mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
Can secondary half way rectification over heat an industrial control transformer? And is so what conservative multiplier is typically used for de-rating? 480:120
Can secondary half way rectification over heat an industrial control transformer? And is so what conservative multiplier is typically used for de-rating? 480:120
191016-0815 EDT
mbrooke:
Yes. Don't run much DC in either the primary or secondary. This is why you must use special phase shift dimmers into a transformer.
Unbalanced DC thru any coil on a transformer unbalances the hysteresis curve increasing saturation in one direction, and increasing magnetizing peak current in one direction. Thus, increasing RMS current. A full wave center tapped rectifier balances the two half wave rectified waveform DC components.
Run experiments. You need an RMS meter that also simultaneously measures the DC component along with the AC component.
Do temperature rise experiments.
.
This is what I was thinking.
I know when the diode is on the primary the thing gets HOT (had to unplug it)- tried that experiment long time ago. But I'd imagine the secondary would not be as bad... right?
191016-1021 EDT
mbrooke:
Generally I would judge just about as bad, possibly worse. It will depend on what the primary and secondary currents are, and how all circuit elements balance out.
I think if the ampere-turns of DC are the same, that secondary loading maybe worse. The DC ampere-turns produces the flux unbalance, and if both primary and secondary have current flow, then power is dissipated in two places.
.
On the primary, it will cause saturation. Don't do it.
One problem with half wave is that all the power that needs to be delivered to the load is taken in very brief pulses that only happen once every cycle. So the peak current through the transformer is much higher than the output current of the rectifier/filter.
The better the filtering (assuming you are using a capacitor across the output), the shorter and larger the spikes are. So a 80 ma current may flow for 2 ms, and that allows the filter to supply a 10 mA load for a full 16 ms. In both cases the energy is the same 80*2 = 16*10.
With no filtering at all, the diode is allowed to conduct for a full half cycle, and the output current also only flows for one half cycle. So 10 mA out is only 10 mA in. That is probably your hair dryer.
An inductor at the input of the filter changes everything, and can get rid of all those nasty harmonics and spikes.
191016-1646 EDT
winnie:
No matter where the DC ampere-turns come from on the transformer, the same number of DC ampere-turns produce the same saturation effect. This assumes a single magnetic core path.
.
Quite true, but please keep in mind that because the magnetizing current comes from only one side, a diode on the primary will not create the same DC ampere turns in the core as a diode on the secondary. And, of course, the magnitude of the diode load current will depend on the load resistance, which we have not really considered so far except to tacitly assume that the transformer will be close to fully loaded.
191016-1452 EDT
steve66:
What you say is true, but it is only part of the story.
The DC flux bias is probably the most important factor, and is most easily observed with an unloaded secondary and a series diode in the primary.
All flux sources in a closed core add.
Not 10ma, but literally 15 amps. Hair dryers have a massive diode that will literally rectify everything when set to low.
https://jimogaoshou.en.made-in-chin...Mica-Electric-Hair-Dryer-Heating-Element.html
I mean a 25kv pole pig is obviously fine with it. But what about a 3000 va transformer converter as an example?
I gets what being said here, but some posts downplay others overplay the hysteresis curve when all else is adjusted for. IMHO