PIC Micro Output I/O

Discussion in 'General Electronics Discussion' started by Rajinder, Sep 21, 2016.

1. Rajinder

535
10
Jan 30, 2016
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,

2. hevans1944Hop - AC8NS

4,630
2,159
Jun 21, 2012
Go revisit Ohm's Law. What is the resistance to ground at the output when the output is low?

Last edited: Sep 22, 2016
3. Rajinder

535
10
Jan 30, 2016
If we have a push pull I/o line

4. BobK

7,682
1,688
Jan 5, 2010
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

5. Colin Mitchell

1,416
314
Aug 31, 2014
No.

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

6. Minder

3,149
680
Apr 24, 2015
The output either switches to +v or common as per the attached dwg.
M.

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7. hevans1944Hop - AC8NS

4,630
2,159
Jun 21, 2012
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.