PC port question

Its a few years since I did any of this stuff, but my advice would be, If you are going to use 9V, to use a buffer between the LED and port. Individual transistors will work fine without risking damage to your computer.

If nothing else, the buffer protects the parallel port. So if there's an accidental short on the output, the parallel port survives. It's easier to put some another buffer IC in a socket than it is to fix the parallel port. In the old days, it was merely tedious (you'd have to desolder the dead IC), but nowadays it may not even be possible (because the parallel port is part of a much larger IC that is unavailable to the end user).

Of course, there is always the risk that the wiring to the buffer IC is done wrong.

Michael

Don't.

Graham

Rats. Back to the parts bins.

Thanks everyone.

Not sure if it will do any damage, but I'm sure it wont work (assuming you want to switch the LEDs on and off?)

When the parallel port pin is low current will flow from the +9v though the LED and the LED will be on. That bit should be OK.

When the parallel port pin is high, it will be 3.3V (unless you have a very old PC). You then have 9V - 3.3V across the LED and resistor, so current will still flow and the LED will still be on, though not quite as bright. You also have a problem that this current is going the wrong way (into the parallel port when the pin is high). This may do damage.

PC parallel ports are NOT open collector.

Are you saying that you want to use 1k resistors just because you don't have 330 ohms? You could try the original schematic with 1k resistors in series with the LED in place of 330 ohm. The LEDs wont be as bright, but probably still visible. Another thing you could try is paralleling up the resistors - 3 X 1k in parallel would give 333 Ohms.

Gareth

I didn't know if they were or not; I was kinda hoping...

That's what I'll be doing. I found an old 7805, and I have lots of 1k resistors. Thanks.

I know its academic, but where did you get 3.3 volts from ?

The parallel port uses TTL compatible signals < 0.7v - > 4.8v

and where did you get those standards if may ask? i think many people have forgotten what the real standards are.

if memory serves, -12 = low, +12 high. and the threshold points of +2.5 = high, -2.5 = low. anything between that is undefined.

or has my years of work failed me?

Yes, they've failed you.

"High" and "Low" is in reference to the logic family, rather than an absolute. so your 12V bits would never apply to the TTL family that can never see more than 5v. (And given those voltages, it seems more like you are thinking of RS-232 that does have something like that.)

The logic family likewise defines the threshold point.

But, once this is all in place, the fact that there is a range of voltages for a "high" and a "low" means that you wouldn't know absolutely what a "high" out of a ttl gate would be. It just has to be above the threshold, and it might vary from IC to IC (within a certain small range) and from function to function (because the internal circuitry varies and that affects the exact voltage when the output is high).

Michael

if you say, so, i was talking about the RS-232 port, not the 5 Volt logic family..

have it your way, i'll stick with what i do. thanks.

You almost described (inverted) RS-232. But not quite. Parallel ports ("Centronics", IEEE 1284 etc.) are a different beast all together.

Tim

You are talking serial ports !

If you want to discuss serial ports then fine ! The OP was talking about a Parallel port device !

yes, your right, i was..

Succinctly put !

You have been playing with serial ports to much :) :) LPT port is ttl, with a pull up to Vcc minus one or two diode voltages(between 1 and 2 volts).

yeah well, at least i have something to play with. :)

Determined experimentally using a multimeter and a few PCs.