| ??? 06/04/10 13:39 Read: times Msg Score: +1 +1 Good Answer/Helpful |
#176387 - Thanks for that typo, More for 8KHz with 50% duty cycle Responding to: ???'s previous message |
Thanks for that typo pointed out.
Calculations for 8KHz T0 ISR tick: 11059200 / 12 cycles = 921600 921600 / 8000 = 115.2 round to 115 65536 - 115 = 65421 or 0xFF8D reload values TL0 = 0x8D; TH0 = 0xFF; Try the above reload values and the Timer T0 ISR will tick aprox at about 8KHz. Achived 921600 / 115 = 8013,91 Hz If the required signal on the specified pin has to be at 8KHz with duty cycle 50% then the frequency required should be twice, at 16KHz: 11059200 / 12 cycles = 921600 921600 / 16000 = 57.6 round to 58 65536 - 58 = 65478 or 0xFFC6 reload values TL0 = 0xC6; TH0 = 0xFF; By these reload values the T0 ISR will tick aprox at 16KHz, that is twice the required frequency and if you change the state of the MCU pin at this rate then you have a signal with 50% duty cycle at 8KHz. Achived 921600 / 58 = 15889.66 Hz K.L.Angelis |
| Topic | Author | Date |
| 80C52 Timer0 | 01/01/70 00:00 | |
| where is the EA initialize? | 01/01/70 00:00 | |
| EA initialize | 01/01/70 00:00 | |
| no reason in what you show | 01/01/70 00:00 | |
| TL0 and TH0 | 01/01/70 00:00 | |
| Calculations for Timer T0 at 8KHz: have you tried 0xFF8D ??? | 01/01/70 00:00 | |
| correction | 01/01/70 00:00 | |
| Yeah - 0x8D | 01/01/70 00:00 | |
| Thanks for that typo, More for 8KHz with 50% duty cycle | 01/01/70 00:00 | |
| Put the compiler to work... | 01/01/70 00:00 | |
| fine, but | 01/01/70 00:00 | |
| T0 overflows at 65535+1, don't overlook this | 01/01/70 00:00 | |
| right answer, wrong premise | 01/01/70 00:00 | |
| Exact calculations require semantics | 01/01/70 00:00 | |
| you are welcome to find out | 01/01/70 00:00 | |
Thanks for the correction... | 01/01/70 00:00 |