NCEES Power Problem 515

[mull982]

In NCEES power exam prblem 515 can someone please explain to me why you have to use 2x the peak voltage for determining the maximum reverse rating across the diode? I understand that you multiply the 120rms voltage by 1.4 to get a peak voltage of 170V but I dont understand why you must use 2x this voltage for the reverse voltage across the diode?

Can someone please provide me with a quick explanation.

On another note, for problem #518 why is the DC voltage on the secondary of the 3-phase diode bridge 1.2x the L-L Vrms voltage of the input. I dont understand where the 1.2 multiplier is coming from. I thought that DC voltage was the same as AC rms voltage?

 

[Flyer_PE]

For 515, you use double the peak value since the capacitor will charge to a value of +170 Vdc. When the AC waveform will then move to a negative 170 Volt value. The voltage across the diode at that point will be around 340 Volts.

For 518, you are trying to limit the peak charging current. This peak value will be determined by the peak voltage which is sqrt2 * VL-L rms.

 

[mull982]

For 515, you use double the peak value since the capacitor will charge to a value of +170 Vdc. When the AC waveform will then move to a negative 170 Volt value. The voltage across the diode at that point will be around 340 Volts.

For 518, you are trying to limit the peak charging current. This peak value will be determined by the peak voltage which is sqrt2 * VL-L rms.

Thanks Flyer

For 515 would the answer be the same for the diodes arranged is a full wave bridge instead of a half wave bridge. Or in a full wave bridge doesn't the cap carge and discharge every other cycle?

For 518 why would there be a peak current. Wouldn't the current be determined by the DC voltage which is esentially Vrms? Aren't the peaks of the DC voltage esentially equal to the Vrms voltage peaks at 208V?