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# How do I calculate the voltage drop of a full wave bridge rectifier?

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This should be fairly easy to do, I just don't know how to do so, I was hoping someone here could help me out.

I built a full wave bridge rectifier using 4 diodes. I put 116V AC for my input, and I got 103.8V DC out. I'm trying to figure out what my output is when my input is, for example, 125V AC or 110V AC...

I could just break out my variable transformer to some 220V and take a few measurements but, as it turns out, I don't have one. lol :'(

The alternative, of course, is math! If I knew what the output was when the input is at 100V and 120V, I could probably figure out how to calculate the output at 110V. Unfortunately, or perhaps fortunately, my power company doesn't usually fluctuate that much. Well, that doesn't work so well.  ::)

In case you where curious:
I have here a transformer with three outputs, 8V, 16V, and 24V. The input was fairly consistent 115.4V AC.
This is the output:

```At the Transformer	At the Rectifier	Voltage Drop
25.7V			22.7V			3.0V
17.5V			15.4V			2.1V
08.3V			07.1V			1.2V
```

My data suggests the their is a correlation between the input voltage and the voltage drop at the rectifier. If you're good at math, you should be able to determine the correlation and create a formula. I'm not one of those people, so carrying on...

The voltage drop on a diode, as I recall, is .7V but, that doesn't explain the 12.2V drop that I am seeing. However, on the one side I am reading full AC, where as on the other chopped DC. I am assuming it is this choppiness (Ha, choppiness is a real word!) that is giving me the voltage loss. However, I currently have no way of testing this theory.

If you have a formula for calculating this, please share!

Does anyone have any other ideas?

Thanks!

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When you rectify an AC voltage with a rectifier then you probably want DC. But the output of a rectifier without a filter capacitor is pulsing DC but an ordinary DC voltmeter reads the average voltage.

A pretty high value filter capacitor will produce a smooth DC voltage that is 1.414 times higher than the transformer's AC voltage minus 2V if the rerctifier is a full-wave bridge and has its max allowed load current. The DC voltage will be higher if there is no load.

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Thanks, that's easy enough!

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