I'm trying to program a pic16f877a using PICKIT3 and MPLAB without success.
I'm powering the pic with the pickit.
I verified the connections almost 20 times, everythings plugged at the right place , every times when I try to burn the pic I get :
"Too much current has been drawn on VDD. Please disconnect your circuit, check the CLK and DATA lines for shorts and then reconnect." and "target device id (0x0) is an invalid device id. please check your connections to the target device".
I'm using a RC Oscillator with an arduino since the pic does not have an internal oscillator and I don't have any quartz or battery.
I made a quick schematic using paint to show you my connections using an image that I found on another similar thread(How to connect a dsPic30f2010 to pickit3 icsp).

Thanks !


  • \$\begingroup\$ Maybe this? 2.4.4 Debugger Powered The internal debugger power is limited to 30 mA. \$\endgroup\$
    – Tyler
    Commented Jul 19, 2017 at 19:08
  • \$\begingroup\$ The PIC16F877A has only two Vdd and two Vss connections. Your diagram is showing three for both. Can you update your schematic with pin numbers? \$\endgroup\$
    – Glenn W9IQ
    Commented Jul 19, 2017 at 19:15
  • \$\begingroup\$ Sorry, did not see it, it's done. \$\endgroup\$
    – Leosa99 _
    Commented Jul 19, 2017 at 19:19
  • \$\begingroup\$ What is powering the Arduino? \$\endgroup\$
    – Glenn W9IQ
    Commented Jul 19, 2017 at 19:39
  • \$\begingroup\$ An ac/dc adaptator \$\endgroup\$
    – Leosa99 _
    Commented Jul 19, 2017 at 19:41

1 Answer 1


Power the PIC from the same power supply as the Arduino. Remove the Vdd connection to the PIC from the PICKIT3. It may also help to add a 10k resistor from PGM on the PIC to ground. Make sure you have your PICKIT3 set to the correct processor.

You should then be able to program the PIC.


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.