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(First posting)

I am building a simple PSU with a torroidal transformer and bridge rectifier. I'd like to know the output DC voltage from the rectifier while under load. e.g. charging a battery. I see that the formula is in here but these things do my head in (sometimes) so if I quote the numbers some kind souuratingl may help.

The transformer is labelled as being 22 VAC at 16.9 amps but at a less than constant duty cycle I fear. It is centre tapped and grounded. At least there's a green etc. ground wire along with the red and yellow secondaries. However, my DVM is registering 28 VAC and the DVM if not Fluke, isn't rubbish no worse than 2-5% I'd say.

Connected to a bridge rectifier rated at 1000 V and current rating > 20 A, what DC voltage should I expect to see with/without a resistive load?

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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.

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  • \$\begingroup\$ Thanks a lot. Will work on that info and report progress. Regards TT \$\endgroup\$ – Experimenter Sep 26 '16 at 11:49
  • \$\begingroup\$ It is unclear what @Experimenter is doing with that centre-tap. A bridge rectifier suggests the centre-tap could be left un-used. Or two diodes of the bridge could be left un-used. You get far different DC voltage out for those two cases. \$\endgroup\$ – glen_geek Sep 26 '16 at 23:08
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If you add 40% for the conversion from RMS to sine peak or x root(2)

Then add x% for conduction losses from full load to no load.

You get 30V and you measured 29V so your conduction losses are less than typical.

This because instead of average rectified Vac with x% ripple , you are seeing peak sine out with zero ripple at no load.

We call this an "unregulated" Dc out for a good reason.

Even with a 10M input impedance DVM there will be enough capacitance to store the most of the peak voltage. A simple 5% dummy load will reduce the Vdc swing greatly

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  • \$\begingroup\$ Thanks a lot. Will work on that info and report progress. Regards TT \$\endgroup\$ – Experimenter Sep 26 '16 at 11:49
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Yes, glen_geek, you are quite prescient in raising this. Fact is the centre tap earth is purely accidental to the thing as it was there when I got the transformer and I was unsure whether or not to ground it to the metal chassis where also is grounded the mains input.(It's like the guy who climbed the mountain -why- because it was htere!) I guess I can leave it unconnected as I'm going with the bridge rectifier wired in the usual way. I vaguely remember the significance of it OK with old 'C type ' power supplies in ye-olde valve radios but the present 'project' is v. un-ambitous and only came about because I was impressed with the quality and spec of the chunky torroidal - of which I got a few cheaply. All a bit of entertainment really. Would like to consider some very simplest form of output voltage/current control/feedback circuit like a heavy duty transistor that responds to for instance a rising charging battery voltage. (if you're thinking he 'doesn't know what he's talking about' here then you're most likely right.............. Regards TT

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