How to power a 12V6A peltier device with a 14.4V battery?

Question

How to power a 12V6A peltier device with a 14.4V battery.

My son is working on a school science project and unfortunately I have very little knowledge on electrical engineering. Please be patient with my questions.

We need to power a 12V6A peltier device to show the cool down process. We already have the heat sink and peltier theory in place, so this question is more related to the circuit handling the peltier.

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1st Question

1st Test: We performed a simple test using a 6V battery pack (4 AA's) using a breadboard. We applied the lowest resistor we have of 10 ohms (to get as much from the battery pack and still control the current), in an arrangement + >> R10 >> Pelt >> -.

Results: The peltier worked at some extent, but the resistor got really hot and started burning up.

Q1.1?? Why does a small 10 ohms resistor get really hot (to the point of smelling and turning black) when supplying just 6V (4 AA's pack)?

Q1.2?? Does it make a difference to put the resistor closer to the source or closer to the ground? Our thought now is if we put the resistor close to the ground, it will handle less heat because the current is passing through the peltier first and thus the peltier consuming it. If we put it before the peliter, all the current passes through the resistor first, heating up.

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2nd Question

We are going to upgrade the source to a rechargeable 14.4V 6A (that's what we have at our disposal), so we can hopefully run the test for 1 hour.

Using what we learned from ohms law, we need to drop the current from 14.4V to 12V, and limit the amps to 5, so we need a resistor of 0.48 (0.48r = (14.4v - 12v) / 5i).

Q2.1?? What kind of resistor do I need to get this right without burning the resistor?

Q2.2?? With such small resistance, does it even make sense to put a resistor? Should we just connect the peltier directly to the battery source and assume it can handle the extra voltage and current?

Thanks in advance for the help.

Answer 1

Q1.1: The smaller the resistance, the higher the current passing through it. The resistor heated up so much, because the current was around I=6/10 = 0.6A P = VI = 6*0.6 = 3.6W which is probably much higher than the (assumed 1/4 or 1/2W) rating of the resistor. This resulted in the overheating and burning. (The calculation is estimate. I'm not sure what is the voltage drop on the peltier if there is any. The calculation assumes it is a short circuit, but probably it has some resistance.)

Q1.2: It doesn't make any difference. In a circuit where all components are in series, the passing current in all components will be the same. Basically all current goes into the circuit will come out at the end. The current will not be consumed. (Kirchhoff's current law).

Q2: Assuming that the peltier has its internal resistance, it should be generally no problem to connect it directly. However, if too much current flows through it, the peltier's hot side will heat up quickly faster than the heat sink can cool it and will heat up eventually the cool side as well. Should that happen, you will either need to control the current, reducing it or improve the cooling method e.g. increase the heat sink, put a fan on it or use water cooling.

To control the current you may want to either introduce a big resistor, but given the relatively high current, it should be a big power resistor which are not too cheap, or you should get a power supply which allows you to control the output current.

Answer 2

I would suggest that you don't use any resistor. The Peltier element can be connected directly to the 14.4 volt battery. 12 volt Peltiers are usually meant to be used in car refrigerators and the car battery runs at 14.4 volts when the car is running. Car and motorcycle batteries work best with the Peltier element, because they can deliver the 6 amperes the element wants to take with no problems.

When discharging lead-acid batteries, note that they suffer (age) if you take more than half of their capacity before charging them again.