Risk to damage my computer with this circuit?

[J_Nichols]

I'm building a circuit to control a relay through the parallel port of my computer.
Basically, the parallel port sends bits to drive a transistor. That transistor allows a 12 VDC to flow in the coil of the relay. So when the relay is energized, the relay closes the AC contacts and a light bulb can flow.

My question is if I can damage the parallel port, since I know parallel port work from 5 VDC to 9 VDC (depending the model) and the battery voltage is 12 VDC.

Is like this:

A software controls the light bulb through a transistor and a relay.

 

[Dragon Tamer]

It would help to see the circuit diagram in question, but if you have the NPN transistor on the negative end of the relay coil with a 10k resistor on the base, a diode to suppress inductive spikes form the relay coil, and the ground of the parallel port is connected to circuit ground then I see no harm that can be done to the computer.

 

[mdanh2002]

Parallel port output are strictly TLL level (0V or 5V) and can draw only a limit amount of current (max 20mA if am not wrong). To be sure you need to try on an old computer (so that damaging it would not cost you much), or use an ISA (not PCI) parallel port card to facilitate replacement. To be further safe, you need an optoisolator to provide some sort of protection between the port and your circuit. Always triple check your connections before connecting anything to the port.

 

[MikeMl]

The YOU TUBE thing is worthless. Show a schematic of what you propose to do.

I wouldn't have 120Vac anywhere close to a computer unless (pick one):
a. I knew what I was doing,
b. used an opto-isolator, an external power supply, a transistor to turn on/off a low voltage lamp.

 

[crutschow]

You could probably drive an AC solid-state relay directly from the parallel port.

 

[Dragon Tamer]

That's true, that completely skipped my mind. Solid state would be better because quiet, and will last longer than a relay. Even though relays can operate for a very long time, they do wear out if used frequently.

As for what mdanh2002 said, you could add a replaceable parallel port card, to your computer so you have some sort of isolation.

 

[Reloadron]

My experience with the old parallel ports is they can be anywhere between 3.3 and 5 volts for a logic high. They also do not source / sink much current at all. If you are worried use a low power opto-coupler or use a circuit similar to the one found in this PDF file.

Ron

 

[J_Nichols]

**broken link removed**

Uploaded with ImageShack.us
there you have the schematics. Some improvement?

The transistor is IRF-840
The relay is a solid state relay

 

[ericgibbs]

hi,
I guess you realise that image is for the serial port.?
The IRF Vgs threshold is at ~4V, I suspect thats why the have chosen serial port drive.

Also there is no suppression diode across the relay coil.

 

[crutschow]

Also there is no suppression diode across the relay coil.

Eric, I guess you missed that it is a solid-state relay and thus needs no diode.

 

[ericgibbs]

Thanks Carl, I just looked at the schematic.

 

[J_Nichols]

I forgot to say that the DC voltage of the battery is 12 VDC. The serial port could be around 5 VDC.

For that I'm asking if I can damage the serial port due:
1) I'm working with 12 VDC and the serial port is aroun 5 VDC
2) In the "output" of the relay there is 250 VAC

 

[Reloadron]

The attached schematic is from the link I posted earlier. The relays used are 12 volt coil relays that can switch 250 VAC @ 10 Amps.

Another solution would be the use of an SSR which would afford isolation for the parallel port from the load using opto coupling of the parallel port signal. The problem with this is the parallel port can't sink or source current very much.

The guy with an excellent handle on the parallel port and has written some great software to control it is Eric Gibbs who is active in this thread.

To answer your question....No. Just make sure you wire it correctly. I would place a snubber across the relay contacts.

Ron