1
\$\begingroup\$

I'm paranoid right now because I made another circuit meant for the parallel port of the PC. While all my previous experiments involving the port were successful (except for one experiment I made a long time ago), I feel I need to make some sort of protection circuit just in case my circuits in the future decide to blow the parallel port.

So my question is, if I made a miniature parallel port extension cable that's about a couple inches long and replaced all of the wires in the cable with fuses, would this be sufficient enough protection for all computers?

I'll explain by example:

If my circuit accidentally sends a logic high output to a data pin on a parallel port hardwired in SPP mode that outputs a logic low, I can see something blow up, but I'm thinking adding a fuse in series with this connection so that this "accidental" connection will blow up the fuse instead of the electronics in the computer. Is my theory correct on this?

If fuses work, what ratings and values would be best?

My circuitry runs on 5V maximum and will likely draw less than 150 mA current.

Also, can I apply this idea to the serial port (or any other port) as well but with fuses of different ratings to match the serial port?

After all, its much cheaper to blow up fuses than it is to blow up my circuit or the PC itself.

\$\endgroup\$
1
  • \$\begingroup\$ In the past we saw protection units that were placed between expensive cash ATMs and other bank equipment and RS232 serial modems connected to telephone lines. They had on each signal pin a pair of 25V Zenner diodes and a 50mA fast blow fuse. The fuse was on the modem side and protected the diodes in the event of voltage spikes. RS232 is able (A, B and C versions at least) to accept +-25V on any pin (input or output). TTL is a bit fussier and most PC ports are CMOS or HCT type input that do not like spikes much. \$\endgroup\$
    – KalleMP
    Commented Mar 25, 2019 at 17:04

2 Answers 2

Reset to default
2
\$\begingroup\$

The classic parallel port (SPP) used to use 74LS374 TTL IC, which can drive out about 2.6 mA at TTL-level (2.4-3.2 V, it has a built-in 100-Ohm resitor), and sink up to 24 mA (to TTL 0.4 V level). If shorted to 5 V and turned low, the max current is no more than 130 mA. Since early ages all integrated parallel port interfaces use the equivalent drive capabilities. Therefore it is very difficult to burn down the parallel port, one has to have a very special skill to do so.

But if you are really paranoid as you say, you can use any low-current fuses that would keep your load running, there are 10-20-30-50 mA etc. rated fuses.

\$\endgroup\$
1
  • \$\begingroup\$ I think a long time ago I wrecked a parallel port because two wires were accidentally shorted together on my board leading into the parallel port, but that was also the time when my PCB making was sub-par. So I guess a 50mA fuse is ok then. \$\endgroup\$
    – Mike -- No longer here
    Commented Jun 23, 2018 at 19:05
2
\$\begingroup\$

If you really want to protect the parallel port, which already has a lot of protection, it may be wiser to use diodes and a buffer (although not as prototype friendly). That way if you do have a over current the diodes will protect it, if it is really bad then it will blow out the diodes and not the port.

enter image description here Source: http://www.iq-technologies.net/projects/pc/007/

\$\endgroup\$
3
  • \$\begingroup\$ Don't tou mean over voltage? \$\endgroup\$
    – Oskar Skog
    Commented Jun 22, 2018 at 7:49
  • 1
    \$\begingroup\$ This is if you want to read something. What if you want to drive something? :-) \$\endgroup\$
    – Ale..chenski
    Commented Jun 24, 2018 at 0:03
  • 1
    \$\begingroup\$ @OskarSkog In my view it's the current that will kill the port, the port already has over-voltage protection and can tolerate some abuse. If it's a low impedance source that can source current, then thats when the real problems begin. A 10k overvoltage would have much different results than a 10 Ohm source. \$\endgroup\$
    – Voltage Spike ♦
    Commented Jun 25, 2018 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.