Re: [CAD_CAM_EDM_DRO] Re: Getting Parallel port to run relay

Hello,

> Coil current 89.3mA resistance 56 ohms

That is 89.3ma * 56 ohoms or 5 volt coil.

As another respondant has suggested, use a PNP transistor as an emitter
follower. A 2n2907 may be adequate. Coil between transistor emitter and
+5 volts. Diode parallel with coil and reverse biased. Transistor
collector to power supply ground and to parallel port signal ground.
Parallel port output to transistor base. No resistors needed.

When the parallel port output is high the transistor is non conducting
so the relay coil will have no current. When the parallel port output is
low the transistor will conduct and the coil will have current. The
voltage drop across the coil will be about 4.3 volts. That should be
sufficient to activate a 5 volt relay coil.

The load to the parallel port output is the transistor current gain
times the coil resistance. A 2n2907 can have a gain of 100 so that would
mean that the load appears as a 5600 ohm resistor tied to +5 volts. The
parallel port output would only have to sink 0.893 ma. That should not
be a problem for the parallel port output which can likely sink on the
order of 1ma to 2ma.

Tom Hubin
thubin@...