Reuse a 800VA microwave oven transformer for a power electronics project

Written by Guy Fernando

Created Sep 2019 - Last modified Sep 2024

Microwave ovens can kill. Please know what you are doing before taking one apart, and be very careful throughout.

If you have removed a microwave oven transformer (MOT) from an old microwave oven, it is possible to reuse it as a low
voltage power source for an electronics project requiring hefty power. The method described here keeps the mains primary
winding intact, and removes the high voltage (HV) secondary and the filament windings without having to cut the
transformer laminations. A new secondary winding can then be wound around the gaps in the laminations, providing a
stepped down voltage but high current AC source.

The magnetic shunts first require removing to assist the removal of the secondary winding. The shunts are removable
laminations that are fitted between the primary winding and the filament winding. Use a 4mm punch and a hammer to push
out the shunts, one at a time.

The shunts are used in a microwave oven to limit the flux density in the secondary and so limit the maximum secondary
current. Unless current limiting is required for your project, then there is no further use of the shunts.

The HV secondary used to supply the 2kV for the microwave oven magnetron, requires removal. Next using a hack saw
carefully from the top cut through the secondary coil. The secondary coil can be further identified as it uses thinner
wire. Be careful not to nick or damage the primary coil below.

Once the secondary has been cut through, use a 10mm punch and hammer out the copper windings. Alternate between both
sides to draw out the coil from the other side of the transformer. You may decide to keep the paper former depending
on how well it has survived or it can just be removed if no longer required. Using wire cutters, remove the filament
windings (the orange wire shown in the photos).

Using a brush clean away any copper filings and debris, and you should be left with something like the photo shows.
Bear in mind an MOT will get very hot when run continuously at the nominal primary voltage, even with no load due to
core losses. They are after all cheaply manufactured and designed to only operate for short periods as in a microwave
oven. However these core losses can be reduced by running the transformer primary at a lower voltage than the nominal
voltage, or even by using two MOTs with their primaries connected in series.

To quickly test the MOT a new temporary secondary winding was wound using multi-core cable which happened to be laying
around in the junk box. After winding 40 turns and attaching two crimp terminals the mains was connected to
the primary winding.

The RMS voltage on the secondary was measured to be 35.9V, giving
0.9 secondary volts per turn.

When full wave rectified and smoothed with a suitable capacitor, the maximum DC voltage is approximately given by:

Maximum DC voltage,
\( \begin{aligned}
V_{DC} = \sqrt{2}.V_{RMS}
\end{aligned} \)

\( \begin{aligned}
V_{DC} = 1.414 \times 35.9 = 50.8V
\end{aligned} \)

By testing the transformer under open circuit and short circuit conditions it is possible to find the transformer losses,
including iron-core, magnetising, and eddy-current / hysteresis losses.

The test equipment required to perform the test include a voltmeter, ammeter and wattmeter. The voltmeter and ammeter
measures "apparent power" in VA while the wattmeter measures "true power" in watts. The short circuit test also requires
a variac or some other means to reduce the voltage across the primary winding.

A transformer with an open circuit secondary can be modelled as shown in the circuit below. For the open circuit test case,
R_{1}, R_{2}, X_{1} and X_{2} can be ignored leaving just R_{w} (iron-core loss
resistance) and X_{m} (magnetising loss reactance). The open circuit test is conducted by leaving secondary winding
open, while the full rated primary voltage is applied to the primary winding. At this point E_{o}, I_{o}
and P_{o} readings are taken.

Iron-core loss as measured by the wattmeter,
\( \begin{aligned}
P_o = E_o . I_o . cos \phi_o
\end{aligned} \)

No load power factor,
\( \begin{aligned}
cos \phi_o = \frac{ P_o }{ E_o . I_o }
\end{aligned} \)

Iron-core loss current component,
\( \begin{aligned}
I_w = I_o . cos \phi_o
\end{aligned} \)

Magnetising loss current component,
\( \begin{aligned}
I_m = I_o . sin \phi_o
\end{aligned} \)

Iron-core loss resistance,
\( \begin{aligned}
R_w = \frac{ E_o }{ I_w }
\end{aligned} \)

Magnetising loss reactance,
\( \begin{aligned}
X_m = \frac{ E_o }{ I_m }
\end{aligned} \)

No load impedance,
\( \begin{aligned}
Z_o = \sqrt{ R_w^2 + X_m^2 } = R_w + jX_m
\end{aligned} \)

Mains voltage is applied to the primary, with the secondary open circuit.
Readings of E_{o}, I_{o}, and P_{o} are taken.

\( \begin{aligned}
E_o = 240V, \text{ } I_o = 2.72A, \text{ } P_o = 53W.
\end{aligned} \)

\( \begin{aligned}
cos \phi_o = \frac{ 53 }{ 240 \times 2.72 } = 0.081
\end{aligned} \)

\( \begin{aligned}
I_w = 2.72 \times 0.081 = 0.221A
\end{aligned} \)

\( \begin{aligned}
I_m = 2.72 \times 0.997 = 2.711A
\end{aligned} \)

\( \begin{aligned}
R_w = \frac{ 240 }{ 0.221 } = 1087 \Omega
\end{aligned} \)

\( \begin{aligned}
X_m = \frac{ 240 }{ 2.711 } = 89 \Omega
\end{aligned} \)

\( \begin{aligned}
Z_o = \sqrt{ 1087^2 + 89^2 } = 1090 \Omega
\end{aligned} \)

A transformer with a short circuit secondary can be modelled as shown in the circuit below. For the short circuit test case,
R_{w} and X_{m} can be ignored. R_{1} and R_{2} are combined together as R_{c}
(copper loss resistance), and X_{1} and X_{2} are combined together as X_{i} (eddy-current /
hysteresis loss reactance). The short circuit test is conducted by shorting the secondary winding, while the primary voltage
is gradually increased using the variac until the maximum rated primary current is reached. At this point E_{s},
I_{s} and P_{s} readings are taken.

Copper loss as measured by the wattmeter,
\( \begin{aligned}
P_s = I_s^2 . R_c
\end{aligned} \)

Copper loss resistance,
\( \begin{aligned}
R_c = \frac{ P_s }{ I_s^2 }
\end{aligned} \)

Full load impedance,
\( \begin{aligned}
Z_s = \frac{ E_s }{ I_s }
\end{aligned} \)

Eddy-current / hysteresis loss reactance,
\( \begin{aligned}
X_i = \sqrt{ Z_s^2 - R_c^2 }
\end{aligned} \)

Full load power factor,
\( \begin{aligned}
cos \phi_s = \frac{ R_c }{ Z_s }
\end{aligned} \)

The variac is increased until the maximum primary current is reached, with the secondary short circuit.
Readings of E_{s}, I_{s}, and P_{s} are taken.

\( \begin{aligned}
E_s = 88.7V, \text{ } I_s = 3.33A, \text{ } P_s = 186W.
\end{aligned} \)

\( \begin{aligned}
R_c = \frac{ 186 }{ 3.33^2 } = 16.8 \Omega
\end{aligned} \)

\( \begin{aligned}
Z_s = \frac{ 88.7 }{ 3.33 } = 26.6 \Omega
\end{aligned} \)

\( \begin{aligned}
X_i = \sqrt{ 26.6^2 - 16.8^2 } = 20.7 \Omega
\end{aligned} \)

\( \begin{aligned}
cos \phi_s = \frac{ 16.8 }{ 26.6 } = 0.63
\end{aligned} \)

Transformer efficiency is given by,

Efficiency,
\( \begin{aligned}
\eta = \frac{ output\text{ }power }{ input\text{ }power } = \frac{ output\text{ }power }{ output\text{ }power + losses }
\end{aligned} \)

The value of transformer efficiency will be maximum when the copper losses, P_{c} are equal to the iron-core losses,
P_{w} in the transformer. This will be somewhere in between no load and full load conditions.

Efficiency,
\( \begin{aligned}
\eta = \frac{ E.I.cos \phi }{ E.I.cos \phi + P_c + P_w }
\end{aligned} \)

Assuming that the copper losses are lumped in just one winding we can approximate copper losses.

Efficiency,
\( \begin{aligned}
\eta \approx \frac{ E.I.cos \phi }{ E.I.cos \phi + I^2.R_c + I_w^2.R_w }
\end{aligned} \)

\( \begin{aligned}
E = 240V, \text{ } I = 3.33A, \text{ } cos \phi = 0.63.
\end{aligned} \)

\( \begin{aligned}
\eta \approx \frac{ (240 \times 3.33 \times 0.63) }
{ (240 \times 3.33 \times 0.63) +
(3.33^2 \times 16.8) +
(0.221^2 \times 1087) }
\end{aligned} \)

\( \begin{aligned}
\eta \approx 0.68 \text{ } (68 \text{%})
\end{aligned} \)

In the MOT described here, the secondary winding could of course be wound more efficiently by using single core copper wire with
thinner insulation and a specific gauge to minimise air gaps in the core. These changes would help to improve the transformer
efficiency, calculated to be approximately 68%. If there is a need for continuous operation,
then it is advisable to use two MOTs with the primaries connected in series. This method reduces heat generated and noise
caused by the MOT's high magnetising current.

Owing to the microwave oven being a mass market appliance, and so the requirement for minimal manufacturing costs, the MOT will
not be the most efficient of transformer designs. However the MOT can still be effectively used as a means of producing a cheap
high power mains step down power supply.

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