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

1. Feb 13, 2013

1. The problem statement, all variables and given/known data

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. Relevant 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.

2. Feb 13, 2013

### Staff: 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...

3. Feb 13, 2013

4. Feb 14, 2013

### Staff: 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:

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

5. Feb 14, 2013

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.

6. Feb 14, 2013

### Staff: Mentor

Yes, it has to be the secondary peak voltage since the formula subtracts the diode drop which occurs in the secondary circuit.
Good luck with that

7. Feb 14, 2013