Hi, I am currently looking into LC filter to control a Peltier element in cooling mode (I know that is super inefficient, needed a huge heatsink with fan to get a tiny petri dish to 4°C, but other factors dictate it and efficiency doesn't matter).
Empirically from simulation, this seems to do the trick: http://tinyurl.com/y7s73yt4
I also have the formula f = 1/(2pisqrt(L*C)) for the cutoff frequency.
What I don't understand, though ... If I remove the capacitor in the simulation above, not a lot changes. Slightly more ripples, but the inductance can handle the issue by itself quite well. From physics, I can also see why, that is why I placed the diode and only later found out that that is also how it is done in other sources.
Only: How does that that comply with the formula above? Is that only applicable for filter with real HIGH/LOW, not LOW/[high impedance] input, like my PWM-MOSFET?
The capacitor controls the voltage ripple, but with a mainly resistive load that's proportional the current ripple
when the capacitor is omitted. In other words with a large enough inductor the ripple is all under control,
but normally capacitors are a lot cheaper so it pays to use more capacitance and less inductance for
a specific ripple budget, assuming you avoid actual resonance (which typically you do by keeping the
Q value low enough, for a resistive load).
With a variable load you need the final capacitor to handle load-induced current changes.
I get that, but the formula for the cutoff frequency doesn't seem to check out. For no capacitance, the cutoff frequency should be infinite (super high due to parasitic capacitance, but I am not sure if those play a role in the simulation), but practically, the inductance does filter a lot.
I guess your formula works only with no load applied to output of the filter? RC filter will have also reduced ripple when you connect heavy resistive load after it and output peak to peak voltage will be less than input even without the cap.
Try adding in the apparent resistance of your Peltier devce to your sinulation - eg if it takes an amp at 12 volts , use 12 ohms.
If you're using PWM and eg a MOSFET to drive it, try also addiing a fast eg schottky flywheel diode between the inductor in connection and ground or Vdd depending whether your MOSFET is P or N channel.- that makes it more like a conventional buck switchmode PSU and will have better efficiency and much smaller switching spikes .
Typically for this sort of filtering you calculate the ripple directly from the defining equations for
L and C. And you also want to check its not resonant at the frequency you are driving it at (nor the
harmonics for PWM).
L dI = V dt
C dV = I dt
[ The resonance point if it happens should be significantly lower in frequency than the
PWM, and its worth calculating the Q at that point given an estimate of the load's resistance,
high Q is bad - its probably a reasonable thing to make the impedances of L, C and R roughly
equal at the resonant point (L and C are by definition equal there). ]
allanhurst:
Try adding in the apparent resistance of your Peltier devce to your sinulation - eg if it takes an amp at 12 volts , use 12 ohms.
The Peltier element IS in the simulation, it is the resistor, calculated just as you suggested. I also have a diode there, I am not sure I get where you suggest one.
Basically you are designing a buck converter. I've never approached a design considering the "low pass" frequency. I've used either ripple current or ripple voltage. I know you can find many guides on buck converters on the internet.
One note in your simulation. I'm guessing the flow of the "dots" indicates relative current flow. You can see the capacitor is carrying very little current compared to the load. So it makes sense that it will have a small effect.
You can make a converter with no capacitor, many china based constant current LED drivers have no output capacitor. However LED's arn't sensitive to ripple currents.
I believe Pelitier junctions do not like pulsing current, however I don't know their sensitivity to ripple.
Given that they are sensitive I would suggest you look at 100 to 200 µf to start. Also note your simulation does not include the component losses. You have to particularly watch the capacitor as a "non switching" capacitor can have quit a large internal series resistance that will reduce its effect and reliability.
By removing the capacitor you change a second order filter into a first order filter. What this means is that it will not drop off as quickly. If you plot the output against the frequency you should see this. Better is to plot the log of the output against the log of the frequency to give you a better view as to what it will sound like.
Grumpy_Mike:
The OP said:-
A Peltier element is much less reliable using PWM as it is using DC, hence the need for a filter.
Thanks for that. For my education could you or the OP post a reference relating the reliability issue?
Edit: I have just found Application Tips | RMT Ltd which on the surface supports the efficiency issue. I will have to spend some time trying to understand the issue.
It is not just efficiency it is the fact that the ON / OFF of a PWM signal puts more mechanical strain on the device and so can weaken the junctions that make it up, hence resulting in an earlier failure. Some manufacturers claim it does not but I suspect that is only their marketing department and not the engineers.
The mechanical strain comes about through the sudden magnetic field created by the large current passing through the closely packed junctions.