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How does a half wave rectifier or phase control dimmer affect the triggering of a circuit breaker? For example, if a 20A resistive load was on a 15A breaker with a half wave rectifier. Now average RMS current is only 10A even though peak RMS is still 20A.

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  • \$\begingroup\$ You worked it out yourself ….the breaker is rated at an RMS current so it's the RMS load current that will mainly cause the trip. However many breakers have a mag element which will trigger at high peak currents and even the thermal element will trip at high peak currents. \$\endgroup\$ – Jack Creasey May 20 '19 at 0:15
  • \$\begingroup\$ No. a 20A (full wave)resistive load run at 50% duty cycle is 14.1A RMS \$\endgroup\$ – Jasen May 20 '19 at 1:51
  • \$\begingroup\$ average RMS current what does that even mean? \$\endgroup\$ – Jasen May 20 '19 at 2:51
  • \$\begingroup\$ A link to the breaker's datasheet would be a help as we could check the trip times. \$\endgroup\$ – Transistor May 20 '19 at 6:26
  • \$\begingroup\$ @Jasen I did not know that the RMS of a 50% duty cycle sine wave was not RMS(100%)/2...though I see it now after playing around on Wolfram. Since I thought it was a simple relationship I was doing [RMS(1st half)+RMS(2nd half)]/2 which is were I got "average RMS current" from. \$\endgroup\$ – chew socks May 22 '19 at 23:12
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Now average RMS current is only 10A even though peak RMS is still 20A

No, it's 14.14 amps and quite close to the breaker's limit.

If the current taken is normally 20 amps RMS for a full AC scenario, you can project this to power by squaring. So 20 amps becomes 400 amps squared. It is the power that is halved by introducing a half wave rectifier so, 400 amps squared becomes 200 amps squared and, the effective RMS current is the square root of 200 i.e. 14.14 amps.

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