"Kelvin switching" on programmable gain TIA?? - Page 1
Forum section: electronics / beginners

author: msat
author_rank: Regular Contributor
timestamp: November 30, 2020, 05:03:37 am
content: Hello!I'm wanting to design a low frequency transimpedance amplifier with a selectable feedback network. I came across the following articlehttps://www.analog.com/en/analog-dialogue/articles/programmable-gain-transimpedance-amplifiers.htmland was intrigued but confused by the circuit in Fig. 10 which implements what the author is referring to as Kelvin switching. He describes the reasoning for it, but the explanation turned my brain to mush. How would the errors which are present in Fig 9 be avoided in Fig 10? I trust that it's true (AD even have an OP AMP with such circuitry built-in) and it's easy enough for me to implement with a DPxT switch, but I do want to have a better grasp of what's going on here. Can someone explain this in a way that's somewhere between the article's explanation, and what would be written in a For Dummies book?Cheers,Mark

author: exmadscientist
author_rank: Frequent Contributor
timestamp: November 30, 2020, 07:03:20 am
content: The trick in that circuit isn't the first (left) switch, it's the second (right) switch. The left switch presumably makes sense to you: it's used to select the gain resistor. The problem is, then, that the on-resistance and parasitics of the switch are added to the gain resistance.The key idea is that we can choose which node we call "output". Instead of taking the output signal from the op-amp output, we can take it instead from the junction of the switch and gain resistor. This has two drawbacks: we need twice as many switches, since that point is no longer common among all selectable gains, and the output impedance of the op-amp is increased by as much as twice the switch resistance, depending on how you look at it.This trick may or may not be a good tradeoff, depending on how large your gain resistance is in comparison to the switch resistances, and on how much you care about bandwidth (=parasitic capacitance).

author: pwlps
author_rank: Frequent Contributor
timestamp: November 30, 2020, 10:02:24 am
content: Quote from: exmadscientist on November 30, 2020, 07:03:20 amJust to add for clarity - the first sentence isn't quite correct : the trick actually IS the first switch since we now take the output voltageafterit so that the switch is not in the feedback path anymore and its resistance does not enter into the gain equation. But this topology implies having the respective outputs on different nodes therefore we need a second switch .

author: Marco
author_rank: Super Contributor
timestamp: November 30, 2020, 06:12:49 pm
content: Figure 14 seems overly complex.Why couldn't you put a SPST and a SPDT in series on each feedback path? When the feedback path is disconnected you open the SPST and the bottom of the SPST gets connected to ground through the SPDT. When the feedback path is connected you close the SPST and the bottom of the SPST gets connected to output through the SPDT.

author: msat
author_rank: Regular Contributor
timestamp: December 01, 2020, 02:04:54 am
content: Thanks for the replies! These responses are in line with what I was thinking while lying in bed last night. The output side of the first switch is common to both the feedback network and measurement equipment (ADC, voltmeter, etc)and thus it's resistance is effectively negated and merely affects the output impedance of the op amp. However, if the switch resistance is in the milliohm range while the feedback resistors are in the kOhm+ range, the benefit of going with the circuit in Fig 10 is likely negligible for all but the most accurate of circuits (and then there's probably plenty of other design issues that also would need to be addressed), which exmadscientist pointed out.The second switch is nothing more than a selection circuit necessity due to the first switch being common to the measurement output and feedback network. The second switch adds to the output impedance to the measurement side, but assuming that the measuring equipment has a high impedance input, the switch's effect is negligible. What I particularly like about this arrangement is that it's much easier to ensure that the switch's contact whetting current is met.Using a DP3T switch, I can implement a 3 selectable gain circuit as described in Fig 10. If my feedback resistance ends up being sufficiently high, I can use the same switch, but I can switch out both ends of all feedback networks except for the one selected. I don't know for sure, but I suspect this would reduce some of the stray capacitance. Then again, I'm primarily interested in DC, so the "Kelvin switching" is likely more desirable.I agree that 14 gets a little crazy. There's probably a handful of ways you can arrange your switches for the same outcome, but I suspect there's better ways to minimize stray capacitances.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 01, 2020, 06:09:57 am
content: Keep in mind that much of the literature on transimpedance amplifiers is interested in maximizing bandwidth (and usually is designing for photodiode applications). If you are interested in LF or DC performance, many of the extreme lengths people go to (such as that Figure 14) start to look kind of silly. But when your capacitances are best measured in femtofarads, they become necessary!Accordingly, many such circuits, if they use switchable gain at all, need something nicer than a standard mechanical switch or relay. Sometimes RF switches are used, but more often you'll see a CMOS analog switch. These can have quite high on-resistances, hence the need for elaborate schemes to work around switch on-resistance.Transimpedance amplifiers can have extremely good performance, but they are really, really sensitive to stray capacitance. Even if you don't actually need much bandwidth, it can still be very difficult to compensate them if there's any kind of cabling between the current source and the transimpedance circuit. This is what sunk the transimpedance-amplifier prototype I was working on a year ago: it worked beautifully at DC and met all gain performance and stability requirements... then we stuck a cable on the front and watched it oscillate. Unstoppably.(It should have been possible to compensate it for any given cable, but in the end application, the cabling was necessarily going to be highly variable. So the whole thing got scrapped. That's why you build prototypes early!)

author: msat
author_rank: Regular Contributor
timestamp: December 01, 2020, 11:49:45 pm
content: The AD part I mentioned in an earlier post which had integrated switches did have on resistance in the realm of 300 ohms or so, and therefore I can certainly see the benefit of the Kelvin switching circuit.I'm interested in the issues you had with your circuit, as it might be applicable to me as well. My current sourcing/sinking DUT won't be on-board, so I'll need some cabling, though I'm thinking well under a meter in length. How are oscillations addressed? Is it by capacitor selection in the feedback circuit? If so, I could probably just keep it socketed and change it out at any sign of oscillations. I'd like to understand this better before designing a PCB.

author: David Hess
author_rank: Super Contributor
timestamp: December 02, 2020, 09:01:09 pm
content: The simple explanation is that moving the output to a point after the switch rather than before the switch moves the switch's resistance so that it is in series with the output resistance of the amplifier rather than in series with the feedback network.

author: msat
author_rank: Regular Contributor
timestamp: December 03, 2020, 12:35:36 am
content: Or rather that you're now measuring at a point that's within what's effectively a voltage divider, and thus eliminating what would be a measurement error caused by the unknown or variable resistance of the switch. Technically, the switch's resistance is still in series with the feedback network, but the point of measurement makes that mostly a non-issue.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 03, 2020, 06:14:39 am
content: Quote from: msat on December 01, 2020, 11:49:45 pmThat... is probably going to be an issue. You don't say how much gain and bandwidth you want, but I'd expect any decent length of cable will be a problem unless you are quite slow or low-gain.I know of two ways to attack the problem. Trading bandwidth for stability by increasing the feedback capacitance, as you mention, is one. But that can only take you so far. The other method is to directly cancel the capacitance using a couple of clever circuit tricks. The attached paper by Hoyle and Payton is a good starting point for those. These tricksdo not work at allif your capacitance is variable, such as an uncontrolled (user-controlled) cable. You can probably dial in one unit on the bench, but I would not want to rely on cancellation in production without some very good benchtop evidence.(I also would not want to socket a capacitor. That does work for large values, but it's too erratic for smaller capacitances or higher frequencies. It's another thing that depends on your gain and bandwidth.)Ultimately the reason we abandoned the TIA was that it couldn't work for us with random cabling separating the measurement circuit and the sourcing device. We already had a Plan B circuit (actually, the TIA was the "hey-maybe-we-can-improve-this" Plan B...), so there was no strong incentive to fix that. Perhaps there is some clever solution; I'd love to learn about it if so. I'd also love to hear a good intuitive explanation of the Hoyle paper; its techniques do work, but my brain never quite wrapped itself around them.

author: Marco
author_rank: Super Contributor
timestamp: December 03, 2020, 06:37:43 pm
content: Quote from: exmadscientist on December 01, 2020, 06:09:57 amIt's no longer compensating for the switch resistance, so borrowing the Kelvin topology stops making sense. From the output point of view it has only one switch in series with the feedback, rather than 2 with my suggestion, but it needs more switches and more traces (with parasitic capacitance of their own). It's not worth the headache any more.

author: msat
author_rank: Regular Contributor
timestamp: December 05, 2020, 12:44:48 am
content: Quote from: exmadscientist on December 03, 2020, 06:14:39 amI'll be working with DC. For my specific purposes, it'll be acting as a low-impedance sink/source current meter. I'm currently thinking of 2 feedback loops, one with 1k resistor, and the other with 1M and calculated 159uF and 159nF caps (almost definitely on the larger side), for a measurement range of roughly 100nA-10mA and 100pA - 10uA, respectively using my 6.5 digit voltmeter. I've only accounted for meter errors, so bottom-end accuracy of each range is probably a bit optimistic. Though I'm thinking (or hoping) that for the specific experiments I have in mind, I'll be expecting currents that reach the upper end of each range, and am not all too concerned with the accuracy of the tail end of the measurement values.Also, since my planned usage is quite specific for now, and I have full control of the cabling I use, my problems are hopefully much less of an issue than yours. I skimmed through the paper you linked, and this active capacitance control business made my eyes glaze over. I think it's probably unnecessary for my needs.I'm slowly getting my ducks in a row. I think I should be able to start drawing my PCB here shortly.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 05, 2020, 05:43:55 am
content: That sounds quite realistic. You should be able to get something working nicely. For the low end of the range you will need a very good op-amp, but there are plenty of those available to choose from. Let me know if you need suggestions, as I've got a table somewhere (which I unfortunately cannot simply post as it's marked up with all kinds of proprietary stuff).

author: Marco
author_rank: Super Contributor
timestamp: December 05, 2020, 04:41:34 pm
content: Quote from: exmadscientist on December 03, 2020, 06:14:39 amI disagree, the unity gain buffer to force the TIA input voltage to the photodiode bias voltage works just fine with variable capacitance, the value of the capacitance is not a factor in the design after all. With or without the bootstrap, the value of the capacitance will only affect the feedback capacitance you need for stability.Whether the bootstrap works with the delay of a transmission line rather than a pure capacitor, that I'm doubtful about.

author: msat
author_rank: Regular Contributor
timestamp: December 05, 2020, 07:50:01 pm
content: Op amp choice is obviously important, and I have a few that I'm going to try out. I'm looking for a footprint common among all of them. Some of the amps I want don't have the footprint & pinout I really want. Ah well. I think my biggest issues, especially on the low end of the range (which, as I've said, I'm not super concerned about) will be the quality of my passives in the feedback loop, and my power supply. I'm going to try my bench PSU, but I'll go with batteries if I have to.I might take you up on op amp suggestions, but I think I got it for the time being. Thanks!

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 05, 2020, 09:04:11 pm
content: Quote from: Marco on December 05, 2020, 04:41:34 pmThat is only possible if you have a floating source and can drive its other terminal, which is the case with photodiodes. If you have a "ground referenced" or otherwise independent source, as you do when trying to build a more general-purpose ammeter type circuit, you need something more complicated and less flexible. (Compare Figures 3/5 against 4/6 in the Hoyle paper above.) At least, that's true as far as I know. I'm very welcome to be proved wrong here!Quote from: msat on December 05, 2020, 07:50:01 pmIf you have a free choice, for any given part, take the SO-8 over the SOT23-5 or MSOP-8. You'll find the SO8 will generally have better performance, due to obscure reasons involving packaging (something about die stresses). Of course, other factors can and should take priority.

author: msat
author_rank: Regular Contributor
timestamp: December 05, 2020, 11:40:21 pm
content: Quote from: msat on December 05, 2020, 07:50:01 pmIf you have a free choice, for any given part, take the SO-8 over the SOT23-5 or MSOP-8. You'll find the SO8 will generally have better performance, due to obscure reasons involving packaging (something about die stresses). Of course, other factors can and should take priority.Huh. I was indeed thinking of going with a SOT23 package, so I'll reconsider the SO-8. Thanks for the tip!

author: msat
author_rank: Regular Contributor
timestamp: December 06, 2020, 03:21:15 am
content: exmadscientist,Looking at various op amp datasheets, besides some not listing input impedance at all, others sometimes show differential and common mode impedances. For example, the TLV915x series show 6Tohm common mode and 100Mohm differential. I haven't come across any articles/papers talking about the effect of differential impedances on TIAs. I'm assuming that, given the above part as an example, if I'm using 1Mohm in the feedback loop, then 1/100th of the input current would be going straight through to the non-inverting input and therefore create an equivalent error at the output. Does that sound correct?

author: David Hess
author_rank: Super Contributor
timestamp: December 06, 2020, 09:01:49 am
content: Quote from: msat on December 06, 2020, 03:21:15 amThe listed differential impedance is before negative feedback is taken into account so it can be ignored, which is why nobody writes about it.

author: Marco
author_rank: Super Contributor
timestamp: December 06, 2020, 03:35:03 pm
content: Quote from: exmadscientist on December 05, 2020, 09:04:11 pmMan the paper is needlessly terse in describing those, those amplifiers are no longer really bootstrap amplifiers ... they are part of a negative capacitance circuit.That said, you don't need negative capacitance to offset cable capacitance. You can simply use coax and drive the shield with a buffer, again the exact capacitance of the cable becomes irrelevant. Though that won't fix transmission line effects.

author: msat
author_rank: Regular Contributor
timestamp: December 06, 2020, 07:15:29 pm
content: Quote from: David Hess on December 06, 2020, 09:01:49 amThanks for the response, David. Let me see if I understand this: The source current to be measured terminates at the inverting input of the TIA which is basically seen as a roughly 0 ohm source/sink to whatever potential the non-inverting input is at. That "resistance" is parallel to the op amp's differential impedance, and thus since the ratios between the two resistances are extremely large, the effect of the differential impedance is negligible. Is that correct?

author: msat
author_rank: Regular Contributor
timestamp: December 07, 2020, 01:39:23 am
content: I have a further question, and while it's starting to get a little off-topic, I think it makes more sense to ask here than to start a new thread.My intent is to feed a variable biasing voltage to the non-inverting input, which will therefore be reflected at the non-inverting input. If I simply measure what this voltage is at the inverting input during set up of my experiment (rather than at the inverting input), do I have to give any consideration to op amp input offset voltages? To my understanding, the answer is no, but I'd like to know if that's correct.

author: David Hess
author_rank: Super Contributor
timestamp: December 07, 2020, 04:51:40 pm
content: Quote from: msat on December 06, 2020, 07:15:29 pmThe resistive component can be ignored because the differential resistance operates between the inputs, and the negative feedback prevents any change in voltage at the inverting input. The excess gain of the feedback loop falls as frequency increases so the input resistance and input impedance have a greater effect as frequency increases.There may be some extreme applications where common mode and differential impedance matter, and they also have a much greater effect in open loop applications like with a comparator.Incidentally it has been a while since I looked at this, but I think the common mode impedance is so much higher because it represents the impedance of the current source which feeds the input differential pair times the current gain of the input transistors. The differential impedance comes from the emitter resistance of the differential pairs which is much lower but irrelevant for the reasons given above.

author: David Hess
author_rank: Super Contributor
timestamp: December 07, 2020, 04:58:19 pm
content: Quote from: msat on December 07, 2020, 01:39:23 amI think you mean "which will therefore be reflected at the inverting input".That is exactly right; if you measure the voltage at the inverting input, then you can ignore the input offset voltage, and some automatic zero circuits effectively do exactly that. (1) But keep in mind that when measuring that voltage, any current into or out of the inverting input will create an apparent offset voltage by shifting the output, so there will be a premium on a low input bias current measurement.(1) Chopper stabilized amplifiers work this way, with the error amplifier first measuring and correcting its own offset and then measuring and correcting the offset of the main amplifier.

author: msat
author_rank: Regular Contributor
timestamp: December 08, 2020, 02:23:17 am
content: Quote from: David Hess on December 07, 2020, 04:58:19 pmWhoa! I mixed up inverting and non-inverting in my post several times, but you understood what I meant.I am a little bit confused (unsurprisingly) about current into or out of the input causing an offset voltage, given that, to my understanding, the inverting input side basically acts as a constant voltage source. Of course, inherently some sort of offset has to be generated so that a correction via the feedback path could be made, but I'm assuming this will be quite small, and thus not a huge concern for my needs, as I'm not aiming for utmost accuracy or performance. I'm thinking that being able to set the voltage at the inverting input down to the microvolt range would be more than good enough.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 08, 2020, 06:14:49 am
content: Quote from: Marco on December 06, 2020, 03:35:03 pmTell me about it! I banged my head against it off and on for longer than I'd have liked before we cut bait on this poor idea....QuoteThat only works if you control the cable. We couldn't even guarantee it would be coax of some sort. Cutting bait was the right choice, I think! But, hey, now I've got a decent gain-selectable 10-decade ammeter box that only I can coax to get working...Quote from: msat on December 08, 2020, 02:23:17 amI believe David isn't talking about the op-amp input by itself, but rather what happens when you try to measure the voltage at that node. If you use a meter (or meter circuit) with insufficiently high input impedance, you will disturb the operating point of the circuit. The definition of "sufficient" here is up to you and your required precision. Someverygood parts are available these days and while they can get expensive, if you only need one, who cares if it's $25 instead of $0.25?

author: msat
author_rank: Regular Contributor
timestamp: December 08, 2020, 11:56:53 pm
content: Quote from: exmadscientist on December 08, 2020, 06:14:49 amThat's what I was thinking he meant too, but even if my meter places a load on the circuit (just to clarify, I would only use my meter to check the voltage at the inverting input, and once set, I would remove the meter and attach the DUT), as long as it doesn't sink/source more current than the feedback loop can handle, the effect should be minimal. After all, isn't that really the point of a TIA?

author: David Hess
author_rank: Super Contributor
timestamp: December 09, 2020, 05:53:17 pm
content: Quote from: msat on December 08, 2020, 11:56:53 pmI misunderstood exactly what you wanted to do, however what you suggest has another problem.Since the operational amplifier has finite open loop gain, a change in output voltage does result in a difference between the inputs. If you make the measurement and current into the inverting input drives the output to 10 volts, and the open loop gain is 100,000, then the input will have shifted 100 microvolts.As an issue of practicality, probing the inverting input also tends to cause oscillation because any added capacitance combined with the feedback resistor creates phase lag in the feedback. A capacitor can be added across the feedback resistor, which is usually a good idea anyway, or sometimes a resistor can be placed in series directly at the probe tip.Input offset is usually measured using the non-inverting configuration with the input grounded, and measuring the output voltage, so the operational amplifier amplifies its own input offset voltage to a convenient level.

author: msat
author_rank: Regular Contributor
timestamp: December 10, 2020, 01:08:36 am
content: I have yet another question (I know, I know...):Measurements at the output of the amp will be made with my voltmeter which has a 10G Ohm input impedance in the range I'll be using. Do I need to terminate the output of the amp to ground (through a resistor, of course), or is leaving it floating ok? If either is acceptable, is any particular way better? The amp will be powered with a double ended power supply, with the mid point referenced to ground.Quote from: David Hess on December 09, 2020, 05:53:17 pmI do intend on having a cap parallel to the feedback resistor even though I'm not interested in AC. From what I've read, it just seemed like good practice for stability purposes.You specify open loop gain, and intuitively it seems like the same would apply to a closed loop system, is that correct?Ultimately, I just want the simplest way of setting a bias voltage at the inverting input , so if I can do that without having to determine the amp's offset voltage, that would be preferable. As I mentioned above, my meter's input impedance is 10G Ohm, so I assume any bias currents from its inputs would be fairly small.

author: msat
author_rank: Regular Contributor
timestamp: December 10, 2020, 02:55:01 am
content: Quote from: David Hess on December 09, 2020, 05:53:17 pmAfter thinking about this comment some, I believe it's starting to make sense.So if I want to reduce the deviation from my target voltage at the inverted input, either I need to find an amp with a higher open loop gain, or cascade amps together to multiply their gain (and allow for a smaller feedback resistor on the leading amp.)

author: Marco
author_rank: Super Contributor
timestamp: December 10, 2020, 05:49:55 pm
content: It's unlikely to be a relevant error source.

author: David Hess
author_rank: Super Contributor
timestamp: December 10, 2020, 06:08:41 pm
content: Quote from: msat on December 10, 2020, 01:08:36 amUsually there is no reason that the output needs to be terminated with additional load. Sometimes it is done to force the output stage into a particular mode of operation or to aid stability by controlling the output stage current.QuoteThe added feedback capacitance is good practice in TIAs to compensate for the parasitic capacitance at the inverting input which would otherwise compromise stability.QuoteIn an open or closed loop system, the output voltage is equal to the open loop gain times the difference between the inverting and non-inverting input. In a closed loop system, external networks control the inverting input but the output still follows this difference.QuoteThe usual method is to measure the output voltage, which is a convenient low impedance point, and trim the voltage at the non-inverting input. When the output is zero, then the input offset voltage should have been compensated for by the non-inverting adjustment. The offset null pins could also be used and might be preferable for parts where offset null adjustment also trims the the offset voltage drift, assuming that offset current errors are small; the offset null pins should never be used to compensate for offset current errors because this would increase offset voltage drift.It has been a while since I worked with this sort of thing but I seem to remember using a relay or switch to connect a lower value resistor between the inverting input and ground configuring the TIA as a non-inverting amplifier amplifying its own input offset, and then the offset adjustment was performed.

author: David Hess
author_rank: Super Contributor
timestamp: December 10, 2020, 06:15:37 pm
content: Quote from: msat on December 10, 2020, 02:55:01 amThat is exactly right, and in high performance low frequency TIAs, I have seen those methods used.There is also another method to increase open loop gain. Since open loop gain is primarily limited by thermal feedback from the output stage to the input stage, unloading the output using an external buffer to drive the feedback network and the following stage will maintain the highest possible open loop gain. This is the primary reason that open loop gain falls with increasing load.Precision amplifiers do not have higher open loop gain than general purpose only because of more gain stages or higher gain per stage. They have higher open loop gain primarily because they use techniques to lower the effects of thermal feedback from their output transistors to their input transistors.

author: KT88
author_rank: Frequent Contributor
timestamp: December 10, 2020, 08:09:18 pm
content: Reading through this thread, I didn't find the correct answer to the initial questionQuote. The answer is: It isn't an improvement - it even makes things worse as now two switches are in parallel... The only purpose of fig. 10 is the introduction of the concept of Kelvin switching. Figure 14 describes the practical improvement one can get with Kelvin switching: It allows to (almost) completely eliminate the influence the parasitics have on the circuit. The goal here is not better (oscillation-) stability - it's gain stability. Leackage of MOS transistors inreases roughly by the factor of two every 10°C. This means that the disconnected branch contributes more or less to the gain, depending on the temperature.The solution is to conncet the open node to ground - first, zero voltage contributes zero current to the feedback loop. Second, leackage capacitance is shunted to ground.That way one can get at least close to the isolation performance of the legendary Coto relays and alike...CheersAndreas

author: msat
author_rank: Regular Contributor
timestamp: December 11, 2020, 03:30:40 am
content: David,Thank you for your helpful responses. I'm quite exhausted by the time I get home from work, so it often takes a while to get technical concepts to click in my head.I really want to avoid complicating what is otherwise a fairly minimal circuit to null out offsets. I did consider breaking out those pins on op amps that have them, but on their own they wouldn't do any good if the rest of the circuit requirements weren't met. I might just accept these errors, as they probably wouldn't pose an issue for my needs. At this point it's hard for me to say, but I can go the easy route, play with biasing a bit and if I see big differences, I'll try to design a more precise circuit. This thought also extends to the effects to deviations from target bias voltage due to input currents. It may not be worth the effort to design a more sophisticated multi-stage amp, but time an experimentation will tell. If I have some data rather than just some hunch that more gain will help, then I'll address it when the time comes.I did come across an interesting circuit in the LTC6244 datasheet in which the current source is fed into the non-inverting input of the first amp with a gain set to 3 and the output of that went to the inverting input of a 2nd amp whose feedback path looped back over to the first amp's input. The point of that was specifically to reduce input capacitance somehow, but I doubt it's just limited to that.Lastly, trying to wrap my head around the effects of increasing gain some more, if I'm not mistaking, for a voltmeter with a given resolution (which is used as the measurement device), for any increase in open loop gain of the amp, while allowing smaller input deviations for a given output, has the effect of reducing the range of measurable input currents, correct? Hopefully that sentence makes sense.

author: msat
author_rank: Regular Contributor
timestamp: December 11, 2020, 03:38:03 am
content: KT88,The original question was about how did the Kelvin switching work, but, at least for me, I believe is resolved. Without Kelvin switching, the resistance of the switch (which is potentially unknown, or variable) would add to the errors also caused by the other components in the feedback network. But as stated in the article, Kelvin as well as standard switching networks have a detrimental effect by creating additional capacitances in the feedback network.

author: David Hess
author_rank: Super Contributor
timestamp: December 11, 2020, 11:22:36 pm
content: Quote from: msat on December 11, 2020, 03:30:40 amThat example circuit shows what is required to increase open loop gain without compromising stability. Increased open loop gain would lower the voltage change between the inverting and non-inverting input lowering the effect of the differential input capacitance, but I am not sure if that is what they are referring to unless it is that the common mode input capacitance is lower than the differential mode input capacitance.QuoteI do not know why that would be; were you thinking of closed loop gain?A higher open loop gain will result in a smaller deviation in offset voltage at the input, but usually 10s of microvolts of change is insignificant for a current input. The change in the current is divided by the impedance of the source which is high for a current by definition, or it would not be a current. Some applications of course do require the best possible virtual ground at an inverting input.

author: msat
author_rank: Regular Contributor
timestamp: December 13, 2020, 02:18:43 am
content: Quote from: David Hess on December 11, 2020, 11:22:36 pmApparently it is a way of negating differential capacitance. Why that is is well beyond my comprehension ability. Still, I find two amp series circuit interesting. I'll probably avoid it for the time being. I don't know how I would go about biasing the inverting side as would be done in a standard single amp TIA. Would the bias voltage have to be applied to both amps?QuoteHonestly, I have only a vague idea of why I thought that. The basic transimpedance gain equation is not all that revealing, and the more complicated one considers the various capacitances which is difficult for me to predict. Plus open loop gain of a given amp varies quite a bit. I'm better off just accepting this as fact and proceed with the project rather than bog myself down with details that aren't necessarily critical for me to understand (though I would like to). There's only one way for me to find out whether microvolts of offset from the target will be an issue. If it is, that would be quite enlightening.Thanks again for taking the time to answer my questions even though sometimes they just bounce off of my head.

author: David Hess
author_rank: Super Contributor
timestamp: December 13, 2020, 03:18:36 am
content: Quote from: msat on December 13, 2020, 02:18:43 amIt is the same except the inverting and non-inverting inputs are swapped because the output driving the feedback network is inverted.