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Half Wave Diode Rectifier Circuit
how to calculate voltage and current in half wave rectifier circuit?
Circuit has 120Vrms @ 60Hz. V secondary is 20Vrms and the load resistor is 1k ohms. Can you show me the formulas to calculate the load current and diode current, volts peak, Vdiode peak inverse voltage.
If this is truly a resistve load, it's all pretty simple. First, for a half wave rectifier, the rms voltage is not just half the rms of a full wave. Full wave rms is 1/sqrt(2) of the peak voltage, half wave rms is 1/2 the peak voltage. This all comes from the definition of rms being the square Root of the Mean of the Square of the voltage. The rms current and voltage can then be used to calculate the power. SInce this is a simple series circuit, the diode current is the load current. The rms voltage is just 20*0.5 = 10 volts. The rms current is 10 volts/1000 ohm = 10 mA.
The peak current is the same for full or half wave rectifiers. For a sine wave Vpeak = Vrms*sqrt(2) so the peak voltage is 20 V * sqrt(2) = 28.3 volts. This is also the peak reverse voltage of the diode.
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Animation of Diode Rectifier
Why would there be a difference in peak Voltages of a half- and full-wave rectifier circuit?
My teacher told me there should be a difference, but didn't tell me whether it would be higher in the full-wave or half-wave rectifier. He also didn't explain why. I am just working on an assignment a little ahead of time, and he hasn't explained that yet. Does it have something to do with the fact that the full-wave rectifier circuit is going to have 4 diodes as opposed to the 1 diode in the half-wave rectifier?
In theory, the "peak" voltage of the raw rectified signal should be the same in either one. After filtering, obviously, the half-wave rectifier puts out only half the power of a full wave rectifier.
Maybe your teacher misspoke. Ask him what he means.
====edit====
I think that this is what he was talking about:
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...An aspect of most rectification is a loss from peak input voltage to the peak output voltage, caused by the threshold voltage of the diodes (around 0.7 V for ordinary diodes and 0.1 V for Schottky diodes). Half wave rectification and full wave rectification using two separate secondaries will have a peak voltage loss of one diode drop. Bridge rectification will have a loss of two diode drops. This may represent significant power loss in very low voltage supplies. In addition, the diodes will not conduct below this voltage, so the circuit is only passing current through for a portion of each half-cycle, causing short segments of zero voltage to appear between each "hump"....


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