# Building a power supply, Mysterious voltage jump?

I have been building a 28V linear power supply to power a vintage military radio. I want it to be as simple as possible,o I built it with a 120-30VAC transformer, it then runs through bridge rectifier diode.

The voltage coming out of the diode is 26VDC which is close enough for those old radios. But when I wire in a capacitor to smooth out the signal the voltage jumps up to 40VDC regardless of the value of the capacitor I use. I have tried a 50V 3300 microfarad capacitor and a 28V 6600 microfarad capacitor. I am not using any kind of voltage regulator IC because I didn't think I would need one if I could just get the power supply to output the voltage I want. My main question is why does the voltage jump so high after the capacitor and how can I fix it? I have included a wiring diagram for the circuit I made.

• Because VAC is a sinusoid and the rating has been given in RMS voltage (not peak voltage or peak-to-peak voltage), and RMS is not defined the way you assume it is. May 17, 2020 at 18:35

My main question is why does the voltage jump so high after the capacitor and how can I fix it?

In simple terms, the capacitor acts as a peak voltage hold component thus charging to the top of the peak of the rectified secondary voltage.

Without the capacitor, the average voltage measured from the rectifier (using a standard DC multimeter) would be about 0.637 x 30 volts x $$\\sqrt2\$$ = 27 volts.

The 0.637 factor comes from here: -

Here's the proof: -

With the capacitor present, the DC becomes "smoothed" at the peak value from the bridge rectifier which is around 42 volts (30 volts x $$\\sqrt2\$$). This is where the 0.707 value arises in the top picture i.e. $$\\sqrt2\$$ = 1.4142 = $$\\dfrac{1}{0.7071}\$$.

To get a more accurate result we should remove about 1.2 to 1.4 volts from the numbers above to account for the volt drop caused by the diodes in the bridge rectifier.

To restore the smoothed and rectified voltage to something like 26 volts DC requires a transformer change with a secondary voltage that is more like 18 volts RMS.

In other words, 18 volts RMS x $$\\sqrt2\$$ = 24.5 volts minus a couple of diode drops to result in a final voltage of about 23.2 volts. If you can find a secondary voltage rating of 20 volts then it would be more suitable.

Alternatively use a regulator but be prepared for issues such as excessive warming if the current draw is too much.

• Ok that makes a lot of sense now. I will replace the transformer and see where that gets me. Thank you very much! May 17, 2020 at 18:35