18M2 issue

magnustullock

New Member
Hi,
I have an 18M2 with a script which has run successfully in the past. HOwever, after not looking at it for a number of years I think I may have fried my chip and wondered if anyone could advise?

the 18M2 chip has b.0 to b.7 as output, c.0, c.1 and c.06 and c.07 as output
c.4 is input with an interrupt on high
c.5 is input with an interrupt on low

I could not understand why my script was not working until I realised my interrupt on low was did not have an input as default. I then stupidly connected my c.5 to a 9V input and not the 5V from the voltage regulator I use to drive the chip. Now the chip does not see an interrupt at all on c.5 (output pins still seem work). will the 9V have damaged the input pin?

thanks
Magnus
 

inglewoodpete

Senior Member
From my experience (and embarrassment:rolleyes:), PIC chips do not survive voltages over about 8v on their power supply or input pins. Anything over about 6v is high risk.
 

magnustullock

New Member
thanks everyone. I was a bit confused as I thought a 9V input on c.5 would have killed the whole chip. However, the output pins are all working. Does that mean that each pin essentially works in isolation when it comes to input voltage?
 

Buzby

Senior Member
.... Does that mean that each pin essentially works in isolation when it comes to input voltage?
It depends on the voltage !.

A slight over-voltage ( maybe 6 to 12v ) might just destroy one pin, but a bigger over-voltage is more likely to destroy the chip.

Even more over-voltage, like 240v, will explode the chip, melt the PCB tracks, and stink the room out.

Guess I how know that !

Cheers,

Buzby
 

hippy

Technical Support
Staff member
I was a bit confused as I thought a 9V input on c.5 would have killed the whole chip.
You might have got lucky because C.5 (leg 4) on an 18M2 is the MCLR pin on the underlying PICmicro. That doesn't have a clamping diode to V+ and is, I believe, used for delivering a programming voltage greater than V+ when being production line programmed.

The damage done may therefore be more limited than if 9V had been applied to any other pin.

One of the problems with breaking something is that one can never tell exactly what damage has been done and what effect it may have. That doesn't mean you must stop using it but it does mean you should take care and have to consider it could catastrophically fail at any time, worst case would probably be if it were to short all its pins to V+. So it might be wise to stop using it rather than use it in a situation where failure would cause cascade failure.

Used in a project where 'what's the worse which could happen?' doesn't actually matter it is probably fine on the grounds it might fail but there will be no damage done.

For projects where damage to other components could occur it's probably best to move to a new chip.
 
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