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12th December 2010, 06:19 PM  #1 
diyAudio Member
Join Date: Aug 2005

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. 
12th December 2010, 07:19 PM  #2 
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Join Date: Jul 2005
Location: Montréal QC

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 cathodeside of the DC network.
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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? 
12th December 2010, 07:31 PM  #4 
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Location: Montréal QC

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 http://macao.communications.museum/i...6_0_12_eng.png
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12th December 2010, 07:37 PM  #5 
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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.

12th December 2010, 07:39 PM  #6  
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Quote:
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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 
12th December 2010, 08:16 PM  #8 
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Average and rms are two different things, but I'll let Psychobiker sort things out.

12th December 2010, 08:18 PM  #9 
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Join Date: Feb 2009
Location: Greater Seattle Area

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 fullwave 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 
12th December 2010, 08:20 PM  #10 
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