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Diode Applications
REC 101: Basic Electronics Unit 1
PN junction diode: Introduction of Semiconductor Materials Semiconductor Diode: Depletionlayer, V-I characteristics, ideal and practical, diode resistance, capacitance, Diode EquivalentCircuits, Transition and Diffusion Capacitance, Zener Diodes breakdown mechanism (Zenerand 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 1REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
Diode Application: Series, Parallel and Series, Parallel Configuration
• The forward resistance of diode is so smallcompared to the other series elements of networkthat it can be ignored (if not specified)
• In general, a diode is in the ON state if the currentestablished by the sources is such that, its directionmatches to the arrow of diode symbol, and VD VKotherwise it is in OFF state
9/11/2017REC 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 diode1. 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
VKRf
OR
Diode Equivalent circuitVK knee voltage
Rf diode forward resistance
Diode Application: Series , Parallel and Series, Parallel Diode Configuration,
9/11/2017REC 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
mA 93A 093.0100
7.010
I
0.7 V
10 V 100
I
Diode Application: rectifier
9/11/2017REC 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
Diode Application: Half Wave rectifier
9/11/2017REC 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 ONSo 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 toestablish a dc level is calledhalfwave 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 OFFSo Vo(t) = 0
+
Vo(t)=0
-
TtTfor
TtfortVtV
i
2/ 0
2/0 )()( period ofoutput combined Therefore 0
Diode Application: Half Wave rectifier
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
6
t
Vi(t)
T
Vm
-Vm
V0(t)
T
Vm
t2T
2T
Vdc= 0.318Vm
mmT
mdc
T
m
TT
i
T
dc
VCos
TCos
VtCos
T
VV
dttSinVT
dtdttVT
dttVT
V
)0(22
1.0)(
1)(
1
20
2
0
2
0
2
00
0
VDC: is periodic average of output wave
mdc VV 318.0
TtTfor
TtfortVtV
i
2/ 0
2/0 )()(0
2
04
2sin42
2cos12
.0)(1
)(1
2
20
2
20
22
0
2
2
0
222
0
2
0
2
0
2
mmT
mT
m
T
mdc
T
m
TT
i
T
orms
VVt
T
Vt
T
Vdtt
T
VV
dttSinT
VdtdttV
TdttV
TV
Vrms: square root of periodic average of squared output
mrms VV 5.0
Diode Application: Half Wave rectifier
9/11/2017REC 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
21.1157.112
12
factor form is k where
11factor Ripple
2
2
2
f
2
222
222
m
m
f
dc
rms
dc
dcrms
dc
ac
dcacrms
V
V
kV
V
V
VV
V
V
VVV
Diode Application: Bridge type Full Wave rectifier
9/11/2017REC 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 soD2 and D3 is OND1 and D4 is ONSo Vo(t) = Vi(t)
+
Vi(t)
-
R
- Vo(t) +
D1 D2
D3 D4
The process of utilizing fullperiod of ac signal to establisha dc level is called fullwaverectificationA bridge with 4 diodes asshown in figure is mostfamiliar full wave rectifier.
Diode Application: Bridge type Full Wave rectifier
9/11/2017REC 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 andD2 and D3 is OND1 and D4 is ONSo Vo(t) = - Vi(t)
m
T
m
T
i
T
dc
VdttSinV
TdttV
TdttV
TV
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
TtfortVtV
i
i
2/ )(
2/0 )()( period ofoutput combined Therefore 0
Vdc= 0.636Vm
Diode Application: Bridge type Full Wave rectifier
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
10
22
2)(
2)(
1 22
0
222
0
2
0
2 mm
T
m
T
i
T
orms
VVdttSin
T
VdttV
TdttV
TV
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
48.01109.1122
12
21
2
2
2
2
m
m
dc
rms
V
V
V
V
Diode Application: Centre tapped Full Wave rectifier
9/11/2017REC 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
tR
1:2
+Vi(t)- - Vo(t) +
+Vi(t)-+Vi(t)-
Positive Half cycle(0 t< T/2)Vi(t) >0 soD1 is ON & D2 is OFFSo V0(t) = Vi(t)
+Vi(t)-
R
- Vo(t) +
+Vi(t)-+Vi(t)-
1:2
D1
D2
Another type of fullwaverectifier is Centre tappedIt uses 2 diodes with a centretapped 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 soD1 is OFF & D2 is ONSo V0 (t) = -Vi(t)
Diode Application: Centre tapped Full Wave rectifier
9/11/2017REC 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
TtfortVtV
i
i
2/ )(
2/0 )()( period ofoutput combined Therefore 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
Diode Application: Clippers
9/11/2017REC 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)= 0If Vi(t) 0, Diode ON, Vo(t) =Vi(t)
Negative Series ClipperPositive Series Clipper
Diode Application: Clippers
9/11/2017REC 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
-Vt
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T
-(Vm+V)
If Vi(t) -V0, Diode ON, Vo(t)= Vi(t) -VIf Vi(t) -V<0, Diode OFF, Vo(t) =0
If Vi(t) -V>0, Diode OFF, Vo(t)= 0If 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+VV
V
V
-Vt
Vi(t)
T
Vm
-Vm
+
V0(t)
-
+
Vi(t)
-
R t
Vo(t)
T-(Vm-V)
If Vi(t) +V0, Diode ON, Vo(t)= Vi(t) +VIf Vi(t) +V<0, Diode OFF, Vo(t) =0
If Vi(t)+V>0, Diode OFF, Vo(t)= 0If Vi(t) +V 0, Diode ON, Vo(t) =Vi(t) +V
T
Negative biased Series Clipper Positive biased Series Clipper
Diode Application: Clippers
9/11/2017REC 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)= 0If 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)= VIf 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
Diode Application: Clippers
9/11/2017REC 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)= -VIf 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)
TV1
V2-V2
-V2
D1 D2
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
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
17
Diode Application: Clampers
9/11/2017REC 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
Diode Application: Clampers
9/11/2017REC 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)-Vm0, 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 soDiode is ONSo V0(t) = 0
Negative part of cycle(T/2 t< 0)Vi(t) =-Vm soDiode is OFFSo V0(t) = Vi(t)-Vm= -2Vm
Diode Application: Clampers
9/11/2017REC 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
Diode Application: Clampers
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
21
t
Vi(t)
T
Vm
-Vmt
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
-VmV
+V0(t)
-
+
Vi(t)-
R
C
(Vm+V)
+ -
tVo(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
-VmV
+V0(t)
-
+
Vi(t)-
R
C
(Vm-V)
+ -
t
Vo(t)
T
V
-2Vm+V
Diode Application: Zener diode as shunt regulator
9/11/2017REC 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
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
23
+
V0
-
R
+VZ
-
IZ
RLVi
+
V0
-
R
+V-
RLVi
L
iL
RR
VRV
If V VZ Then Zener is ON (breakdown region)
If V < VZ Then Zener is OFFIf diode is ON, then circuit shall be
+
V0
-
R
RLVi VZ
IRIZ IL
ZZZ
L
Zi
L
ZZiZ
LRZ
ZO
IVP
RRV
R
V
R
V
R
VVI
III
VVV
11
Diode Application: Zener diode as shunt regulator
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
24
L
iL
RR
VRV
If V VZ Then Zener is ON (breakdown region)If V < VZ Then Zener is OFFIf diode is ON, then circuit shall be
sheet dataper ascurrent diodemax is where
so constant, is as
breakdown in ON be todiodefor resistance LoadMin
thus
min
minmin
max
min
min
ZMZMRL
R
L
Z
L
LL
L
Zi
ZL
L
iLZO
IIII
I
R
V
R
VI
R
VV
RVR
RR
VRVVV
+
V0
-
R
+VZ
-
IZ
RLVi
+
V0
-
R
+V-
RLVi
+
V0
-
R
RLVi VZ
IRIZ IL
Diode Application: Zener diode as shunt regulator
Diode Application: Voltage-Multiplier Circuits
9/11/2017REC 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
Diode Application: Voltage-Multiplier Circuits
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
26
Half wave Voltage doubler
t
Vi(t)
T
Vm
-Vm
D1C1
+
Vi(t)
-
-
2Vm
+
D2C2
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
Diode Application: Voltage-Multiplier Circuits
9/11/2017REC 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
Diode Application: Voltage-Multiplier Circuits
9/11/2017REC 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
doublerQuadrupler
Tripler
Thanks
9/11/2017REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
29
Contact Dr Naim R KidwaiProfessor & DeanJETGI, Faculty of Engineering Barabanki, (UP)Email: [email protected]