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I'm an electronics hobbyist and I decided to make my own electric heater/kiln thingy. It works fine for a couple of minutes, but then the breaker trips. This is my first project involving AC, so I'm a bit clueless as to what could be happening here.

The circuit is very simple:

  • I have a coil of wire that is switched on/off by a SSR which is controlled by a PID controller.
  • The PID and the heater coil each have their respective fuses.

  • I'm run it off a wall outlet and since I live in Canada, it operates at at 120V AC 60 Hz.

  • I have it grounded to the heatsink of the SSR, which is housed close to all the other wiring, but definitely not touching.

Edit: I hope I didn't botch this circuit drawing, but here it is.

schematic

simulate this circuit – Schematic created using CircuitLab

My thoughts:

  1. The breaker cannot be tripping from over-current because the fuses are never blown, and the wire itself can only handle a maximum of 5 amps before melting. (I think the breaker trips at 10 or 15 amps).
  2. Is it possible that the coil is generating a voltage spike that is arcing over to ground? Like how in DC, the voltage builds up as the field collapses.
  3. If 2. is a possibility, how would prevent that from happening? Like how in DC you would have a fly back diode.

That's all I could think of, and honestly I don't know what to look up so I decided to ask here before turning it back on again.

Thanks.

PS: I decided not to draw a circuit because I don't know what the exact values would be for all of the components. I'm hoping that the circuit is simple enough to understand from the description in words. I can attempt to go calculate everything if it's really necessary, just hope I don't have to haha.

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  • \$\begingroup\$ a wire under water will handle hundreds of times more current than the same wire in air. if your breaker pops, it's because you're using too much current, unless you have a brand-new breaker with AFCI, but that's not likely the case, or cause even if true. \$\endgroup\$ – dandavis Aug 21 '19 at 20:53
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    \$\begingroup\$ Fuses don't actuate instantly. Circuit breakers are generally faster, depending on the type. \$\endgroup\$ – Hearth Aug 21 '19 at 20:56
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    \$\begingroup\$ Measure the resistance of the heater, calculate the current flow and post the information into your question. \$\endgroup\$ – Transistor Aug 21 '19 at 20:59
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    \$\begingroup\$ Draw a schematic, and note where you don't have values. The drawing is in any case better than words describing a circuit. Just make sure the drawing correctly represents the circuit. \$\endgroup\$ – JRE Aug 21 '19 at 21:19
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    \$\begingroup\$ As said above, picture(s) would be very helpful. There may not be a 'problem' with your circuit, but a problem with the design. As transistor rightly said, surely you have an idea of the current drawn from your heater, and you are asking this question as the current is much higher than you anticipated. If not, you will probably need to measure / post some component / circuit values. \$\endgroup\$ – David777 Aug 21 '19 at 21:32
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The breaker is almost definitely tripping from an overcurrent.

As to why it would only happen after a certain period of time, the resistance of your heater is liking decreasing with temperature. So you are actually drawing more current once the heater is warmed up.

If you dig deeper into the data sheet for the fuse you might find that it will not trip that quickly for 15A. Check out the current versus time curves on the data sheet.

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  • \$\begingroup\$ I find this a little dubious. Heaters are almost always made of high-resistivity refractory metal alloys like nichrome, which increase in resistance with increasing temperature. I see little point in making a heater out of a negative temperature coefficient material, not least because it would be difficult to control its current. \$\endgroup\$ – Hearth Aug 23 '19 at 12:16
  • \$\begingroup\$ @Heath Good point, I am now skeptical of my own answer! I guess less materials decrease resistance with temperature than my own experiences made me think. After looking at some charts online it seems most metals will increase resistance as they get hot. \$\endgroup\$ – MaxwellEE Aug 28 '19 at 20:27
  • \$\begingroup\$ Semiconductors decrease in resistance with temperature because the increased thermal energy results in increased carrier concentration. All other materials, as far as I know, increase in resistance with temperature as more energy in the crystal lattice means more potential to scatter electrons. \$\endgroup\$ – Hearth Aug 28 '19 at 20:55
  • \$\begingroup\$ I believe graphite and some piezoelectric materials also have decreasing resistance. \$\endgroup\$ – MaxwellEE Aug 28 '19 at 21:07
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The short is happening in your PID or SSR and they are not connected correctly. If the diagram you drew is correct, then this is the only current pathway that could trip the breaker and leave the fuse 'unblown'. Check the wiring of the PID and SSR and the pinouts. Make sure the SSR is rated for 110VAC and can handle 10A. Also there is no current in an SSR, so if it has a coil, then it's a relay. Keep in mind that Relays or SSR's need to have proper spacing (look for IEC61010 certified components or equivalent certification). Using improper components could lead to arcing and shorts.

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