How much current can a coil take?

Hello guys,

I'm building the light electro-mechanical shutter from a Hard disk drive mentioned here:

**broken link removed**

So I opened a hard-disk (different kind), and took the shutter part (as shown in the pictures), and by putting current in the coil, we can control the shutter. So, to drive the shutter effectively, there's a circuit to be designed (also available in the same page):

**broken link removed**

The coil of the hard-disk I have has a resistance of 6 Ohms, but I don't know whether this circuit is valid to drive it. There are big capacitors in the circuit, and I'm afraid I'd burn stuff, where the circuit drives almost 3 Amps.

How should I look at this matter? should I just design the circuit blindly and just use it? or should I do some relevant calculations before building the circuit?

Thank you for any efforts.

It should be OK, you need the big capacitors to ensure it can handle the brief high currents needed to get the actuator moving. The differences between actuator characteristics between drives is quite small so what works for one should work for others.

Brian.

Thank you so much for your fast reply. I still have another question though.

I couldn't find a 2500 uF capacitor, so I'm using 3300 uF. Would this difference introduce any problems? How much different could the capacitors be?

Thank you.

The capacitor is just there as a local reservoir of power so a bigger value will actually work better although 2500uF is already quite big so the difference will be very small. I would try not to go much lower than 2500uF because capcitors of that value can have very wide tolerances, sometimes as much as -50% to +150% of the value and if you were unlucky enough to use one at the lowest end of it's range it might cause problems.

The reason it is fitted it that at the instant the coil is told to move, the current needed to shift it from a standing start is quite high, once it's moving the current drops. If the power wires were long or the power source was struggling to produce enough current, the voltage would drop and that could prevent normal operation. The capacitor, being close to where the power is needed can release energy quickly and right where it's needed.

Brian.

Thank you so much for your help. I'll design the circuit and contact you on this post if I have any further question. I hope you'd get some notifications when I do.

Cheers