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

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

3. 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.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. 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

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

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

12. 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

13. 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

14. 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.

15. 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

16. Vac Vdc Vac Vdc Ignoring diode loss, the average dc is 90%
of the ac input for
full-wave. ac
Vac
Vdc

17. 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

18. Three-phase rectifier output
200
160
120
Vdc = x Vrms = 162 V
Volts 80 40 10 20 30 40
Time in milliseconds

19. 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%

20. 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

21. 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.

22. 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 -

23. 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.

24. 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

25. 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

26. 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

27. Half-wave voltage doublerC2
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

28. 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%

29. 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

30. 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.

31. 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

32. 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

33. An ideal power supply has perfect voltage regulation.
The voltage
does not change.
Vac
Vdc

34. 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

35. 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.

36. 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

37. 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

38. 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

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