If I have a 2.4A 5V Max Output Battery Pack, and I buy a 12V, 60 Watt Peltier Element and a 5V .4A computer fan (leaving room for comfort, item is actually 5v .36A), could I run both off of that one battery pack?
For Further clarification, I will split the USB cable that powers both objects two ways (Y Split connection). As referenced by someone else, using Ohm's law a 12V, 5A 60Watt Peltier would run at 5V 2.1 Amps when plugged into a 5V power supply, and added with a 5V .4A computer fan would run at a total of 5v, 2.5A. The Peltier I am using is the TES1-12706 or TEC1-12706 (12v 60Watt)
Thank you so much, I mainly want to see if you guys think this would work, or I need to get a lower powered Peltier to be on the safe side, also I want you guys to tell me if that Ohm's Law conversion is correct.
Best Answer
Here is a much more readable datasheet for the TEC1-12706
What can be easily said is that you are unlikely to be able to run both the Peltier and a fan on a power supply that can only deliver 2.4 A. This will depend on how accurately the battery pack shuts off on overcurrent of course, but assuming it's accurately measuring the current it won't work. You need a power supply that will deliver at least 3 A if you don't want to regulate the voltage going to the Peltier.
The effective resistance of the Peltier module depends on temperature (both absolute and differential). The Peltier effective resistance for this module has a positive temperature coefficient, as the absolute temperature goes up, so does the effective resistance. The datasheet shows performance data using a constant current with variable voltage with the Hot side at a specified temperature.
You are wanting to feed the module with a constant 5 V ...so I marked it on the performance curve.
This graph shows T(hot) as 27 degC and we can consider this to be ambient temperature for this exercise.
Now consider the points I marked:
When you first turn the unit on there is no temperature difference between the hot side and the cold side, both are at 27 DegC (a balmy 80 Deg F). From the graph we can derive that you may draw about 2.5 Amps (halfway between the green and purple line). The effective resistance is about 2 Ohms.
At point 2, as the temperature on the cold side is reduced by 20 degC (and assuming our fan on the hot side keeps it at ambient, 27 DegC), the cold side temperature is about 7 DegC ...this is 20 DegC less than the hot side. The current at this point is about 2 Amps. The effective resistance about 2.5 Ohms.
At point 3, the cold side is at about -3 DegC (30 DegC less than the hot side). The current is about 1.8 Amps. The effective resistance about 2.8 Ohms.
From the above you can see that the effective resistance varies with temperature, and how great a differential temperature can be maintained depends on the thermal losses in your mechanical design.
It's unlikely that you will be able to maintain the hot side at 27 DegC. If the ambient temperature varies and the hot side temperature varies then you have to work out what the effective resistance might be from the curves given.
Notice in the datasheet that it shows a separate performance curve where the hot side is at 50 DegC (about as hot as the hand can tolerate). Notice also that the current flows have gone down (with a constant 5 V input) compared to those at 27 DegC.
The bad news is that as the ambient temperature drops to say 15.5 DegC (60 F) you can expect the current when you first turn the unit on (hot = cold side temp) to be MORE than at 27 DegC. Projecting the data given, I'd expect the turn on current to be very close to 3 A @ 15.5 DegC (60 DegF) ambient.
The only really viable solution is to measure the current and adjust the voltage to hold the current maxima at your desired level.
Datasheets for modules
Thermonamic have a full set of datasheets and the TEC1-12704 does seem closer to a reasonable choice for you.