"Kelvin switching" on programmable gain TIA??

[exmadscientist]

Figures 3/5

Man the paper is needlessly terse

Tell 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....

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.

That 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...

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.

I 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. Some very good 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?

[msat]

I 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. Some very good 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?

That'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?

[David Hess]

That'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?

I 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.

[msat]

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.

I 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.

I 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.

[msat]

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.

After 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.)

[Marco]

It's unlikely to be a relevant error source.

[David Hess]

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.

Usually 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.

I 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.

The added feedback capacitance is good practice in TIAs to compensate for the parasitic capacitance at the inverting input which would otherwise compromise stability.

You specify open loop gain, and intuitively it seems like the same would apply to a closed loop system, is that correct?

In 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.

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.

The 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.

[David Hess]

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.

After 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.)

That 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.

[KT88]

Reading through this thread, I didn't find the correct answer to the initial question

How would the errors which are present in Fig 9 be avoided in Fig 10?

. 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...

Cheers
Andreas

[msat]

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.

[msat]

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.

[David Hess]

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.

That 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.

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.

I 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.

[msat]

That 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.

Apparently 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?

I 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.

Honestly, 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.

[David Hess]

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?

It is the same except the inverting and non-inverting inputs are swapped because the output driving the feedback network is inverted.