Hello,

Recently I bought a Peltier element for use in a temperature controller application. They didn't provide any ratings for the peltier, and it doesn't have any ratings specified as a code either. The brand is "Supercool", with dimensions of 30 mm X 15 mm X 4 mm.

The shopkeeper told me to 'try' 12 V, but I'd like to be sure about the ratings. So can someone please help me to identify the peltier? I've linked the pictures.

http://i1322.photobucket.com/albums/u569/abhaymv/Photo1773_zps89ed85e0.jpg

http://i1322.photobucket.com/albums/u569/abhaymv/Photo1770_zpsd60681d2.jpg

http://i1322.photobucket.com/albums/u569/abhaymv/Photo1772_zps2a4f4e2b.jpg

http://i1322.photobucket.com/albums/u569/abhaymv/Photo1771_zpsfb4103e9.jpg

 

Good question, I just bought some also.

I know they are used in car and truck coolers, so I'd say 12-24 Volts.

 

Good question, I just bought some also.

I know they are used in car and truck coolers, so I'd say 12-24 Volts.

But won't the peltier break if I drive it at a voltage greater than maximum? How can I ensure that this won't happen?

 

You will need some type of control for this device; PID would be ideal, but can be expensive. A DC Burst (SCR) control; adj duty cycle ON---OFF; typically used in heating applications would be a possibility. You may not need a feedback element, if the environment can be controlled by other means. See KB Electronics, Coral Springs, FLA. for economical controls solutions.

Cheers, DPW [Everything has limitations...and I hate limitations.]

 

What will damage a TEC is the maximum temperature on the hot side.
Try running at 6 to 12VDC and monitor the current.
Do not operate without proper heat sinks on both sides of the module.

Edit: Judging by the size of that TEC I would suggest not exceeding a current of 3A.

 

What will damage a TEC is the maximum temperature on the hot side.
Try running at 6 to 12VDC and monitor the current.
Do not operate without proper heat sinks on both sides of the module.

Edit: Judging by the size of that TEC I would suggest not exceeding a current of 3A.

Thank you for the reply. But why do I need a heat sink on both sides? I thought only the hot side would need a heat sink...

EDIT: I was able to get the specs of the peltier after digging through the internet. Apparently I had to use the keyword 'TEC' instead of peltier. Thank you for giving me the idea!

http://www.ge.infn.it/ATLAS/Electronics/Humidity/Peltier/peltier%20RS.pdf
The specs are 3.4 A, 7.8 V and 16.5 W.


So do I operate this at 7.8 V? Do I need heat sinks on both sides?

 

Put heat sinks on both sides for testing the current/voltage/power/temperature characteristics.
After that you are free to do as you wish, at the risk of killing the TEC.

 

Put heat sinks on both sides for testing the current/voltage/power/temperature characteristics.
After that you are free to do as you wish, at the risk of killing the TEC.

I am planning to use a heat sink on the hot side, but how can I use the TEC at all if I need heat sinks on both sides?

My application requires both heating and cooling. Would this be a problem if I reverse the polarity of the voltage and use the same peltier for both? While heating, my heat sink would be attached to the cold side of the peltier and I'd be heating an aluminium block. While cooling, the heat sink would be attached to the hot side, as it should be...

 

In your actual application your load will take the place of heat sinks.

I am suggesting putting heat sinks on both sides for testing purposes.

 

Firstly they don;t run at "volts" as they are current operated devices.

The max current for that package will be similar to any semiconductor, it depends on how big the heatsink is on the hot side vs how much work it is doing (how much heat it is transferring).

You need to rate it within the final hardware structure. What I do is to assemble with the heatsink/fan etc, slowly increase current and measure the temperatures in many steps.

Then do a quick chart, and you will see the cold temperature gets colder in proportion to current, but it soon gets to a point on the chart where more current does not give the expected increase in "coldness". That is showing that the limit of its heatsinking and/or current capability is being reached, and it is becoming much less efficient and risks damage.

 

I've never seen a Peltier, so correct me if I'm wrong. This is just basic thermodynamics:

You have to think of heat as a flow. If you put a radiator (and I use the term loosely) on the hot side and apply power, the cold side will get cold. It doesn't have enough surface area to collect the heat that the Peltier might move, if it could. The cold side might even become coated with frost and frost is an insulator. As less heat can get into the cold side, the flow of heat to the hot side will be reduced. This will wreck your performance graph because you will be measuring the insulating ability of still air or frost, not the performance of the Peltier. That would be like trying to measure a hair dryer with your hand over the air intake holes.

After you get done finding out what to expect, install it where you want to. If it doesn't live up to what you measured with good radiators on both sides, you will know where to look and adjust things to get it working well.

 

Firstly they don;t run at "volts" as they are current operated devices.

The max current for that package will be similar to any semiconductor, it depends on how big the heatsink is on the hot side vs how much work it is doing (how much heat it is transferring).

You need to rate it within the final hardware structure. What I do is to assemble with the heatsink/fan etc, slowly increase current and measure the temperatures in many steps.

Then do a quick chart, and you will see the cold temperature gets colder in proportion to current, but it soon gets to a point on the chart where more current does not give the expected increase in "coldness". That is showing that the limit of its heatsinking and/or current capability is being reached, and it is becoming much less efficient and risks damage.

Thank you! I'll try this procedure ASAP!

I've never seen a Peltier, so correct me if I'm wrong. This is just basic thermodynamics:

You have to think of heat as a flow. If you put a radiator (and I use the term loosely) on the hot side and apply power, the cold side will get cold. It doesn't have enough surface area to collect the heat that the Peltier might move, if it could. The cold side might even become coated with frost and frost is an insulator. As less heat can get into the cold side, the flow of heat to the hot side will be reduced. This will wreck your performance graph because you will be measuring the insulating ability of still air or frost, not the performance of the Peltier. That would be like trying to measure a hair dryer with your hand over the air intake holes.

After you get done finding out what to expect, install it where you want to. If it doesn't live up to what you measured with good radiators on both sides, you will know where to look and adjust things to get it working well.

Again, thank you for this post. It helps understand peltiers a bit better.

 

Believe me, you can kill these little things very quickly.

With 24W being generated think of a 24W soldering iron. They get hot very quickly.
My point is, while you are in the experimentation stage, use a very large heat sink.

Why use heat sinks on both sides?
Because you don't know which side is the hot side and which side is the cold side.

And don't exceed 3A.

 

Believe me, you can kill these little things very quickly.

With 24W being generated think of a 24W soldering iron. They get hot very quickly.
My point is, while you are in the experimentation stage, use a very large heat sink.

Why use heat sinks on both sides?
Because you don't know which side is the hot side and which side is the cold side.

And don't exceed 3A.

Right. Thanks. I'm planning to use CPU heat sinks, which should have a thermal resistance of 0.5W/K if the datasheets are to be believed...