I am running a PLC and a 2Kw 24V dc motor from the same acumulator. but when the motor starts the plc trips to stop. I believe the easiest way to over come this problem is to mount a capacitor but I cannot figure out the specs for it. how can ı calculate what my need is.or have any better ideas or any solutions other than this?

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G.DEMIRURAL

 

2Kw at 24V is 83 amps!, assuming that's running current?, then the start current could easily be a few hundred amps - how big (VERY BIG!) is your accumulator?.

A capacitor to try and provide the current required would be absolutely HUGE - a better way would be to diode isolate the supply to the PLC, and fit a large capacitor on the PLC, to keep that running for a while when the 24V drops way down!.

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NOT only that but the capacitor will add its own in-rush current to the system which will compound the situation

Two option

1) in-line inductor to limit the current. This will effect dynamic performance
2) Since it is a startup problem, has a resistor in series with the DC-machine and after a set period (controlled by some other means) have a relay short out the resistor, thus bypassing it

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The acumulator is a 280 amp/hour array of 2V acumulators ( the ones that they use in tow trucks) I will try the series resistor which is some what easier to apply for me. I am still open to suggestions though.

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G.DEMIRURAL

 

As I suggested before, diode isolate the power to the PLC, and hang a big capacitor on the PLC (which will keep the PLC powered when the batteries drop briefly).

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http://www.winpicprog.co.uk

 

I agree with that suggestion.

JimB

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Experience is directly proportional to the value of the equipment ruined.