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

On Sunday 30 January 2005 07:08 pm, Tom Hubin wrote:

> 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.

Hm? Coil connected between transistor emitter and +5 supply, base is pulled
toward ground, tends to turn transistor on but won't go any lower than the
minimum output voltage specified by the parallel port, which may be as much
as several tenths of a volt. I have some 5V relays around but haven't really
tested to see how low a voltage it takes to pull in. I'd put the relay in
the collector leg and use a base resistor.

> 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.

Definitely a point in favor of using a PNP transistor, anyhow.