# LC damping for controlling Peltier element

I am trying to control my Peltier element with a H-bridge motor driver, namely this DRV8873.
My TEC is rated at 128.7 W, 13.1 A, 15.7 V.
My controlling signal is PWM at 100 kHz.
I have a pair of LC filter connecting from the output of the driver to my TEC element. My inductors have about 300 mΩ series resistance.
The problem: When I turn on the board, nothing happens. It seems like tripped the over current protection of the IC with this LC filter. Without the LC filter, the circuitry works but with terrible noise.
What can I do to tackle this problem? Any help is really appreciated :)

simulate this circuit – Schematic created using CircuitLab

• What PWM% are you setting the outputs to, initially? Commented Mar 22, 2023 at 21:46
• @TimWilliams: it's supposed to be 30% or 50%. I also tried slowly ramping up the Duty Cycle from 0% but it also didn't help. Commented Mar 22, 2023 at 21:48
• In which mode? Are you using the outputs independently or bridged? Commented Mar 22, 2023 at 21:49
• @winny Presumably for efficiency, or EMI. Running a TEC at full voltage maximizes ohmic losses, minimizing efficiency. Commented Mar 22, 2023 at 21:57
• @Michael Do you mean OUT2 is set to 0V, or it is open circuit? What is the minimum duty cycle or pulse width you are able to produce, and does that still cause faulting? Do you have any waveforms? Commented Mar 22, 2023 at 21:59

It's probably tripping the overcurrent protection because when you turn on the device, the LC circuit you designed has an under-damped response and creates a huge current spike, in my experience low-pass filters that have a cutoff frequency of at around 1/100 part of the PWM frequency works very well to convert the PWM signal to DC, if the PWM frequency is 100kHz then F3dB = 1kHz or a bit higher, but this is not a rule of thumb you should test, to reduce the current peaks, try placing a inductor with a higher value, you could try first with a 330uH one and calculate the value of the capacitors with the formula:

  f_s = 1/(2*pi*sqrt{LC}}


The larger the value of the inductor, you can place capacitors with lower value, you could place a 330uH inductor and 4.7uF capacitors without any problem, but you should also consider the ripple current, which you can approximate with:

V = L * (dI/dT)

where:

dI = current ripple, dT = you find it with the period of the PWM, V = maximum voltage.

I would say that this is enough, the only problem that could arise is that the inductors are too big, but there is always some inductor, you could also do some simulations in LTSpice and read this article that will help you with your problem.