If the secondary voltage is 30 volts RMS, I am led to believe that DC voltage of the rails is around 42 volts based on a book I am reading.

I don't understand why the DC voltage would be equivalent to the peak AC voltage.

- Home
- Amplifiers
- Power Supplies
- Having trouble with PS math

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter MichaelJHuman
- Start date

- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.

If the secondary voltage is 30 volts RMS, I am led to believe that DC voltage of the rails is around 42 volts based on a book I am reading.

I don't understand why the DC voltage would be equivalent to the peak AC voltage.

V_DC = (VacRMS * 1.414) - Diode Drop.

100V RMS AC signal, .7V diode drop...

VDC = (100*1.414) - .7

VDC = 140.7

http://macao.communications.museum/images/exhibits/small/2_16_0_12_eng.png

Let the OP understand rectification and then go from there, average and peak power calculations come later.

I know what the answer should be if there's no capacitor, or at least I think I do.

If the sine waveform was equivalent to water flow through a pipe I would know the average flow. Admittedly that's probably a bad analogy, because voltage is probably more analagous to water pressure.

Anyway, I appreciate the answers, thanks

RMS is more than a mathematical convenience. The RMS voltage is proportional to the amount of energy dissipated in a resistive load. Note that for sine waves the average and the RMS values are the same. This is not the case for all waveforms. A square wave, for example, has Vavg = Vpeak * D and Vrms = Vpeak * sqrt(D), where D is the duty cycle.

~Tom

Average and rms are two different things, but I'll let Psychobiker sort things out.

Calculus agrees with you. The average of the function ABS(sin x) is not identical to RMS. It's 2/PI, or about .64. The average of sin x is 0 of course

I think you are confusing average voltage with average power.

I think you are confusing average voltage with average power.

You can calculate average value of abs(sin x) using calculus. Integrate from 0 to PI and then divide the result by PI, as memory serves.

2 / PI I think or .637. Isn't that what I said above?

The integral is -

(-cos PI - (- cos 0 ) / PI = 2 / PI, as memory serves...been awhile.

No, I was not confusing average voltage with power, least I don't think so...of course there's also a simple equation using integrals for RMS power, which is the integral of RMS voltage times RMS current divided by delta time I believe.

Last edited:

Note that for sine waves the average and the RMS values are the same. This is not the case for all waveforms. A square wave, for example, has Vavg = Vpeak * D and Vrms = Vpeak * sqrt(D), where D is the duty cycle.

~Tom

No they aren't (I assume you're speaking about the absolute value of the waveforms).

In this case, the rms is Vpeak/sqrt2, and the average Vpeak/0.5*Pi.

The difference is not huge, but significant.

I understand RMS.

What I did not understand, and someone explained above somewhere, was how the reservoir cap allows the DC voltage to be as high as it is. In my brain, the capacitor would have caused the DC signal (ignoring ripple,) to be some sort of average voltage value from the rectifier.

I understand, sort of, what he was saying.

Thanks for all the answers

The 2/pi figure appears in choke input supply theory, but you were asking about capacitor input supplies.

The 2/pi figure appears in choke input supply theory, but you were asking about capacitor input supplies.

I follow you. I was speaking imprecisely in post 7. RMS voltage != average voltage.

What always gets me is that average power is calculated from RMS voltage and RMS current not average voltage and average current. But the math works out, I believe.

- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.

- Home
- Amplifiers
- Power Supplies
- Having trouble with PS math