| ??? 04/13/05 21:51 Read: times |
#91528 - Always an input, no matter how you treat Responding to: ???'s previous message |
Joseph Hebert said:
... write a one to a pin in order to treat it as input. It's not really a matter of treating it as an input - it's always an input. The thing is, if you write a '0' to a pin there is no way that any external stimulus can ever force that pin to anything other than zero[1] - therefore the only thing that input will ever see is a '0'. It is effectively shorted to ground. But if you write a '1', an external signal can force the pin high or low - and thus the input can read the state of that signal. [1] Without destroying the open-drain driver. |
| Topic | Author | Date |
| ports input to ouput | 01/01/70 00:00 | |
| Fundamental questions | 01/01/70 00:00 | |
| Not relevant | 01/01/70 00:00 | |
| Input/Output | 01/01/70 00:00 | |
| Top of pullup?? | 01/01/70 00:00 | |
| Yes, the top. | 01/01/70 00:00 | |
| now you even confused me | 01/01/70 00:00 | |
| A good representation of 8051 I/Os here | 01/01/70 00:00 | |
| What do you think, Karthik? | 01/01/70 00:00 | |
| Always an input, no matter how you treat | 01/01/70 00:00 | |
| I concede | 01/01/70 00:00 | |
| Pull down | 01/01/70 00:00 | |
I am clear now thanks to everybody | 01/01/70 00:00 |