| ??? 08/26/08 19:54 Read: times |
#157779 - Detecting Both Edges - A very old trick Responding to: ???'s previous message |
One way to detect both edges of an input signal is to put an XOR gate between the interrupt source signal and the INTR_N input of the MCU. The gate output goes to the MCU. The other input of the XOR comes from another port I/O pin set as an output high or low. If this port pin is low then the MCU's negative sensitive interrupt input will trigger from a H-->L of the interrupt source signal. If the port pin is high then the MCU's negative sensitive interrupt input willl trigger from a L-->H of the interrupt source signal.
The scheme is now to toggle the port pin output bit each time an MCU interrupt comes. Now an interrupt will be generated for each transition of the source signal.......as long as the source signal is not faster than the MCUs interrupt response time and the time your code spends in the interrupt service routine. Michael Karas |
| Topic | Author | Date |
| pls help....8051 external interrupts | 01/01/70 00:00 | |
| show definition of "a" variable | 01/01/70 00:00 | |
| definiton : sbit a=P3^2; | 01/01/70 00:00 | |
| Need more information | 01/01/70 00:00 | |
| the entire code | 01/01/70 00:00 | |
| keyscan | 01/01/70 00:00 | |
| How the LED are connected? | 01/01/70 00:00 | |
| One tip | 01/01/70 00:00 | |
| P3.2= iNT0 | 01/01/70 00:00 | |
| P3.2= INT0 | 01/01/70 00:00 | |
| pin 3.2 | 01/01/70 00:00 | |
| P3.2= INT0..1 | 01/01/70 00:00 | |
| Obvious Reason | 01/01/70 00:00 | |
| time duration | 01/01/70 00:00 | |
| mistake | 01/01/70 00:00 | |
| ok thanks alot | 01/01/70 00:00 | |
| ? | 01/01/70 00:00 | |
| GATE0/1 ? | 01/01/70 00:00 | |
| no | 01/01/70 00:00 | |
| try | 01/01/70 00:00 | |
| Debounce maybe? | 01/01/70 00:00 | |
| Solution | 01/01/70 00:00 | |
| interrupt flag bit clear | 01/01/70 00:00 | |
Detecting Both Edges - A very old trick | 01/01/70 00:00 |