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Help with a keypad connection to GPIO pins

I am hoping someone in this group can explain why, the below, is ok to do. I'm a little cautious in implementing it because I already did something similar (without pulling up the 4-lines high with resisters) and it appears to have shorted the ports on my development board. I don't want to make another costly mistake! If you can respond to this...I'll be very indebted!

I'm attaching a schematic of what Zilog recommends in terms of how to connect a keyboard to the GPIO ports. The problem I have is that I don't understand it. Specifically:

Looking at the schematic - If you imagine that the GPIO pins PB4 - PB7 are output and set to high...and that the pins PB0 - PB3 are set to input. And that by pressing a key you make a direct connection from PB7 out through line 1 of the keypad then it would see that you would have a problem. with the current going through PB0. Because it is being sent through PB7 through the keypad and out of line 1 and into PB0 directly (there is no resister on that end) you would have a problem, wouldn't you?

Initially I think I burnt out my development board because I had half of the GPIO pins tied to the input of the other half of the pins with no current limiting resister. It appears that this schematic does the same thing, doesn't it?

Some further information about the pins when they are set to output and input modes is in the other attachment, GPIO modes.jpg.

The problem I'm having is that - based on the table you see in that attachment it appears that there should not be any problem connecting the output of one pin to the input of another. Because it is 'high impedance' it limits the current and, therefore, I don't need a limiting resister. If that's the case - how did I destroy my original development board? I didn't use resisters but I did tie the output to the input of another pin.

Links to the two examples I'm talking about can be found: ftp://

formatting link

Thanks very much for your help!

Reply to
dave.shattuck
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If PB0 is an input, it won't act like it's an output. It acts like a 1MEG resistor, unless you have pullups enabled, in which case it'll act like a resistor of whatever value the pullups are. I don't know which processor you are using (your schematic didn't make it) so I can't help there.

because I had half of

It may, or it may not. Again, no schematic.

This is true

I dont' know. However, it's possible that the processor comes up with all outputs. I know the PICs don't do this, for this very reason; they all start out as inputs.

If you have a software bug where you've made them outputs by mistake, they'll pass as much current as they can.

Another possibility is that the input was whacked by SCR latchup. If you put too much current into an input, and it is higher or lower than the power pins, it can get stuck into a state where the pin is sinking large amounts of current. If you don't catch it and stop it, it'll destroy the input port.

One simple thing you can do is read the datasheet, and find out what the maximum current is into a pin, and put a resistor there that will keep it from getting to that maximum current. So, if the max current is, say,

20mA, and your power rail is 5V, you put a 5/.02 = 250 ohm resistor in front of the port. That way, no matter what you do, you won't be able to hurt it.

another

Reply to
Bob Monsen

Forgetting all else, if PB4 - PB7 are output and set high, you can't get anything useful like this. You must ripple a low (0) from PB4 through PB7 with your keyboard routine. 0111, 1011, 1101, 1110 rapidly. Between each increment, you read PB0 through PB3 for a low input. Pull-up resistors wouldn't hurt anything and is recommended.

Knowing which output bit is low and which input bit read that low allows you to know which key was pressed. Any low that is read should stop scanning and re-read the input to eliminate bounce. Scanning should resume when the inputs are again all high.

The tricky part comes when more than one key is simultaneously pressed.

Reply to
Lord Garth

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