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#1
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| My Compaq Evo N600c laptop has some of it's parallel port pins @ 5.0v some @ 3.3v and the rest are grounded @ 0v. The question is will it do any damage to my parallel port by using pullup resistors on the breakout board so that pins 1 to 17 are at 5.0v ? I also noticed that pin 20 is @ 5.0v too will it be OK to ground that one as the design of my breakout board requires that pins 18 to 25 are commoned to ground? |
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#3
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| Although I cannot say specifically how a "Compaq Evo N600c laptop" parallell port works, I maybe can help you with some hints anyway. The thing that might hurt a parallell port is current. I guess you should be safe sinking or sourcing 5 mA from/to each pin on the port. If you use 1 kOhm or higher valued resistors to pull up to 5 V or down to 0 V, the current will not exceed 5 mA and you should be safe. This is supposing your breakout board does not load the pins further. (Use 2k resistors to be sure, perhaps?) I'm not sure how effective this will be however since if the parallell port can source this current, all the potential will be over the resistor, meaning you will still have 3.3 V at the pin... ??? It seems very strange to me that pin 20 is not at ground potential. The parallell interface is specified to use pin 18-25 as ground. Many parallell printer cables simply short these pins together - so the port should be able to take a short to ground on that pin. No guarantees, however! Arvid |
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#4
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| The reason you are seeing 5V on pin 20 is because your meter has a high input impediance. If pin 20 was shorted to ground through a amp meter I bet you will only see milliamps and this is probably some kind of loop to let the PC know something is connected. I know you are concerned about blowing your laptop LPT but I can assure you this pin gets connected directly to pins 18-25 on all printers I have seen. Darek |
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#5
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| Thanks Arvid and Darek, you're both right by the way. Pin 20 is connected directly to pins 18 - 25 and the plug shield casing in my printer cable so I'll be doing the same in my breakout board with 4K7 pullup resistors, though I'll probably go the opto route anyway. I was measuring the voltage at pin 20 with a high impedance oscilloscope, hence the voltage, but I'm still a little confused as the parallel port is supposed to be able to source 2.5mA and sink 20mA on any pin not connected to ground. I'd really like to know what else pin 20 can be used for, as 1 gound pin and the shield would be more than enough for my use. |
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#6
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| You source through the pull up resistor that is why you are limited to 2ma. Above 2ma the voltage drop across the resistor would make the output to low. You sink through the transistor junction and I limit is determined by the device. Darek |
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#7
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| Got it, thanks Darek. I'm an ex-army radio telecoms tech but came out 25 years ago and have been a computer engineer, off and on ever since, but trying to get the little grey cells back into componant level electronics again is more difficult than I first imagined. It's surprising just what you can forget in 25 years... but I'm trying, mostly other peoples patience |
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