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I am building a DSLR cooler (something I actually started last year, and ended up shelving when winter hit, and which I've picked up again for this coming summer), and I'm trying to regulate the power to one or two TECs via PWM outputs on an Arduino. I'll be using a few sensor inputs to the arduino as well as a pot to control the PWM level to the TEC(s).

The TECs require 12V power, however depending on the exact Arduino unit I end up using, I may be limited to 5V for the Arduino (i.e. Arduino Mini). So I've put together a circuit diagram that uses an LM7805 to regulate the voltage down. I am trying to build a pretty compact cooler, and while I am not going to be drawing a lot of amperage (130mA tops I think, given the various sensors and LEDs I'm going to be using, and accounting for the 12-15mA the Arduino itself will draw) , I am worried that a 7V voltage drop in the regulator will produce a lot of heat. It's just shy of 1W energy dissipated in the regulator itself.

The datasheet for the LM7800 series regulators states that the maximum thermal resistance junction-ambient for a TO-220 package is 50°C/W (interestingly, a TO-3, which seems a good deal larger in terms of footprint, is only 35°C/W). Assuming I experience the maximum, that would mean a 1W dissipation would increase the temperature of the unit by 50°C? If I'm running at 25°C, then I'd have a 75°C regulator temperature?

Or is there a lower actual operational temperature, below 50°C, that I should take into account? I've scoured various forums, mostly audio forums, and I've seen members throw out junction-ambient temps of 10°C or 25°C for TO-220 packages before, and I'm curious why. It seems like a 50°C increase for just 1W thermal dissipation is a lot, but maybe that's just the reality of things.

Thanks!

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  • \$\begingroup\$ In comparison, the IR datasheet for the IRF9540N gives 62degC/W. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 29 '15 at 19:20
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    \$\begingroup\$ How about using a cheap buck converter? An SMPS, switched mode power supply. Ebay and Amazon have loads of these modules and convert your 12V to 5V probably at a lower price than you can buy the heat sink for your 7800 regulator for. \$\endgroup\$ – jippie Dec 29 '15 at 19:24
  • \$\begingroup\$ Watch out for cheep smps and noise though if your doing analog measurements. \$\endgroup\$ – MadHatter Dec 29 '15 at 19:29
  • \$\begingroup\$ DSLR = digital single lense reflex (camera)? TEC = ? \$\endgroup\$ – Transistor Dec 29 '15 at 20:07
  • \$\begingroup\$ @transistor a Peltier element: ThermoElectric Cooler \$\endgroup\$ – jippie Dec 29 '15 at 21:11
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It seems like a 50°C increase for just 1W thermal dissipation is a lot

This is accurate for many TO-220's, a TO-3 would only rise 35°C for 1W of thermal dissipation because it has a much larger body that acts like more of a heatsink. Simply add a heatsink to keep regulator 'cool'.

More help on this is addressed well here.

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  • \$\begingroup\$ So there really isn't a "typical" temperature, when these things dissipate, they just dissipate at the max? This is going to be a part of a cooler, and that cooler will use a fairly large 98mmx40mm heatsink, cooled by two fans. The electronics, at least in theory, will go along one side of the cooler. So I could just bolt the regulator, and probably the MOSFETs I'm using to deliver 12V power to the TECs, right to that same heatsink. Any reason not to do that? \$\endgroup\$ – jrista Dec 30 '15 at 1:50
  • \$\begingroup\$ I guess I have the same question about the IRLB8748PbF MOSFET I'm using to deliver the full 12V power to the TECs. The junction-ambient for those is 62°C. The package, also TO-220, is rated for 30V/78A. I'm probably going to be using them at 12V/5.8A...is the package still going to heat up to 62°C, or is there a way to calculate actual operating temperature at the power level I'll be using it at? \$\endgroup\$ – jrista Dec 30 '15 at 1:56
  • \$\begingroup\$ @jrista if my math is correct, you should be fine not heatsinking your FET. It has an RdsON of 4.8mOhm. Take this times your current -> 4.8mOhm * 5.8A = 27.84mV. V*I=P, right? so 5.8A*27.84mV=161.47mW. \$\endgroup\$ – SolderFumes Dec 30 '15 at 14:44
  • \$\begingroup\$ ...so with 62°C/W j-a, then 62*0.16147= ~10°C rise above whatever its ambient temp is, if my thinking is correct. (Great FET BTW, a 78A switch for ~$1!) \$\endgroup\$ – SolderFumes Dec 30 '15 at 14:52
  • \$\begingroup\$ Thanks Solder, I calculated the same. I just wasn't sure if multiplying the j-a by the power dissipation was valid or not. I agree, the FET is nice! Although I got em for a little over $2 each (although I think most of the cost was shipping, actually.) \$\endgroup\$ – jrista Dec 30 '15 at 18:05

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