I know the basics of how to find these values from a graph, but not sure how to prove that average voltage of half-wave rectifier is peak/pi and rms is peak/2. Same for full-wave rectifier where average is 2peak/pi. I understand that the full-wave rectifier equation for rms, peak/root 2, is a standard equation, would you derive the other equations from this one?
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If you accept that the RMS value of full wave rectified sine is \$ \frac {V{pk}} {\sqrt2} \$ (exactly the same as the unadulterated AC signal) then you should see that the heating effect (power) for any given resistor as a load will be halved on half wave rectified voltage.
So, let's assume a 1 ohm load resistor (for mathematical convenience) and square the full wave RMS voltage to get power: -
Power = \$ \frac {V^2} 2 \$ and half this power (half wave rectifier) is \$ \frac {V^2} 4 \$.
So convert back to RMS voltage by taking the square root and you get the RMS value for half wave rectified voltage is \$ \frac V 2 \$.
For average value of a FWR sine wave with peak \$V_P\$: -
$$ V_{AVE} = \frac {1} {\pi} \int_{0}^{\pi} V_P \ sin\theta \ d\theta $$
$$ V_{AVE} = \frac {V_P} {\pi} ( -cos \theta )_0^{^\pi} $$
$$ V_{AVE} = \frac {2V_P} {\pi} = \frac 2 \pi V_P = 0.637 V_P $$
It's all about finding the area under one half cycle so, the integral finds the total area under the sine wave and the division by \$\pi\$ divides area by length to get average height. Pi is used because it's convenient to work in radians for \$ \theta \$.
Because we are talking averages, for a HWR sine wave, it's half the average value of the FWR sine wave.
"Average" formulas taken from here.
Andy aka
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\$V_{DC}=\frac{V_{peak}}{\pi}=\frac{1}{\pi}\int_{o}^{\pi}V_{peak} \ sin(t)dt\$
Tony Stewart EE75
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2Link-only answers become useless if the link ever dies. Better to summarize the content here, and then provide the link for further details. – Dave Tweed Sep 10 '17 at 16:40