# Active Integrator Parallel Capacitor in Feedback

I am designing an integrator circuit, looking to perform in the range of 10-10kHz - and I had a thought and I cant find any information online about it.

For low frequencies of 10Hz, the reactance of the 0.01uF capacitor on its own is in the 10's of Megaohms such at low frequencies the capacitor struggles to integrate and causes an undesired phase shift. My thinking is by adding a larger capacitor in parallel, with lower impedance at low frequency, it will help the low end integration. Is my thinking correct, and what are the implications of doing this, if any?

simulate this circuit – Schematic created using CircuitLab

• What is the purpose of R3? ("Current limiting"?? ). It destroys the integration capability. – LvW Oct 19 '18 at 10:29
• The circuit has a DC gain of 1E4. So - 1mV offset will produce 10V at the output! – LvW Oct 19 '18 at 10:31
• Yeah, ignore R3. I don't know what I was thinking. – Fat Diode Oct 19 '18 at 13:08
• Forgetting offset effects, you have the classical inverting integrating device. However, you need a low-offset opamp (Voff in the µV range). – LvW Oct 19 '18 at 14:18
• What's the budget for this? And just back of the mind calcs suggest you are talking about handling what could be as high as $100\:\frac{\text{kV}}{\text{s}}$. Am I way off on that? What's the range of currents? (You've given the sampling range.) Are you intending to add a switch to reset the integration? Or are you depending on $R_2$ for reasons I'm not sure I'd understand right now? – jonk Oct 19 '18 at 17:50

All capacitors can integrate charge. An arbitrary value of x Coulombs of charge on a 10 nF capacitor results in a much larger potential across the capacitor compared to a 1 uF capacitor. The relation is $$V = \dfrac{q}{C}$$