Peltier PWM Filtering problem

Discussion in 'Power Electronics' started by ozlow, Jun 19, 2017.

  1. ozlow

    Thread Starter New Member

    Apr 13, 2016
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    Hi everyone!

    I need to control a Peltier module (TEC1-12709) in heating and cooling mode. Browsing similar topic i decided to control the Peltier with a LC filtered 50kHz PWM signal with an H bridge configuration. I simulated the prototype in LTSpice (the project is attached).

    - The H bridge will be modified into a 4-NMOS configuration with proper gate driver, the one in the schematics is just for simulation purpose;
    - Vdd is 12V;
    - Frequency isn't a constraint, i've chosen "high" frequency PWM to reduce the inductors values;

    With the LC filter the current is pretty smooth but the voltage at the non conducting PMOS source is acting strange, causing the MOS turning on in the opposite way (i suppose) and conducting something like 50A to the power source.
    How should i modify this circuit (or rebuild it from scratch)?

    Thanks for the help.
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Why do you want to filter the PWM signal?
     
  3. ozlow

    Thread Starter New Member

    Apr 13, 2016
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    Because in many threads i found that a full on/full off control on Peltier module will reduce the efficiency, due to increasing loss for internal Joule effect.

    Thank you for reply.
     
    MrChips likes this.
  4. MrChips

    Moderator

    Oct 2, 2009
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    Thanks. I didn't know that.
     
  5. MrChips

    Moderator

    Oct 2, 2009
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    That would make sense for DC.
    Does it still hold true for PWM since it is the RMS value that matters?
     
  6. wayneh

    Expert

    Sep 9, 2010
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    Yes, the loss due to PWM occurs at the "on" peaks. It's less efficient every time it's on, than the equivalent (lower) DC voltage.

    It's true with LEDs too, but much less pronounced.
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    I will have to see a mathematical analysis to be convinced.
    Is this because power is proportional to the square of the current?
     
  8. wayneh

    Expert

    Sep 9, 2010
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    No, it's fairly simple. Consider the LED intensity versus current curve. At higher currents it's not linear and bends over, meaning you get less intensity per amp at higher amperages. So if your PWM peak on periods are at that higher, and thus less efficient current level, all of your 'on' time is at that lower efficiency. An equivalent time-averaged DC current will operate the LED at a current where the efficiency is higher. The LED geeks also worry about the effect of current on color.

    I've seen LED drivers that somehow exploit this by lowering the PWM voltage? (maybe they are pulsing a constant-current?) in addition to lowering the duty cycle. I don't recall how they accomplished that without a linear device.
     
  9. crutschow

    Expert

    Mar 14, 2008
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    Exactly.
    The resistive loss in the Peltier is proportional the the RMS current value, which is much higher than the average current without an inductor to losslessly smooth the peak current value.
    The Peltier cooling, on the other hand, is proportional to the average current.
    So you want to keep the RMS current as close to the average current as is practical.

    The same effect occurs in LEDs if you are using a resistor to limit the peak current for a PWM signal.
     
  10. ozlow

    Thread Starter New Member

    Apr 13, 2016
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    1
    Yup! To achieve this current control - because as noted the most efficient way to drive the Peltier is not with Imax current (the one that you achieve applying 12V to the cell) but with a lower current, obtained by some sort of technique. DC is unconvient because a MOS in linear region will dissipate too much, so usually the trick is to filter the PWM signal with an LC filter (not RC because of dissipative loss, it's like 12V 6A PWM).
     
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