Elerion, here is the promised simple analysis of your active integrator circuit (AIC):
The diagram above shows the same circuit performing two functions: low pass filter (LPF) and active Integrator Circuit (AIC)
You have already derived the formulas for the Shunt Feed Back Low-Pass filter correctly, for a sine wave input, so no need to describe that further. The characteristics for the low-pass filter are given anyway in the box below the schematic.
ACTIVE INTEGRATOR CIRCUIT DESCRIPTION
(1)
Caveats
(1.1) The 1M Ohm feedback resistor has been omitted for clarity; it plays no part in the fundamental operation of the active integrator
(1.2) Assume supply voltages for the AIC of -5V and 5V with 0V being the circuit reference.
(1.3) R2= 1K Ohms, C1= 1uF
(1.4) The the input square wave changes state when the AIC output reaches -2V and 2V
(1.5) The AIC input square wave has excursions of +2V and -2V with 0 sec rise and fall times.
(2)
Theory
(2.1) The relationship between charge, voltage, time, and current for a capacitor is given by:
Q = CV = IT ...(2.1.1)
Where Q is the charge in C in Coloumbs (1 Coloumb = 6.2415 *10↑18 electrons)
C = the capacitor value in Farads
V= the Voltage across the capacitor
I = the current (constant) flowing into the capacitor in Amps
T = time in seconds
Take an example:
C = 0.01F (10mF)
V = 5V
I = 0.2A (200mA)
T = unknown
From Q = CV = IT you get T= CV/I ...( 2.1.2)
Then substituting you get, T = (0.01* 5)/0.2 = 0.25 seconds (250mSec)
This tells you that if you fed a 10mf capacitor with a constant 200mA for 250mSec the voltage across the capacitor would increse linerarly, in a ramp, to 5V.
Q = CV = IT is an extreemly useful formula and you can use it for numerous situations involving capacitors.
By the way, the inverse is the relationship for an inductor:
Q = LI = VT (where V is a constant Voltage) ... (2.1.3)
(3)
Theoretical Operation of the Active Integrator Circuit
(3.1) Assume the the following initial conditions:
(3.1.1) AIC IPV = 0V
(3.1.2) AIC OP = 0V
(3.1.3) Thus the voltage across C1= 0V
(3.2) The AIC IPV Jumps to 2V in zero time
(3.3) Thus, the current through R2 jumps to R2/VIN = 2V/1K = 2mA
(3.4) In order for the AIC opamp to be at ease, it must remove an equal current current from the virtual earth point. It does this by making its output fall and generate a matching current through C1. To keep the current through C1 flowing the opamp output voltage must keep falling at a linear rate (dV/dT)
(3.5) By definition, the current flowing into the virtual earth point is constant so it follows that the current flowing out of C1 has exactly the same magnitude.
(3.6) The left end of C1 is at virtual OV and the right side of C1 has a constant current flowing out of it so from formula (2.1.2) a linear negative ramp is generated. Its aiming point is - infinity. And that is where it would go if nothing stopped its progress. Note that the opamp has no idea what voltage is on the right plate of C1, all it knows is that it must extract a constant negative current from its virtual earth point to keep its two inputs at equal voltages.
(3.7) So, practical aspects aside, the right plate of C1 could be at -100V and the AIC opamp would have no idea. It would continue making its output voltage more and more negative to extract exactly the same current from its VEP. It would do this forever if left to its own devices.
(3.8) You are probably thinking, that's all very dramatic, but you have removed the 1M Ohm feedback resistor; that would let the AIC opamp know what the output voltage is. No it wouln't: at -100V output the 1M resistor would have 100V across it so the 1M resistor wolud only remove 100uA from the VEP. The other 2mA-100uA = 1.9mA woud still be required to keep the AIC opamp inputs at 0V. In, fact the 1M resistor would only stop the ramp negative progress at -2,000V when it would exactly drain an equal curren out of the VEP as in flowing in. At that point, ignoring the fact that the 1M resistor would have arked over at about -200V, it would be dissipating 4W.
(3.9) The curent through the 1M resistor is proportional to ramp voltage and this illustraytes the exponentional function that it adds to the ramps pure linear shape. Not only does the 1M resistor do no nothing in practical terms, but it also distorta the ramp.
(3.10) Back to reality; of course, the ramp would stop abruptly when the AIC opamp output saturated, probably around -3V5 for an LM358.
(3.11) But the AIC opamp output would never saturate because of caviat (1.4)
(3.12) Assume that when the ramp reaches -2V the ISTO switches over and produces -2V at the AIC input. Now there is 2mA flowing out of the VEP so the opamp out voltage output starts rising linearly at a rate sufficient to ensure an exact 2ma flowing into the VOP and so the cycle will continuue for ever.
(4) Practical Aspects
(4.1) The AIC will operate indefinately given perfect components. As this cannot be the case, because real-life components are not perfect, the circuit is fraught with problems and almost impossible to characterise. The normal outcome, even with the 1M feedback resistor, is for the output to slowly drift towards the positive or negative supply rail and stay there.
(4.2) Your latest circuit takesd care of all the prolblems- that is why, as you say, it works so well.