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Question about MCLR in PIC16F88

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coolchip

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how to connect ads on pic16f88

If the MCLR is internally tied to Vdd by changing CONFIG1 bit 5 ( MCLRE) , then is there still a need to pull up the MCLR to 5v ?
thanks:D
 

Code Warrior

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Re: MCLR in PIC16F88

Hello,
Instead of leaving the pin to float it is better to use resistor.

Bye
 

    coolchip

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coolchip

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Re: MCLR in PIC16F88

Code Warrior said:
Hello,
Instead of leaving the pin to float it is better to use resistor.

Bye
If the MCLR pin is internally tied to Vdd (+5V) then is there a need for pulling it up externally?
 

Code Warrior

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MCLR in PIC16F88

Hello,
If MCLR is internally tied to VDD then I think there is no need for pull up resistor.

Bye
 

C-Man

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Re: MCLR in PIC16F88

It is always a good idea to have no floating pins in your design.

So if a pin is not used program it as input and connect it to GND or +5V this can prevent a lot of ESD problems later.

Just my experience ...

best regards
 

VVV

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Re: MCLR in PIC16F88

Whenever you do not use pins, you can either leave them as inputs and connect them to GND of Vdd, or you can make these pins outputs and so they do not need to be connected to anything.

Personally, I favour this second approach, since the pins are not connected to anything, so if I need one more pin to do something, then I just run a wire and use the pin, without having to disconnect anything.
Plus, many times during the debugging process I find it useful to toggle a bit at a certain poin in the code, so that I know when a certain event takes place, I can tell where the code is, so again, an unused and unconnected pin comes in handy.

Now to answer your questiom: when you make the /MCLR internal (connected to Vdd), the actual pin of the PIC becomes available as a general purpose I/O pin. This is nice, since in an 18-pin package, pins are scarce.
So, with a pin that is a general-purpose I/O now, you have to do one of the things I described above. Personally, i would make it an output and leave it unconnected.
But you cannot just leave it as an input and unconnected.
 

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