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The capacitor C2 keeps the value of a slow varying signal for some time. The capacitor discharges through the two resistors R3 and R4 with a time constant of \$\tau = RC = 10\mu F \times 1M\Omega = 10s\$. The voltage across resistor R4 should be half of the voltage across the capacitor. This works as expected when the op amp (NE5534N) isn't connected to the circuit.

When V_R4 is used as reference to the inverting input of the op amp the voltage changes to about 80% of the voltage across the capacitor C2. I have not been able to find a solution to this. Am I missing something here?

Op amp comparator

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  • \$\begingroup\$ what's the current that flows into a non-ideal opamp like the NE5534? \$\endgroup\$
    – Marcus Müller
    Commented May 9, 2019 at 17:49
  • \$\begingroup\$ Why do you need a diode if the input voltage lies between 0 to 5V? Actually a PI controller is a better solution if you want to achieve an offset cancellation. \$\endgroup\$
    – Marko Buršič
    Commented May 9, 2019 at 18:26
  • \$\begingroup\$ The diode is to force the capacitor to discharge through the resistors \$\endgroup\$
    – opamp
    Commented May 10, 2019 at 7:14

1 Answer 1

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I don't see any negative feedback on the op-amp. You are exceeding the differential input voltage of the op-amp. According to page 2 of its datasheet the maximum is 0.5V and it has protection diodes so it is drawing a large current to protect the device.

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    \$\begingroup\$ 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 \$\endgroup\$
    – Marcus Müller
    Commented May 9, 2019 at 17:52
  • \$\begingroup\$ 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 \$\endgroup\$
    – opamp
    Commented May 10, 2019 at 7:11
  • \$\begingroup\$ Can you elaborate on what your goal is? \$\endgroup\$
    – Charles H
    Commented May 10, 2019 at 15:40
  • \$\begingroup\$ 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? \$\endgroup\$
    – opamp
    Commented May 12, 2019 at 16:15
  • \$\begingroup\$ 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 \$\endgroup\$
    – Charles H
    Commented May 13, 2019 at 16:24

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