rec101 unit 1 (part iii) diode applications

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,


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


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


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



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



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


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.



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/ )(


Vdc= 0.636Vm



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) +


48.01109.1122

12

21

2

2

2

2

m

m

dc

rms

V

V

V

V

Diode Application: Centre tapped Full Wave rectifier


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


12

VDC: equal to that of bridge type full wave rectifier

mdc VV 636.0

TtTfortV

TtfortVtV

i

i

2/ )(


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 +


48.0

Diode Application: Clippers


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



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



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



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


17



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

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



20

The output is drawn

t

Vi(t)

T

Vm

-Vm

t

Vo(t)

T

-2Vm

-2Vm

-2Vm



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


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

+

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



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: Voltage-Multiplier Circuits


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

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



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

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


29

Contact Dr Naim R KidwaiProfessor & DeanJETGI, Faculty of Engineering Barabanki, (UP)Email: [email protected]

[email protected]

mailto:[email protected]