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**Principles & Applications**

Electronics Principles & Applications Sixth Edition Charles A. Schuler Chapter 4 Power Supplies (student version) © Glencoe/McGraw-Hill

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**INTRODUCTION The System Half-Wave Rectification**

Full-Wave Rectification RMS to Average Filters Multipliers Ripple and Regulation Zener Regulator

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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 you to view that segment again, if you want to.

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**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.

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**The power supply energizes the other circuits in a system.**

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.

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**Most line operated supplies change ac to dc.**

Power Supply Circuit A Circuit B Circuit C dc dc ac dc Most line operated supplies change ac to dc.

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**the positive end of the load.**

The cathode makes this the positive end of the load. + Half-wave pulsating dc + ac - - A series rectifier diode changes ac to dc.

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**the positive end of the load.**

The cathodes make this the positive end of the load. + Full-wave pulsating dc + - ac - Two diodes and a transformer provide full-wave rectification.

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**VLOAD is equal to one-half the total secondary voltage.**

Only half of the transformer secondary conducts at a time. VLOAD is equal to one-half the total secondary voltage. C.T. VTOTAL ½ VTOTAL

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**The bridge circuit eliminates the need for a transformer.**

+ + ac - Full-wave pulsating dc - The bridge circuit eliminates the need for a transformer.

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**Reversing the diodes produces a negative power supply.**

+ + ac Full-wave pulsating dc - - Reversing the diodes produces a negative power supply.

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**Power Supply Basics Quiz**

Most line-operated power supplies change ac to ________. dc A single diode achieves ________ -wave rectification. half Two diodes and a center-tapped transformer provide ________ -wave rectification. full A bridge rectifier uses ________ diodes. four The positive end of the load is the end in contact with the diode ________. cathodes

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**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

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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 3f full-wave rectifiers.

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**Vac Vdc Vac Vdc Vac Vdc Ignoring diode loss, the average dc is 45%**

of the ac input for half-wave. Vac Vdc ac Converting rms to average

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**Vac Vdc Vac Vdc Ignoring diode loss, the average dc is 90%**

of the ac input for full-wave. ac Vac Vdc

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**Three-phase rectification is used in commercial, **

industrial and vehicular applications. Full-wave, 3 f bridge 3 f 120 V 60 Hz Vdc = x Vrms = 162 V 0 V -200 V +200 V 20 ms 40 ms 0 ms

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**Three-phase rectifier output**

200 160 120 Vdc = x Vrms = 162 V Volts 80 40 10 20 30 40 Time in milliseconds

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**Average dc Quiz The average dc voltage with half-wave is**

equal 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-wave rectifier is _________ of the effective ac voltage. 90% The average dc voltage with a bridge rectifier is equal to ________ of the ac voltage. 90% The average dc voltage with a 3 f bridge rectifier is equal to ________ of the ac voltage. 135%

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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 3f full-wave rectifiers. Repeat Segment

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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.

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**VP Filter capacitor Discharge + ac Charge -**

A relatively large filter capacitor will maintain the load voltage near the peak value of the waveform. + Charge ac -

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**VP Discharge time is less. + ac -**

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

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**Adding a filter capacitor increases the dc output**

Vac Vdc Ignoring diode loss and assuming a large filter, the dc output is equal to the peak value for both half-wave and full-wave. Adding a filter capacitor increases the dc output voltage. Vac Vdc ac

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**Vac Vdc Vac Vdc Ignoring diode loss and assuming a large**

filter, the dc output is equal to the peak value for both half-wave and full-wave. ac Vac Vdc

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**Vac Vdc Vac Full-wave doubler Vdc Ignoring diode loss and**

assuming large filters, the dc output is twice the peak ac input. ac Vac Vdc Full-wave doubler

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**Half-wave voltage doubler**

C2 The charge on C1 adds to the ac line voltage and C2 is charged to twice the peak line value. C1 is charged. C1 ac Half-wave voltage doubler

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**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%

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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

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**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.

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**An ideal dc power supply has no ac ripple.**

Vac Vdc An ideal dc power supply has no ac ripple. ac ripple = 0% Vac Vdc

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**Vac Vdc 1.32 Vac 12 Vdc Vac Vdc ac x 100% ac ripple = = 11 %**

= 11 % Real power supplies have some ac ripple. Vac Vdc

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**An ideal power supply has perfect voltage regulation.**

The voltage does not change. Vac Vdc

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**Vac Vdc Vac Vdc DV x 100 % Voltage Regulation = VFL 1 V = x 100 % 12 V**

= % The output of a real supply drops under load. Vac Vdc

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**The voltage across a conducting zener is relatively constant.**

Reverse Bias in Volts 12 10 8 6 4 2 5 I 10 Reverse Current in mA 15 Vrev 20 25 30 35 V The voltage across a conducting zener is relatively constant.

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**be used to regulate voltage.**

Vac Vdc If the zener stops conducting, the regulation is lost. Vac Vdc Vac A shunt zener diode can be used to regulate voltage. Vdc ac

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**Power Supply Quality Quiz**

The voltage regulation of an ideal power supply is ___________. 0% The ac ripple output of an ideal power supply is ___________. 0% V in a real power supply should be as ___________ as is feasible. small The ac component of an ideal dc power supply should be as ________ as is feasible. small A device that is commonly used to regulate voltage is the ________ diode. zener

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**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

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**REVIEW The System Half-Wave Rectification Full-Wave Rectification**

RMS to Average Filters Multipliers Ripple and Regulation Zener Regulator

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Principles & Applications

Principles & Applications

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