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Old 12th December 2010, 06:19 PM   #1
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Default Having trouble with PS math

I am trying to understand the math behind a typical amp power supply.

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.
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Old 12th December 2010, 07:19 PM   #2
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Think of it this way - the RMS of the AC signal is 70.7% of its peak. A simple rectifier and capacitor will build voltage to 100% of the wave (it's peak value), and then become reverse biased...ergo it will be the 100% value ('peak') that you see on the cathode-side of the DC network.
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Old 12th December 2010, 07:26 PM   #3
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That seems counter intuitive, but ok

Would that require a sufficiently large smoothing cap to be true?
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Old 12th December 2010, 07:31 PM   #4
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If you wished to see a very stable DC level after teh diode, yes. If not, you'd simply see a ripply wave, peaks being at the AC peak voltage minus the diode drop :

V_DC = (VacRMS * 1.414) - Diode Drop.

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

VDC = (100*1.414) - .7

VDC = 140.7

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Old 12th December 2010, 07:37 PM   #5
Elvee is offline Elvee  Belgium
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Remember that rms value has no relation with anything in the waveform. It is purely a mathematical process yielding convenient results. The equation of a sinusoidal voltage is something like Vpeak*sin wt. Vrms appears nowhere, and it is normal that you get Vpeak as an output when the waveform is passed through a peak rectifier.
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Old 12th December 2010, 07:39 PM   #6
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Quote:
Originally Posted by Elvee View Post
Remember that rms value has no relation with anything in the waveform. It is purely a mathematical process yielding convenient results. The equation of a sinusoidal voltage is something like Vpeak*sin wt. Vrms appears nowhere, and it is normal that you get Vpeak as an output when the waveform is passed through a peak rectifier.
Let the OP understand rectification and then go from there, average and peak power calculations come later.
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Old 12th December 2010, 08:06 PM   #7
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Thought I did understand rectification. And having a minor in math, I thought I did understand RMS. It's just an average value of the waveform.

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
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Old 12th December 2010, 08:16 PM   #8
Elvee is offline Elvee  Belgium
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Average and rms are two different things, but I'll let Psychobiker sort things out.
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Old 12th December 2010, 08:18 PM   #9
tomchr is offline tomchr  United States
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If you have an infinitely large reservoir cap, the voltage across the cap will be the peak voltage of the incoming sine wave (less a diode drop -- or two diode drops in case of a full-wave rectifier). However, this requires the conduction angle (the amount of the cycle the rectifier is conducting) to be zero, which is not the case in practice. Hence, in practice you get about 1.2~1.3 times the RMS voltage (less diode drop(s)) across the reservoir cap rather than the theoretical sqrt(2) times.

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
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Old 12th December 2010, 08:20 PM   #10
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Quote:
Originally Posted by Elvee View Post
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
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