| ??? 11/06/08 22:00 Read: times |
#159858 - Pins are physically tied to Vcc Responding to: ???'s previous message |
So that yields:
11101110 > 0xEE So isn't that the final hardware address? Or do I shift it over by one and that becomes the address? and if I do shift it, doesn't the upper 1 drop off?..... All of my previous I2C hardware was already spec'd, this is the first time I've had to design in a new device and come up with its address. I read it as if I want to write, it would be 0xEE and if its reading then its 0xEF When it shifts left one time, doesn't the upper bit get dropped? That is where my confusion lies I guess.....thanks for the fast reply Chris |
| Topic | Author | Date |
| I2C Addressing Oddity I'm not quite sure | 01/01/70 00:00 | |
| From the data sheet Slave is 0xExxx0 | 01/01/70 00:00 | |
| Pins are physically tied to Vcc | 01/01/70 00:00 | |
| Also forgot to mention | 01/01/70 00:00 | |
| it depends how do you see it | 01/01/70 00:00 | |
| 7-bit addresses are confusing | 01/01/70 00:00 | |
Got it now, it makes sense | 01/01/70 00:00 |