Minimum reverse angle
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My brain is about fried, what I meant was the peak reverse voltage PRV (or minimum reserve voltage rating) for a full wave rectifier. In NCEES 515 they have a half waver rectifier, I'm wondering what is the answer for a full wave.
if the input is a 120 Vrms, on a half waver rectifier its PRV is 2 * sq rt 2 * 120.
So for a full waver rectifier would it be twice that of a half waver?
if the input is a 120 Vrms, on a half waver rectifier its PRV is 2 * sq rt 2 * 120.
So for a full waver rectifier would it be twice that of a half waver?
iahim
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The peak inverse voltage is the maximum reverse voltage, between anode and cathode, that the diode can withstand.
If you have a simple circuit with a single diode connected to a 120 Vrms AC circuit, during the positive half of the AC wave, the diode will conduct (act as a closed switch). During the negative half, the diode will block (act as an open switch). The maximum voltage that the diode has to be able to block occurs when the AC wave reaches the minimum (negative peak). The cathode will be at zero volts, because the diode is blocking. The peak for a sin wave is Vrms x sqrt (2). So, in this example the PIV is 170V.
In the NCEES problem, in addition to the diode, the circuit includes a capacitor. The capacitor will charge when the diode conducts (positive half of the AC) and discharge while the diode is blocking. So in this circuit, the cathode voltage is no longer at zero. It will be a positive value (whatever voltage is left across the capacitor). If the capacitor is very large and the load very small, the capacitor will not discharge much during the negative cycle, so the voltage across the capacitor will remain at +Vpeak. As before, the minimum voltage of the anode will be during the negative cycle, at -Vpeak. So the PIV is 2 x Vpeak, or 2 x sqrt(2) x Vrms.
A full wave rectifier is difficult to explain without a picture and waveforms, but if you apply the same concepts as above, you will see that the maximum will occur when the cathode is at zero and the anode at -Vpeak. So PIV is equal to Vpeak. Adding a capacitor in this circuit will not affect the PIV.
If you have a simple circuit with a single diode connected to a 120 Vrms AC circuit, during the positive half of the AC wave, the diode will conduct (act as a closed switch). During the negative half, the diode will block (act as an open switch). The maximum voltage that the diode has to be able to block occurs when the AC wave reaches the minimum (negative peak). The cathode will be at zero volts, because the diode is blocking. The peak for a sin wave is Vrms x sqrt (2). So, in this example the PIV is 170V.
In the NCEES problem, in addition to the diode, the circuit includes a capacitor. The capacitor will charge when the diode conducts (positive half of the AC) and discharge while the diode is blocking. So in this circuit, the cathode voltage is no longer at zero. It will be a positive value (whatever voltage is left across the capacitor). If the capacitor is very large and the load very small, the capacitor will not discharge much during the negative cycle, so the voltage across the capacitor will remain at +Vpeak. As before, the minimum voltage of the anode will be during the negative cycle, at -Vpeak. So the PIV is 2 x Vpeak, or 2 x sqrt(2) x Vrms.
A full wave rectifier is difficult to explain without a picture and waveforms, but if you apply the same concepts as above, you will see that the maximum will occur when the cathode is at zero and the anode at -Vpeak. So PIV is equal to Vpeak. Adding a capacitor in this circuit will not affect the PIV.
