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Old 18th June 2004, 09:55 AM   #1
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Question Basic theory question: effective AC voltage

I got an exam fysics today and there's a part about AC. Now, we've proven that the 230V AC is actually 325V p-p AC and that this 230V is the effective value of that 325V (325/sqr(2)). So far, no problem.

However, every time I read something about toroids, it is said that the voltage after full wave rectification is sqr(2)*AC voltage. Does this mean that the voltage written on a toroid is actually the effective AC voltage and that the "real" AC voltage is sqr(2)*effective voltage? I guess so, but I just wanted to know.

I don't know all the rigth terms, so if you don't get my question the first time, reread it please .
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Old 18th June 2004, 10:19 AM   #2
Jennice is offline Jennice  Denmark
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Hi Devil...,

You're on the right track.
The voltage on the transformer secondary (regardless if torroid or other core) is specified as AC RMS (effective) value at rated load.

First, note that the unloaded output voltage (idle, no load) is a little higher, depending on the output impedance of the transformer.

Secondly, what happens is, that the storage capacitor which is usually placed after a (full) bridge rectifier, is charged up to the transformer's peak output voltage -- NOT the RMS value. As you concluded yourself, the peak voltage V_peak = sqr(2) * V_rms.

Both voltages are "real" voltages. You just need to remember the difference between RMS and PEAK.

Furthermore, remember to de-rate the output current capability of such a configuration due to core saturation and higher losses during the peak charging currents. Recall.. The rectifier only conducts in short periods, but at higher-than-rms currents.
Example: A transformer with secondary specification of 18V 36VA (2 Ampere), conected to a load through a full-bridge rectifier, will charge the capacitors to approximately sqr(2)*18V = 25V (ideal components). However, you can not connect a load that is rated 25V, 2A. That would be 50 Watts. (Well, yes you can, but your components will be stressed over their rating).

Hope this explains things.
Jennice
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Old 18th June 2004, 10:24 AM   #3
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The voltage across the smoothing capacitor WITH NO LOAD will approach the peak value of the AC signal, so it will approach 1.414 x V.

Actually it will usually be even higher than this, as the secondary rated voltage of the transformer will assume some load current and the open-circuit AC voltage will be higher than the nominal rating.

Once you start drawing current from the smoothing capacitor, the voltage will start to drop.

I suggest you download the PSU simulator from the Duncan Amplification website. This will let you see exactly what is happening.
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