| ??? 02/15/09 13:53 Modified: 02/15/09 13:54 Read: times |
#162432 - Mistake? Responding to: ???'s previous message |
The transfer function of your mentioned circuit is Vout = V2, assuming that the OPamp can handle the much too high input common mode voltage.
What do you intend? Vout = V2 - V1 ? Kai |
| 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 |