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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:

schematic

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?

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  • \$\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\$ Commented Sep 8, 2016 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
    Commented Sep 8, 2016 at 20:45

6 Answers 6

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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\$ Commented Sep 8, 2016 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\$ Commented Sep 8, 2016 at 19:36
  • \$\begingroup\$ industrial-electronics.com/image/p_split.gif \$\endgroup\$ Commented Sep 8, 2016 at 19:38
  • \$\begingroup\$ Or perhaps just add 20-100 ohm resistance in the first second and take it out with a relay? \$\endgroup\$ Commented Sep 8, 2016 at 19:41
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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\$ Commented Sep 8, 2016 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\$ Commented Sep 8, 2016 at 20:45
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Something that is being overlooked is the freon side of the circuit. Most likely the freon system has a cap tube as the metering device. When it is running there is at least 100 PSIG difference between the input and output of the compressor depending on the refrigerant. The system when shut down needs to sit for five to ten minutes, depending on the system to allow both sides (High side & Low side) to equalize. If you do not allow for this your start load will be extremely high and it may not start at all. Most systems have some type of protection to protect for this, one of the most common is a thermal disk usually mounted on the compressor. I expect the compressor is cooled by the return saturated gas from the evaporator so changing compressor speeds other then momentary can be a very bad thing. The compressor needs a given amount of energy to start under all normal conditions. Trying to circumvent that can cause compressor burnout or spoiled contents in the refrigerator as it will not maintain temperature. My suggestion is to simply get a bigger inverter, it will be the most cost effective in the long run.

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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\$ Commented Sep 8, 2016 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\$
    – user80875
    Commented Sep 8, 2016 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\$ Commented Sep 9, 2016 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\$
    – user80875
    Commented Sep 9, 2016 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\$ Commented Sep 9, 2016 at 14:45
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That is a permanent split capacitor (PSC) motor. The starter wire in the case of an electromagnetic or electronic relay, which disconnects it after starting the engine, always has a greater resistance because it is built with a very thin wire, so as not to take up space, so that the aggregate comes out with the smallest possible dimensions . When the starting coil remains connected through the capacitor, a balance is found between the cross-section of the wire and the size of the capacitor, so that the current that will pass through it does not burn it. The currents at the beginning of the pump are the largest, not only from the moment of the start, but during the entire time that the refrigerator has a large evaporation of gas, which returns to the suction pipe of the compressor. when it reaches low temperatures, the compressor works more easily, because the amount of gas in suction is much lower. The refrigerators are built by groups of engineers and scientists in this field and they have found the balance of minimal electricity consumption. Your inventions will cost you enough to buy two new refrigerators and again you will achieve nothing.

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use a $20 SSR to start your fridge or AC it switches only ON when the AC sinus wave is at the 0 crossing point, explained in the youtube channel from GreatScott! titled how to make an AC softstarter ...

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    \$\begingroup\$ Welcome to EE.SE, John. You may need to revise your answer. Zero-cross is the worst time to switch on an inductive load. \$\endgroup\$
    – Transistor
    Commented Jul 4, 2021 at 17:04
  • \$\begingroup\$ ok, then buy an easy start for $300 to use with your inverter, there are some very secret cheap components inside to make this thing work. \$\endgroup\$
    – John
    Commented Jul 4, 2021 at 17:09
  • \$\begingroup\$ Use a bigger inverter is is the best solution. \$\endgroup\$
    – Gil
    Commented Jul 4, 2021 at 20:21

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