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I would like to know how the Peltier device would act if you heat sink the hot side WITH the cold side. Would it just sit there and consume power?

I need something to consume ~15 watts or more at 14V. Was thinking of other possibilities than a power resistor.

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  • \$\begingroup\$ "Was thinking of other possibilities than a power resistor." Is this just an exercise, or is there a reason you want to avoid using a power resistor? \$\endgroup\$ – Adam Davis Feb 28 '14 at 16:30
  • \$\begingroup\$ I was wanting to see if I could find a solution to dissipate 4 times 15 watts without a sizeable heat sink. So I thought maybe flexible heat sink film connected to both sides of a Peltier device may work? \$\endgroup\$ – brett s Feb 28 '14 at 17:01
  • \$\begingroup\$ Or, use 4 peltier devices with cold side of one connected to hot side of the next, and so forth until the last one only needed 1 heat sink. But I have no experience with these so it was just thoughts. \$\endgroup\$ – brett s Feb 28 '14 at 17:02
  • \$\begingroup\$ The energy has to go somewhere. Whether you use a power resistor or a peltier device as you suggest, you're going to be generating heat. In fact the Pletier device will, overall, create more heat than the power resistor, so if the point is to get rid of the excess heat output while still consuming the same amount of power, you'll need to try another route. Here's a similar question which might give you some ideas: electronics.stackexchange.com/questions/85520/… \$\endgroup\$ – Adam Davis Feb 28 '14 at 18:13
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Yes, and it will get hot, just like any power resistor. Remember, Peltier devices are not all that efficient — the heat out is much greater than the heat in. The difference is the electrical power you're putting into it.

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  • \$\begingroup\$ I'm new to thermoelectric devices, so based off what I'm trying to do, do you have a rough spec that I should look for in a Peltier for testing? What internal resistance/voltage rating should I look for? \$\endgroup\$ – brett s Feb 28 '14 at 15:48
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    \$\begingroup\$ Just use a power resistor. It'll be cheaper and more robust. \$R = \frac{V^2}{P}\$, so to dissipate 15W at 14V, you'll need a resistance of about 13 ohms. \$\endgroup\$ – Dave Tweed Feb 28 '14 at 15:55
  • \$\begingroup\$ Can you elaborate on how it will be more 'robust'? \$\endgroup\$ – brett s Feb 28 '14 at 16:03
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    \$\begingroup\$ Well, for one thing, if a Peltier device gets too hot, it will fall apart because the solder that holds it together will melt. Low-value power resistors are generally just wire and ceramic, and don't suffer from that problem. \$\endgroup\$ – Dave Tweed Feb 28 '14 at 16:18
  • \$\begingroup\$ I can't imagine a Peltiier device being happy in water.... But power resistors can be so robust that I have seen special Metal resistors cooled in a bucket of water. I don't know the type of resistor , it was being demonstrated to me, needless to say I stood well back when it was switched on... \$\endgroup\$ – Spoon Feb 28 '14 at 17:11

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