Timeline for Reference voltage changes when connected to inverting input of op amp comparator
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 13, 2019 at 16:24 | comment | added | Charles H | You should look up the basics of op-amps, you're not using them correctly. This task would be easier accomplished with a comparator or (maybe) a differential amplifier. I don't really understand why you would need to do it though or why the amplitude would be changing | |
| May 12, 2019 at 16:15 | comment | added | opamp | I forgot to mention that the signal will be a square wave that is between 0 and 5v. The max amplitude of the wave can change. My goal with the diode, cap and two resistors is to make a "dynamic reference", such that if the signal changes in amplitude, I will always compare it to half of the recent max amplitude. Is this possible if the max differential input voltage is +/-0.5v? What if I change the two resistors in the voltage divider to 100k, and place an additional resistor of higher value (lets say 500k-1M) at both inputs? | |
| May 10, 2019 at 15:40 | comment | added | Charles H | Can you elaborate on what your goal is? | |
| May 10, 2019 at 7:11 | comment | added | opamp | Thank you so much for the answers! I'm a bit new to working with opamps. Is there a way to configure the current opamp to work as I intend? or should I maybe change it up for a dedicated comparator IC like farnell.com/datasheets/… ? I see that this one has a 15v max differential input voltage | |
| May 9, 2019 at 17:52 | comment | added | Marcus Müller | also it has an input bias current even under proper differential input voltage of up to 1.5 µA – and that's around 1.5e-6 A · 5e5 Ω = 0.75 V that drop over either part of the voltage divider | |
| May 9, 2019 at 17:49 | history | answered | Charles H | CC BY-SA 4.0 |