Rewinding a microwave oven transformer (MOT) for general purpose use

Introduction

In this article I am going to present my struggling about rewinding a microwave oven transformer (MOT) in order to make a low voltage (20-30V) but high current (15-30A) transformer. I have done these experiments in order to make a cheap but a very handy transformer for a bench supply. These types of transformers 500-100 VA transformers cost hundreds of USD where I live, but these microwave oven transformers can be purchased for a few dollars used, because they can be only used for special applications (like where it comes from, microwaves). They usually get shorted or interrupted on the secondary winding due to the high voltage, poor insulation, thin aluminium cross section wire. Many times the primary winding remains intact.

Before starting with the practical work, we have to settle a few theoretical consideration, and proving them in small scale.

Inductive circuit

In a pure inductive circuit the current lags by 90º or Π/2 in radians. In figure 1 is presented the schematic a practical inductive circuit (fig. 2) for measuring the phase shift. As you can see on both of the figures there is a 1 ohm resistor, which is used for shunt, to measure the current through the inductor.

#1 – inductive circuit
#2 – practical inductive circuit with current measure shunt

Current equals to I = U/R, where I is the current, U is the voltage, R is the resistance, in our case 1 ohm.

#3 – current lags behing voltage

On the scope capture (fig. 3) the P1 marked with yellow is the voltage, and P2 marked with blue is the current, and the probes are connected as on fig. 1.

Capacitive circuit

Now lets see a what we can learn about a capacitive circuit. In this case the voltage is behind by 90º in an ideal situation. The R is the same like before used for shunt, and on fig. 6 you can see the relation between current and voltage.

#4 – capacitive circuit schematic
#5 – practical inductive circuit with current measure shunt

We don`t have to make the exact calculations for these experiments. This study was about proving experimental these theorems.

#6 – voltage lags behing current

LC tank circuit resonance

Now let`s see what happens when we connect a capacitor with an inductor in parallel like on fig. 7. What we should know about LC tank circuit? An ideal LC tank circuit at resonant frequency consumes no current. As the inductor causes phase shift in current by 90º, the capacitor makes the exactly opposite, the total impedance (impedance is the term used for resistance, but in alternative current) increases to infinite, so no power is consumed from the source.

#7 – LC tank circuit

In the real world, this is not true, because the capacitor has leakage and the inductor has resistance, so a part of the power consumed is just simply lost in them, in Joule – Lenz effect (in heat).

#8 – schematic diagram of the LC circuit used by me on fig. 9

In order to prove resonance, we have to keep in mind a few formulas:

#9 – reactance of inductor and capacitor

Where:

  • XL – reactance of inductor
  • XC – reactance of capacitor
  • jω – pulse

Basically every LC circuit has got it`s resonant frequency. If we want to create a tuned LC tank we have to keep in mind that the reactance of the inductor and the reactance of the capacitor has to be equal.

For simplifying the calculations we consider that we are going to work with a sinusoidal signal.

#10 – sinusoidal waveform

Doing a little bit of more math, we can calculate the resonant frequency of the circuit.

#11 – step by step determination of the resonant frequency

I have created my own LC circuit (can be seen on fig. 12), tuned to aproximately 4414 Hz, where the value of  the capacitor is 10uF and the inductor 0.13mH. Obviously a signal generator was used to test the circuit.

#12 – practical LC circuit

On the following screen capture you can see the current flowing through the circuit, the blue line, which is almost equals to zero.

#13 – current through an LC circuit at resonant frequency

I know this introduction for a microwave oven transformer rewinding is quite long, but – if you continue to read the article you will find out why I have did it.

Rewinding the transformer

In order to rewind it, we have to cut of the secondary coils. There is one for high voltage (2000V) and one for the heater (aprox 6V). But how can you make difference between the secondary and primary winding? Simple, the secondary is made out of thinner wire that the primary and the winding for the heater has got a thick (in most of the cases red colored) insulation and only a few turns. And also you can`t see directly the secondary winding, it is covered in a thick insulating paper. You can cut of the secondary simply with a hacksaw, but pay attention not to damage the primary.

#14 – a re-winded MOT

On fig. 14 you can see my rewinded MOT, I used a 2mm section wire for the secondary and as many turns as it was possible to wrap around.

After powering up from the mains ( 230V ), (fig. 15. A – without capacitor) with no consumer at the secondary,  the current was 3.5 A, so the power dissipation was around 600-800 W, and of course the transformer was warming up slowly.

#15 – power factor correction

I have attached a resistive consumer to the secondary (a metal bowl with salty water), and I got a secondary current between 10 and 30A at significant 22-25V. Unfortunately the primary current does not change. It remained the same.

My first thought was to insert a capacitor in series with the transformer, because basically a transformer is made out of two inductors on the same core. I only had to measure the inductance of the coil and then buy a suitable capacitor to reduce the free running current consumption. The inductance was 723mH, so at 50Hz If we want to achieve resonance we have to connect a 14uF capacitor In parallel with the transformers primary winding (fig. 15 – B). I have not found a 14uF only 15uF, but thats close enough.

After testing it, I found out that the primary current was 3.6A, almost the same without the capacitor so I have made no improvements. I have tested it with different consumers but the results were exactly the same like the previous one.

Because I had no any other ideas I tried connecting the capacitor in series with the transformer, in order to double the impedance of the circuit. This time I have not obtained a better, but different results, on fig. 16 you can see them. As I increased the secondary current the primary current dropped, in a non-liniar way.

#16 – in / out current ratio

The good thing was that I was able to reduce the primary current to around 1A, but the secondary voltage immediately followed it. At 9A secondary current, secondary voltage dropped to 14-15V.

Even in the last case, the transformer dissipates around 100W which is a quite large amount of power.

Conclusion

Conclusion
So what next? Yes, the whole project was a failure but I have learned a few practical things what I have not found in this form exactly on the web. In order to use this transformer as it is (with capacitor in parallel) I would submerse it in a metal case and fill it up with transformer oil in order to cool it down. An other option would be, with the capacitor in series to make a 12-13.5V linear power supply for HAM radio. In my opinion the best choice would be to use the classic formulas to calculate the necessary winding and then rewinding both the primary and secondary coil according to that. All in all, in my opinion this project was useful because it taught me what I can and what I can`t do with a microwave oven transformer.