| ??? 02/26/03 05:20 Read: times |
#40101 - thanks everyone Responding to: ???'s previous message |
The data sheet says the input current for an 80c552 is less than 1 uA, and typically about 100nA. For less than 0.5LSB error, the formula is to take 0.5 * 1024 * Vcc / I
and to keep the resistance LESS than the calculated amount. The example was with 5.12V time 0.5 times 1024 divided by 1uA, which equals 2500ohms. |
| Topic | Author | Date |
| A dumb question | 01/01/70 00:00 | |
| thanks everyone | 01/01/70 00:00 | |
| RE: thanks everyone | 01/01/70 00:00 | |
RE: thanks everyone | 01/01/70 00:00 |