I'm thinking of getting an auto transformer and rectify it to make mains isolated 240V 8A DC power supply. I would like to know if its possible to rewind an old microwave transformer to 1:1 isolating transformer with maximum output of 2kW. I know that the waveform wouldn't be smooth so i'll be using beefy 450V 2000uF capacitor for rectification.
Electronic – How to rewind microwave transformer to isolation transformer
microwavetransformer
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Best Answer
The following is largely complementary to Dave Tweed's answer
The welds are generally along the surface and can be removed with an angle grinder "fairly easily". The laminations can be rebuilt using a clamp to hold the core together. Epoxy helps prevent lamination chatter but DO NOT introduce any additional air gap (or "epoxy gap") between laminations.
The magnetic shunt must be removed if the transformer is to have anything like normal regulation. Otherwise the transformer has an intentionally designd "droop" in its load characteristic.
Microwave ovens tend to run the iron rather hard - well up the saturation curve.
Adding a relatively small number of extra turns greatly reduces the magnetising current.
It may not be true in all cases but in at least some transformers the mains primary on its bobbin comes off easily as a complete unit. If the secondary is then removed - destructively or otherwise the available space left from the secondary & shunt would allow two identical primaries to be added.
To get 2 kW you'll need two typically sized transformers - or primaries from 4 identical ones installed on two cores. These can then be connected with primaries in parallel and secondaries in parallel.
If you add extra turns to the primaries and not the secondaries you'll get a degree of stepdown which will reduce the DC level that youd otherwise get. A sinewave will peak rectify to 1.414 the RMS ac value or in this case 240 x 1.414 =~ 340 V. Under load this will be somewhat lower and mean DC depends on the degree of filtering.
With 2000 uF filtering at 8A you'd probably get 10-20V of ripple. V drop across a hald cycle at 8A =
Vdrop = t x i / c = 0.01s x 8A / 0.002F
= 40V.
That assumes instantaneous peak charging of the capacitor at the start of each half cycle - which would not be the case. But it gives you a feel for the order of ripple voltage involved.