Ok, so for the past few days, i've been getting more and more interested in Peltier Modules, and i've finally decided to make a portable mini fridge with one. After MANY long hours of research, part selection, crying, and cussing, I finally came up with a circuit diagram that seemed pretty solid.

That is, until I showed it to my chemistry teacher. (He is VERY experienced with computing and circuits)

He says that if I hook up this circuit, it will most likely overload and die out, but he doesn't know why, because his teachers didn't explain it in great detail. He said that I should hook up a resistor and a fuse, but didn't tell me the specifics: how many Ohms, voltage, etc...

Keep in mind, this isn't my first time playing around with circuits. I have grasped the basics and I understand some stuff. However, this is my first time messing around with batteries in a major way. I.E: I have used batteries, like 2 AA batteries for my small breadboard projects; NEVER have I used a 12V, 20 AMPHOUR battery before! So when it comes to the big stuff, i'm a total noob.

I would appreciate it if someone can help me fix my circuit before I turn it on for real! I really would not like to throw away $160 in parts!

A picture of my diagram is below: NOTE: Some stuff has changed, like the temperature sensor now being one unit, not a probe and main unit.



If anyone can provide me a diagram or design schematic of my fridge that works without overloading or not working in general, you will receive my eternal respect and gratitude!!!

Circuit Diagram

  • 3
    \$\begingroup\$ First study what happens when you connect devices in series vs in parallel. Hint: You should not be connecting your three devices in series. \$\endgroup\$
    – The Photon
    Mar 27 '17 at 18:15
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    \$\begingroup\$ Fan2 should be in parallel with the 12V unit containing the Peltier module and Fan1. You should also determine what current Fan2 needs and add that to the 5A for the other circuitry, then add a fuse appropriately rated. Other than that I don't see any major issues, although its possible I'm missing something. \$\endgroup\$
    – Redja
    Mar 27 '17 at 18:20
  • \$\begingroup\$ TEC element may need 12V at 5A direct to Battery. For variable cooling and reliable efficient operation, a linear control supply is much more efficient and not PWM. Ideally an adjustable current LED driver. \$\endgroup\$ Mar 27 '17 at 22:13
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    \$\begingroup\$ Not related to your circuit, but I hope you have a big heat sink on the hot side of your TEC. \$\endgroup\$ Mar 28 '17 at 18:34
  • \$\begingroup\$ @GeorgeHerold Yeah, that's already taken care of. It's a all-in-one Peltier TEC Assembly. Certified not to overheat with the fan and heatsink that's already built in. \$\endgroup\$ Mar 28 '17 at 19:06

While @peufeu has given a perfectly good answer, I'll give you my own take on it.

I think you may have misunderstood your chemistry teacher's answer a bit. Your circuit as shown will basically have both fans being driven at half their rated voltage, and almost no voltage across the TEC. So you'll get the fans blowing slowly and no cooling. You won't actually damage anything, but it won't exactly be useful, either.

Assuming your temperature switch is a simple mechanical on/off switch which turns on for high temperature, your circuit should look like


simulate this circuit – Schematic created using CircuitLab

You'll note that this does not incorporate some of peufeu's safeguards, but it may serve you as a starter, just to assure yourself that the idea is sound. It can also be used to get an idea if your TEC is up to the task, and if your heat sink/fans are properly functioning.

Wiring should not be that big an issue. 5 amps is not a great deal, and 20 gauge wire should do you fine. Also, the TEC is the big current hog, and the fans will run on much less (which is why they would limit the TEC current in your original design).

Over-discharging a lead-acid battery is not a good idea, but one or two incidents probably won't be an issue. It's not uncommon, for instance, for car batteries to get drawn down, and they usually are usable afterwords for quite some time. That's not to say that it's a good idea, but it's not as catastrophic as peufeu may have led you to believe. It's worth pointing out that discharging to less than 50% of capacity will typically limit battery lifetime to several hundred charge/discharge cycles, as opposed to several thousand for 10% levels.

Precision control of the voltage is not necessary. A fully-charged lead-acid battery under load will drop to about 12 volts almost instantly. For the sort of load you're proposing, you can expect the battery to be at about 11.5 volts when the battery is 50% discharged, and that is an excellent target as a cutoff point. Note that this cutoff is 10 amp-hours, and at 5 amps (more or less) this will only take 2 hours. Even if you're willing to run the battery flat, you'll only get about 4 hours of cooling.

If you experiment with your cooler, you'll see that you want to get your contents cold BEFORE you put them in the cooler. It simply will not cool things quickly. What it will do is keep them cool.

  • \$\begingroup\$ The switch actually closes the circuit when it gets too warm: about ~43-45°F, which activates the TEC so it cools the inside back down to 34°F, which is the temperature at when the switch opens the circuit, shutting everything off. It's sorta a low-power mode: keeps the battery from running itself down in 4 or so hours. Ideally, if the refrigerator only ran for 30 minutes every hour, then theoretically the battery would last for 8 hours on a single full charge. Of course, this is a lead acid battery though; you should never run it lower than 50% capacity, EVER. \$\endgroup\$ Mar 28 '17 at 19:18
  • \$\begingroup\$ One more thing, though: According to your circuit example, are you saying that the battery will automagically limit the current to each component? Or will it really overload everything? \$\endgroup\$ Mar 28 '17 at 19:24
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    \$\begingroup\$ @gnimmargorP - Oops. You're right about the switch description. I've edited. Thanks. Pesky brain-farts. And you are correct about battery life - if the cooler contents are already cool, and the TEC doesn't have to run continuously to maintain temperature, the battery will last longer. As to discharging below 50%, the dangers are, in my opinion, overstated. A single complete discharge of a lead-acid battery will not, in my experience, normally kill the battery. And I've done it repeatedly. And yes, if the fan and TEC are rated for 12 volts you do not need to do anything to limit current. \$\endgroup\$ Mar 28 '17 at 21:26

Wire your loads in parallel instead of in series.

Make sure your temperature switch can handle the current required.


Batteries are excellent for starting fires. A 12V lead acid battery will happily put out enough current to heat your wires red hot, melt them, and set everything on fire.

Add an overtemperature safety. When one fan fails, or someone covers the "do not cover" heat sink, or dust accumulates, etc... the Peltier will overheat and destroy itself. It is not cheap.

Add an undervoltage lockout. Discharging lead-acid batteries too deep will very quickly render them unusable. Check minimum allowable voltage for your battery.

Now, you have 3 inputs to your switch:

  • Fridge temperature: run
  • Peltier too hot: stop
  • Battery low: stop

It would be a bit complex to wire all this with relays, so I suggest keeping your mechanical temperature switch (if you already bought it...) and wire a dead simple safety for the rest: a low-RdsON MOSFET, a LM339 comparator, TL431 for low-battery check, and your choice of thermistor.

Test the setup using a current-limited bench power supply, not your beefy battery.

EDIT: I upvoted your question to compensate for the -1 you had because, although you clearly gonna need lots of help, you took the time to explain your problem clearly...

  • \$\begingroup\$ I really like that you gave a very constructive answer. Newcomers sometimes get some tough love when asking questions in here :) \$\endgroup\$ Mar 27 '17 at 20:45
  • \$\begingroup\$ Would using a resistor rated @ 60W, 2.7Ohms limit the current to a safe level and, more importantly, would it overheat the battery? \$\endgroup\$ Mar 27 '17 at 22:23
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    \$\begingroup\$ It's OK for testing, put it in series with your battery. Remove it while using your fridge of course! You can also use a 12V 100W lightbulb (or several smaller ones in parallel) \$\endgroup\$
    – bobflux
    Mar 27 '17 at 22:32
  • \$\begingroup\$ So, ok. Is there really any way to get a steady 12V, 5A without a horrifying waste of energy in the process? \$\endgroup\$ Mar 28 '17 at 0:00
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    \$\begingroup\$ @gnimmargorP Some other things you need to be aware of. Despite their function, peltier modules do NOT LIKE to be thermally cycled very often. If that thermal switch is turning on and off every few seconds the module will not last very long. Peltier modules should never be allowed to get wet and sealing them is not so simple. The fan control needs to be separate from the peltier control. Turning the fan off with the peltier will cause it to overheat due to thermal inertia. GOOD thermal contact between the entire surfaces of module and heat-sinks is IMPERATIVE. \$\endgroup\$
    – Trevor_G
    Mar 28 '17 at 18:25

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