# Input and Output Voltage of a Full bridge rectifier

I know If we provide a sinusoidal AC Voltage across the Inputs of a full bridge rectifier we get a fluctuating DC Voltage (more or less) with the peaks of source AC Voltage. But what would be the peak voltage after smoothing the Output DC with Capacitors if we hypotactically provide 220VAC RMS?

• Depends whether you're talking about an ideal case, or not (but in both cases it's about the same, peak). And of the "hypotactical" question. :-) (did you mean "hypothetical"?) Apr 11 '21 at 12:34
• It would be the peak voltage of the 220 VAC. Do you know how to calculate that? Apr 11 '21 at 12:35
• @Transistor Unfortunately, No. Apr 11 '21 at 15:45

Figure 1. Relationship between sinusoidal peak, RMS and average (after rectification) values. Image source: Learn About Electronics.

The peak voltage of a sinusoidal voltage waveform is $$\ \sqrt 2 V_{RMS} \$$. Conversely $$\ V_{RMS} = \frac 1 {\sqrt 2} V_{peak} \$$.

The RMS voltage can be considered that DC voltage which will give the same heating affect as the AC being considered. Since the power in a resistive load is given by $$\ P = \frac {V^2} R \$$ the RMS value is not the same as the average value.

Without smoothing capacitors the fullwave rectified DC will have the same RMS value as the AC waveform (ignoring losses in the rectifier).

Figure 2. Smoothing capacitor ripple. Image source: Electronics Notes.

The smoothing capacitor charges up on each peak of the rectified AC and supplies power to the circuit when the voltage falls away. See the linked article for how to size the smoothing capacitor.

• I disagree. The average value of an AC is always equal to 0. Apr 11 '21 at 16:13
• @Miss, fixed, thank you. Apr 11 '21 at 16:22
• @Transistor So the peak Voltage after smoothing is also 230V or root 2 x Vrms ? I'm confused!. Enough with the hypothetical, what would voltage reading I'm supposed to get in a normal multimeter if I were to pull my probes across the outputs? would it be an average value or just peaks? Apr 13 '21 at 5:32
• Figure 2 should make it clear enough that the capacitors will hold the peak voltage. With no load on the power supply you should see $230\sqrt 2 \ \textrm V$. Your capacitor voltage rating must be greater than this. Apr 13 '21 at 7:05