# Left-over AC voltage in a half-wave rectifier

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1. Homework Statement

My professor gave us a formula for the left-over output AC voltage in a half-wave rectifier. This formula was given without any calculus or explanation as to how it was obtained.

2. Homework Equations

$$V_{out(AC)}\simeq (0.385)(V_m-V_{th})$$

3. The Attempt at a Solution

I don't know if Vm is the peak input voltage or the peak secondary voltage (see image below). I do know that Vth is the threshold voltage of a diode; 0.7V for Silicon and 0.3V for Germanium. Does anybody know how this formula was obtained? I don't have any information in my textbook on it and I've read the chapters twice.

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#### gneill

Mentor
Vm is most likely the secondary voltage that's being rectified. If I had to guess, I'd suggest that the formula would come from some tinkering with the Fourier series of the rectified signal. One can pull out the DC component and the various AC harmonics...

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#### gneill

Mentor
FYI, curiosity overcame me this morning so I took the Fourier series of an ideal half-rectified sinewave and extracted the total RMS AC component of the signal. I found an exact expression for the constant:

$\frac{\sqrt{\pi^2 - 4}}{2 \pi} ≈ 0.385589...$

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So if a transformer were to be involved, this RMS AC voltage would be based on the *secondary* peak voltage, not the primary peak voltage, right?

My professor didn't say whether the AC was peak or RMS but you have clarified that. Thank you. I'm going to have a discussion about the lack of specificity in this class.

#### gneill

Mentor
So if a transformer were to be involved, this RMS AC voltage would be based on the *secondary* peak voltage, not the primary peak voltage, right?
Yes, it has to be the secondary peak voltage since the formula subtracts the diode drop which occurs in the secondary circuit.
My professor didn't say whether the AC was peak or RMS but you have clarified that. Thank you. I'm going to have a discussion about the lack of specificity in this class.
Good luck with that   