op-amp DC AC integrating behavior explained

[Circuit prepared in 2. state, TOGGLE THE SWITCH TO START INTEGRATING (read below)] Transfer function: Vo=-Vin (Rf/R1)(1/sRfC) 2 possible behavior according to Rf values: 1. Rf=100kOhm, Cf=10nF (random values): LPF behavior, cut off frequency ~159Hz (low freq. gain =-Rf/R1 W.R.S. to 2.8V which is the biasing voltage); 2. Rf=+inf (not connected): pure integrator behavior, unity gain freq. ~318Hz. DC response is indeed a ramp. Why so? From a math point of view it's of course his integral but from an electronic pow what happens is that the power mosfet is forcing a constant current into Cf in order to compensate the DC applied voltage variation (w.r.t. 2.8V = bias) across R1 and maintaining - at 2.8V. How to bias the circuit? In discrete electronics Rf has this role and it must be chosen big enough to force the circuit in 2. (performing as integrator indeed and not as an inverting amplifier): 2MOhm is good here. In CMOS technology the switch (a Mos) may be used in order to force the circuit in buffer cfg (mathematically speaking it forces the starting condition from which integrating) and then switched off when integrating.

published 7 years ago