"Kelvin switching" on programmable gain TIA??

[msat]

Hello!

I'm wanting to design a low frequency transimpedance amplifier with a selectable feedback network. I came across the following article https://www.analog.com/en/analog-dialogue/articles/programmable-gain-transimpedance-amplifiers.html and 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

[exmadscientist]

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).

[pwlps]

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.

Just to add for clarity - the first sentence isn't quite correct : the trick actually IS the first switch since we now take the output voltage after it 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 .

[Marco]

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.

[msat]

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.

[exmadscientist]

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!)

[msat]

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.

[David Hess]

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.

[msat]

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.

[exmadscientist]

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.

That... 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 tricks do not work at all if 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.

[Marco]

many of the extreme lengths people go to (such as that Figure 14)

It'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.

[msat]

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.

That... 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 tricks do not work at all if 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.

I'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.

[exmadscientist]

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).

[Marco]

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 tricks do not work at all if your capacitance is variable

I 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.

[msat]

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!

[exmadscientist]

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 tricks do not work at all if your capacitance is variable

I 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.

That 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!

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.

If 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.

[msat]

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.

If 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.
[/quote]

Huh. I was indeed thinking of going with a SOT23 package, so I'll reconsider the SO-8. Thanks for the tip!

[msat]

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?

[David Hess]

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?

The listed differential impedance is before negative feedback is taken into account so it can be ignored, which is why nobody writes about it.

[Marco]

Figures 3/5

Man 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.

[msat]

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?

The listed differential impedance is before negative feedback is taken into account so it can be ignored, which is why nobody writes about it.

Thanks 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?

[msat]

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.

[David Hess]

The listed differential impedance is before negative feedback is taken into account so it can be ignored, which is why nobody writes about it.

Thanks 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?

The 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.

[David Hess]

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.

I 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.

[msat]

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.

I 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.

Whoa! 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.