NCEES PE Power Prep Exam Problem 515

Could anyone explain why Vpeak is calculated as sqrt(2)*Vrms on this problem? I have understood that for a half-wave rectifier Vpeak=2*Vrms, and for full wave rectifier is that Vpeak=sqrt(2)*Vrms

I appreciate your help in advance.  

Guest Darian said:

I have understood that for a half-wave rectifier Vpeak=2*Vrms, and for full wave rectifier is that Vpeak=sqrt(2)*Vrms

Click to expand...

This isn't exactly correct.

For a sine wave with RMS value V_rms, the peak value will be V_peak = sqrt(2)*V_rms. This is regardless of half/full wave rectification.

For the problem, we first need to calculate the peak value of the voltage signal. This is sqrt(2)*120 = 169.7 V. This is also the voltage that will be present on the capacitor at full charge. Then when the voltage input signal is at its negative cycle peak (-169.7 V), and the charge on the capacitor is +169.7 V, the voltage difference across the diode is 339.4 V.

Chattaneer PE said:

This isn't exactly correct.

For a sine wave with RMS value V_rms, the peak value will be V_peak = sqrt(2)*V_rms. This is regardless of half/full wave rectification.

For the problem, we first need to calculate the peak value of the voltage signal. This is sqrt(2)*120 = 169.7 V. This is also the voltage that will be present on the capacitor at full charge. Then when the voltage input signal is at its negative cycle peak (-169.7 V), and the charge on the capacitor is +169.7 V, the voltage difference across the diode is 339.4 V.

Click to expand...

But, the half wave rectifier does not follow a full sine wave, that is why calculations shows Vrms=Vpeak/2. What am i missing?

Darian said:

But, the half wave rectifier does not follow a full sine wave, that is why calculations shows Vrms=Vpeak/2. What am i missing?

Click to expand...

Not sure where you're not following. 

I don't see where Vrms = Vpeak/2 in the calculations.

Take a look at this graph. Vsource is the source voltage (120 V rms), Vcap is the voltage across the capacitor when it is initially charged, and after it retains the charge. The yellow plot shows what the voltage would be if the capacitor wasn't holding a charge. Notice on the negative cycle of the source voltage, the voltage across the diode is Vcap - Vsource. Since the peak is +/- 169.7 V, the voltage across the diode is Vcap - Vsource = 169.7 - (-169.7) = 339.4 V.

The problem is not asking for the output voltage of the half wave rectifier.

It's asking for the reverse minimum voltage rating of the diode. The voltage rating of the diode is how much voltage it can be subjected to without failing or becoming damaged.

You have to calculate what the maximum voltage is across the diode during the entire period of the input AC voltage signal (from zero to +Vpeak, back to zero, to -Vpeak, and back to zero).

In order to use a diode in this circuit, the diode will have to be rated for a "minimum" of this value, meaning as long as the voltage rating of the diode is equal to, or greater than, the most voltage it will see during the entire period of the input signal, the diode can be used in the circuit without being damaged.

During the positive AC voltage input cycle, the diode is forward biased, closed, and the capacitor is charged to a voltage equal to the peak voltage of the AC voltage input. 

During the negative AC voltage input cycle, the diode is reverse biased, open circuited, and the capacitor is now discharging a voltage amount equal to the same peak voltage value of the AC input signal that it was previously charged to. 

During this time when the diode is reverse biased and open circuited, the voltage across the open circuit of the diode will be equal to the sum of the peak voltage of the AC input signal in series with the peak voltage discharge of the capacitor:



The peak voltage of the AC input signal is √2 times the RMS value. 

The peak voltage discharge of the capacitor is equal to the same value because it was charged to the peak voltage of the AC input signal during the previous positive input cycle of the AC input when the diode was forward biased and closed.

The AC input shown on the left has the voltage polarity reversed with the negative reference on top and the positive reference on the bottom because the circuit it shown during the second half the AC input period when the voltage is going from zero, to the negative peak value, and back to zero which in turn results in the diode being reverse biased and the capacitor discharging.

This is the most voltage (in absolute value terms) that the diode will see during the course of one period of the AC input signal. After one period, the process just repeats itself.