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Peltier voltages? Is there a 'max'

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Killa-X

Fri Apr 17 2009, 01:36AM

Registered Member #1643 Joined: Mon Aug 18 2008, 06:10PM
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Posts: 1039

say we can do 20V 4A for sample, and 10V 4A. I found the 20V 4A is a lot colder than the 10V 4A, but both ways the heatsink doesn't get very hot. This is because its a 12-15V 9A peltier.

Do peltiers have a max volt/amps? Or is the max when it over heats and it burns the case? Or does it simply over-ride the cold into hot.. I want to know if theres a true limit to a peltier as long as it doesn't over heat on my heatsink. My big 4X3X2 inch heatsink doesn't even get that warm with 20V ~4A

Proud Mary

Fri Apr 17 2009, 01:57AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

All thermoelectric devices have a figure of merit which depends upon the Seebeck Coefficient, and the thermal and electrical conductivities of the materials used.

To answer your question in a more general way, Peltier-Seebeck devices are sometimes used in superconductive experiments at very low temperatures, though not perhaps the basic devices you can buy for computer cooling!

Killa-X

Fri Apr 17 2009, 02:05AM

Registered Member #1643 Joined: Mon Aug 18 2008, 06:10PM
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I just got these to mess with.. I was just wondering if it was bad to use 20V on a 12V 15V MAX peltier

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Fri Apr 17 2009, 03:22AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

It should be fine, it might not last as long as it would at 15v (especially with lots of thermal cycles) due to increased thermal stresses, but for most hobby applications I would keep turning up the power until it doesn't get any colder

Killa-X

Fri Apr 17 2009, 03:39AM

Registered Member #1643 Joined: Mon Aug 18 2008, 06:10PM
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the issue is that both my 10 & 20v on that transformer is so lower amp that it wont even ice water (most peltiers can frost) . mines max rated 15v 9A and it gets cold but no frost. i think 4AA is almost cold equal to my 10v if not colder >_>. id just like to see frost on one sometime. ive seen small frosted ones before with no heat sinks oddly as that was....

Dr. Slack

Fri Apr 17 2009, 07:28AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

A good model for the thermal side of a Peltier is a thermal current source from cold to hot proportional to the current flowing, in parallel with the fixed thermal conductivity of the device from hot to cold, with a resistive heat generator proportional to current squared flowing into both hot and cold (the I2R losses of the device's intrinsic resistance). Figure of merit engineering on better peltier matierials is concentrated today on reducing the ratio of thermal to electrical conductivity

At any given bias, there will be a load line - type behaviour, with exactly the same maths as for a resistor from a battery. The temperature difference will be max when pumping no net heat (thermal open circuit if you will), the heat pumped will be max when both junctions are at the same temperature (thermal short circuit!), neither of these extremes is actually useful. The max "pumping power", or "useful performance" in watts.degrees (WK) will be when pumping half this maximum flow against half this maximum temperature difference, WK(useful)max = 0.25Wmax.Kmax. There is a small complication in that the temerpature difference actually generates a reverse voltage which changes the effective input voltage, but Peltiers are so inefficient that it doesn't change the numbers much, and doesn't change the conclusions at all.

At zero bias, WK = zero. As bias increases, the pumping power increases quickly, the losses as the square of a small current increases slowly, so WKmax increases. It passes through a maximum at some current. As bias increases further, the self-heating now increase quickly as the square of a large current and comes to dominate the pumping capacity, and the WK falls again.

Given that manufacturers are trying to present their products in the best light, at what point on that curve do you think they specify their wares?

scroll down for answer

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Those who answered "at the bias that gives the maximum WK, duh!", get a mark.

So yes, you can operate a Peltier at higher bias, but no, there is absolutely no point unless you want it to behave as a very expensive resistive heater. The V and I that the data sheet gives is at the best output of the device.

Steve Conner

Fri Apr 17 2009, 10:32AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I used to design TEC controllers for cooling laser diodes. TECs are often specified by two parameters:

Maximum Delta-T, which occurs at zero heat pumping power. (The TEC is using all its pumping power to overcome its own thermal leakage from hot side to cold, so there is no power left to transport any external heat load.)

Maximum heat pumping power, which occurs at zero Delta-T. (both sides at the same temperature)

Neither operating point is any practical use, but they are the two end points of a load line. They're also the two most impressive figures you can get from the device for marketing purposes.

If you overdrive a TEC too much, it will go into thermal runaway. Both sides get sizzling hot and your laser diode gets cooked.

Proud Mary

Fri Apr 17 2009, 11:12AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Steve McConner wrote ...

If you overdrive a TEC too much, it will go into thermal runaway. Both sides get sizzling hot and your laser diode gets cooked.

There you go, Killa! More is not always better!

Dr. Slack

Fri Apr 17 2009, 12:00PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

If you overdrive a TEC too much, it will go into thermal runaway

Is that a real runaway Steve, as in apply constant volts (or current), and the dissipation increases changing the temperature which changes the effective ressitance which furrther increases the dissipation?

Or is it an engineer-induced runaway, as in

while "cold junction not cold enough"
         increase the applied power
end while

Steve Conner

Fri Apr 17 2009, 04:35PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I'm not sure to be honest, whether a TEC will run away if it's not inside a control loop.

All I know is that as you increase the TEC current, the cold side gets colder and the hot side gets hotter at first. But beyond a certain current level, the cold side stops getting colder and starts getting hotter, while the hot side gets "frickin' hot".

I believe it's because the TEC's internal heating (which goes to both sides equally, hence must be pumped back from the cold side) is proportional to the square of current, whereas the heat pumping power is just linearly proportional to current. So if you turn the current high enough, the maximum Delta-T stops increasing and gets smaller, as the TEC can't pump its own I2R losses away from the cold side.

Also, in a practical application, when the TEC is on a heatsink, the heatsink temperature will increase with drive, and that increases the Delta-T and hence heat leakage to the cold side, if a controller is holding the cold side temperature constant.

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