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I am attempting to run a Molybdenum heat coil. The best I can tell it is 0.3-0.4 ohms per rack and I would like to run say two racks at 230V at 3.9KW (so it doesn't trip my 230V 20 amp breaker).

Does that mean, assuming a value for two racks is a total of 0.8 ohms, can I add a ~12.7641 ohm resistor for a total resistance of 13.5641 to run it at 3900 watts at 230 volts?

If not, how is it possible to run these? Do I need a lower voltage supply with more current?

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    \$\begingroup\$ Are those heat coils rated for 230V? I assume AC? With 230 VAC and 12.8 ohm resistor in series with your 0.8 ohm coils you would be burning some 3.7kW in your resistor. No heat coil needed! ;) \$\endgroup\$ – Dejvid_no1 Oct 8 '15 at 17:42
  • \$\begingroup\$ I don't know much about them unfortunately. But I have tried plugging them in using a SCR and it starts conducting around 50v or so and shortly trips my 20amp breaker. It seems like I need more resistance, but I am no electrical engineer. \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 17:45
  • \$\begingroup\$ I am sure they are intended for an industrial furnace, perhaps a tube ceramic sintering furnace. \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 17:45
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    \$\begingroup\$ Does that mean I need to run them at lower voltage and higher amperage? \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 17:49
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Using resistors is Very Bad Idea. Look at datasheet of Your heating element, this one is for example: http://heatingelements.isquaredrelement.com/Asset/Moly-D-technical-brochure.pdf On page 4 You can find a graph showing resistivity in function of temperature of heating material. At normal operating temperature (about \$1600^{\circ}C\$) molybdenum elements has resistance 7 times larger than at room temperature! If You have an element with \$0.3\Omega\$ resistance at room temperature, it will rise up to about \$2.1\Omega\$ at \$1600^{\circ}C\$. Connect them in series and You've got more than \$4\Omega\$ in operating condition.

I know, it's still too much to run it with 20A circuit breaker - only reasonable solution for such power is to use SCRs with proper driver. Such regulator gives You another two important features:

  1. Soft-start - limit current during start-up, when molybdenum elements are cold and have low resistance.
  2. Regulation of temperature - use more sophisticated regulator with feedback from some temperature sensor and you'll have an automatic regulation.
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  • \$\begingroup\$ Very interesting. I currently have a (manually adjustable) SCR in place and also have it hooked up to a SSR controlled by a PWM temperature controller. So if I gave it the proper voltage with enough amps, would that work as a proper driver? \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 19:08
  • \$\begingroup\$ For clarification: as the resistance increases, do I need to increase the voltage? It looks like it would run decent at 4.0Ω at 150V but I would need a 40 amp breaker (doable I assume). \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 20:11
  • \$\begingroup\$ @JeshuaLacock You may use Your temperature controller to automatically regulate the temperature, but don't forget about some kind of softstart! \$\endgroup\$ – Jakub Rakus Oct 11 '15 at 12:36
  • \$\begingroup\$ Thank you. I keep hearing that term - does that just mean I start the volts low and ramp up? \$\endgroup\$ – Jeshua Lacock Oct 11 '15 at 16:28
  • \$\begingroup\$ @JeshuaLacock You're right, it's the simplest softstart. More sophisticated method is to control the current, to not exceed desired value. And as You noticed earlier to get 3.9kW at 4Ω You need about 31Amps. Take a look at the datasheet from link in post - it briefly explains all what You need to know to use Your molybdenum coils. \$\endgroup\$ – Jakub Rakus Oct 12 '15 at 16:26
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  1. You have not provided enough information about the heater elements to answer your first question properly. Your 230V, 20A power source provides 4600W of power. If you provide the recommended voltage type & rating for your heater element, you will probably get a better answer. If you have no information about your heater element, you could start here for possible clues: http://heatingelements.isquaredrelement.com/category/oly-d-molybdenum-disilicide-mosi2-heating-elements
    Note: Adding a 12.8 Ohm resistor in series with your heater elements won't work because the majority of the energy will be dissipated as heat through your largest resistance--which is your 12.8 Ohm resistor. A Moly-D 3-6 would use 61A@16V and consume 975W (so, to produce ~2,732° F in a 1 ft^3 oven, you would need 4 heater elements, which would consume 3.9kW), just as an example.

  2. Yes, you would probably need a lower V & higher current to use the heating elements that you have. Most likely you will have to use some kind of step-down transformer that can provide a large A output--which will probably be pricey.

HTH

Best Regards!

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You have three options:

  1. Find more of the coils an add them in series until you get down to your target power. Sounds like you will need 30-40 of them though so that seems unreasonable.

  2. Use a transformer to reduce the voltage and therefore power. Still, looking for a 3.9KW transformer is going to be difficult or expensive.

  3. Use PWM to reduce the average current draw. This is going to need a separate question about creating a driver capable of powering a load this big.

Unfortunately I think you will just have to find alternate heating coils that are of a lower power if you are not able to use the ones you have without tripping your breaker.

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  • \$\begingroup\$ Would welding supply(s) (low voltage high amperage correct) perhaps in parallel as needed to achieve the operating voltage work? \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 17:59
  • \$\begingroup\$ I can ask this as a separate question if that makes more sense, but if you could amend your answer I would appreciate it, thank you. \$\endgroup\$ – Jeshua Lacock Oct 8 '15 at 18:04

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