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Not a duplicate question: this question is about reducing start current, not about identifying the motor type.

I have a refrigerator that has a permanent split capacitor (PSC) motor, which I am powering from solar panels/battery/inverter.

My inverter is rated at 1000w (2000w peak), but unfortunately, the motor occationally trips the overload protection of the inverter, with the risk of spoiling the contents of the fridge if it goes unnoticed.

I am considering replacing the motor with a DC compressor (Danfoss BD35F), but it's quite expensive (about $350).

Before I do that, I want to experiment first, to see if I can get it to work with a consistently lower start current.

I realize it might break the motor, that it might not start consistently, but it's worth a try.

I've read about the following options:

  • a) use a start capacitor, 150uF, for the first 1000ms
  • b) use an NTC thermistor
  • c) power resistors, for the first 1000ms (similar effect as NTC)
  • d) voltage transformation to a lower voltage, e.g. from 230V to 200V

but:

  • a) I'm having a hard time finding a start capacitor to try with, and random experimentation could be expensive, since they seem to cost about $10-15 a piece and can break if powered for more than a few seconds.
  • b) I've tried a 10ohm NTC, but that made no difference at all.
  • c) I'm going to try with various resistors at 50, 70, 100 ohms (rated at 50w).
  • d) I'm not sure which type, which voltage to try etc..

ALSO:

I tested the motor with the following results:

  • Disconnected for 24h, ~850w
  • Connected for 20min, disconnected for 5min, ~600w
  • Connected for 45min which included a self-restart, ~1100w
  • Disconnected, then reconnected after 15min, ~800w

I'm puzzled by these results, because it seems that the motor draws more if left connected. But the thermostat disconnects the circuit. The only way this makes sense to me, is if the thermostat somehow causes an arc that draws an additional 300w when reconnecting. But I tested it by disconnecting for 15min, turning the thermostat to 0, reconnecting, then turning the thermostat to 3. The max power draw was only 782w.

Very strange... Any ideas?

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  • \$\begingroup\$ Note that the motor really needs that power when starting up, so not giving it will possibly make it fail to start up. Also it is very hard on the motor to start up shortly after stopping, due to the way the refrigerant moves around,which is the reason why a lot of fridges have measures to prevent it from starting up within a couple of minutes after being stopped. \$\endgroup\$ – PlasmaHH Sep 9 '16 at 10:19
  • \$\begingroup\$ Add any soft starter for that Refrigerator, but it will cost you compare to other methods. You need to find the starting current and other parameter accordingly you have to select soft starter. \$\endgroup\$ – Photon001 Sep 9 '16 at 10:21
  • \$\begingroup\$ The motor is a Donper AG100CY1, seemingly without available specification. The resistance is 43ohm + 26 ohm for the two windings. I guess the start current is 1100w/230V = ~ 4.8 A. How do I determine the right soft starter? \$\endgroup\$ – user95482301 Sep 9 '16 at 10:24
  • \$\begingroup\$ 26Ω is your starter winding //43Ω. Req=16Ω Surge current is V²/R=3304 Watts \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 9 '16 at 11:03
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    \$\begingroup\$ Possible duplicate of Refrigeration compressor circuit, help trying to reverse engineer it \$\endgroup\$ – Charles Cowie Sep 9 '16 at 13:57
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I have a similar situation, off grid PV / battery storage system with a 1Kva 12-volt inverter and I run a small 230-volt standard fridge. It used to knock out my TV & Satellite at start-up and hit the battery bank voltage quite hard. I connected a VDR (voltage dependant resistor) in series with the live feed. This just takes the sting out of the start-up inrush current and overcame the problem for me

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I solved this with a 2 farad 16v super capacitor on the input of the inverter> Although it doesnt reduce the start current it provides enough electrons to start the compressor without shutting down the inverter

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Clearly, if your inverter cannot provide enough power for the fridge to start, the proper solution is to get a more powerful inverter. That said, if you know your inverted can provide the extra current for a short time during startup, you may just be able to replace its protection circuit by a slow-blow circuit breaker, which is specially designed for loads like motors and such.

Soft-starter can be another solution, but only if your compressor supports it (many big compressors do).

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  • \$\begingroup\$ The problem with bigger inverters, e.g. 1500w, is that they also have a higher standby current, which can be significant. Also, the load is usually not more than 200w total. \$\endgroup\$ – user95482301 Sep 9 '16 at 10:59
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    \$\begingroup\$ Well yeah, there's no free lunch. If it was possible for the same inverter to provide more current, or of the same compressor to consume less at startup, that would have already been done. \$\endgroup\$ – Dmitry Grigoryev Sep 9 '16 at 11:24
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Reducing startup power is not needed for on grid use. The measures would only increase the cost of the fridge. To reduce the startup power some form of unloading is required. For big compressors this is done by lifting the suction valve(s). Sometimes a softstarter is also included. A better way in your situation would be to change to a socalled absorption type of fridge. They work with a small heating element and have no inrush current at all. It might be possible to work even without an inverter if you select the right type (12 or 24 VDC also needed for the Danfoss compressor).

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    \$\begingroup\$ Thanks for your answer. Unfortunately absorption fridges tend to have awful energy efficiency, burning about 2.5kWh/24h compared to a normal modern fridge, which uses about 0.4 kWh/24h. In addition, I am actually using this in an offroad motorhome which will basically never be level enough to use absorption. Good point nevertheless :) Absorption: typically 120w electrical x 24h = 2.88 kWh/24h (duty cycle may be between 50-100%). When burning gas, the energy used is double, if I remember correctly. \$\endgroup\$ – user95482301 Sep 9 '16 at 19:11
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I think your best bet would be one of those power factor correction capacitors. A large motor run cap might be a suitable alternative. The only method I have of calculating the required size is from someone else (I haven't tried it) and is really hit and miss.

It goes like this: connect the capacitor in parallel with a filament bulb and connect both of these in series to the appliance while it is running. Keep increasing the capacitor size till the bulb goes out. Or you know you could use the saying: the bigger the gob the better the job.

Alternatively you could use a larger inverter to start it and use a change over switch to go to the correct size.

Finally. Use a DC inverter compressor refrigerator. Obviously you'll have to buy a new one.

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  • \$\begingroup\$ Actually the DC fridge isn't necessarily a solution -- mine causes enough of a sag in the supply voltage that the fridge's own low battery protection circuit kicks in, even when the battery is rather full. I'm looking into putting an NTC thermistor in series with the winding, and/or a big capacitor across the control circuit input (I haven't opened it up yet to check the possibilities). \$\endgroup\$ – Chris H Jul 11 '18 at 14:55
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You need to limit inrush current, it seems. Perhaps do a search on "mains inrush protection". One such item from the search is this

This device is useful where the current drawn from the mains supply by a piece of equipment at switch-on (eg. an audio power amplifier) is considerably higher than the steady state current consumption of the equipment. This current surge can be troublesome in that it may blow fuses or trip circuit breakers. The unit has IEC input and output connectors and limits the current at switch-on to an acceptable level for 200mS–400mS. Two LED’s on the front panel indicate the status of the unit. A voltage dependent resistor is included for transient suppression.

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Something to keep in mind with all AC motors, reducing the voltage actually causes motors to draw more current. Remember that the only thing that reduces the current when a motor is running is the back voltage that is generated by the motor spinning. If the voltage is too low you will burn out the motor, just as will happen if it is on too high a voltage. All AC motors have frequency and voltage specifications that it is designed to run properly. Also there are some motors that are designed specifically for inverter use, and can be speed adjusted over a range to increase and decrease the power consumption.

Just keep in mind that the higher the frequency the faster the motor will spin, and the higher the operating voltage. When the frequency goes down so must the voltage. This is how variable frequency drives work in the real world. Also with a soft start it is reducing the voltage for a very short time so that the motor can build up to speed, and then a set of contacts close to bypass the voltage reduction part of the circuit.

One of the problems that I have personally encountered is that some people connect motors and other things without even bothering to make sure the motors are connected for the correct voltage. Also one of the more important things to do when doing any connections is give all the connections a pull test before closing up anything (power off obviously for safety). That is to make sure all the connections are good and the wires have not opened on the inside between the ends. There is nothing more annoying than intermittent connections, and when you find a wire that stretches you found the problem most of the time.

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    \$\begingroup\$ This is interesting general discussion, but it's not clear that there is an actual answer to this nearly two year old question within. The answer form is strictly limited to being used for actual answers; it may not be used for commentary. \$\endgroup\$ – Chris Stratton Jun 27 '18 at 5:03
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Here is a simple fix for your problem that you can do for very little investment. Buy a surge protector suitable for the device and pinch off the noise maker. When your fridge draws hard the surge protector will supply the extra voltage then return to normal pass-through. Cheap easy fast available almost everywhere.

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    \$\begingroup\$ Welcome to EE.SE, COTS, but this answer is technically incorrect. Surge protectors do not store energy and can not provide energy. Instead, when the incoming voltage exceeds a certain threshold, they provide a partial short-circuit to protect the load. You can edit your question to fix it using the edit link. \$\endgroup\$ – Transistor Aug 23 '18 at 18:08

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