circuit analysisoperational-amplifierpowerpower supplyvoltage
With regard to the picture below, it says with an AC signal as an input, the output is a square wave. I am familiar with the uses of op-amps like in differentiator, integrator, inverting and non-inverting, active filters etc. But those I can memorize their configuration. But this I could not understand.
These are all RC time constants, let's call them tau. (freq. = 1/(2*pi*tau))
For differential circuit tau = R1*C3, (with R1 giving you some gain.)
For the integrator tau = R6*C3, (R5 is resetting the cap with yet another time constant.)
What they mean is that a passive R/C filter can only approximate a differentiator/integrator so long as the time constant is much slower than the signal. The reason for this is that the true behavior of an R/C and R/L circuit is exponential in time e.g. from basic circuit theory, the general response of an RC circuit is $$V=V_0 (1-e^{(-t/RC)})$$ If RC is large, e^x is close to linear for small values of t, yielding behavior close to an ideal integrator/differentiator.
Another way to think about it is to consider the integrator case in 1.15 with a constant voltage input (e.g. Vin = 10 V). We expect the output to be a linear ramp of constant slope (integration of a constant = straight line). However, if RC is too small, what happens is that after a time of integration, V will increase due to the capacitor C charging up. This will decrease the current through R, which in turn decreases the "slope" the output voltage. At some point, when V = Vin the integrator stops working completely. This is how the behavior of the passive integrator deviates from the ideal integrator. Conversely if RC is large enough, the voltage across capacitor C will never get large enough to reduce the current through resistor R, and so the current through resistor R will be approximately constant, behaving as an integrator.
Note that you can use op-amps and other active components to force the current through the resistor (in 1.15) to be constant, this is how the "ideal integrator" circuit below works:
Best Answer
This operates like a comparator except rather than open drain or collector is a push-pull to bipolar supplies (Vcc~Vee) minus some dropout voltage. The output is slew-rate limited to the component spec [V/us].
Remember that the (open loop) output inverts when the inputs are not equal (ignoring Vio offset) .
Other
This is not how you draw a sine wave which is more like triangular but peaks are smooth about 25% below the apex. You often see people draw sinewaves this wave with far too much "risetime". When we measure risetime of a pulse from 10 to 90% there is a formula to estimate the bandwidth BW(-3dB) BW= 0.35/Tr (10~90%)
Here is how you can verify a hand drawn wave using 3/4 of a triangle.
Since an open loop Op Amp may saturate internally after the internal compensation , then the linear GBW does not apply and the slew rate is limited by the output stage current limit defined in the datasheet. There may be reverse recovery times specified for overdrive unlike a good comparator IC.