Basic PIC circuit is not working

Question

I have just begun in the world of microcontrollers, and the first microcontroller. I chose to work with was the PIC from Microchip. The PIC I am using is the PIC16F877A. I am using MPLAB IDE and HI-TECH C. I am trying to make an extremely simple program that turns on an LED. This is the code I am using:

#include<htc.h>
#define _XTAL_FREQ 8000000
__CONFIG(UNPROTECT & PWRTDIS & WDTDIS & HS & LVPDIS); 

int main()
{
    TRISB0 = 0;
    RB0 = 1;
    while(1);
}

When I hook up the PIC to my circuit the LED does not turn on. Here is my circuit diagram and a picture of my circuit:

A few other notes about my circuit:

• I am using a 9 volt battery hooked up to a 7805 regulator for the power supply.
• I have measured the voltage coming from pin RB0 with a multimeter and it measures 0.0 V.
• If there is no problem with my circuit I could have programmed the chip wrong.
• My capacitors hooked up with my crystal are 22 pf, not 22 µf as in the schematic.
• I have put 100 µf capacitors between pins 11 and 12 and between 31 and 32.

Answer 1

As it's impossible to put code in comments, I'll put my suggestion for code that sets the Config bits correctly here. The issue is I don't have this compiler so I cannot be certain of the names (LVPDIS as opposed to, say, LVP_OFF in assembly) but if I'm wrong, someone could comment(?)

#include <htc.h>

#define _XTAL_FREQ 8000000

__CONFIG(UNPROTECT & PWRTDIS & WDTDIS & HS & LVPDIS);

int main()
{
    TRISB0 = 0;
    RB0 = 1;
    while(1);
}

Answer 2

Have you tried checking the clko pin to see if you are getting a clock signal? Hook a scope up to it and make sure the chip is even running. If you don't have a scope or usb logic analyzer, you might be able to see an LED glow slightly from that pin.

Have you measured the Vdd to make certain you are supply the correct power level (between 4.0 and 5.5)?

As RB0 is also an external interrupt pin, try moving the LED to pin34 and changing your code to use RB1 instead:

TRISB1 = 0;
RB1 = 1;

Answer 3

1. There is no bypass cap in sight. Put 100 nF to 1 µF ceramic directly between each power pin and ground.
2. Showing only the pin numbers doesn't show us the pin function. It's impossible to see what you have connected to what pin without looking up the pinout in the datasheet. This information should be in your schematic.
3. 22 µF is about 6 orders of magnitude too much capacitance for the crystal and its drive circuit. This will definitely keep the oscillator from running, and therefore the PIC from running.
4. There is no resistor in series with the LED, so it will draw too much current. That can load down the supply, damage the LED, and/or damage the PIC pin output driver.
5. Note that any pin that can be a analog input wakes up that way by default. I don't remember off the top of my head whether the pin you are trying to use is one of them, but it's a good idea to go to the A/D and/or comparator modules and make sure all pins are configured as digital.