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I have a Peltier (TEC) module, with diodes in series with each lead. (see image below)

Image of Peltier Leads

I'm not quite sure what the specs are for the diode, but it says 130 on one line and SB560 on another, so I did a google search and found this.

How can I figure out how what the actual voltage on the Peltier element will be?

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  • \$\begingroup\$ A schematic would help. Is that the exact Peltier module you have? \$\endgroup\$ – W5VO Jul 23 '14 at 15:58
  • \$\begingroup\$ Yes, that's the exact Peltier. And I'm currently testing by hooking the Peltier up straight to a power supply. \$\endgroup\$ – M.Y. Jul 23 '14 at 16:12
  • \$\begingroup\$ The diodes are in parallel with the TEC? (one pointing one way and one the other?) Or something else. \$\endgroup\$ – George Herold Jul 23 '14 at 16:17
  • \$\begingroup\$ I believe so? i.imgur.com/msBRTf3.png \$\endgroup\$ – M.Y. Jul 23 '14 at 16:43
  • \$\begingroup\$ OK I can't see the TEC... but I assume it is at the end of the white and black wires. So those look to be in series, and both pointing the same way. I have no idea why there are two of them, but they will allow current to only flow in one direction. (That will allow either heating or cooling of you TEC depending on which side is connected to the heat sink. Perhaps the diodes are there just to add a bit more voltage drop so you don't damage the TEC by an over voltage. (?) \$\endgroup\$ – George Herold Jul 23 '14 at 17:49
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If I wanted to know the voltage across the Peltier, I would measure it directly with a multimeter.

The Peltier datasheet you linked a says it's rated for a maximum current of 3 A DC. If you push 3A DC through it, then the graph at the bottom of that one-page datasheet seems to indicate that the voltage across the Peltier should be about 0.8 V. That's the voltage directly across the Peltier at 3 A DC no matter what other stuff you have in series with it -- diodes, resistors, inductors, whatever.

From the description, it sounds like one Fairchild SB560 Schottky diodes b is in-line in each of the two wires that come out of this Peltier cooler. (I'm a little mystified what they are for, but I've already speculated on a few possibilities).

Like all diodes, it has a datasheet c. On the first page of the datasheet, it says each SP560 has a forward voltage drop ("VF") of 0.67 V at 5 A. But at the 3 A you're probably trying to use, the graph on page 2 of the datasheet shows it has a forward voltage somewhat less than 0.6 V -- maybe 0.56 V?

So I expect the total voltage across the entire series chain when 3 A is going through the chain -- two diodes at 0.56 V each and a Peltier at 0.8 V -- to be roughly 1.92 V.

Honestly, that's an unusual power supply voltage. The Peltiers I'm using this week are 12 V 5 A (similar to d e ); and 12 VDC power is more convenient than 1.92 V or 0.8 V.

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  • \$\begingroup\$ Good answer, just measure it! (Is this typical, someone who doesn't know how to measure something?) \$\endgroup\$ – George Herold Jul 24 '14 at 1:03
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    \$\begingroup\$ (input voltage -> voltmeter reading): 4V -> 2.55V, 3.5V -> 2.4V, 3.0V -> 2.1V, 2.5V -> 1.8V. Thanks! I'm wondering how the Peltier isn't frying when I'm supplying more than its Vmax of 0.8V... \$\endgroup\$ – M.Y. Jul 24 '14 at 14:28
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The diodes are there so you would not wire it wrong. When you invert the + and the - the heating side switches with the cooling side. The ceramic plates have a different structure for each side and if you invert the heating/cooling it would make them less resistant over time.

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