NCEES Problem #515 Diode Rating/Capacitor

[Spark-E]

Sorry for posting again on this question but I'm having trouble understanding the answer provided by NCEES and the other postings.

The input signal is 120VACrms. I would think that the cap would charge until it reached +60VACrms, to the positive peak, and it would start discharging. I would think this would repeat back and forth, charging and discharging the cap.

This would lead me to see that the output of the diode sees a max/min of +60/-60 VACrms. Same with the input.

Using the maximums, I would think that the rating would be sqrt(2)*(abs(+60)+abs(-60)), basically sqrt(2)*120.

I feel like I'm missing something fundamental here. Any ideas?

Thanks.

 

[iahim]

Have you seen this post (you could have missed it because the title was not very accurate): http://engineerboards.com/index.php?/topic/22179-Minimum-reverse-angle/page__view__findpost__p__7061917?

Let me know if you still have questions and I will try to help.

 

[EEpowerOK]

Have you seen this post (you could have missed it because the title was not very accurate): http://engineerboards.com/index.php?/topic/22179-Minimum-reverse-angle/page__view__findpost__p__7061917?

Let me know if you still have questions and I will try to help.

Your link did not work

 

[iahim]

Sorry about that. I just copied the post below.

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.

 

[Spark-E]

The proper link:

http://engineerboards.com/index.php?showtopic=22179&hl=minimum-reverse-angle

So I still have an issue... why would the full 120VACrms be the maximum and not half seen on the anode and capacitor? Since it is an AC wave I'd expect the positive peak to be 60 and the negative to be -60. So on the output the maximum (with charged cap) will be +60. And the greatest possible difference would be when the -60 occurs (negative peak) making the difference 120 VAC maximum.

Using 120/0 is the same, Say the cap charges to 120 VAC. If the anode is 120 (and the cathode is 120 via the cap) the net difference would be 0. If the anode is 0 (and the cathode is 120 via the cap) the net difference would be 120. Seeing this, I would think 120 would be the rating?

 

[iahim]

I'm not sure I understand where the 60 is coming from. The diode conducts for half a cycle, and blocks the other half, so the PIV is calculated during the negative half, while the diode is blocking. During the negative cycle the minimum voltage is -170V.

 

[Spark-E]

I figured it out. I was treating the RMS as a peak-to-peak voltage and not referenced from 0 volts. Oops. Thanks for the help.