| ??? 08/23/10 18:04 Read: times |
#178249 - It's a correct feedback loop Responding to: ???'s previous message |
M said:
It mentions that the LED current If increase with the Vin (dc). However, the input opamp is basically in an open loop configuration (the Vin is dc). This would mean its output would swing to either +ve or -ve rail. If that is the case, how can the If current depend on the Vin? The feedback forces the inverting input of left OPamp to 0V. To do this, the current flowing through R1 must flow into pin 3 of LOC110. To allow this, the fotodiode must become low ohmic enough, which is provided by increasing the current through the LED. By this is a full loop with negative feedback is formed. The cap is used to compensate for the stray capacitance at inverting input of OPamp (junction capacitance of fotodiode). It provides a phase lead and restores the phase margin of OPamp. Kai Klaas |
| Topic | Author | Date |
| Wrong concept yields wrong simulation? | 01/01/70 00:00 | |
| GIGO indeed | 01/01/70 00:00 | |
| COntrol loop | 01/01/70 00:00 | |
| Simulation circuit | 01/01/70 00:00 | |
| Bias needed | 01/01/70 00:00 | |
| OP07... | 01/01/70 00:00 | |
| It's a correct feedback loop | 01/01/70 00:00 | |
| Biasing and dual supply for opamp | 01/01/70 00:00 | |
| Find out Why.. | 01/01/70 00:00 | |
| Not required | 01/01/70 00:00 | |
Drop of photodiode? | 01/01/70 00:00 | |
| You don't need the additional parts | 01/01/70 00:00 |