Transfer function of a current integrator - Page 1
Forum section: electronics / beginners

author: Atomillo
author_rank: Regular Contributor
timestamp: February 06, 2023, 02:59:36 pm
content: Hello!I'm designing and simulating a current integrator, and I have to know it's transfer function H(f) = Vout/Iin because I plan to reset it by placing it inside a control loop that when activated will dispense current until it's output reaches zero.I've been having some difficulties understanding the results obtained and would greatly appreciate some help. First of all using the test circuit of the first image I get the results shown in the second picture.Now, mathematically, I understand that for an ideal integrator the gain at f=1Hz is 1/2*pi*C, which seems to agree with the results, but the open loop gain of the used AD712 at this frequency is just 115dB (verified with another Spice simulation so it is correctly modelled). How can the gain of the integrator be greater than the open loop gain of the op amp?Furthermore, how can it be that the gain of the integrator reaches 0dB at 10GHz (can you believe it!)? I'm assuming I screwed up something in the simulation, but I've not been able to find what.Any help is greatly appreciated!PS1: In order to obtain this result I've had to specify that in the initial condition of the capacitor it's charge is zero. If not the simulator gives a completely different result.PS2: In a transient simulator I verified the circuit works properly in the time domain (which it does) and as expected with currents high enough the output doesn't slew fast enough.

author: JohnGi
author_rank: Newbie
timestamp: February 06, 2023, 05:19:55 pm
content: "AC 1" on your current source input means 1 amp used as the input reference for the AC simulation.It's a small signal analysis, and doesn't take account of clipping etc - you would expect an enormous value at the output.The 115 dB open loop gain you mention is Vout/Vin for the opamp.Your simulation plot db(Vout) is showing you Vout/Iin.

author: Atomillo
author_rank: Regular Contributor
timestamp: February 06, 2023, 07:58:02 pm
content: JohnGi:Many thanks for your answer! I understand that we are not seeing any large signal behaviour, but even then how can the bandwidth of Vout/Iin supposedly extend to the GHz range if we only have poles and no zeros to cancel them?Furthermore, what can be the upper limit for the low frequency gain? If we are not limited by the Aol of the op-amp I'm at a loss

author: Benta
author_rank: Super Contributor
timestamp: February 06, 2023, 09:09:30 pm
content: You're way outside the limits of your opamp model. No one knows what's happening.Get your signal levels right first.

author: MrAl
author_rank: Super Contributor
timestamp: February 07, 2023, 02:03:56 am
content: Quote from: Atomillo on February 06, 2023, 02:59:36 pmHello,It looks like your transfer function for AC is:Vout=(5.0e10)/(pi*f)but that is in pure theory, where there are no limitations on voltage or current and no limitations on what the components can do. That would, for one example, mean the op amp can take an infinite input voltage without blowing out and can source an infinitely high output voltage.We all know this cant work in the real world and looking at the output for even 100Hz we can see the output is way above anything even remotely reasonable.To add to that, during the first instant the output can not respond at all. That puts 1 amp through a 10pf capacitor for the first instant of a cosine wave input, and the voltage due to that appears at the inverting terminal of the op amp. Note that this is a current drive not a voltage drive, so there is no delay. Because it is an electrical current drive it MUST be inside a closed loop (barring any non conservative losses) and that means it must immediately pass through the capacitor. If you look at what voltage this would produce at the inverting input you will find it will be very very high. In fact, it will reach up to 1/C volts for the very first instant in theory, which is way too high for any op amp i know of.So maybe you need to rethink this project test setup and see what you can change to make it a more reasonable circuit and/or a more reasonable test setup.