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/images/exhibits/small/2_16_0_12_eng.png
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
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.
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
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
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
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
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
If you can do calculus than you know enough maths to understand RMS for sine waves. Remember that power goes as V^2/R, so you can calculate the average power produced from Vsin(x) = (V^2/R) integral(sin^2(x)) - the answer is (V^2/2R) (by inspection or trig). What DC voltage would give you this power? Answer: V/sqrt(2).
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.
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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.
The average of abs(sin(x)) is 2/pi, as you said. I wasn't disputing that, but it has absolutely nothing whatsoever to do with the RMS value of a sine wave! You were asking about RMS vs. peak for a sine wave so I explained where the factor of sqrt(2) comes from. It may be that your confusion is not about maths but physics.
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
Your post #7 suggested that you didn't understand RMS, as you spoke about waveform average. RMS is about waveform^2 average. Then you seemed to confirm this in post #10. Sorry if I misunderstood.
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.
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