Peltier PSU haxoring
 

I had an idea ... if I was to take a pelt rated at 15.2V/24A and hook it to a 15V/20A PSU and run a .2 Ohm resistor in series between the neg terminal and the pelt, I should mod the power going to the pelt to 11V/20A.

r1 = (15V-11V)/20A = .2 ohm

Would something like that allow the pelt to run at approx. 75% an not blow the PSU? Just curious, cos I think I need one in my system.

Okay, calcs are done.

R = (15.2V - [15.2V*.75])/20A = .19 ohm

15.2V/24A -> 11.4V/18A @ 75% power (max efficiency point)

Setup would be like the pic below. Does anybody know of a reason this wouldn't work on the 15V/20A PSU I mentioned? It seems to be cut and dry, but it's been a long time since I've done this type of stuff. This forum seems to be full of engineer types or people so inclined ... lets wrap our head around this.

Make sure the resistor is capable of handling that amount of wattage.

Small resistors only handle 1/4 watt.

yea looks about right, you'll need a high power resistor(or multiple resistors in paralell) since it'll be dissapating around 50-60w.

10x 2ohm 5w ceramic resistors would do nicely.

Um, high power resistors? I suppose that makes sense. I'm trying to keep this as simple as possible, so I'd like to avoid a line of resistors in parallel. Where can you get the high power variety? I've only dealt with the little ones.

5w and 10w should be readily available from most electronics stores, larger wattage ones(50-250w) I've only seen in 4 and 8ohm versions,mainly used for testing audio amplifiers.

in this case 10x 2ohm 5w or 5x 1ohm 10w *is* the simple and easy option :cool:

oh paralell is splitting the "electron flow" so it goes through all the resistors at the same time, not in a long line.

Here's some math:

P=VI (power=voltage * amps)

V=RI (voltage=resistance * amps)

Knowing these two equations is the fundamentals of electronics.

You can then extrapolate P=RI^2. Since you know that R=0.19, and that I=20, P will be 76 Watts. Most resistors are available in 1/4 (most common) and 1/2 Watt (common), so you'll have to do a little digging, and I can tell you off hand, it's going to be hard to find. I'd start with suppliers of automotive audio equipment.

Allelectronics sells a 1 Ohm, 75 Watt, flat ceramic type resistor, for $3.50 .

You might consider figuring a way to alter the PSU instead, it'd probably be cheaper than trying to dissipate 76 Watts.

Here's some more math for parallel resistors:

1/R(total) = (1/R1) + ... (1/Rn)

The guy above was right in saying that .2 Ohms could be fudged with 10x 2 ohm resistors in parallel. I'll check that site for that size.

BTW, I don't want to crack the PSU ... too many things could go wrong ... I'd rather make an "expansion card" for it. If I remember how big that size of resistor is, it'll be a good size card, too (maybe slip it in a pci slot? Naw, too cheesy ... ). At any rate, it would be an easy board to build.

Then your only option is that 1.0 Ohm, 75 Watt resistor, qty 5, all in parallel. at 3.50 each, that's $17.50, plus tax and shipping.

Do the math, you won't find any other combination of resistors that will work, unless you get into a mix of parallel and serial.

I actually found some 3 Ohm 10W resistors at MPJA (12518 RS) that might work. 16 (!!!) in parallel would give 16/3 = 5.3 or .188 Ohms, giving me darn close to a .19 Ohm resistance, and I'm getting 160W of heat dispersal (overkill, but that wont hurt) at it'll only cost $5. Think that'll do it? Oh, and these are designed for high heat use in power supplies.

http://www.mpja.com/product.asp?product=12518+RS

That would give me 11.44V/18.1A on the peltier, at 75.3% capacity, btw. I think that would be managable on the PSU (in my lay opinion, heh). Also under consideration is to put a high wattage potentiometer on the card and slip that whole assembly in a drive bay on the front of my box with the dial sticking out. That would allow me to up the resistance (lower the pelt power) during the winter when I don't need -20C temps (it gets 10-15F here on a regular basis for a few months per year) and want to save electricity. If all this works, I just have to figure out a way to waterproof my socket. With my current block, that might be a biznatch.

Yes, that'll work!

No it won't give you 18.1 amps on the pelt. You'll still have over 20 amps. The amps don't go down, only the voltage is split.

Since the voltage is split, the pelt also, will react differently, because its resistive quality is not linear with voltage.

What I mean is that max capacity of the pelt will be proportianally lowered with the voltage ... it won't try to pull more amps than it can use. My whole goal with this project was to lower the voltage of the loop forcing the pelt to draw lower amps. (Args, I don't know how to express this right). I want the max amp load of the pelt to be lower than the max output of the PSU. I guess that'll work.

BTW, do you know of anyplace that carries high power pots? Most places I have looked only carry the weenie .5W variety.

I see what you're getting at, but you should note:
The PSU will always supply the same voltage, so whatever is not going to the pelt, goes to the resistor. That being said, if you lower the voltage to the pelt, it will draw less amps (so I was wrong) but what I was trying to say is that it's not linear, in that if you lower the voltage to the pelt by 25%, the amp draw from the pelt will NOT be 25%.

A power pot huh? Try searching for a rheostat.
I found something here

I think I read on tetech.com or its affiliate site that it is directly proportional (V/A) when raising or lowering voltage .... If you have any ideas of a way to ensure that my amp draw is lower than 20 amps, I'd be happy to take it under consideration. I just don't need caps blowing in my machine (heh) and then my cpu/board burning cos my tec goes offline. I want to cool it, not cook it, and I just want to find a safe, cost effective way to do it.