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I need to make a small heater. I was looking at this project:

http://www.makeuseof.com/tag/make-your-own-temperature-controller-with-an-arduino/

but I need to design the heating element. I was considering taking 120VAC from the wall, running the hot side through a 1 amp fuse, running that into a relay, then running the relay output into a 300 ohm power resistor for a total of 50 watts. Looking at this part currently:

http://www.mouser.com/ProductDetail/Vishay-Dale/HL05006Z300R0JJ/?qs=sGAEpiMZZMtbXrIkmrvidHh8qJ0vS%2f1J%252bhAYDyLahp8%3d

Questions:

  1. Do I need an AC isolation transformer? If so, why? Isn't the 1 amp fuse enough for safety?

  2. I am not planning on using a fan or heat sink, only ambient air. What is duty cycle of such a power resistor? Generally I would expect that it can dissipate 50 watts 100% of the time (taking into account the ambient temp derating of course), and I've also seen the 1000h component life spec. Any thoughts?

  3. Max voltage = (P*R)^1/2, which in this case comes out to 122V. Obviously this is close to wall voltage, and peak to peak is much higher than 122V. This thing looks like it's built like a high voltage component though, will this really be an issue? (If it is I can adjust the dissipated wattage downwards a bit, while still using a 50 watt rated resistor)

Thanks!

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  • \$\begingroup\$ The need for an isolation transformer is in case someone touches any of the 120V wiring from the input through the resistor. If the wiring is protected from being touched, then the transformer is not needed. the fuse makes no difference, 1A is much more than is required to kill. \$\endgroup\$ – DoxyLover Jun 1 '15 at 21:26
  • \$\begingroup\$ The wall voltage is reported using Vrms so that you can treat it as if it was a DC voltage for power calculations. You don't need to worry about the peak vs. rms values. V=(P*R)^1/2 is correct if you use the rms value for V. Using V=120 Vrms only gives 24 W. You need to reduce the resistor to get 50 W. \$\endgroup\$ – Austin Jun 2 '15 at 3:06
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  1. No, you don't need an isolation transformer provided the low voltage power supply, relay and temperature sensor are isolated from the resistor and mains circuit. If any of those isolation barriers is not present or fails, even for an instant, bad things may happen (such as a shock, fire, destruction of your computer, etc.)

  2. The power the resistor can dissipate is given in the datasheet.

enter image description here

If you are using it to heat something, say to 100°C, then you can dissipate about 80% of 50W or 40W. The actual power will depend on how insulated the resistor is as well as the ambient temperature (it's the air immediately around the resistor that matters).

  1. That voltage rating is a power dissipation limit. There is also a 1000V test. That part should be fine on mains voltage barring any huge surges.
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  • \$\begingroup\$ I understand the power rating, but is that power rating for a 100% duty cycle? \$\endgroup\$ – cat pants Jun 1 '15 at 21:36
  • \$\begingroup\$ Average power,however that comes about, assuming the resistor temperature is fairly steady. \$\endgroup\$ – Spehro Pefhany Jun 1 '15 at 21:52
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    \$\begingroup\$ The power rating is practically always for 100% duty cycle, unless otherwise stated. So yes, the resistor can handle 50W continously at 25 degrees centigrade, and up to 500W for five seconds according to the datasheet. \$\endgroup\$ – jms Jun 1 '15 at 21:53
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Answers:

  1. Yes. Because 1A through you = still VERY BAD! Mains voltage is related to Ground, you are connected to ground, thus current can flow. A 120VAC to 120VAC transformer is less bad, but possibly only slightly, because you can still touch two ends of 120VAC which is still at the very least painful.
  2. First: Then how are you going to transport the air to where you want it? Waiting for convection to do it's work on its own is going to make for a "watching paint dry" exercise unless it's inside a tiny box. Then: if you DO heat something with it, the ambient air temperature will rise, so no, you can no longer use it continuously on at 50W anymore, because it'll become only 45W at some heating of the environment (approx 50degC), 40W at some more, etc etc etc.
  3. This looks like a linearly wirewound resistor, so you can very likely put well above 120V RMS across it. The point to 120V being close is not the voltage, but the fact that if the resistor is off by a little (5% specification + aging effects of the materials) or for a while the voltage happens to be 125VAC you are actually overloading the resistor. Take at least 10% margin.

Then, to make suggestions:

If you are intent on using 120VAC, why not use a 100W (or more) halogen lightbulb in an official mounting? If you don't want the light, add a black metal box around it that you connect to power ground for safety.

If you are not: Why not just go to 24VDC with a cheap system power supply from mouser ($20 from Meanwell for about 70 to 100W) and using that to put through whatever you want. It'll cost you $20 and you will at least not kill anyone, plus you can use a high power MOSFET to control the load and actually use PWM later if you want even better control.

Oh and you have power right away for a fan, as well, which makes moving the heat around the space you want it in much easier.

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  • \$\begingroup\$ The 100 watt bulb is a good idea! Aren't those products all EOL though? \$\endgroup\$ – cat pants Jun 1 '15 at 21:34
  • \$\begingroup\$ Also: how does an isolation transformer help if you get shocked? IMO it doesn't do anything, 1 amp will kill you either way...(or even less) \$\endgroup\$ – cat pants Jun 1 '15 at 21:35
  • \$\begingroup\$ @catpants I don't think the halogen tubes are, and I'm sure you can still find normal halogen bulbs. 100W+ Workman's lamps also still use the halogen tubes, because LED is waaaaaay too expensive, or too fragile for builders. \$\endgroup\$ – Asmyldof Jun 1 '15 at 21:36
  • \$\begingroup\$ @catpants The difference is, as I state, what comes from the wall only needs 1 wire to kill you, granted has to be the right one, but none the less, after isolation the only path is through the transformer, so water and such will be much less risky. But as said, I'm not a fan of using 120VAC on items that are not designed for that use without a box around them. Adding a box would defeat the heater purpose, so your isolation solution will always be fiddly and risky. \$\endgroup\$ – Asmyldof Jun 1 '15 at 21:37
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    \$\begingroup\$ @catpants No, if it's an isolation transformer, it isolates. The reason the phase can conduct into ground is that the source of that phase is related to ground itself. The isolation transformer turns 120VAC into a magnetic field, which it then turns back into 120VAC, so current cannot flow from the secondary into ground, because ground is no longer electrically connected to its secondary... unless you connect them, but that would be quite stupid. But again (cannot say it often enough): Touching both ends of secondary will always still be 120VAC. \$\endgroup\$ – Asmyldof Jun 1 '15 at 22:26

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