When i build this program it built successfully. Then, i programmed it into a 16F84a but when i place an led in RB0 with correct resistor nothing happen!

I am trouble shooting my circuit

Q>how i know if the crystal i bought 4MHz is working or not?

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Hesham Ismail Mohammed Sharif
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It would appear that you have the wrong oscillator type selected in your __CONFIG line. A 4MHz crystal should have XT selected, not LP. Your config line
should read:
The program can't get much simpler. If the LED is wired correctly, the MCU has 5V power and ground connected and your MCLR pin is pulled high with a 10K to 33K resistor then it has to work.

Here's a basic 16F84 breadboard you can have a look at to compare with how yours is set up.

Your crystal needs two (typically) 22pF loading capacitors to work properly. Wire them like this (not a 16F84, but same idea):
bleh.jpg

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I change the codes you suggested still not working.

I will post the schematic that i am following

Attached Images

LEDwith16F84a.JPG (82.8 KB, 10 views)

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Hesham Ismail Mohammed Sharif
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There's the problem. Like I said previously, MCLR must be pulled high with a resistor. Your schematic shows it pulled low. When MCLR is held low the PIC is in RESET, and halted. No program runs with MCLR pulled low like that.

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Q> What you mean by pulled high and pulled low and when do i use pulled high or pulled low?

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Hesham Ismail Mohammed Sharif
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Wow it worked i connected MCLR to 5V and it worked i guess that was the problem.
Thanks futz for giving me guide

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Hesham Ismail Mohammed Sharif
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A pull up or pull down resistor is used to create a default value or state for a signal. Now you have MCLR wired to +5V. If you connect an ICSP it will not work because the signal is stuck at +5V. If you were to put a 10K or 20K resistor between MCLR and +5V then MCLR would be +5V when no other signal was present. But because of the resistor a signal from an ICSP connector could change the value to VPP or GND as needed.

A pull down works the same way but the default become GND.

Most often the an input will have a pullup or pulldown and one or more other signals that will be tristated (turned off or made an input) when not in use.

HTH

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In case you didn't understand 3v0's pullup/pulldown explanation I found another good explanation on the interweb.

And, like 3v0 says, don't connect MCLR directly to VDD. Use a pullup resistor between the MCLR pin and VDD. 22K or 33K are good, but it isn't critical. Anything from 10K to 33K is fine.

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This is really just personal choice, PIC's are designed for a direct MCLR connection if you wish, but obviously it prevents ICSP - if you don't use ICSP, then you don't need a resistor - if you do use ICSP, then you need a resistor and a diode.

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I was under this assumption as well but it seems that later pics shouldn't have MCLR connected direct to Vdd.

From the 16F876A data sheet,
Mike.

 

Sorry for my basic question

Q> what is ICSP?

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Hesham Ismail Mohammed Sharif
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In Circuit Serial Programming

Means that you connect a 5-wire cable to the target circuit to reprogram it in place, rather than pry the PIC out and take the chip to the programmer.

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Figure 14-5 in the datasheet gives an upper bound for the value for one of the resistors and a lower bound for the value of the other, but doesn't seem to indicate what the capacitor value should be. Any ideas? or have I just not looked carefully enough?

 

I think setting a resistor <40k is neccesary between MCLR and vdd now.

I searched the datasheet of 16F84A as david said a capacitor must be included and also i found out that another resistor and diode is needed.

Q> What is the value of C and diode that i must used?

Attached Images

MCLR.JPG (15.3 KB, 3 views)

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Hesham Ismail Mohammed Sharif
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The diode and second resistor isn't really necessary. Just put a 22K or 33K (or thereabouts) resistor between MCLR and VDD.

A 0.1uF capacitor between VDD and VSS is a good idea. Place it as close to the chip as possible.

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