capacitorcircuit analysisintegratorpassive-networks
I'm reading the first chapter of the AoE. I've come across this section on differentiator/integrator circuits and couldn't understand the math behind it.
For the first picture, it says small RC means dV/dt being much lower than dVin/dt, but I don't understand how it does this. Similarly, I'm not sure how large RC means Vin >> V. I know this may be a petty question, so please be patient with me.
The easyest way should be to ask me directly through my laserharp website ;) i'm the designer of this schematic. It's an output stage with a balanced/unbalanced output driver. if not used as balanced, you must connect the negative output to the ground to get full unbalanced signal. It's explained in the user manual of the laser harp. "ILDA wiring"
The node voltage equations are not so difficult. But it's really a tedious work to type the equations. I got the same result with you, maybe we are both wrong :).
Best Answer
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: