Week 3 Diode Applications

8/12/2019 Week 3 Diode Applications

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1. Half-Wave Rectifier

A Rectifieris a circuit that converts ac to pulsating dc.

Half-wave rectifier: A circuit that eliminates either the positive ornegative alternations of its input signal.

The basic half-wave rectifier

shown is a diode that is

connected in series between

an ac source and a load

resistance.

DIODE APPLICATIONS


2/26

Half-Wave Rectifier Operationwith Resistive Load

For the positive half-cycle of the

input, the source forces positive

current through the diode, the

diode is on, and vO = vS(for anideal diode).

During the negative half cycle,

negative current cant exist in the

diode. The diode is off, current in

resistor is zero, and vO =0 .


3/26

Half-Wave Rectifier Circuit withResistive Load (cont.)

Using the offset model, during the on-state of the

diode vO = vS- Von = (VP sinwt)- Von. The output

voltage is zero when the diode is off.

Often a step-up or step-down transformer is used

to convert the 220-V, 50-Hz voltage availablefrom the power line to the desired ac voltage

level as shown.

Time-varying components in the rectifier output

are removed using a filter capacitor.


4/26

Peak Detector Circuit Operation

As the input voltage rises,the diode is on, and the

capacitor (initially

discharged) charges up to

the input voltage minus the

diode voltage drop.At the peak of the input

voltage, the diode current

tries to reverse, and the

diode cuts off. The

capacitor has no dischargepath and retains a constant

voltage providing a constant

output voltage:

Vdc = VP Von


5/26

Half-Wave Rectifier Circuit with RCLoad

As the input voltage rises during the first

quarter cycle, the diode is on and thecapacitor (initially discharged) charges

up to the peak value of the input voltage.

At the peak of the input, the diode

current tries to reverse, the diode cuts

off, and the capacitor dischargesexponentially throughR. Discharge

continues till the input voltage exceeds

the output voltage which occurs near the

peak of next cycle. This process then

repeats once every cycle.

This circuit can be used to generate

negative output voltage if the top plate of

capacitor is grounded instead of bottom

plate. In this case,

Vdc = (VP Von)


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Half-Wave Rectifier Circuit withRCLoad (cont.)

The output voltage is not constant as in an ideal peak detector, but has a ripple

voltage Vr.

The diode conducts for a short time DTcalled the conduction interval during

each cycle, and its angular equivalent is called the conduction angle c.

P

rc

onPonPr

V

VT

C

T

R

VV

T

T

RC

TVVV

2

)(1)(

D

D

w


7/26

EXAMPLEHalf-Wave Rectifier Analysis

Problem: Find the dc output voltage, output current, ripple voltage, conduction

interval, and conduction angle for a half-wave rectifier.

Given data: secondary voltage Vrms = 12.6 (60 Hz),R = 15 W, C = 25,000

mF, Von = 1 V

Analysis:

Using discharge interval T = 1/60 s,

VC

T

R

VV

V

AVR

VVI

VVVVV

onP

r

onPdc

onPdc

747.0

)(

12.115

8.16

8.16)126.12(

W

cwDT

2Vr

VP

0.290 rad16.6o

DTcw

c

2f

0.29

1200.769 ms


8/26

Peak Diode CurrentIn rectifiers, nonzero current exists

in the diode for only a very small

fraction of period T, yet an almost

constant dc current flows out of the

filter capacitor to load.

If the repetitive current pulse is

modeled as a triangle of height IP

and widthDT,

using the values from the previous

example.

IPI

dc

2TDT

48.6 A


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Surge Current

In addition to the peak diode currents, there is an even larger current through the

diode called the surge current that occurs when power is first turned on.

During first quarter cycle, current through diode is approximately

The peak value of this initial surge current occurs at t = 0+:

using values from previous example.

Actual values of surge current wont be nearly as large as predicted abovebecause of the neglected series resistances associated with both the rectifier

diode and transformer.

)cos(sin)()( tVCtVdt

dCtiti

PPcd www

ISCwCV

P168 A


10/26

Peak Inverse Voltage RatingThe peak inverse voltage (PIV) rating of

the rectifier diode is the diodebreakdown voltage.

When the diode is off, the reverse-bias

across the diode is Vdc - vS. When vlreaches its negative peak,

PPonPldc VVVVvV 2)(PIV min


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2. Full-Wave Rectifiers

Full-wave rectifiers cut capacitordischarge time in half and require

half the filter capacitance to achieve

a given ripple voltage.

All specifications are the same as

for half-wave rectifiers.


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Full-Wave Rectifier Equations

P

r

P

onP

onPronPdc

V

V

V

VV

RC

T

T

C

T

R

VVVVVV

21)(1

2

)(

ww

D

PVPIV

V

VT

P

rc

2

2

Dw


13/26

3. Bridge Rectification

The requirement for a center-tapped transformer in the

full-wave rectifier is

eliminated through use of 2

extra diodes.

All other specifications are

the same as for a half-wave

rectifier except PIV = VP.


14/26

Rectifier Topology Comparison andDesign Tradeoffs

Rectifier Parameter Half-Wave Rectifier Full-Wave Rectifier Full-Wave Bridge Rectifier

Filter CapacitorC

VPV

on

Vr

T

R C

VPV

on

Vr

T

2R C

VPV

on

Vr

T

2R

PIV Rating

2VP 2

VP

VP

Comments Least Complexity Smaller Capacitor

Center-taped Transformer

Two Diodes

Smaller Capacitor

No Center-taped Transformer

Four Diodes


15/26

DESIGN EXAMPLE

Rectifier Design Analysis

Problem: Design a rectifier with given specifications.

Given data: Vdc = 15 V, Vr< 0.15 V,Idc = 2 A Assume: Von = 1 V.

Analysis: Use a full-wave bridge rectifier that needs a smaller value of filter

capacitance, smaller diode PIV rating, and no center-tapped transformer.

VVPIVACVI

Ams

sAT

TIIms

V

V

V

VT

FV

sAV

TIC

rmsVVVVV

V

PPsurge

dcP

P

r

rdc

ondcP

17=|711)17)(111.0(120

7.94352.0

60/12

22|352.0

17

)15.0(2

120

121

111.015.0

1

120

12

2/

0.122

215

2

2

2

D

D

w

w


16/26

4. Diode Limiters (Clippers)

Limiter: A diode circuit that limits or clips off the positive (or negative)

part of its input signal.

The circuit shown is a

positive limiter.

Because the cathodeis at ground potential

(0 V), the anode

cannot exceed 0.7 V

(assuming a silicon

diode). Thus, pointAis clipped at 0.7 V

when the input

exceeds this value.


17/26

A Limiter Applications | 5 V SupplyAn example shows a circuit that limits the signal into a computer.

The Diode D2 will conduct when the voltage V1 > 4 V.

When D2 is ON, V2 = 4 V + Von.

When D2 is OFF, V2 = V1.

(V1) (V2)


18/26

5. Diode Clampers

A diode clamper adds a dc level to an ac signal. They are often referred to as dc

restorers.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-

-

im


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Zener DiodesZener diode: A siliconpnjunction device that differs from the rectifier diode in

that it is designed for operation in its reverse-breakdown region.

SPECIAL PURPOSE DIODES


20/26

Zener Diode Reverse Characteristics

Ideally, the reversebreakdown has a

constant breakdown

voltage.

This makes ZenerDiode useful as a

voltage reference,

which is its primary

application.


21/26

VOLTAGE REGULATOR USING THE ZENERDIODE

The Zener diode keeps the voltage across load resistor RL constant. ForZener breakdown operation,IZ > 0.

For proper regulation, Zener currentIZ must be positive.

If the Zener current < 0, the Zener diode no longer controls the voltage

across the load resistor and the voltage regulator is said to have dropped

out of regulation.

min1

|011 R

V

V

RR

RRV

R

VI

Z

S

LL

ZS

Z


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ISVSV

Z

R

(205)V

5kW 3 mA

ILVZ

RL

5V

5kW

1 mA | IZI

SI

L2 mA

EXAMPLEVoltage Regulator using the Zener Diode

Problem: Find the source, load and Zener

diode currents for a Zener diode regulator.

Given data: VS= 20 V, R = 5 kW, V

Z= 5 V


23/26

EXAMPLEVoltage Regulator Including Zener Resistance

Problem: Find the output

voltage and Zener diode current

for a Zener diode regulator.

Given data: VS = 20 V,R = 5

kW,RZ = 0.1 kW, VZ = 5 V

Analysis: The output voltage is

now a function of the current

through the Zener diode.

0mA9.1100V5V19.5

100

V5

V19.5

05000100

V55000

V20

W

W

W

W

W

L

Z

L

LLL

V

I

V

VVV


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Line and Load Regulation

Line regulation characterizes how sensitive the output voltage is to input voltagechanges.

mV/VRegulationLine

S

L

dV

dV

Load regulation characterizes how sensitive the output voltage is to changes in

load current withdrawn from regulator.

W RegulationLoad

L

L

dIdV


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EXAMPLE: Line and Load RegulationCalculations

For the circuit shown in the figure the lineregulation can be calculated from the nodal

equation for VL

RZ

RR

ZR

RZ

R

LdI

LdV

RZ

R

ZR

SdV

LdV

L

I

ZR

ZV

R

SV

RZ

R

RZ

R

V

IVV

RZ

R

RZ

R

VIVV

ZRR

V

IVVVV

IVVVV

L

LR

Z

R

S

LLR

Z

R

S

L

LR

Z

R

L

R

S

R

L

LR

ZL

R

SL

ZZ

ZZZ

//|

11

00


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Examples of Practical DiodeApplications

Rectifier Circuits Conversions of AC to DC for DC operated circuits

Battery Charging Circuits

Simple Diode Circuits Protective Circuits against

o Overcurrent

o Polarity Reversal

o Currents caused by an inductive kick in a relay circuit

Zener Circuits

Overvoltage Protection Setting Reference Voltages

Week 3 Diode Applications

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