| ??? 10/09/08 16:44 Modified: 10/09/08 16:46 Read: times |
#158942 - What I meant... Responding to: ???'s previous message |
 It's correct that the voltage divider output is connected to the non-inverting input of OPamp. But the output of OPamp should be connected to its inverting input. So, the inverting input shouldn't be grounded. The mistake with the input buffer is, that the input signal shouldn't be connected to the inverting input, but to the non-inverting input, which shouldn't be grounded either. What you have drawn is a current summer, with the input held on virtual ground. The input impedance is nearly zero, because the inverting input is always kept on ground potential by the OPamp. Kai |
| Topic | Author | Date |
| Analog input signal conditioning | 01/01/70 00:00 | |
| voltage follower | 01/01/70 00:00 | |
| Good link that.. | 01/01/70 00:00 | |
| Wont work | 01/01/70 00:00 | |
| It should have been +/- 15V | 01/01/70 00:00 | |
| Excel automates resistor calculation | 01/01/70 00:00 | |
| Thanks. | 01/01/70 00:00 | |
| I would eventually omit the input buffers... | 01/01/70 00:00 | |
| Now that is worrying.. | 01/01/70 00:00 | |
| What I meant... | 01/01/70 00:00 | |
| Inputs swapped | 01/01/70 00:00 | |
| Extremely sorry for the goof up. | 01/01/70 00:00 | |
| Some hints... | 01/01/70 00:00 | |
| Just in time... | 01/01/70 00:00 | |
| As long as... | 01/01/70 00:00 | |
| Yes I understand. | 01/01/70 00:00 | |
Active clamp... | 01/01/70 00:00 |