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ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition
Chapter 4Power Supplies
©2003 Glencoe/McGraw-Hill
Charles A. Schuler
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• The System• Half-Wave Rectification• Full-Wave Rectification• RMS to Average• Filters• Multipliers• Ripple and Regulation• Zener Regulator
INTRODUCTION
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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.
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ac Power Supply
Circuit A
Circuit B
Circuit C
dc
dc
dc
Most line operated supplies change ac to dc.
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+
-
0
+
-
0ac
Half-wave pulsating dc
The cathode makes thisthe positive end of the load.
A series rectifier diode changes ac to dc.
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+
-
0
Full-wave pulsating dc
+
-
0
ac
The cathodes make thisthe positive end of the load.
Two diodes and a transformer provide full-wave rectification.
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VLOAD is equalto one-half thetotal secondary
voltage.
C.T.
½ VTOTAL
VTOTAL
Only half of the transformer secondary conducts at a time.
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+
-
0
Full-wavepulsating dc
+
-
0
ac
The bridge circuit eliminates the need for a transformer.
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+
-
0 Full-wavepulsating dc
+
-
0
ac
Reversing the diodes produces a negative power supply.
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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
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Vac
Vdc
ac
Vac
Vdc
Ignoring diode loss,the average dc is 45%
of the ac input forhalf-wave.
Vac
Vdc
Converting rmsto average
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Vac
Vdc
ac
Vac
Vdc
Ignoring diode loss,the average dc is 90%
of the ac input forfull-wave.
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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
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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
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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%
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Filtercapacitor
+
-
0
Charge
Discharge
VP
A relatively large filter capacitor will maintain theload voltage near the peak value of the waveform.
ac
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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
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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.
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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.
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ac
Vac
Vdc
Full-wave doublerVac
Vdc
Ignoring diode loss andassuming large filters,
the dc output is twice thepeak ac input.
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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.
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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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Vac
Vdc
Vac
Vdc
An ideal dc power supply has no ac ripple.
ac ripple = 0%
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Vac
Vdc
Vac
Vdc
acac ripple =
12 Vdc
1.32 Vac x 100% = 11 %
Real power supplieshave some ac ripple.
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Vac
Vdc
An ideal power supply has perfect voltage regulation.
The voltagedoes not change.
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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 %
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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
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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.
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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.
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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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ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition
Chapter 4Power Supplies
©2003 Glencoe/McGraw-Hill
Charles A. Schuler
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• The System• Half-Wave Rectification• Full-Wave Rectification• RMS to Average• Filters• Multipliers• Ripple and Regulation• Zener Regulator
INTRODUCTION
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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.
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ac Power Supply
Circuit A
Circuit B
Circuit C
dc
dc
dc
Most line operated supplies change ac to dc.
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+
-
0
+
-
0ac
Half-wave pulsating dc
The cathode makes thisthe positive end of the load.
A series rectifier diode changes ac to dc.
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+
-
0
Full-wave pulsating dc
+
-
0
ac
The cathodes make thisthe positive end of the load.
Two diodes and a transformer provide full-wave rectification.
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VLOAD is equalto one-half thetotal secondary
voltage.
C.T.
½ VTOTAL
VTOTAL
Only half of the transformer secondary conducts at a time.
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+
-
0
Full-wavepulsating dc
+
-
0
ac
The bridge circuit eliminates the need for a transformer.
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+
-
0 Full-wavepulsating dc
+
-
0
ac
Reversing the diodes produces a negative power supply.
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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
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Vac
Vdc
ac
Vac
Vdc
Ignoring diode loss,the average dc is 45%
of the ac input forhalf-wave.
Vac
Vdc
Converting rmsto average
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Vac
Vdc
ac
Vac
Vdc
Ignoring diode loss,the average dc is 90%
of the ac input forfull-wave.
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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
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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
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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%
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Filtercapacitor
+
-
0
Charge
Discharge
VP
A relatively large filter capacitor will maintain theload voltage near the peak value of the waveform.
ac
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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
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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.
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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.
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ac
Vac
Vdc
Full-wave doublerVac
Vdc
Ignoring diode loss andassuming large filters,
the dc output is twice thepeak ac input.
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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.
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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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Vac
Vdc
Vac
Vdc
An ideal dc power supply has no ac ripple.
ac ripple = 0%
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Vac
Vdc
Vac
Vdc
acac ripple =
12 Vdc
1.32 Vac x 100% = 11 %
Real power supplieshave some ac ripple.
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Vac
Vdc
An ideal power supply has perfect voltage regulation.
The voltagedoes not change.
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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 %
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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
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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.
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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.
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REVIEW
• The System• Half-Wave Rectification• Full-Wave Rectification• RMS to Average• Filters• Multipliers• Ripple and Regulation• Zener Regulator