Cooling Integrated Circuits with Peltier Cells

[EvIl_DeViL]

Homework Statement

Hi all,
I have a question about using peltier cells for both cooling and reading temperature in the same moment.
I need to cool down a couple of Integrated Circuits so I'll put put a 20mmx20mm peltier cell over them but I cannot read the cold side of the cell because it's sticked to the IC; I hope physics land me a hand here, providing I put a thermocouple to the hot side...

Homework Equations

S = - (Δv/Δt)
where:
S is the Seebeck coefficient (which I'll need for later)
Δv is the voltage I'll read with a multimeter
Δt is thermodynamic temperature which I'll read with a thermocouple. If I understood correctly it should be the difference between the hot and cold side expressed in kelvin's degrees

I'll take all these measurements before placing the peltier over the IC to avoid experimental errors

The Attempt at a Solution

Now that we found S we place the cell on the IC with some thermal compund and we get:
Δt = - (Δv / S)
hence
Δt = - ((Vin - Vcc) / S)
because I expect Δv across the peltier cell should be Vcc + the voltage generated by the seebeck effect because the cell is in series.

then I convert °K to °C

Now that I have the temperature difference between the hot and cold side I can measure with a thermocouple the hot side, subtract the difference I just found and get the temperature of the cold side sticked to the IC.

Are my assumptions correct? will it work?

I know peltier are dangerous with electronics because of the condense but I recently used a dht11 which is a sensor which can read temperature and humidity.
I'll use it to know how much voltage I can provide to the cell before generating condense using a table like that http://cdn2.i4memory.info/reviewimages/cooling/memory/CorsairDHX_TEC/CorsairDHX_TEC_dewpoint_table.png (I'll drive it with a transistor) to keep it above the dew point (maybe 1 °C with a 5% tollerance).

I also know peltier are cooling better if the hot side is helped in dissipation, my thermocouple will find place between the peltier cell and a fan

Thank you all!

 

[Hesch]

It's the IC you want to cool ( not the peltier cell itself ). The IC is mounted on some PCB ( I guess ), so why don't you drill a hole in the PCB, so that you can measure the temperature of the IC from beneath through this hole? ( I assume that you have not made the layout of the PCB yet ).

It will be much more accurate and simple.

 

[EvIl_DeViL]

No way; the PCB isn't mine. It's a commercial router (It's called "banana pi bpi-r1" which is a cool piece of hardware but cool in the wrong way eheheh...) so it's multilayer (don't know how many but surely 2 at least: top and bottom). Also some ICs are BGA so drilling is a no go :( It's an official fact that using sata + wifi + external usb + high cpu load would result in hardware fault and shortened life (his, not mine luckily) because of the high current walking around (almost 3 Amps) and the poor AXP209 (PMU) is heavy loaded!

Nice advice though! If I'll ever project a board I'll consider a hole under the hottest ICs for this purpose.

returning to physics: I can't believe nobody didn't ever tried something similar before. There must be something on the web!

 

[berkeman]

Alternately, attach a thin piece of aluminum between the IC and the Peltier cell, and have it stick out a little past the cell on one side. Attach your thermocouple to that tab of Al that is sticking out...

 

[SteamKing]

Homework Statement

Hi all,
I have a question about using peltier cells for both cooling and reading temperature in the same moment.
I need to cool down a couple of Integrated Circuits so I'll put put a 20mmx20mm peltier cell over them but I cannot read the cold side of the cell because it's sticked to the IC; I hope physics land me a hand here, providing I put a thermocouple to the hot side...

Homework Equations

S = - (Δv/Δt)
where:
S is the Seebeck coefficient (which I'll need for later)
Δv is the voltage I'll read with a multimeter
Δt is thermodynamic temperature which I'll read with a thermocouple. If I understood correctly it should be the difference between the hot and cold side expressed in kelvin's degrees

I'll take all these measurements before placing the peltier over the IC to avoid experimental errors

The Attempt at a Solution

Now that we found S we place the cell on the IC with some thermal compund and we get:
Δt = - (Δv / S)
hence
Δt = - ((Vin - Vcc) / S)
because I expect Δv across the peltier cell should be Vcc + the voltage generated by the seebeck effect because the cell is in series.

then I convert °K to °C

Now that I have the temperature difference between the hot and cold side I can measure with a thermocouple the hot side, subtract the difference I just found and get the temperature of the cold side sticked to the IC.

Are my assumptions correct? will it work?

I know peltier are dangerous with electronics because of the condense but I recently used a dht11 which is a sensor which can read temperature and humidity.
I'll use it to know how much voltage I can provide to the cell before generating condense using a table like that http://cdn2.i4memory.info/reviewimages/cooling/memory/CorsairDHX_TEC/CorsairDHX_TEC_dewpoint_table.png (I'll drive it with a transistor) to keep it above the dew point (maybe 1 °C with a 5% tollerance).

I also know peltier are cooling better if the hot side is helped in dissipation, my thermocouple will find place between the peltier cell and a fan

Thank you all!

For some reason, people seem to forget that a temperature difference of 1 degree Kelvin = 1 degree Celsius.

If you measure temperatures in degrees Celsius, no conversion is necessary when calculating the temperature difference.

 

[EvIl_DeViL]

For some reason, people seem to forget that a temperature difference of 1 degree Kelvin = 1 degree Celsius.

If you measure temperatures in degrees Celsius, no conversion is necessary when calculating the temperature difference.

I know but the library of the thermocouple sensor throw out celsius values so we still have to convert so I can sum. I can also convert my sensor values from °C to °K that's not the problem :)

@berkeman: I'd like something more professional/accurate and which make me feel a genius :D I think my idea could work and I came here just to have a confirmation before even wasting time on it :P