PIC Micro Output I/O

[Rajinder]

Hi,
I have a question regarding the PIC18F4520 micro. If i connect a 10K pull up to one of the pins and make this as an output. I believe that this will be a logic '1'. If i now make this (still as an output) but output a 0, how can the output be a 0 since it is connected to VCC via a 10K Pull-up.

Best regards,

 

[hevans1944]

Go revisit Ohm's Law. What is the resistance to ground at the output when the output is low?

 

[Rajinder]

If we have a push pull I/o line

 

[BobK]

Connect a 10K resistor from the + to the - terminal of a 1.5V battery. Is the voltage at the - terminal now 1.5V? If not, what is it?

Bob

 

[Colin Mitchell]

No.

The micro makes the output HIGH or LOW. The 10k has absolutely no effect what-so-ever.

 

[Minder]

The output either switches to +v or common as per the attached dwg.
M.

 

[hevans1944]

how can the output be a 0 since it is connected to VCC via a 10K Pull-up.

The output is shorted to common by the lower MOSFET when the output is low. The upper MOSFET in the totem-pole (NOT push-pull) output is turned off when the output is 0. The 10 kΩ "pull up" resistor is in parallel with the upper MOSFET (which is NOT conducting) and therefore provides a small current (about 0.5 mA with Vcc = 5 V) through the lower MOSFET (which IS conducting) when the output is 0. This current is NOT sufficient to raise the output to a logic 1 level. In essence, when the output is 0, the 10 kΩ resistor and the conductive resistance of the lower MOSFET form a voltage divider at the output producing an output voltage equal to (Vcc) (Ron/Ron + 10,000). The value of Ron is very low, about a tenth of an ohm or less, so the effect of Vcc acting through the 10 kΩ resistor is negligible.