Overpowering a TEC

Krow said:

Just a quick Side Question:

Is Vaseline pretty much the same thing as Die Electric grease or is there a benefit to using the latter over it.

Click to expand...

Vaseline sort of works, but I don't think it's the best if the TECs hotside heats up too much.
Please note, you are only suppose to use it to seal the TECs sides if you have an open TEC device. If it is already sealed then you are fine.

Combining the 12v and 5 v rail will give you 7v not 17v. (differnce between the two)

Combining 12v and -5v rail will give you 17v.

Also..


NORMALLY Increasing voltage reduces the amount of current needed to perform hte same task. However a peltier is a resistive load device. Increasing the voltage means increasing the current as well at least from what I can surmise from some googling.

TEC voltage's specification provided by the factory is 'usually' the sweet spot, especially at the voltage, so it is not recommended to increase the voltage. If its not toasted, I believe you will get a really bad efficiency (assuming with the right pressure, the right dissipator and the right load).

Regarding measuring the current consumed by the TEC, use this method HERE.

Wow guys thanks for all the info! I kinda grasp the relationship between Voltage and Current. And that method for using the TEC is awesome, but you think it woould just be best to get the TEC as close to 12V as possible?. So why a need for resistors? Just get some higher guage wire so as not to make a resistor out of it.

Neur0mancer said:

Combining the 12v and 5 v rail will give you 7v not 17v. (differnce between the two)

Combining 12v and -5v rail will give you 17v.

Also..


NORMALLY Increasing voltage reduces the amount of current needed to perform hte same task. However a peltier is a resistive load device. Increasing the voltage means increasing the current as well at least from what I can surmise from some googling.

Click to expand...

Here are the calculations to find out what happens

Power=volts x current

Voltage = current x resistance

given a 400w resistive load at 12v the resistance would be

c=v/r

so

p=v*v/r or r=v^2/p or r=12^2/400=.36

Given that the resistance will pretty much stay the same for what we are dealing with

increaseing the voltage to 17v will give you p=17^2/.36=802W and c=17/.36= 47 amps or much higher than what you wanted.

Given a 5v voltage you will find that p=5^2/.36=69.44W and c=5/.36=13.9a.

This is one of the reasons why it makes sense to buy the pelt with the highest voltage rating so you can and undervolt it whenever possible. Since no one really does that much with pelts anymore if you keep your eyes open you can find some great deals on 24v PSUs for running them.