How would you control up to 10A @ 12VDC with fixed resistance using a microcontroller?

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Daemach

Joined Nov 30, 2023
2
I want to control a 12715 TEC using a microcontroller (ESP32). I have a regulated 12VDC power supply, and I know I will need to deal with ripple current, etc, but I need advice on how best to control 1 to 10A depending on feedback from a thermistor, which I can handle.

Current control is the most important thing.

I have the microcontroller end covered; I just need guidance on an elegant/cheap way to control current (ideally as easy as using PWM), as oversaturation leads to thermal runaway, negating any cooling effect. How would you do it, or can you point me to some existing schematics?

https://peltiermodules.com/peltier.datasheet/TEC1-12715.pdf
 

Irving

Joined Jan 30, 2016
3,794
Is it current or voltage you need to control? Voltage is a lot simpler and this translates to current via the 'fixed' resistance of the device. You can use a current sense feedback to optimise the PWM to address the resistance change with temperature. Something like this, though I'm not sure how effective PWM'd current is with a peltier device.
:1701372328973.png

An alternative is a true constant current source, the downside being the power dissipated in the MOSFET, requiring a reasonable heatsink.

1701436816091.png
 

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BobTPH

Joined Jun 5, 2013
8,661
but they are also sensitive to ripple
If they are sensitive to ripple, the PWM is out because it is ALL ripple.

I read up on this a while ago, and there seems to be no consensus over whether or not PWM should be used on Peltier devices.
 

crutschow

Joined Mar 14, 2008
34,044
The problem with using straight PWM is that PWM will dissipate more power in the resistance of the TEC, as the cooling is proportional to the average TEC current (since the TEC junction voltage, which generates the cooling, is relatively constant) whereas the power dissipated in the parasitic resistance portion of the TEC cooler, which reduces efficiency, is proportional to the RMS current.
Thus, for example, with a 50% PWM (current) duty cycle, the power dissipated in the resistance would be twice the value of using 50% DC current.

You can use PWM and avoid that difference if you use a series inductor and free-wheeling diode to power the TEC with smoothed DC (i.e. a switching regulator).
 
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Irving

Joined Jan 30, 2016
3,794
So, here's a 1st attempt at a buck converter driving the Peltier device... The device is modelled from a 2016 paper by V. Kubov that I found on ResearchGate, attached below - they used a smaller TEC device but I've modified the model to represent the parameters of the bigger unit, with one exception - I've reduced the capacitance Cq representing the thermal capacity of the device & therefore the rate of change of temperature to allow the simulation to run to stable state in a reasonable time! Note that this simulation is with the the hot side (Th) tied to ambient with a big heatsink of <0.2K/W thermal resistance (represented by R2) and no power input to the cold side. For the thermal (lower) section of the device, voltage = temperature (°C) and current = heat flux (W). Obviously, with a power input to Tc (cold side) the temperature at Tc would be higher than the -70°C shown here (top pane RH scale, blue/green trace)

R1 is a current sense resistor that provides feed back to raise the supply voltage as the current drops to maintain a constant current - as the device cools the Seebeck voltage on the cold side increases forcing the current lower. I've not modelled any feedback and my PWM is fixed here at 90% (18uS/20uS, 50kHz). What we can see from this is that 12v input doesn't give sufficient headroom - you'll need 15v or more into the buck converter to achieve a constant 10A Peltier current with heat input to the device.

1701543211511.png°
 

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MrAl

Joined Jun 17, 2014
11,263
So, here's a 1st attempt at a buck converter driving the Peltier device... The device is modelled from a 2016 paper by V. Kubov that I found on ResearchGate, attached below - they used a smaller TEC device but I've modified the model to represent the parameters of the bigger unit, with one exception - I've reduced the capacitance Cq representing the thermal capacity of the device & therefore the rate of change of temperature to allow the simulation to run to stable state in a reasonable time! Note that this simulation is with the the hot side (Th) tied to ambient with a big heatsink of <0.2K/W thermal resistance (represented by R2) and no power input to the cold side. For the thermal (lower) section of the device, voltage = temperature (°C) and current = heat flux (W). Obviously, with a power input to Tc (cold side) the temperature at Tc would be higher than the -70°C shown here (top pane RH scale, blue/green trace)

R1 is a current sense resistor that provides feed back to raise the supply voltage as the current drops to maintain a constant current - as the device cools the Seebeck voltage on the cold side increases forcing the current lower. I've not modelled any feedback and my PWM is fixed here at 90% (18uS/20uS, 50kHz). What we can see from this is that 12v input doesn't give sufficient headroom - you'll need 15v or more into the buck converter to achieve a constant 10A Peltier current with heat input to the device.

View attachment 308923°
Hi,

I am not so sure a 0.1 Ohm sense resistor is the right size for sensing currents up to 10 amps, or even if a simple sense resistor is the right approach at all due to the higher current range. After all, 0.1 Ohms drops a full 1v at 10 amps. There are better ways to sense current albeit with a little more complexity.

Also, the Rds ON resistance of the MOSFET would have to be very low, and the 68uH inductor ESR very low also.

It would be nice if he had a 13v or higher power source, but he says he's got a 12vdc power source.
 

Irving

Joined Jan 30, 2016
3,794
I am not so sure a 0.1 Ohm sense resistor is the right size for sensing currents up to 10 amps, or even if a simple sense resistor is the right approach at all due to the higher current range. After all, 0.1 Ohms drops a full 1v at 10 amps. There are better ways to sense current albeit with a little more complexity.
As I said, a first stab to illustrate concepts - its not intended to be a manufacturable design. I'd probably use 0.01Ω or maybe an ACS712 hall effect sensor. 5 and 10milliΩ sense resistors are very typical in such circuits when used on low-side sensing.

Also, the Rds ON resistance of the MOSFET would have to be very low, and the 68uH inductor ESR very low also.
Again, its not intended to be a manufacturable design. However, the devices weren't just picked at random. The MOSFET, picked from the LTSpice list is 2.3mΩ @ 20A, & isrepresentative of many suitable devices, and the inductor is a KEMET MPX series, often used for 10-15A @ 12 - 24v smps, it has a DC resistance of 30mΩ max,

It would be nice if he had a 13v or higher power source, but he says he's got a 12vdc power source.
Then he won't get the full range of the cooler - you cant change the laws of physics.
 

MrAl

Joined Jun 17, 2014
11,263
As I said, a first stab to illustrate concepts - its not intended to be a manufacturable design. I'd probably use 0.01Ω or maybe an ACS712 hall effect sensor. 5 and 10milliΩ sense resistors are very typical in such circuits when used on low-side sensing.


Again, its not intended to be a manufacturable design. However, the devices weren't just picked at random. The MOSFET, picked from the LTSpice list is 2.3mΩ @ 20A, & isrepresentative of many suitable devices, and the inductor is a KEMET MPX series, often used for 10-15A @ 12 - 24v smps, it has a DC resistance of 30mΩ max,


Then he won't get the full range of the cooler - you cant change the laws of physics.
Hi,

Are we talking about the same circuit here :) Maybe I am not understanding you with that last statement.

If we have a buck circuit with a low Rds ON mosfet and an inductor with low ESR around 30mOhms and the switch is turned 'on', we have about 33mOhm between the input voltage source and the load of 1.2 Ohms. That means with a 13v voltage source we get 12.65 volts across the load of 1.2 Ohms.

With that 33mOhm and the 1.2 Ohms load, to get 12v output at 10 amps the extra voltage needed would be 0.033*10=0.33 volts, so a 12.33 volt source would do it.

With a 12v input, the output would only be 11.68 volts, and the current would be around 9.73 amps which would be a loss of around 3 percent.
 
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