# Input current filter

I am trying to design an LC input filter with the following requirements:

-The input current peak to peak value should be limited to about 1A because it will be connected to a PV panel

-The output of the filter will be connected to a full bridge converter (with output transformer for isolation with the load). I want the converter to be voltage fed, so this is why I need the capacitor.

The topology is shown in the attached figure. Because of the full bridge, I know that the output current of the filter will be some kind of square wave, so a lot of higher order harmonics compared to the rather flat DC current that I want to draw from the solar panel.

I managed to determine the value of L and C via simulation but I would like to under build this more mathematically. So I am looking for the transfer function I_out/I_in. It's easy to find if the full bridge is replaced with an impedance, because than you can use a current divider. But I cannot find it in case of a complex switching network, such as the full bridge. Any help or clues are appreciated.

• Don't. Whatever complicated choppy current the bridge takes will be reduced to a wobbly voltage at the cap, which will draw a yet smoother current from the panel. If you have got suitable values from a simulation, be happy. Mathematical solutions to this would start with approximating the waveforms, or making assumptions about them to simplify to the point where they can be analysed, and are only really applicable for professors setting students questions, and people who want to publish learned papers. Empericism FTW! – Neil_UK Sep 26 '16 at 7:25
• @Neil_UK And people in aerospace, low-noise measurement and drive systems, etc etc. It's not just restricted to "university use". I should know, it's why I'm paid the big bucks for not sticking to just plonking in a million parts to find the right one. – Asmyldof Sep 26 '16 at 7:30

Consider the output capacitor (say 100 uF for convenience) that feeds the converter. Assume it is charged to some typical voltage level (maybe 10 volts again for convenience). If the converter input current (I) is 10 amps (during a peak) then, the change in voltage on that capacitor is: -

$\dfrac{dv}{dt}$ = I/C = 100 kV per second or 0.1 volts per micro second.

If the duty cycle of the converter is (say) 50% and the average capacitor voltage remains at 10 volts then, the current from the solar cell must be 5 A. A higher duty cycle means the panel supplies a higher average current.

So, the real dv/dt reduces to 0.05 volts per micro second.

Next, consider that the converter switches "on" for 10 us and "off" for 10 us (as per the previously mentioned duty cycle of 50%). During the "on" period, the capacitor voltage will reduce by 0.5 volts and, for the "off" period, it will rise back by 0.5 volts hence capacitor ripple is 0.5 volts p-p.

If the input voltage to the inductor (from the panel) is constant and the inductor is (say) 100 uH, it will see a change in voltage of 0.5 volts over a 10 us period.

Now here's a little cheat - assume that that voltage change is instantaneous (it's a triangle in reality) and, using V = L di/dt you can "estimate" what di/dt will be into the inductor.

I "calculate" di to be 0.5V x 10 us / 100 uH = 0.05 amps.

Like I say, this is just an estimation based on me cheating by saying the change in voltage across the inductor is instantaneous (rather than triangular shaped). So, I would expect to see the panel feed an average current of 5 A with a superimposed ripple of no more than 50 mA p-p. Might be worth simulating......

Hmmm, not far off - output ripple voltage is spot on and input ripple current (barely visible) is one-quarter of my somewhat exaggerated calculation.

That's the power of simulation.

The simulation also tells you that there might be a significant transient on power up that might exceed the input supply limit on the converter - watch out for this.

• ... and logical though and years of experience and ability to write clearly ... +1. I try to teach the younger guys where I work to do general engineering questions in their head to get an order of magnitude result before they start using software to work them out. It allows them to test their own understanding and should flag any gross errors in their detailed calculations or software inputs, etc. This answer is approached in a similar vein. – Transistor Sep 26 '16 at 8:26
• @transistor thanks but the +1 I guess was virtual LOL. – Andy aka Sep 26 '16 at 8:50
• AJAX doesn't work so well on the train. I don't know if this internet thing is going to catch on. Fixed. – Transistor Sep 26 '16 at 8:59
• Thank you very much for the good and detailed explanation, Andy. I also think that it's better to have an idea of what to expect from a simulation before just starting and trying. I will work it out for my application and redo the simulations this week. I'll post it afterwards. Anyway, your answer is really a great help! – Simon R Sep 26 '16 at 9:22

Ic = C * dVc/dt

I_out=I_in-Ic = I_in-C*dVc/dt

Then I_out is pulsed current (MPPT reg?) so I_out2 * duty cycle is related to I_out

like this?

I prefer to work in transfer functions using Z(PV) , ZL(f), Zc(f), Z_bridge) where PV is a quasi-current source with a V limit, Voc but when supplying power has an effective minimum Source impedance of Vmp/I... then choose L, C based on Z(f) to create a low source impedance for better load regulation of current pump to battery

Then bridge becomes a duty cycle modulated LC-LC "T to Pi" filter