# How does Multisim calculate voltages?

If I rectify 10Vac (rms) by diodes without a voltage drop to a full wave pulsating dc, the peak voltage would be Vrms x square root of 2 = 14.142135V The dc (average) voltage would be

2V peak/pi = 9.00316V,

And the ac voltage would be 4.35236V (the rms value of the ripple)

I have adjusted the supply voltage to get 9.003V on the Vdc voltmeter.

To get rid of the voltage drop over the diodes, I am measuring the AC through 2 diodes on each polarity. The Multisim measurements are different from mine, and since they are also calculated, I wonder what formulas they use.

edited: I have made the actual circuit and have measured with 3 different multimeters. They came closer to the calculated values than the multisim.

I have scaled it up to reduce error. For 100Vac (after diodes voltage drop) sine wave, the Vdc (average) was 90.5, and AC measured 42V. (for 10V rms,.. Vdc = 9.05, .. vac = 4.2)

multisim for 10Vrms had: Vcd = 8.83, and for Vac =4.62

• Why do you think getting adding more diodes to the signal path will get rid of diode drops? Feb 15, 2015 at 4:09
• The full wave rectified voltage goes through 2 diodes. To find an equivalent AC voltage, I measure the AC through 2 diodes also. This should give me an ac voltage without the diode drop. Feb 15, 2015 at 4:50
• Forgive me if I'm wrong, but wouldn't the bridge only conduct through one diode on each leg at a time? So it would be more accurate to get rid of D7 and D8? Feb 15, 2015 at 5:45

Taking an educated guess here, I'd start from the origin of this $V_{RMS} = V_{pp} \frac{1}{\sqrt{2}}$