1
C1
+
Vi(t)
-
-
2Vm
+
D2
C2
t
Vi(t)
T
Vm
-Vm
C1
+
Vi(t)
-
+
Vo(t)
-
C2
+ -
Vm+
Vm
-
-
Vm
+
+ -
C1
+
Vi(t)
-
-
2Vm
+
C2
Vm
Half wave Voltage doubler operation:
Positive half cycle
Half wave Voltage doubler operation:
Negative half cycle
t
Vi(t)
T
Vm
-Vm

27. Diode Application: Voltage-Multiplier Circuits
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
27
Full wave Voltage doubler
Full wave Voltage doubler operation:
Positive half cycle
Full wave Voltage doubler operation:
Negative half cycle
t
Vi(t)
T
Vm
-Vm
C1
D1
+
2Vm
-
C2
+
Vi(t)
-
D2
t
Vi(t)
T
Vm
-Vm
C1
D1
+
-
C2
+
Vi(t)
-
D2
Vm
+
-
t
Vi(t)
T
Vm
-Vm
C1
D1
+
-
C2
+
Vi(t)
-
D2
Vm
-
+
+
-
Vm
+
-
Vm
+
-
Vm
+
-
Vm

28. Diode Application: Voltage-Multiplier Circuits
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
28
Voltage Tripler and Quadrupler
t
Vi(t)
T
Vm
-Vm
Vm
+
Vi(t)
-
D1
C2
D2
C3
D3
C4
D4
C1
+ -
+ -
2Vm
+ -
2Vm
+ -
2Vm
doubler
Quadrupler
Tripler

29. Thanks
9/11/2017
REC 101 Unit I by Dr Naim R Kidwai,
Professor & Dean, JIT Jahangirabad
29
Contact
Dr Naim R Kidwai
Professor & Dean
JETGI, Faculty of Engineering Barabanki, (UP)
Email: naim.kidwai@gmail.com
Naim.Kidwai@jit.edu.in