Realistic Op-Amp bandwidth - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 05, 2020, 03:29:51 pm
content: I have a question about some calculations I'm performing as a last-minute check. I have a circuit here which converts a current to a voltage. The current is very small (pico-Amps), thus the gain is very high (108). Now, based on this alone, I was trying to run the numbers on the needed Gain Bandwidth Product (or Unity Gain Bandwidth) on a real op-amp to see what kind of op-amp would match the closed-loop bandwidth of 30 kHz.So, we are given the minimum bandwidth needed in this closed-loop configuration, 30 kHz, and we are given the gain, 108. I calculated that the necessary GBP is 3000 GHz (30,000 * |108|). That number looks incredibly unrealistic, so I was wondering what am I not taking into consideration that makes this wrong? Is it because of the gain that I'm using?

author: Paul Rose
author_rank: Regular Contributor
timestamp: October 05, 2020, 03:37:40 pm
content: Quote from: LoveLaika on October 05, 2020, 03:29:51 pmYes. That isveryhigh.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 05, 2020, 04:01:54 pm
content: Thanks for your reply.Given your response, it seems that my rough calculations check out. If this is the case though, with the numbers being so unrealistic, why would a product description give such numbers? Sure, the gain is very high, but this gain is what they specify here. Something seems very off with what they're saying and with what is realistic.

author: jdutky
author_rank: Regular Contributor
timestamp: October 05, 2020, 04:16:15 pm
content: fromhttps://www.electronics-tutorials.ws/opamp/opamp_1.htmlit looks like you multiply by the log of the gain at your desired frequency by the bandwidth, so I think your GBWP should be something more like 240 KHz. That said, I'm not entirely following the math here, and I'm not sure how a gain of 20dB becomes the value "10" in the example given (but if we were just considering a square wave at your desired frequency of 30 KHz, then we would expect that the amplifier should be able to handle at least the 3rd and 5th harmonics at 90 KHz and 150 KHz, and maybe the seventh harmonic at 210 KHz too, in order to faithfully reproduce the signal, so the 240 KHz value seems to at least be in the right ball park).It also looks like you really need to examine the specs for actual op amps to see what their open loop gain graphs look like and where your desired gain slots in given the required bandwidth.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 05, 2020, 04:31:21 pm
content: The GBW applies to the voltage gain, not the current to voltage conversion or current amplification.For the TIA the required / wanted gain depends on the input resistance / input capacitance. The higher the OPs GBW, the lower the AC voltage at the input and thus less effect of the capacitance at the input. The details can be founds in articles on TIAs used for photo diodes. AFAIR there is something like the square root of the ratio of input capacitance to feedback capacitance (likely just parasitic) that comes in. So I would expect that a few MHz may be needed, but nothing super fast.

author: dmills
author_rank: Super Contributor
timestamp: October 05, 2020, 04:39:53 pm
content: I would guess that that is an equivalent circuit, not what they have really done.You would typically carve this up into a couple of stages, say 10^2 or 10^3 as a front end (and the only bit where noise matters) followed by a second amplifier stage, and maybe even a third.You can even nest opamps within the feedback loop as long as the later ones are MUCH faster then the early one so as not to hurt the phase margin, but work out how little capacitance equals 100M ohms at 30kHz before thinking of going there.Composite opamps are an interesting area, and sometimes quite useful.Regards, Dan.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 05, 2020, 06:04:14 pm
content: One could get a similar response with a TIA with a smaller resistor and than some voltage gain. However this would increase the noise from the resistor. 100 M may still be a real resistor. There is no voltage gain of 10^8 but a transimpedance gain of 10^8 V / A. For this circuit it can help to just run a simulation.

author: rstofer
author_rank: Super Contributor
timestamp: October 05, 2020, 07:07:31 pm
content: In the end, you're going to have several stages of amplification. First I get the GBW for a device, solve for allowable gain for my required bandwidth and then figure out how many stages it takes.Dave did a video:

author: magic
author_rank: Super Contributor
timestamp: October 06, 2020, 09:39:43 am
content: Two minutes in LTspice to get a feel of how it behaves.As I said before in the original thread, about 50fF feedback capacitance is required to meet the bandwidth spec. Input capacitance up to about 10pF is tolerated by this opamp model (10MHz GBW and 90° phase margin). Reducing to 1MHz causes gain peaking and requires Ci<1pF.Open loop gain is the ratio V(out)/V(in), of course. Here it's set to 100k at DC and falls towards AC. It doesn't need to be astronomically high, but low gain increases sensitivity to Ci (see above) and increases the voltage swing at the input node, which may or may not affect a real world current source with less than infinite output impedance. Input impedance of the TIA can be seen on this plot as 1kΩ at DC and approaching 300kΩ at 30kHz and I'm sure it would go down with a faster and/or higher gain opamp.

author: radiogeek381
author_rank: Regular Contributor
timestamp: October 06, 2020, 01:47:49 pm
content: pico-amps? Is that going to be swamped by the noise from the various sources? (Op amp input noise, diode noise, noise from the various passives? )

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 06, 2020, 03:24:18 pm
content: My prototype circuit will not have the protection diodes at all, though there will be some pads to add it on if need be. Going by the approximate capacitance of an 0603-sized resistor, there will be some natural feedback capacitance of around 40 fF. The datasheet for my resistor didn't say anything, so I'm just working off of an assumption. At the very least, I've also added an additional pad to add a feedback capacitor as needed (0.1-0.15 pF). All that I can really do now is build the circuit and test it.https://www.vishay.com/docs/60107/freqresp.pdf

author: Kleinstein
author_rank: Super Contributor
timestamp: October 06, 2020, 03:32:11 pm
content: Quote from: radiogeek381 on October 06, 2020, 01:47:49 pmThe circuit would probably get relatively close to the physical limit with the noise. So chances are the Johnson noise if the feedback resistor would dominate the noise over much of the bandwidth. 1/f noise from the OP will come in at low frequencies. So the expected noise would be around some 12 fA/sqrt(Hz) or some 2 pA_RMS over the 30 kHz bandwidth.