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I would like to cool a laser diode, which has maximum thermal power of 1 W (actually according to datasheet should not be higher than 830 mW). The operating temperature range is (25-30) Celsius.

I think that the small 15x15 Peltier 2 V should be sufficient.

I have good contact of the laser diode with an aluminum heatsink. I will also use a heatsink with the warm side of the Peltier.

I have a feeling that the power consumption of the Peltier wont be high. So my question is what could be the estimated power consumption of the 15x15 2V Peltier.

Thanks in advance.

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  • \$\begingroup\$ Have you tried looking at the datasheets here: mouser.ca/Thermal-Management/Thermoelectric/… \$\endgroup\$ – DKNguyen Mar 2 at 20:36
  • \$\begingroup\$ that depends on the Peltier element. The voltage doesn't tell us how much current it draws – and it's power is voltage × current, assuming you can keep it at the specified temperature. Generally, Peltier elements make the energy transport problem easier, but the cooling problem harder, due to their low efficiency; my guess you're not actually solving anything by using one! \$\endgroup\$ – Marcus Müller Mar 2 at 22:22
  • \$\begingroup\$ Have you read much material on peltier coolers yet? Power consumption can be quite significant, and they suffer from multiple fully different types of losses. Your overall setup will determine how hard the peltier has to work for the size/quality that it is, and even with a good setup they are not super efficient coolers. At any rate, the power the peltier uses to move 1W of heat plus its own losses could be significant, or with a bad setup it could be more than the 1W load. Would you like to know more? \$\endgroup\$ – K H Mar 2 at 22:22
  • \$\begingroup\$ @MarcusMüller I'm probably fond of peltiers, but I think a laser diode is generally a good application for them, although in this case the load is small. Heavy concentrated thermal load on a device that sheds heat poorly and suffers significantly in performance and lifetime from bad thermals. His desired operating temperature is room temperature, which is quite hard to achieve without active cooling). I don't think he should be greatly discouraged, as long as he is not discouraged by the fact that there are many complications to using a peltier. \$\endgroup\$ – K H Mar 2 at 22:28
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    \$\begingroup\$ Looking at this datasheet for a cp30138h, as long as you keep your temperature differential less than 20\$^\circ\$C(use a large enough heat sink), you can pump 1W from the cold side at a drive of 0.6A and 1V, so 0.6W, and if you keep your temperature differential down to 2\$^\circ\$C, you can pump about 2W of heat at 0.6A and 0.7V, so about 0.42W, so the question becomes partly how big of a heatsink can you tolerate? And... I've started calculating and it appears at face value a 40mmx40mmx20mm extruded plate fin heat sink might do. \$\endgroup\$ – K H Mar 3 at 5:29
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The challenge to get Qmax is to have an infinite heatsink attached to the Peltier junction.

If good thermal resistance and high velocity 35mm fan air over heatsink to push heat away then 15x15 mm size may do with Rja <=1'C/W or 10'C T rise with 10W while a CPU heatsink & fan <0.1'C/W

Qmax: 3.4 W 
Delta Tmax: + 87 C 
Maximum Current: 2.1 A 
Maximum Voltage: 3.8 V   (means 8 Watts applied plus W Laser  9 W)
Length: 15 mm 
Width: 15 mm
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