REWIRING A MOT (Microwave Oven Transformer)

Thank you for the great guide!

I'm planning to build a 13.8V HAM PSU as well, but I have few things to consider. If I use 0,8mm copper strips for secondary, I'l be probable be limited to 17 rounds. If I get for example 14,7V AC out, would this be enough for 13.8V after bridge rectifier & smoothing capacitors (14.7-1.2)*1,41= abt. 19V. Will the bridge rectifier drop voltage more on high currents? Can I use even less turns on secondary in order to keep temps down? Will this affect the capacitor specs (referring your schematics)? Will leaving shunts in lower temps, because 20mm*0,8mm secondary is capable of handling abt. 100A, so I don't necessarily need to remove them.

Keep up the good work and thanks in advance,
Pena OH6PP
 
Proximus said:
Thank you for the great guide!

I'm planning to build a 13.8V HAM PSU as well, but I have few things to consider. If I use 0,8mm copper strips for secondary, I'l be probable be limited to 17 rounds. If I get for example 14,7V AC out, would this be enough for 13.8V after bridge rectifier & smoothing capacitors (14.7-1.2)*1,41= abt. 19V. Will the bridge rectifier drop voltage more on high currents? Can I use even less turns on secondary in order to keep temps down? Will this affect the capacitor specs (referring your schematics)? Will leaving shunts in lower temps, because 20mm*0,8mm secondary is capable of handling abt. 100A, so I don't necessarily need to remove them.

Keep up the good work and thanks in advance,
Pena OH6PP
Click to expand...
Dear Pena,
As far as I know when you rectify the AC to DC the voltage you get is (DC voltage-diode voltage drop)*1,41. In full bridge rectifier the current flow is through two diodes in series for both polarities. Thus, two diode drops of the source voltage are lost (0,7*2=1,4 V for Si regardless of the current) in the diodes.
In your case is (14,7-1,4)*1,41=18,75. I think you are good to go!
If you use less turns you will reduce heat but you might have lower volts/instability in power fluctuations
Petros
 
Sir;
I use several microwave transformers as auxillary power supplies for my model trains. Voltages range from 8-24 volts. I don't know how much current one can draw, I've never overloaded one. I used a "fixed" load and a Variac to load the primary to 7 amps. Which is as high as my Variac will handle. None has ever heated up beyond ambient.(room temp ~80 deg) For low voltage, nominal 12 V-AC, 7 amps on the primary is 70 amps on the secondary. I don't work with those current levels so it is a non-issue. I do produce a "welding current" at a low voltage, ~200 Amps at 2.5 volts. Primary current is well under the 7 Amp limit. This tested with a clamp on ammeter into a bolted short, not lab instruments.

I use the primary winding intact, removing only the secondary. Using AWG 10 salvaged from my house, insulation class curioslly THHN. I make my own secondaries. The AWG 10 is rated at 30 Amps inside a cable. I think probably 35-40 amps free air exposure. I use high current, high voltage bridge rectifiers because they are cheap. For a linear supply, 60 Hz, the 'eyeball' capacitor is 10K uF per amp for <2% ripple. You may can use less, it's a matter of noise in the load.

The rectifier drops <1.5 volts, about ~1.4. At the line load, 1.5 is close enough. This drop is the "Fermi" level for silicon diodes and is constant until the diode is overloaded. That's another reason to use 40 Amp 600 Volt bridges.

It isn't absolutely necessary, but I highly recommend using a dowel that has been split for each side. This to avoid a sharp bend as the wire comes around the core, giving it a radiused bend. I use "fiche paper" (pronounced fish paper) as an insulator. That is from my electrical background, it need only be a good insulator. Even shoe box cardboard will work, as long as the transformer doesn't run hot.

Adjusting final AC voltage can be varied by adding or subtracting a "half turn", where the wires come out the same side or opposite sides. Closer adjustment is a matter of rewinding the primary which I don't bother with. Final DC voltage can be lowered 0.7 volts by adding a diode in series with the load. 0.7 volts per diode. . . More than 3 or 4 diodes gets cumbersome though.

Most microwave transformers use aluminium wire. The secondary is about useless, just cut it free. The primary is run intact, watch for heat, backing off the load if necessary. I don't like aluminium wire, obviously. It is much less conductive than copper.

.
 
Thanks a lot for replies!

petcnc said:
In your case is (14,7-1,4)*1,41=18,75. I think you are good to go!
Click to expand...
Great, so I'll stick at 17 turns, if I get the same voltage/turn.

Bi11Hudson said:
For a linear supply, 60 Hz, the 'eyeball' capacitor is 10K uF per amp for <2% ripple.
Click to expand...

In Finland mains is 230 V ± 10 %, 50Hz, so do I need more capacitance for smoothing? I'd like to have low ripple, so would it be a good idea to invest on this capacitor? I'm not sure what ripple current mean in the specifications, but I think that's probably what I am looking for...

Cheers,
Pena OH6PP
 
Great project, pictures and analysis!

Minor point: Is it not 14.7*1.41-1.4~19.4Vdc? I don't think we multiply the diode drop by the root 2 factor. 14.7 Vac (rms measurement). 1.414=root(2) to convert to Vdc (0 to peak voltage). ~1.4=two diode drops for a full wave rectifier (first order model of a high current Si diode).

Of course there are also losses due to the copper coil resistance which will lower the output voltage. If the magnetic core approaches saturation then the Vac is actually not a sine wave so rms is inaccurate as it becomes distorted as the magnetization response is not linear. Of course the magnetic state is a function of the current being pulled so these effects will be load dependent. The output will be lower when 50 amps are pulled rather than what it will be when 5 amps is all that is being used. So even if we use a true RMS meter we would still need to measure the transformer output voltage while the full 50 amp load is connected.

However, this 50Amp PS is going to have a lot of 120Hz ripple on the output even if it is going on to a large filtering capacitor. I do not know a lot about HAM units, but is this ripple not a problem or is so far down in frequency compared to the modulation frequency that it never shows up? Of course if one is just charging a lead acid battery who cares. So good, low ripple power supplies start out higher voltage and then are regulated down with active pass transistors. Or more modern units use switching supplies like you see in computers.... no large transformers at all. Low loss smaller transformers or flyback circuits that used magnetic materials (ferrites or special amorphous metal alloys that have something called uniaxial anisotropy to minimize hysteresis losses) that have minimized the Eddy currents so that they can be run at 10s of KHz rather than 60Hz. But these are still regulated to give a constant voltage. Some PC power supplies put out a lot of 12Vdc current and many Amps at 5 volts.
 
Proximus said:
In Finland mains is 230 V ± 10 %, 50Hz, so do I need more capacitance for smoothing? I'd like to have low ripple, so would it be a good idea to invest on this capacitor? I'm not sure what ripple current mean in the specifications, but I think that's probably what I am looking for...

Cheers,
Pena OH6PP
Click to expand...
Well, In simple words at startup the circuit will see a LARGE capacitive tank EMPTY and it will have to fill it up instantly (it is like a short circuit for the startup). I think the in-rush current will be too high! So you need some sort of in-rush current limiter to handle it.
Other than that although the capacity is high it is doable
 
Ripple voltage is just the AC change in the DC voltage due to current being removed from the PS storage capacitor and then being replaced.

Current is the flow of charge. I=change in charge/change in time.

When the circuit rectifies the AC to make DC voltage you put the current (charge) on a large storage capacitor. The diodes (rectifiers) carry current in only one direction to the capacitor, and so as noted, the current flows until the capacitor voltage reaches the peak AC voltage less the diode drops. If there is initially no voltage on the capacitor the current "surges" to a very high value to try to charge the capacitors up rapidly. After this initial surge there is some charge on the capacitor and so currents after this are somewhat lest. The charge does not flow back into the AC circuit during the negative half of the AC cycle because the diodes only conduct in one direction. If there is no load (current/charge leaving the capacitor to be used) the DC voltage on the capacitors reaches the peak and is then the DC voltage is constant at this peak from then on. There is no ripple voltage, as the DC is constant and there is no current/charge being drawn from the capacitor. However, as soon as there is a load the charge (current) on the capacitors discharges into the load continuously. During the period of time while the AC voltage is in the positive half cycle the current through the diode tries to charge up the capacitor and also provides current to the load. Hence, the voltage on the capacitor increases and simply follows the AC voltage wave to the peak. However, during the negative half cycle of the AC voltage the current to the load has to come from the capacitor as the diode cannot conduct or provide any current. Hence, the voltage on the capacitor goes down. So the DC voltage on the capacitor is not really DC. It has both a DC level plus it has a smaller voltage on top of the DC voltage which is oscillating. Think of it as the waves on top of a body of water. Sure the water has depth (DC voltage) but there is also a ripple (AC voltage) on top. As the load increases (more current is used) then the waves get higher as the depth of the DC goes down.

Ripple voltage is usually specified as a percentage. Ripple/DC *100%. If the current that is being pulled under full load conditions is large then the ripple can become quite large. If you had this situation in a stereo amplifier you would hear it as a 120Hz buzz. If bad enough in some circuits they simply will not work or work improperly. So decent power supplies are set up so that under the largest (current flow) load conditions, there is worst case DC (water depth) that is going to be reached. The supplies use active transistor circuits (or tube of old) to just allow the DC part to come out and not much of the ripple. Hence, ripple is not as much of a problem.

So think of it this way, in a good PS the peak DC plus ripple on the capacitor under maximum load might be 20 volts, but the ripple is 5 volts of this and the DC is only 15 volts. (The ripple voltage would be 33%.) The active circuits then output (pass) a constant 15 volts (DC) or less and the output has little to no ripple on it. Of course the active circuits take up some head room voltage and so the output is always less than the 15 volts. Now one must power the active circuits and this is commonly done from the DC plus ripple voltage coming directly off the storage capacitor, so some ripple feeds into the active circuits. Hence the output can carry a bit of this ripple and so is not completely pure DC. So now the spec of ripple percentage is much small, but not zero. And of course the improved ripple is still a function of the load. It is not uncommon in a very good PS that the ripple can be specified at <0.05% at full load current and even less when not at full load!

There is a lot more to this technology, but hopefully what I have discussed is not too obscure.

Dave.
 
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