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I have this circuit, where I would have expected vop to be zero, because I have an op-amp behind it. Using an op-amp, the rules said that the voltage drop between the inputs is zero, if it has negative feedback like in the given case. Instead, vop is -5V, the same as V1. Does somebody maybe know, why vop is -5V instead of 0V? Thanks in advance for a response! enter image description here

Is there a way in which I can generally analyse a circuit containing an opamp, like replacing the (ideal) opamp by electric components? Because I don't always know whether the negative feedback exists so that I would be able to use the rules like differential voltage between the inputs going to zero etc.

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    \$\begingroup\$ For a DC, the C1 capacitor will act just like an open circuit. Thus, no negative feedback for DC. So, vop = V1 = -5V \$\endgroup\$
    – G36
    Aug 28, 2022 at 11:19
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    \$\begingroup\$ Well, the transfer function for the opamp is given by: $$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{out opamp}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=-\frac{\text{R}_2+\frac{1}{\text{sC}_1}}{\text{R}_1}\tag1$$ So, when \$\text{s}:=\text{j}\omega\$ we get: $$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\sqrt{1+\left(\omega\text{C}_1\text{R}_1\right)^2}}{\omega\text{C}_1\text{R}_1}=\sqrt{1+\frac{1}{\left(\omega\text{C}_1\text{R}_1\right)^2}}\tag2$$ \$\endgroup\$ Aug 28, 2022 at 16:28

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Does somebody maybe know, why vop is -5V instead of 0V?

Just think about it; you have -5 volts at the input to R1 and expect vop to be 0 volts <-- that implies a current through R1 of 500 mA. Then ask yourself, what component on the op-amp side of the circuit could possibly take 500 mA in order to keep vop at 0 volts?

I mean, you haven't even got proper DC negative feedback because the feedback resistor is interrupted by the series capacitor (C1) so, even if the OP777 (a micropower op-amp) could muster 500 mA at its output (it won't of course), it just ain't gonna happen.

If instead you used 1 kΩ for R1, 500 Ω for R2 and shorted the C1 out then it would work and, the voltage at the op-amp output would be +2.5 volts. This then ensures that vop is a proper virtual ground and, is at 0 volts.


I think the presence of the op-amp in what is a simple potential divider circuit confuses the issue. Here is the basic circuit for achieving 0 volts at the centre of a potential divider: -

schematic

simulate this circuit – Schematic created using CircuitLab

And that is all that the op-amp is doing; it adjusts its output until the same voltage appears at the inverting input as that set at the non-inverting input. It uses the two inputs to measure the voltage difference and, because the op-amp has high open loop gain (several hundred thousand to tens of million), its output closes in on the "right" value to create balance. It's a basic control system.

So, if you analysed the diagram above, there is an end-to-end resistance of 7.5 Ω with a total volt drop of -5 volts minus +2.5 volts (= 7.5 volts). This has to mean (using ohm's law) that the current flow is 0.5 amps. In other words, if the op-amp could supply 500 mA at its output terminal, then the circuit would "balance".

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  • \$\begingroup\$ thank you for the the answer, but integrator also has capacitor in the negative feedback and the rules apply, so I thought it would be the same case here. \$\endgroup\$
    – n328
    Aug 28, 2022 at 10:30
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    \$\begingroup\$ Integrators (as a stand-alone circuit) also need a resistor in the feedback path. Otherwise, they are saturated as well. However, very often integrators are part of a stabilizing DC feedback loop. In this case, such a resistor is not required. \$\endgroup\$
    – LvW
    Aug 28, 2022 at 13:11
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If the capacitor was discharged at t=0 and the op-amp was ideal, you'd have an output of +2.5V @ t=0 (to make the input 0V). Think of the (discharged) capacitor as a short circuit for this infinitesimally short time.

The voltage across the capacitor would subsequently change by 0.5A/100uF = 5kV/s. So, at t=1s the output should be +5002.5V to maintain balance at the inputs. And about double that a second later. And so on, ad infinitum.

A real op-amp is unlikely to be able to supply either the current or the voltage to maintain a differential input voltage near zero for more than a very short time.

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    \$\begingroup\$ @Andyaka Thanks, signs swapped. \$\endgroup\$ Aug 29, 2022 at 11:39
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When an input of -5 V is applied to the input of the integrator it will perform its function and integrate, the op amp's output ramping up. During the ramping up period negative feedback will keep the op amp's inputs at almost the same voltage as each other and so during the period whilst the output is ramping up Vop = 0 V.

Once the op amp's output reaches its saturation limit (close to +15 V) the output can rise no further but the capacitor continues to charge. There is a charge current flowing into the capacitor because there is a voltage across its series resistor R1. So the capacitor continues to charge and its left hand side ramps down until it reaches - 5V. The capacitor is now as charged as it can be, being limited by the power rail and the -5 V input.

If you were to now take the input to +5 V (up from -5 V) then the left hand side of the capacitor would ramp up (charging through R1) until Vop reaches 0 V at which point it would stay there due to negative feedback kicking in and now the op amp's output leaves saturation and starts to ramp downwards.

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