| ??? 02/20/09 13:09 Read: times |
#162594 - Not clear Responding to: ???'s previous message |
Oliver Sedlacek said:
This suggests that the output can be negative. I assume you want to scale the output into the range of 0 to 5V, so when V2 is V1 - 5V then Vout = 0, and when V2 is V1 + 5V then Vout = 5V. When V2 = V1, then Vout = 2.5V. Say V2-V1=Vout . Then as you say "when V2 is V1 - 5V then Vout = 0" which leads to V1-5-V1= -5? .Whereas you seems to convert it right with 2.5V being at the center of the swing. Regards Ap |
| Topic | Author | Date |
| Opamp question | 01/01/70 00:00 | |
| Please specify! | 01/01/70 00:00 | |
| Oops here its | 01/01/70 00:00 | |
| Mistake? | 01/01/70 00:00 | |
| transfer function | 01/01/70 00:00 | |
| Vout = V2 ? | 01/01/70 00:00 | |
| I 'm entirly confused... | 01/01/70 00:00 | |
| I explain it again | 01/01/70 00:00 | |
| So Vout = V2 - V1 | 01/01/70 00:00 | |
| I do have a common GND | 01/01/70 00:00 | |
| So Vout = V2 -15 ? | 01/01/70 00:00 | |
| trying to achieve | 01/01/70 00:00 | |
| Reference | 01/01/70 00:00 | |
| Reference | 01/01/70 00:00 | |
| Much clearer but.. | 01/01/70 00:00 | |
| Not clear | 01/01/70 00:00 | |
What VOut range do you want? | 01/01/70 00:00 | |
| I think this will do the trick... | 01/01/70 00:00 | |
| re: transfer function | 01/01/70 00:00 | |
| re: transfer function | 01/01/70 00:00 | |
| Ohm's law | 01/01/70 00:00 | |
| Oh the internal protection of opamp! | 01/01/70 00:00 | |
| Errr... | 01/01/70 00:00 | |
| Don't confuse "OPamp" with "OPamp circuit"! | 01/01/70 00:00 | |
| Don't confuse "OPamp" with "OPamp circuit"! | 01/01/70 00:00 | |
| If the feedback loop doesn't work... | 01/01/70 00:00 |