Electrical – Rectifier voltage drop under load

1. The sites formula is correct, but only under ideal conditions. The AC input must be a clean sine-wave. Some UPS's and DC-AC inverters put out a choppy sine wave that would make the 1.414 ratio of RMS value to peak value not true.

2. This peak voltage assumes no load, whether a single diode is used or a bridge rectifier, plus capacitor of sufficient value to remove any AC ripple. Even a tiny load of .1% of capacity will drop the voltage by the amount the diodes dropped. So subtract .7 volts or 1.4 volts from the expected peak, and the numbers should match better.

3. With heavier loads a bridge or full-wave rectifier will provide the most current. At high current levels >10 amps the Vdrop across each diode can be 1 volt. The output voltage will drop as the load increases until a full safe load is reached.

4. By now the peaking effect is gone and the DC voltage is more like the AC-RMS value. The RMS-to-Peak formula is correct but only under ideal conditions, and as you might have guessed by now, real-world conditions dominate once you apply a load or use a inverter/UPS for AC power, making the RMS to Peak formula useless, especially under heavy loads.

5. @ElliotAlderson made an important observation;

You are missing the point of the formula if you think it is wrong. The formula tells you the maximum dc voltage you can achieve from a given rms ac voltage. That value is very important when it is time to specify the voltage ratings of your components, such as the filter capacitor. The formula is very useful if you understand what it means.

Note: I am assuming no filter capacitor is connected.

Well, it sort-of makes sense but you won't get a good reading without a resistive load.

The average voltage of a (full wave rectified) sine wave is 0.9 times the RMS voltage, or about 19.8V in your case (minus a volt or so because of two diodes with little loading), so maybe 19V.

However, with no loading the output voltage of a transformer will be higher than the rated (full current) load, typically by 10-20% for a smallish transformer.

There's another 'however'- the input to your voltmeter will actually have a small capacitor across it and it will act to filter the voltage (since the back leakage of the bridge will be small). So, you'll get a higher than expected voltage, but not reliably.

I suggest you try putting a resistor such as 10K across the output and measure again. I would expect a reading in the 20-22V range with a basically negligible load.