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### Topic: Need an advice in choosing power source (Read 2225 times)previous topic - next topic

#### liorcohen5

##### Nov 10, 2013, 04:09 pm
Im using a TEC1-12706 Peltier thermo-electric plate, and Im wonering what power source I could use to operate it.
is it possible to use a 9v batteries with it?
is it possible to make it portable using any kind of normal size batteries?
here is the documents of the TEC1-12706:
http://www.hebeiltd.com.cn/peltier.datasheet/TEC1-12706.pdf
I guess that if im operating the device using 9v I'd need a 4.5A current. could a regular battery supply such a current?
thanks!

#### retrolefty

#1
##### Nov 10, 2013, 05:14 pm

Im using a TEC1-12706 Peltier thermo-electric plate, and Im wonering what power source I could use to operate it.
is it possible to use a 9v batteries with it?
is it possible to make it portable using any kind of normal size batteries?
here is the documents of the TEC1-12706:
http://www.hebeiltd.com.cn/peltier.datasheet/TEC1-12706.pdf
I guess that if im operating the device using 9v I'd need a 4.5A current. could a regular battery supply such a current?
thanks!

Peltier devices are very current hungry and a poor choice for a battery powered arduino project. There are Li-Po batteries that could supply the current as well as larger lead acid gel cells, but again requiring to be portable battery powered is a poor economic design. Any project that requires appreciable heating is a poor choice for portable battery power.

Lefty

#### jremington

#2
##### Nov 10, 2013, 08:31 pm
To roughly estimate the battery life in hours for a project, divide the capacity of the battery (which is usually given in ampere-hours Ah or milliampere-hours mAh) by the current consumption of the entire circuit in amperes or milliamperes respectively.

For example, if you have a 4.5 Ah lead-acid battery, and your Peltier device consumes 4.5 amperes, the battery would last roughly an hour.

Unfortunately there are lots of rules about charging and discharging different types of batteries that complicate this rough estimate considerably.

#### MarkT

#3
##### Nov 10, 2013, 09:26 pm
Also capacities are often for a slow discharge rate of 10 or 20 hours, and ony apply to
new batteries, expect substantial decrease in capacity over a battery's life.   LiPo will have
the high max power drain, but life will be minutes, Peltier devices are 50 to 100W usually...
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### liorcohen5

#4
##### Nov 10, 2013, 10:27 pm
Ok thank you all, maybe I should reconsider my options.
I guess that heating/cooling will require a lot of energy anyway though.

#### polymorph

#5
##### Nov 11, 2013, 08:40 pm
Google is your friend. You can look up the characteristics of a given brand of 9V alkaline battery.

I doubt that an 9V alkaline of any make will put out 4.5A short-circuited. Much less at any meaningful voltage.

Looking at the datasheet, page 2, you'll need more like 12V or more to push 4.5A through it. You'll also need a big heat sink and a fan to carry heat away from the hot side, or you will end up with the cold side getting warmer than ambient!

What and how much are you trying to cool something? If this is just to carry heat away from a semiconductor (like a CPU), it is better to just stick the big heat sink and fan directly onto the CPU.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - tinyurl.com/q7uqnvn

#### liorcohen5

#6
##### Nov 12, 2013, 10:58 pm
Is it possible to place another Peltier device on top of the first one with the hot side facing the other one's hot side, and then when the first one is running the second one turns the heat, or at least some of the heat, back to voltage?
and maybe placing a heatsink on top of the second one to carry excessive heat away and boost the voltage gained by the second plat?

thanks!

#### polymorph

#7
##### Nov 13, 2013, 07:11 am
Why would you do that? Peltier devices are very inefficient, both at moving heat and at creating electricity from heat. Very inefficient.

You cannot get something for nothing. This is akin to turning a generator to power a motor that is turning the generator. Perpetual motion.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - tinyurl.com/q7uqnvn

#### liorcohen5

#8
##### Nov 13, 2013, 12:56 pm
yeah I got that.
the thing is I want to use only the cold side to cool small objects, and I need the whole thing to be as small as possible.
so I though if Im using a heatsink on the hot side maybe I could get something from it, even if its not much..

#### polymorph

#9
##### Nov 13, 2013, 07:36 pm
It is like saying I'll add a wheel on the back of my electric car to turn a generator, to provide power to charge the batteries. You are just going to use more power to move the car at the same speed, and generate -less- power than the extra you expending.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - tinyurl.com/q7uqnvn

#### liorcohen5

#10
##### Nov 13, 2013, 11:38 pm
so is it better to just use a heatsink?

#### polymorph

#11
##### Nov 14, 2013, 01:18 am
Yes. Just use a heat sink. Anything you add between the peltier device and the heat sink just reduces the efficiency even more.

In addition, a peltier device can only sustain a maximum of about 50C of difference in temperature between the two sides, and that drops with heat flow, so you will get much less than that. In addition, the hot side will always be hotter than ambient temperature, and the cold side will always be colder than the thing you are trying to cool.

So in the end, don't expect a huge difference between what you are cooling, and ambient temperature.

http://www.tellurex.com/technology/peltier-faq.php
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - tinyurl.com/q7uqnvn

#### liorcohen5

#12
##### Nov 14, 2013, 09:10 am
Thank you very much!!

#### ElCaron

#13
##### Nov 14, 2013, 11:09 am
@polymorph: I don't think the car example is correct. In your example, the energy is taken from a desired form of energy, the kinetic energy from the the car. His idea is to use the "exhaust heat", which is an undesired side effect. This is done in many technical applications.
It is probably still probably not a good idea in this case, since an effective heatspreader will sink the heat much better and will make the primary peltier element more efficient.

#### polymorph

#14
##### Nov 14, 2013, 02:44 pm
Except in this case, you are impeding the flow of heat, making the cool side hotter.

You -are- taking from a desired form of energy (I know I said that incorrectly), the flow of heat.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - tinyurl.com/q7uqnvn

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