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I built very simple circuit which allows turning on/off using microcontroller low voltage signal. To have it protected I've decided to add circuit breaker. Since it consumes 120W of power (around 0.5A of current) I've chosen a 1A circuit breaker with C tripping characteristic. The reason I chose the C-curve is that the device contains a pump (which is an inductive load) so there might be some spikes. I've measuared these spikes with a multimeter. The maximum instantaneous current would be 1.74A.

The problem with the setup is that when I turn on the circuit breaker it trips. It doesn't always trip always. Sometimes it works as expected. I thought the problem is with the circuit breaker, but replacing one didn't help.

Could anyone suggest what can go wrong in such a setup? For sure current doesn't exceeds 1.74A (AC RMS) which is not enough for instant tripping. Except the multimeter can't measure such a short current spikes, might be worth checking with oscilloscope.

Schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ You need to show a schematic. \$\endgroup\$
    – Andy aka
    Commented Mar 8, 2023 at 14:55
  • \$\begingroup\$ Added. Basically it's relay controlled by a controller. Relay is connected to live wire \$\endgroup\$
    – seeker
    Commented Mar 8, 2023 at 18:02

1 Answer 1

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For the C-characteristic circuit breaker, if the current is 10 times higher than the 1A current if you switch on the device, so around 10 ampere for only 0.005 seconds, then the magnetic release will stop the current flow.

It depends on the situation, if you switch on the motor if the voltage on the AC rail is at the highest point, then the motor will see a higher current flow at the beginning.

The circuit breaker has to protect the powerline, not the device. For the device you should use a fuse.

You can try a 2A circuit breaker or you measure the current flow in exact the situation if the 230V AC -> (325V) is flowing into the motor and figure out what the best matching circuit breaker is.

With a good multimeter you can measure the maximum-current, but the sampling time of the ADC of the multimeter is sometimes not very good.

You should do it with an oscilloscope and a shunt-resistor.

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    \$\begingroup\$ You were right. Having contraption connected to oscilloscope showed totally different picture - 7.6A RMS for ~7 microseconds. Then current gradually falls. Replacing CB with 2A nominal solved the problem. I am still wondering whether circuit breaker is rated 1A DC or 1A RMS though (datasheet says it's for AC use only) \$\endgroup\$
    – seeker
    Commented Mar 8, 2023 at 18:05

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