I'm programming a pic 16f870 with a pickit2. all has been working fine till today. I now get a message "programming failed at program memory address oxoooo5c) (or 0x000033). I've checked everything and recompiled and reloaded the hex but still gives same message. I do not understand how to read hex or if it even can be read so I don't know where to go from here. Help would be greatly appreaciated
PS Chip is not very old (programmed maybe 30 times)
AGCB!! I had a similar issue last week.. I was trying to program a pic12f675...I have done this a million times, but out of the blue it failed. I tried about 10 times, until I noticed the voltage was sitting at 4.5V instead of the normal 5.0V... I changed it back and now it works... The thing was, I never changed it, and I'm the only one who uses it.
Yes. I checked the voltage and everything I can think of. Shut down everything and reloaded. Searched for what 3VO said but can't find anything. I just programmed this earlier today, several times and all worked great, but now keeps coming up this 0x00033 location problem. I don't know what to do next.
Have you tried the chip? Does it work? Is it really a faulty location.
I get around it by having a different programmer as a back-up and re-programming the chip.
I tried a different chip and it does the same but with a different location. Sometimes 0x0000. Tried a different usb port and troubleshot pickit2. Same thing
No. It's a clone. When I bought it I did not know about such things. Where is the best place to buy a genuine. Microchip denied my credit card for "general" reasons. Whatever that is. Guess I did not spend enough money with them for the "bother"!
Seems to work if I open the programmer and import the hex before inserting the chip and turning on Vcc. It worked for 13 months and programmed at least a thousand times without a glitch till yesterday.
1. You may have worn out the micro.
2. You may have damaged the micro.
3. You may have "locked-up" the micro. - it may not be re-programming.
Try shorting all the pins as the slightest remaining voltage on the chip will keep it locked-up.