ee 21-lecture 3.5 power supply design

8/3/2019 EE 21-Lecture 3.5 Power Supply Design

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12/6/20

Power Supply DesignA little more diode applications…

1

E E 2 1 S l i d e s ( A A M S )

Topics

• Parts of a power supply

• Ripple

•

Voltage Regulation• Capacitor – input filter

• RMS ripple and ripple voltage

• DC voltage output

• Effect of capacitor sizing

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E E 2 1 S l i d e s ( A A M S )

Block diagram of a power

supply

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E E 2 1 S l i d e s ( A A M S )

Ripple Voltage

• A rectified AC signal will have a DC (steady) component

and an AC (varying) component known as the ripple.

• Ripple is given by the equation

4

%100 voltagedc

ms)voltage(rripple )( x

V

Vr r

dc

rms

E E 2 1 S l i d e s ( A A M S )

Example

• The following readings were obtained in

measuring the output signal of a filter circuit:

• DC voltmeter: 25 volts DC

• AC voltmeter: 1.5 volts RMS

• Compute for the ripple of the filter output.

• Ans: 6% 5

E E 2 1 S l i d e s ( A A M S )

Voltage Regulation

• Describes the amount of change between the no-load

and load conditions of a power supply.

• Given by the following equation:

• Where

%V.R. – voltage regulation

VNL = DC voltage at no-load

VFL = DC voltage at full-load 6

%100..% xV

V V RV

FL

FL NL

E E 2 1 S l i d e s ( A A M S )



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Example

• A DC voltage supply provides 60 V when the

output is unloaded. When connected to a load,

the measured voltage drops to 56 V. Calculatethe voltage regulation.

• Ans. 7.1428%

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E E 2 1 S l i d e s ( A A M S )

Ripple factor of a half-wave

rectified signal

8

mdc

V V 318.0

mrmsr V V 385.0)(

%121%100318.0

385.0)( x

V

V

V

V r

m

m

dc

rmsr E E 2 1 S l i d e s ( A A M S )

Ripple factor of a full-wave

rectified signal• For half-wave rectified signal:

9

mdc V V 636.0

mrmsr V V 308.0)(

%48%100636.0

308.0)( x

V

V

V

V r

m

m

dc

rmsr E E 2 1 S l i d e s ( A A M S )

Capacitor Filter

• Capacitor develops voltage from the output of

the rectifier circuit (RC charging)

• The capacitor discharges during “down time”

of rectifier circuit cycle

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E E 2 1 S l i d e s ( A A M S )

Capacitor Filter Operation

• Normal rectifier

output

• With capacitor filter

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E E 2 1 S l i d e s ( A A M S )

Output Waveform Times

At T1:

-diodes conduct

-cap charges to Vm

At T2:

-rectifier V drops

-cap discharges

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E E 2 1 S l i d e s ( A A M S )



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Output Waveform Times

• Charge-discharge

cycle occurs every

half-cycle of rectifier

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• Filtered voltageincludes a dc level Vdc,as well as a ripple

voltage Vr(rms) causedby capacitor chargingand discharging.

E E 2 1 S l i d e s ( A A M S )

Ripple Voltage Vr (RMS)

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C RV

C I

fC I rmsVr

L

dcdcdc 4.24.234

)(

IMPORTANT!

Idc in milliamperes

C in microfarads

RL in kilohms

E E 2 1 S l i d e s ( A A M S )

Example

• Calculate the ripple voltage of a full-wave

rectifier with a 100-μF filter capacitor

connected to a load drawing 50mA.

• Ans. 1.2 V

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E E 2 1 S l i d e s ( A A M S )

DC Voltage Vdc

16

C

I V

fC

I V V dc

mdc

mdc

17.4

4

Where

Idc in milliamperes

C in microfaradsVm is the peak rectifiervoltage

E E 2 1 S l i d e s ( A A M S )

Example

• If the peak rectified voltage for the previous

filter circuit example is 30 Volts, calculate thefilter DC voltage.

• Ans. 27.915 Volts

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E E 2 1 S l i d e s ( A A M S )

Filter Capacitor Ripple

18

%1004.2

%1004.2

%100)(

xC R

I x

CV

I x

V

rmsV r

L

dc

dc

dc

dc

r

Where

Idc in milliamperes

C in microfarads

Vdc in Volts

RL in kilohms

E E 2 1 S l i d e s ( A A M S )



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Example

• Calculate the ripple of a capacitor filter for a

peak rectified voltage of 30 Volts, a capacitor

C = 50μF, and a 50 mA load current.

• Ans. 4.3%

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E E 2 1 S l i d e s ( A A M S )

Diode Conduction Period and

Peak Diode Current • Larger capacitor = less ripple, higher Vdc

• However, capacitor size directly affects the

current drawn by the rectifying diodes

• Larger capacitor = larger peak current drawn

by rectifying diodes

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E E 2 1 S l i d e s ( A A M S )

Capacitor Filter Waveforms

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E E 2 1 S l i d e s ( A A M S )

Current calculations

• Where T1 = diodeconduction time

• T = rectifier period(note that it’s twicefor the fullwave)

• Idc = average current drawn from filter

• Ipeak = peak current through conducting

diodes22

peak dc I T

T I 1

dc peak I T

T I

1

E E 2 1 S l i d e s ( A A M S )

ee 21-lecture 3.5 power supply design

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