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6. Unregulated Power Supply Design
6.1 Rectifier Circuit Classifications 6.2 Half-wave Rectifier Circuits Design 6.3 Full-wave Rectifier Circuits Design 6.4 Bridge Rectifier Circuits Design 6.5 Power Supply Classifications 6.6 Design equation of Power Supply with filter capacitor 6.7 Power Supply Circuits Design
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6.1 Rectifier Circuit Classifications
rectifier = diode is used to convert input ac into output dc half-wave rectifier circuit full-wave rectifier circuit bridge rectifier circuit
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rms and average (dc) values of rectified sine wave
average (dc) value of half-wave rectified sine wave half-wave rectified sine wave rms value of half-wave rectified sine wave
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average (dc) value of full-wave rectified sine wave
rms value of full-wave rectified sine wave full-wave and bridge rectified sine wave
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rms and average (dc) values of rectified sine wave
Vm t Sine wave Half-wave Full-wave rms value of the wave is used to find the heat power such as heater etc. dc (average) value of the wave is used to find the dc power such as running the dc motor etc.
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6.2 Half-wave Rectifier Circuits Design
Peak Inverse Voltage (PIV) of the diode PIV of the diode is found across the diode when the diode is not conducting here in half-wave rectifier, PIV = Vsm
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6.3 Full-wave Rectifier Circuits Design
Peak Inverse Voltage (PIV) of the diode neglected PIV of the diode is found across the diode when the diode is not conducting here in full-wave rectifier, if the voltage drop due to Rs and diode are neglected, PIV = 2Vsm
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6.4 Bridge Rectifier Circuits Design
Peak Inverse Voltage (PIV) of the diode neglected PIV of the diode is found across the diode when the diode is not conducting here in bridge rectifier, if the voltage drop due to Rs and diode are neglected, PIV = Vsm
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Summary of Design Equations
Rectifier Circuits half-wave rectifier circuit PIV = Vsm full-wave rectifier circuit PIV = 2Vsm Vm t Sine wave Sine wave Half-wave Half-wave bridge rectifier circuit PIV = Vsm Full-wave
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Design of half-wave rectifier-1
A half-wave rectifier is to deliver an average voltage of 40.5V to a dc load of RL=90W from an ac supply of 220V, 50Hz. Design the transformer turn ratio “n:1” , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10W.
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Design of half-wave rectifier-2
A half-wave rectifier is to heat a resistive load of RL=90W with a power of 44.1W from an ac supply of 220V, 50Hz. Design the transformer turn ratio “n:1” , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10W.
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Design of full-wave rectifier-1
A full-wave rectifier is to heat a resistive load of RL=90W with a power of 44.1W from an ac supply of 220V, 50Hz. Design the transformer turn ratio “n:1+1” , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10W.
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Design of full-wave rectifier-2
A full-wave rectifier using a full average current rating of the diode of 5A is to deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V, 50Hz. Design the value of RL , transformer turn ratio “n:1+1” , PIV of the diode. Assume that the secondary winding resistance of the transformer is 1W.
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Design of bridge rectifier
A bridge rectifier using a full average current rating of the diode of 5A is to deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V, 50Hz. Design the value of RL , transformer turn ratio “n:1” , PIV of the diode. Assume that the secondary winding resistance of the transformer is 1W.
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6.5 Power Supply Classifications
DC Power Supply = filter C is used to smooth the output dc from the rectifier half-wave power supply circuit full-wave power supply circuit bridge power supply circuit
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6.6 Design equation of Power Supply with filter capacitor
DV= - m1t1 DV= m2t2 VO
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Ripple factor of the dc output voltage
Ripple factor of the dc output voltage is the amount of rms. voltage of the ripple with respect to the dc output voltage
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Summary of Design Equations
Power Supply Circuits half-wave power supply circuit full-wave power supply circuit bridge power supply circuit
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6.7 Power Supply Circuits Design
Design Example 1 Draw the full-wave power supply with filter capacitor. Design the size of the capacitor for the following conditions. Find Ripple factor of the dc output voltage. Transformer turn ratio is 1:2 (1:1+1) with 100V, 60Hz. ac at the primary winding. Load resistor RL=2kW. Required minimum output voltage is 70V.
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Design Example 2 Draw the half-wave power supply with filter capacitor. Design the size of the capacitor for the following conditions. Transformer turn ratio is 1:1 with 100V, 60Hz. ac at the primary winding. Load resistor RL=2kW. The required ripple factor is 10% what is the dc voltage output?
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