# Reverse engineering a power supply characteristics

I have this schematics and the following information: connect the + of the multimeter to test point A and the - to test point B and read

app. +20 VDC, 0.2 VAC

What is the RMS at the secondary of the transformer? You apparently have 18 V on the output of a full wave bridge. Such a bridge has two diode drops. Figure 700 mV per diode, so 1.4 V drop thru the bridge. That means the peaks of the AC wave were 19.4 V. Assuming a sine, the RMS voltage is a factor of sqrt(2) lower than the peaks, so 13.7 V.

Of course the AC line voltage can vary a bit. The 18 V figure may also not be the nominal. Sometimes you put the worst case on a schematic to remind yourself that the rest of the circuit must be able to tolerate that.

Any circuit using a transformer directly from the line, then followed by a full wave bridge, is going to have some tolerance for the resulting voltage. My guess is that the transformer is probably rated for "12 V" output, or maybe "12.6 V". The latter was a common fillament voltage for tubes.

• Thank you, test point A is marked with 18V but the information I have says "you should read 20VDC and 0.2VAC" so maybe I redo your computations starting with 20 instead of 18. – Alessandro Jacopson Feb 25 '17 at 15:53
• @Ales: Yes, you can use the method I show to work backwards to the transformer secondary RMS voltage, starting from any known DC output of the full wave bridge. – Olin Lathrop Feb 25 '17 at 16:06

It should be:

Vrms=Vdc/1.414

Use this equation assuming your capacitor is able yo keep charge with a very low ripple voltage (ideally 0v).

• So 20/1.414=14VRMS? – Alessandro Jacopson Feb 25 '17 at 15:15
• And what about the "0.2 VAC" in my question? – Alessandro Jacopson Feb 25 '17 at 15:16
• I think it refers to the ripple. – Wesley Lee Feb 25 '17 at 15:17
• If it refers to ripple voltage just add half of it to your DC voltage, then I forgot to add your diodes voltage drops, you should add 1.4v to your DC voltage to compensate the difference, so your equation would be: Vrms=(1.4+Vdc+0.5*Vripple)/1.414 – Marcelo Espinoza Vargas Feb 25 '17 at 15:51

What is the RMS at the secondary of the transformer?

the dc output is the peak of the secondary, plus drop on the diode. That means the peak is 20v+0.2 = 20.2v.

add two diodes forward drop, you get 21.6v.

Vrms is 70% of that figure, or 15.3v.