If you replace the feedback resistor with a capacitor, you get an integrating amplifier.
In math, an integration operation is basically the area under a curve. If we have a voltage vs time graph, and the voltage remains constant, the integral of that will be the voltage times the time it stays at that level. As you can see, the longer the time the voltage remains constant, the higher the integral will be.
Back to our integrator, as the input voltage is applied to the inverting input via an input resistor that creates an input current. The Opamp will try to compensate the current by creating a voltage across the feedback element enough to make a current flow equal to that at the input to conform to the current rule: the inputs draw virtually no current.
In the simple inverting amplifier, the feedback resistor developed a constant current at a constant voltage at the output with respect to the inverting input, tied to ground. This time however, the feedback element is a capacitor; an element that can store charge, charge that eventually develops a voltage across it as it gets more and more charged.
If we apply a constant voltage at the input, a current flows through the input resistor. This current the opamp tries to compensate by creating a voltage at the capacitor to induce a current equal to that of the input. If the capacitor is initially completely discharged, the voltage across it is 0v, and its "resistance" is infinite since it is effectively insulating both sides so no current flows.
The gain is initially infinite, since Rfb/Rin tends to infinity by action of Rfb being infinity. This makes the output voltage go down quickly in a small amount of time (remember that the opamp is acting in an inverting configuration). As the capacitor starts charging, the charges entering the out plate of the capacitor push the charges on the other side, effectively creating a current across the capacitor, enough to counteract the input current.
As the charges build up inside the capacitor, a voltage develops across it in opposition of the output voltage, making it seem as if less voltage is applied to it, slowing down the amount of charges getting into the capacitor.
Less new charges going into the capacitor causes less charges being pushed out at the other plate. The Opamp tries to compensate by further lowering the voltage.
As you can see, the charges keep building up and the opamp is always trying to compensate by lowering the output voltage. At one point, the opamp will not be able to lower the output voltage, at which point it is said to be saturated.
The rate of charge of the capacitor depends on the current that is applied to it, and the current depends on the voltage and resistor at the input by ohms law I = V/R. The higher the voltage, the faster the capacitor charges and the output going lower, and the lower the input resistor the more current flows, charging the capacitor faster and resulting in the same faster lower output.
This action is the same as in the integration operation: the higher the value of the graph the higher the integral will be in the same amount of time.
Also if the input goes negative, the capacitor starts discharging and the output will go higher to compensate. If at any point the input goes to 0, the current through the input resistor will be zero, and the opamp will compensate by setting the output voltage at the same level as the capacitor voltage, in order to stop it from being charged or discharged.
Similar to what happens in an integration: if the graph crosses 0 and stays there, the integral will be the sum of areas up until that point and stay there for as long as the graph stays at zero. Also, if the graph goes lower than 0 then the integral will go lower because the area will be negative relative to 0.
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