Circuit idea: Add so much voltage to the input voltage as it loses across the capacitor
Current-to-voltage C integratorEdit
What is the most elementary electrical integrator? Of course, this is the humble capacitor. So, if we drive a capacitor C by a constant current source I, it acts as an ideal current-to-voltage integrator with a current input IIN and a voltage output VOUT = VC. Note that the output voltage changes linearly through the time.
Voltage-to-voltage RC integratorEdit
Only, we need usually an integrator with voltage input and voltage output (voltage-to-voltage integrator). For this purpose, we can build a compound voltage integrator just connecting a voltage-to-current converter (a resistor) before the integrator:
V-to-I converter + I-to-V integrator = V-to-V integrator
Only, a problem arises here - the voltage drop VC across the capacitor C "enervates" the input voltage thus decreasing the input current. As a result, the output voltage changes exponentially through the time.
Active RC integratorEdit
Remember what we do in real life when an obstacle stands in our way - we remove it by an equivalent useful "antidisturbance". Following this recipe, we may remove the "harmful" voltage VC by an "antivoltage" -VC. That means to connect an additional varying voltage source and to make its voltage equal to -VC. As a result, the "harmful" voltage VC disappears and the upper point becomes a virtual ground! The compound current source VIN,R is "fooled": it doesn't "understand" that there is a capacitor connected; it "thinks" that its output is shorted.
But where to take an output from? We have three possibilities.
First, we might use the old output; but we have already destroyed this voltage!
Second, we may use the "original" voltage as an output. It is possible but bad solution to connect the load across the capacitor for two reasons: the load has to have a differential output; the load will shunt the capacitor thus affecting the current.
Finally, we can use the "copy" voltage as an output! What a great idea! First, the load will be connected to the common ground; second, it will consume energy from the supplementary source instead from the input one!
Building an op-amp inverting integrator - go to step 3
Op-amp inverting RC integratorEdit
Finally, we have only to replace the varying voltage source with a real one. Now the op-amp doses the voltage of the power supply thus producing a compensating voltage -VC. In other words, the combination of an op-amp and a steady battery acts as a varying voltage source.
The op-amp "observes" the potential of the virtual ground (the difference between the two voltages) and changes instantly its output voltage so that this point to stay always at zero volts. Doing that, the op-amp compensates the "harmful" voltage drop across the capacitor by copying and adding it to the voltage of the input source; doing that, the op-amp "helps" the input source.
How to make an infinite large capacitanceEdit
It's time to make conclusions. What have we actually done here? How does it operate? What is the final result?
The answer is amazing: we have made a "botomless" capacitor having infinite capacitance. Looking from the side of the input source this "ideal" capacitor acts just... as a piece of wire...?!? What do you think about this speculation? Is it always right?
We are already true magicians as we can transform any imperfect component into an almost ideal one! In this lab, we have transmuted the "imperfect" capacitor into a perfect infinite one having no any voltage drop across it although a current flows continuously through it! But with the same success we can make an ideal diode, zero resistance, etc... For this purpose, we just incorporate a varying battery to the imperfect component that compensate the losses inside the component. Remember: all the op-amp inverting circuits exploit this clever trick.