I am having some problems with an active rectifier (AFE unit) that produces a 8 kHz and a 50 Hz component on my DC.

We have bult in a normal standard Low-pass filter that deals with the 8 kHz, so no problems there.

But we still have the 50 Hz problem (its a software issue in the AFE inverter itself, and therefore not easily solved fast).

Its not much, 600mA rms when charging with 50A@600VDC but apparently still enough to mess up the battery that it is charging.

The problem is that the battery that the AFE is charging can be considered a capacitive load. So the dampening effect on the low pass filter for it to take care of the 50Hz is quite much.

We solved it with putting in a standard power resistor in series before the load, so with a normal power resistor on 1 ohm we can filter the 50Hz signal also. But that is not so economical, just burning off a cupple of hundreds of watt.

So the the next step is to get a 3,2 mH inductor to create 1 ohm of resistance@50Hz and that will we test this week.

But what I am thinking about is the voltage spike from this inductor,

we have capacitors that is sitting between + and – before and after the inductor. Approx 1 mF on each side.

Since a capacitor is a low inductance on rapid voltage changes, they should take care of the voltage spike.

The voltage spike can be simply calculated with E=I x R.

Since the capacitor is the resistance here, it is frequency dependent of course but we can still say that the spike will se the capacitor as a low resistance.

But am I correct in my thoughts here, that capacitors between + and – will take care of the spike?

And how do i calculate the amount of current/energy, that will go into the capacitor?

## Best Answer

A current of 50 amps through an inductance of 3.2 mH implies a total stored energy of 4 joules (\$E = LI^2/2\$). If this energy were wholly dumped into one of your 1000 uF capacitors across the rail, the voltage would rise according to the energy formula for a capacitor (\$E = CV^2/2\$).

So equating 4 joules to the above gives a voltage rise of 89 volts. But this is when the initial charge on the capacitor was zero.

If your capacitor is already charged to 600 volts, it already has an energy of 180 joules and 4 more joules makes 184 joules or, a final voltage of 607 volts after the inductor energy has dumped.

However, if your capacitor has an ESR (equivalent series resistance) of 0.1 ohms, the 50 A that would initially flow into the capacitor would also produce 5 volts across the ESR so the peak of 607 volts could be as high as 612 volts.

Is this value OK? Only you can say. I would also recommend you download a free spice tool (such as LTSpice) as this can help a lot and is worth the up-front learning curve for anyone in electronics.