I have disassembled the motor controlling circuit for my refrigerator. I am trying to understand how it works, so I hopefully reduce the start surge-current and make it work with my 1000w inverter. Currently the fridge sometimes draws 1100w on start, othertimes only 600w.

The fridge is brand new, high-efficiency, "BEKO" with a Donper AG100CY1 compressor.

I've checked the circuit a hundred times, and it really is like this:


simulate this circuit – Schematic created using CircuitLab

But it seems to go contrary to what I've read about refrigerator controller circuits.

  • The start winding is permanently connected (normally it would be disconnected after start, e.g. by a PTC)
  • The run winding is current runs via PTC at start, then via capacitor once hot

My main surprise is that the surprise winding is permanently connected.

(And additionally, I am hoping to add a start capacitor...)

Why isn't the start-winding disconnected? Does this make any sense?

  • \$\begingroup\$ The winding with the higher DC resistance value is usually the RUN winding. The winding with the lower DC resistance is the start winding. \$\endgroup\$ – Dwayne Reid Sep 8 '16 at 20:25
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    \$\begingroup\$ I think you'll find that the high start current is due to a higher mechanical load in the compressor at startup and has nothing to do with the motor windings. The 'start' winding is not only powered when the motor starts - it is powered all of the time. \$\endgroup\$ – brhans Sep 8 '16 at 20:45

That is a permanent split capacitor (PSC) motor.

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  • \$\begingroup\$ Thanks. I was able to find a circuit diagram which seems to show the same... link \$\endgroup\$ – user95482301 Sep 8 '16 at 19:34
  • \$\begingroup\$ But where should I add the start capacitor to reduce start current then? At the voltage source? Or is it not possible to add a start capacitor to this type of motor? \$\endgroup\$ – user95482301 Sep 8 '16 at 19:36
  • \$\begingroup\$ industrial-electronics.com/image/p_split.gif \$\endgroup\$ – user95482301 Sep 8 '16 at 19:38
  • \$\begingroup\$ Or perhaps just add 20-100 ohm resistance in the first second and take it out with a relay? \$\endgroup\$ – user95482301 Sep 8 '16 at 19:41

Why did you check the circuit "a hundred times"? This is a technical site and "a hundred" means 100. ;^)

The 43 Ω winding is the primary one as it is directly across the mains.

The 26 Ω winding is the start winding as it has the series capacitor. I can't guess at the inductance but we could do some guesses about the currents.

The impedance of the capacitor is given by \$ Z = \frac {1}{2 \pi f C} \$. At 50 Hz this will be 1061 Ω and at 60 Hz 884 Ω. Worst case on 230 V 50 Hz is 230 mA and on 120 V is 135 mA. These seem a little low to me but maybe they're enough to give rotational torque to the motor to get it started but low enough not to waste significant power. The start winding does, after all, do some useful work even when up to speed.

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  • \$\begingroup\$ Because I was so convinced that I was wrong, since the circuits I've seen explained all had a start capacitor with PTC. The PSC motor was new to me. \$\endgroup\$ – user95482301 Sep 8 '16 at 20:41
  • \$\begingroup\$ I've read that the start winding is generally the one with the higher resistance? The motor draws about 60w in operation, and 600-1100w at start. \$\endgroup\$ – user95482301 Sep 8 '16 at 20:45

Motors in consumer products are typically carefully designed to do exactly what they need to do at the minimum manufacturing cost with no safety margins. Any alteration of the circuit is likely to result in the motor not starting reliably.

I believe that refrigerator compressor motors often require less starting current when they are started after not running for some period of time. I have seen a refrigerator with a "minimum off time" circuit. If this refrigerator does not have such a circuit, consider adding that. That might result in the motor requiring only 600 watts every time rather than 600 some times and 1100 other times.

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  • \$\begingroup\$ That's a good point. But worst case, I'll probably buy a DC compressor to replace it. I've decided to experiment to see if I can save some money, and perhaps learn something as well. The circuit also has a timer-switch (I think...) which prevents the compressor from being started immediately after being turned off. Perhaps this relates to what you're saying, that the compressor needs to cool down first. And if I remember correctly, it never reached more than 700w when started cold. Perhaps I can run cycles of 1h operating, 1h resting. I'll try it out! \$\endgroup\$ – user95482301 Sep 8 '16 at 20:48
  • \$\begingroup\$ I don't think you need an hour. I would recommend experimenting with a shorter rest time, perhaps 5 or 10 minutes. Then try longer times if that doesn't work. \$\endgroup\$ – Charles Cowie Sep 8 '16 at 21:22
  • \$\begingroup\$ Would there be another way to reduce the start current, as experienced by my inverter, by adding a capacitor to the circuit while starting? E.g. at the power cable \$\endgroup\$ – user95482301 Sep 9 '16 at 6:39
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    \$\begingroup\$ I don't think there is any way to reduce the start current of the motor by any external means without reducing the starting torque capability. A DC motor with an electronic speed controller (ESC) would do the job. The least expensive option may be to find an inverter that has a higher short-term current capacity. \$\endgroup\$ – Charles Cowie Sep 9 '16 at 8:37
  • \$\begingroup\$ if motor stalls when turned on then you risk burning out the motor if limited to run current under high compression load. Theoretical peak surge is 2351 Watts = 230^2/22.5ohm. Would ZCS triac help? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 9 '16 at 14:45

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