My question relates to a video I saw on youtube. In this video, the user is cascading 3 TEC1-12706 peltier modules to achieve temperatures of -50 on the cold side of the cascaded peltier.
The TEC1-12706 is a 12V 6A unit, but in this video the peltiers are supplied with the following voltages as below:
Peltier 1 on the bottom – 12V
Peltier 2 in the middle – 5V
Peltier 3 on top – 3.3V
Unfortunately I don't really understand how to read the data graphs on data sheets. My question is, how does these lower voltages (5V and 3.3V) affect the performance of the TEC's, and wouldn't it be better to replace the 2 12V peltiers with a 5V and 3.3V version respectively.
EDIT
From the answer from @vindarmagnus, I left the following comment to make it more clear what I'm really after
I have an idea how cascading work, and you've confirmed what I thought. However, talking specifically about the TEC1-12706 which is a 12V module, how is its performance affected when you run it on 3.3V and 5V. Does the drop in performance justify the use of a 12V module or will it be better to get a 3.3V and 5V module instead whereby I will get better performance.
Just to add, any type of reference as to why would be highly appreciated and would also help future users
Best Answer
It's hard to say without looking at the datasheets of the 3.3V\5V modules. If you wanted to choose a different module, it would need to do these things in this application:
1) Use less heat then the warmer stage below it.
2) Be more efficient than the 12V module (check the COP at an equal power for both modules)
Typically this is done by using peltiers stacks that get smaller as you get colder (to reduce power being dissipated in the modules)
The problem with peliter modules is they generate heat because they have resistance. For the TEC1-12706 you can see that they have roughly 2Ω (which varies depending on the heat load, the temperature across the module and the power being dissipated in the peltier).
In this application (where you only want to drive to a low temperature with no heat/very small heat load) and get max deltaT (temp between hot and cold side), there is a better advantage to staging peltiers with different voltages because with lower voltages each peltier generates less internal heat.
Shown below are some rough cut numbers (we don't know what the resistance of each cell is or power being dissipated so I chose 2Ω for a rough cut number, to illustrate the effect). Each stage has to move it's own heat, as well as the heat of other stages, the more heat it has to move the more it has to expend energy to do so, so using less energy down the stages gives an advantage over using 12V on all stages.
You can get an estimate of how cold any strategy will work, if we figure in the heats from the picture above we can roughly guess how cold the modules will get:
1) The 3.3V module has almost now heat load on the cold side (blue arrow) so there is rougly 20C of temperature drop (deltaT)
2) The 5V module has roughly 5-10W of heat from the module above for a 30C delta T
3) The 12V module has roughly 17-25W of heat from both modules which would be an upper bound of 40C for a deltaT
This combines for 20C+30C+40C= 90C and when we factor in ambient 25C-90C=-65C
The graph below is also for the Qc side or cold side, so the heat load calculations are done for the cold side.
Keep in mind that there is no heat load in the movie above, only heat that it is drawing from the air. If you were to put any heat load on the peltier it would reduce the stacks ability to move heat drastically as the first module would then have a heat load. There is a good app note here for designing politer modules