Intergrater Question

When you apply a small positive voltage to the input of the resistor, that doesn't mean the input to the op-ap 'sees' the same voltage. There is a voltage drop across the resistor because there is current flowing through it.

Now, an *ideal* op-amp has an infinite input impedance. So there is no current flowing between the inverting input and the junction point (resistor, capacitor and inverting input to the opamp). But there *is* current flowing to the capacitor.

Pretend for a moment the opamp isn't there and the output side of the cap is just tied to ground. When you apply a signal voltage to the input of the resistor, current begins to flow through the resistor to charge up the capacitor. Simple series RC circuit. As the capacitor charges, the voltage at the point between the resistor and capacitor would slowly rise and eventually equal the input voltage.

But now put the opamp back into things. As the voltage at the resistor-capacitor tie *tries* to rise above ground, the op-amp, sensing a rising voltage on its inverting input will slew its output negative. So as the capacitor charges, and the voltage across it rises, the output side goes negative by exactly the same amount as it charges. This keeps the inverting input 'virtually' at ground potential. (in the ideal case anyway, in real opamps it will shift very slightly).

As more charge builds up on the capacitor (owing to a fixed amount of current flowing in from the applied input signal and resistor), the opamp keeps slewing its output negative more and more in order to maintain the inverting input at 'virtual ground'.

All this time, the voltage across the capacitor is rising at a *linear* rate (not the exponential rate often associated with RC networks). Once the output 'hits the rail', the cap continues to charge. But now the voltage at the inverting input can no longer be maintained at 'virtual ground' and it begins to rise. Also now that we've 'hit the rail', the charging current into the cap begins to decay (the voltage between resistor input and the junction point is slowly decaying).

With just this simple a circuit, once the integrator 'hits the rail', and the inverting input voltage rises above 'virtual ground', the circuit is sometimes said to be 'winding up'. Reversing the input signal's polarity will *not* cause the output to start to integrate right away. It will remain at the negative 'rail' for some time while the charge is gradually bled off the cap. Once the charge on the cap has bled down to the point where the junction is at ground potential, *then* the output will move from the 'rail' voltage as the cap's charge continues to bleed down.

Various designs have been used to prevent or at least minimize this delay in reacting to a change in input signal. In the process-control world, the circuits are referred to as 'anti-windup' circuits. One simple method is to place diodes across the two opamp inputs. This prevents the cap from charging excessively and keeps the inverting input within 1/2 volt (okay, one 'diode drop') of the ground potential.

hope this helps

daestrom

Something was really weird with MultiSim, maybe I had too many programs open the night I typed this question causing it to slow my PC down. After the op-amp went to the rail, there was a delay before the RC started rising. That really threw my understanding of an intergrater circuit off.

I tend to get baffled by the "outside" of the circuit and not think about the simpliest thing "the inputs are 'equal'", that explaination you gave made perfect sense.

You never got to the question about putting a cap from the resistor/capacitor junction (the inverting input) to ground. I saw a circuit with this capacitor and wasn't sure how to "add" that into the equation. Per MultiSim, it only changed things VERY little. Keeping in mind I use an ideal op-amp (no offset voltage, bias current, etc..) so when I calculate something and run it in MultiSim, my answers should be just about identical.

The caps aren't in parallel or series because I tried both and my numbers were out in left field.

Thanks again!