# RE: Interfacing a coil to the C64?

ruud.baltissen_at_abp.nl
Date: 2002-09-11 12:46:28

```> Thus: IC2-reg. out---IC1 (high side switch)--coil (16ohms)---gnd
>           \--gnd    \--+C1 (6800uF)---gnd
>                      \--(-) D1 1N4004 (+)---gnd (for coil kickback)

The diode must be parallel to the coil.

Then what puzzles me is the fact that you mention 32 Volt. IC2 can source
500 mA. 500mA * 16 Ohm = 8 Volt. I know coils have a minimum Voltage/Current
otherwise they won't hold whatever they are supposed to hold.

> I need "boost" with C1 to 2A for 100ms, and

That was years ago but my instinct says it doesn't work like this. Let's
theorise. If I'm wrong, anybody please correct me.
F = Cb/V. This means that with 6800uF C1 contains 0.0068 * 32 Cb electrons.
A = Cb/sec.
So the moment you connect 16 Ohms to C1, a 2A current will flow. But because
there ar electrons flowing away, the voltage over C1 will drop and thus the
current will drop. This calculated T value means that the voltage has
dropped to about 30% (????). Keeping the Hold-voltage in mind, you have less
then 0.1 second.

> how long it takes to charge..

0.0068 * 32 / 0.5 = 0.4352 sec. maximum. But the voltage drop won't be 32V
but much less. So the time is shorter.

> and also if this is safe to regulator.

That depends. If it has a build-in current limiter, then the current is
restricted to the 0.5 A used in the calculation. If not, it will supply much
more current. Loading time is less but it could damage the regulator.

> Also, can gnd be shared with C64,

If not using opt-couplers, you have to.

> or does some kickbacks leaks through?

I presume you steer the switch with one of the I/O-pins of the userport
through a resistor. Use diodes to short surpluss Vcc and Vss. Then make sure
the C64 is connected to GND of the circuit at only one place.

--
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