electronics principles & applications sixth edition chapter 4 power supplies (student version)...

ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition

Chapter 4Power Supplies

(student version)

©2003 Glencoe/McGraw-Hill

Charles A. Schuler

• The System• Half-Wave Rectification• Full-Wave Rectification• RMS to Average• Filters• Multipliers• Ripple and Regulation• Zener Regulator

INTRODUCTION

Dear Student:

This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow youto view that segment again, if you want to.

Concept Preview• The power supply energizes the system.

• Rectifiers change ac to dc.

• A half-wave rectifier conducts for half of the ac input cycle.

• The cathode end of the load circuit is positive.

• A full-wave rectifier conducts for the entire ac input cycle.

• Full-wave rectifiers use two diodes and a center tapped transformer.

• A bridge rectifier uses four diodes and provides full-wave performance without a transformer.

Power Supply

Circuit A

Circuit B

Circuit C

The power supply energizes the other circuits in a system.

Thus, a power supply defect will affect the other circuits.

ac Power Supply

Circuit A

Circuit B

Circuit C

dc

dc

dc

Most line operated supplies change ac to dc.

+

-

0

+

-

0ac

A series rectifier diode changes ac to dc.

Half-wave pulsating dc

The cathode makes thisthe positive end of the load.

+

-

0

Full-wave pulsating dc

+

-

0

ac

Two diodes and a transformer provide full-wave rectification.

The cathodes make thisthe positive end of the load.

VLOAD is equalto one-half thetotal secondary

voltage.

C.T.

Only half of the transformer secondary conducts at a time.

½ VTOTAL

VTOTAL

+

-

0

Full-wavepulsating dc

+

-

0

ac

The bridge circuit eliminates the need for a transformer.

+

-

0 Full-wavepulsating dc

+

-

0

ac

Reversing the diodes produces a negative power supply.

Power Supply Basics QuizMost line-operated power supplies change ac to ________. dc

A single diode achieves ________ -wave rectification. half

Two diodes and a center-tapped transformerprovide ________ -wave rectification. full

A bridge rectifier uses ________ diodes. four

The positive end of the load is the end in contact with the diode ________. cathodes

Concept Review• The power supply energizes the system.• Rectifiers change ac to dc.• A half-wave rectifier conducts for half of the ac

input cycle.• The cathode end of the load circuit is positive.• A full-wave rectifier conducts for the entire ac

input cycle.• Full-wave rectifiers use two diodes and a center

tapped transformer.• A bridge rectifier uses four diodes and provides

full-wave performance without a transformer.

Repeat Segment

Concept Preview

• When a meter is connected to the output of a rectifier, it will respond to the average value of the pulsating dc waveform.

• The average dc load voltage is 45% of the rms input voltage for half-wave rectifiers.

• The average dc load voltage is 90% of the rms input voltage for full-wave rectifiers.

• The average dc load voltage is 135% of the rms input voltage for full-wave rectifiers.

Vac

Vdc

ac

Vac

Vdc

Ignoring diode loss,the average dc is 45%

of the ac input forhalf-wave.

Vac

Vdc

Converting rmsto average

Vac

Vdc

ac

Vac

Vdc

Ignoring diode loss,the average dc is 90%

of the ac input forfull-wave.

0 V

-200 V

+200 V

20 ms 40 ms0 ms

3 120 V60 Hz

Three-phase rectification is used in commercial,industrial and vehicular applications.

Full-wave, 3 bridge

Vdc = 1.35 x Vrms = 162 V

10 20 30 400

Time in milliseconds

40

80

120

160

200

0

Vol

tsThree-phase rectifier output

Vdc = 1.35 x Vrms = 162 V

Average dc QuizThe average dc voltage with half-wave isequal to ______ of the ac voltage. 45%

The effective ac voltage in a two-diode, full-wave rectifier is _______ of the secondary voltage.

half

The average dc voltage with a full-waverectifier is _________ of the effective ac voltage.

90%

The average dc voltage with a bridge rectifieris equal to ________ of the ac voltage. 90%

The average dc voltage with a 3 bridge rectifier is equal to ________ of the ac voltage. 135%

Concept Review• When a meter is connected to the output of a

rectifier, it will respond to the average value of the pulsating dc waveform.

• The average dc load voltage is 45% of the rms input voltage for half-wave rectifiers.

• The average dc load voltage is 90% of the rms input voltage for full-wave rectifiers.

• The average dc load voltage is 135% of the rms input voltage for full-wave rectifiers.

Repeat Segment

Concept Preview

• A filter capacitor will charge to the peak value of the ac input.

• The peak value of the ac input is 141% of the rms value.

• The peak value is 141% for both half-wave and full-wave circuits.

• Voltage doublers produce a peak value that is 282% of the rms input value.

Filtercapacitor

+

-

0

Charge

Discharge

VP

A relatively large filter capacitor will maintain theload voltage near the peak value of the waveform.

ac

ac

+

-

0

Discharge time is less.

Full-wave is easier to filter since the discharge time is shorter than it is for half-wave rectifiers.

VP

Adding a filter capacitorincreases the dc output

voltage.

Vac

Vdc

ac

Vac

Vdc

Ignoring diode lossand assuming a largefilter, the dc output is

equal to the peak valuefor both half-wave and

full-wave.

Vac

Vdc

Vac

Vdc

ac

Ignoring diode lossand assuming a largefilter, the dc output is

equal to the peak valuefor both half-wave and

full-wave.

ac

Vac

Vdc

Full-wave doublerVac

Vdc

Ignoring diode loss andassuming large filters,

the dc output is twice thepeak ac input.

C1 is charged.

C1ac

Half-wave voltage doubler

C2

The charge on C1 adds to the ac line voltageand C2 is charged to twice the peak line value.

Capacitive Filter dc Output Quiz(Ignore diode loss and assume a light load for this quiz.)

The dc output in a well-filtered half-wave supply is _____ of the ac input. 141%

The dc output in a well-filtered full-wave supply is _____ of the ac input. 141%

The dc output in a well-filtered half-wave doubler is _____ of the ac input. 282%

The dc output in a well-filtered full-wave doubler is _____ of the ac input. 282%

Concept Review

• A filter capacitor will charge to the peak value of the ac input.

• The peak value of the ac input is 141% of the rms value.

• The peak value is 141% for both half-wave and full-wave circuits.

• Voltage doublers produce a peak value that is 282% of the rms input value.

Repeat Segment

Concept Preview• The ideal dc power supply has no ac ripple.

• The percentage of ac ripple is a measure of how closely a real power supply approaches the ideal.

• The dc output voltage of an ideal supply never changes.

• The percentage of voltage regulation is a measure of how closely a real power supply approaches the ideal.

• Zener diodes are used as voltage regulators.

• The voltage drop across a zener diode remains almost constant, if it is conducting.

Vac

Vdc

Vac

Vdc

An ideal dc power supply has no ac ripple.

ac ripple = 0%

Vac

Vdc

Vac

Vdc

acac ripple =

12 Vdc

1.32 Vac x 100% = 11 %

Real power supplieshave some ac ripple.

Vac

Vdc

An ideal power supply has perfect voltage regulation.

The voltagedoes not change.

Vac

Vdc

Vac

Vdc

ac

The output of areal supply drops

under load.

Voltage Regulation =VVFL

x 100 %

= 12 V1 V

x 100 %

= 8.33 %

0246

5

10

15

20

25

30

35

Reverse Bias in Volts

ReverseCurrentin mAI

V

Vrev

The voltage across a conductingzener is relatively constant.

81012

ac

Vac

Vdc

A shunt zener diode canbe used to regulate voltage.

Vac

Vdc

Vac

Vdc

If the zener stops conducting, the regulation is lost.

Power Supply Quality QuizThe voltage regulation of an ideal powersupply is ___________. 0%

The ac ripple output of an ideal powersupply is ___________. 0%

small

The ac component of an ideal dc power supply should be as ________ as is feasible. small

A device that is commonly used to regulatevoltage is the ________ diode. zener

V in a real power supply should be as ___________ as is feasible.

Concept Review• The ideal dc power supply has no ac ripple.• The percentage of ac ripple is a measure of how

closely a real power supply approaches the ideal.• The dc output voltage of an ideal supply never

changes.• The percentage of voltage regulation is a measure of

how closely a real power supply approaches the ideal.

• Zener diodes are used as voltage regulators.• The voltage drop across a zener diode remains

almost constant, if it is conducting.

Repeat Segment

REVIEW

• The System• Half-Wave Rectification• Full-Wave Rectification• RMS to Average• Filters• Multipliers• Ripple and Regulation• Zener Regulator