Help Deriving an OpAmp Transfer Function - Page 1
Forum section: electronics / projects

author: dreaquil
author_rank: Regular Contributor
timestamp: October 16, 2015, 10:24:07 pm
content: Hi,I'm working on a project which uses an inverting low pass filter and I've introduced some parallel input capacitance to my system so it can be modelled as shown below. Can anyone help me by deriving the transfer function (Vout/Vin) for this schematic? I've gotten the following and wanted to know if I'm correct. If so, I may need help making sense of it tooVo/Vi=-(Rf/Ri)((1+jwRiCi)/(1+jwRfCf))

author: Phoenix
author_rank: Frequent Contributor
timestamp: October 17, 2015, 12:29:02 am
content: The transfer function you've dervied looks good to me. What part don't you understand?

author: dreaquil
author_rank: Regular Contributor
timestamp: October 17, 2015, 07:29:40 am
content: Well the thing is I'm applying a step input for Vi. 0V to 40.96V using a microcontroller. I'm working with extremely high resistances. Here are the values:Ri=10Tohm, Rf=10Gohm, Ci=3nF, Cf=47pF. The output of my opamp is taking a long time to settle because of that input capacitance but I don't have a clear picture in my head of what's the reason. The only thing I am capable of changing is the 47pF but not by much.

author: mikerj
author_rank: Super Contributor
timestamp: October 17, 2015, 08:10:30 am
content: You can't be using something with Teraohms of impedance, the input bias currents and capacitance of the op-amp itself and leakage currents with the caps and even the PCB will have a huge effect.

author: dreaquil
author_rank: Regular Contributor
timestamp: October 17, 2015, 08:26:11 am
content: I've taken that into consideration, using an LPC662 (2fA input bias current) and using special PCB design techniques.Edit: and high voltage caps for very low leakage.

author: Phoenix
author_rank: Frequent Contributor
timestamp: October 17, 2015, 10:04:58 am
content: Quote from: dreaquil on October 17, 2015, 07:29:40 amI think you might need to explain better what you're trying to achieve because it doesnt quite make sense to me.You've setup an amplifier gain of 1000 (using stupidly high impedances)Your opamp supply rails are max +-15V ??You're appling a 40V stepSounds like you're just jamming it from rail to rail, so I'm not surprise it isnt behaving well.

author: SteveLy
author_rank: Regular Contributor
timestamp: October 17, 2015, 10:06:16 am
content: Quote from: dreaquil on October 17, 2015, 08:26:11 amThat's not enough. Your op-amp can no longer be treated as ideal if you're going to use such extreme values for the components. Even if your external circuit was perfect, the op-amp's non-ideal characteristics: input and output impedances, finite gain and bandwidth; need to be figured into the transfer function calculation.Consider that the input impedance of a typical op-amp is a few Mohms, so you cannot expect ideal behaviour with G & Tohm components. You're off by many orders of magnitudes. Try using resistances under ~100k and the ideal transfer function should be reasonably close to reality within the BW limit of the device.

author: dreaquil
author_rank: Regular Contributor
timestamp: October 17, 2015, 10:25:22 am
content: My application does not permit me to do so (picoammeter), also the LPC662 has an input impedance of greater than 1Tohm. Apart from that the signals I am dealing with are DC in nature however there is a step which is causing it to take long to settle.

author: JimRemington
author_rank: Regular Contributor
timestamp: October 17, 2015, 03:48:37 pm
content: QuoteNo surprise there. Calculate the time constants.

author: SteveLy
author_rank: Regular Contributor
timestamp: October 17, 2015, 04:13:36 pm
content: ^

author: Zero999
author_rank: Super Contributor
timestamp: October 17, 2015, 09:57:25 pm
content: Quote from: dreaquil on October 17, 2015, 10:25:22 amThe impedances at each input aren't matched to minimise the effect of bias currents. There should be a 10G resistor from the + input to 0V.What's the input capacitance of the op-amp? It's not specified but is likely to be significant.Quote from: Phoenix on October 17, 2015, 10:04:58 am*snip*QuoteYou've got the gain formula upside down. It has a gain of 1/1000 or a loss of 1000.Yes you're right, at these high impedances, leakage can be a problem.

author: Phoenix
author_rank: Frequent Contributor
timestamp: October 18, 2015, 12:06:00 am
content: Quote from: Hero999 on October 17, 2015, 09:57:25 pmYou're right, my bad.