RE: Interfacing a coil to the C64?
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 
> T=RC=.1088, about 100ms. 

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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