I currently have a 12v 6A Peltier bolted to 2 heatsinks, with 2 separate fans. The old PSU I used for this system went kaput, so I would like to find a temporary replacement before I buy a new power supply. I dug around and found an old laptop brick, supplying 19V at 3.42A. I was wondering since the brick supplies a lower power than the max rating of the Peltier, is it ok if I use it instead? Or will the overvolting cause it damage? Edit: I also have an Arduino board that can output 5V, 20-50 mA. Is this a better choice?
If the peltier module draws 6 Amps at 12 volts, it will attempt to draw significantly more at 19 volts. Since the 19 volt supply is only rated to deliver 3.4 Amp, the supply will likely be damaged.
EDIT: If the supply has overcurrent protection, it will probably shut down, without damage, when the Peltier module demands excessive current. In any case, it will not provide full power to the Peltier, but is unlikely to damage it.
Edit 2: You require a power supply that will deliver at least the 6 Amps that the Peliter requires, at about 12 volts (+/- a volt should be OK). Your Arduino that can only supply 50 mA at 5 volts is definitely not adequate.
The 19.4V supply will probably "sag" until it's maximum output is reached.
This MAY cause overheating or shutdown or damage. Or possibly none of these.
A solution which will treat the supply well is to provide a series resistor such that the supply is maintained within specification. eg such that I = 3.4A and Vp < 12V.
This can easily be determined by experiment (and/or reference to the Peltiers datasheet).
eg if Peltier V/I is linear then Vp at 3.4A is about 12V x 3.4/6 = 6.8V.
So series R = Vr/I = (19.4-6.8)/3.4 = 12.6 / 3.4 = 3.7 Ohms.
Power in R = PR = V x I = 12.6V x 3.4A ~= 43W.
While several x low cost 10W aircooled ceramic R's could be used, a 12V 40W light-bulb sounds about right - say 2 x automobile 12V, 21W "festoon" bulbs in parallel.
Try that or similar and then measure Vpeltier, Vsupply and Isupply. Adding a 1 Ohm 10W resistor allows easy current measurement. (I = V_1_ohm / 1_Ohm)