Transimpedance amplifier maximum current resolution

[ddr_controller]

Hi folks!

Maybe this question is a bit too general but I don't seem to find any solution.

I want to build a current sensor for a current that ranges from 1uA to 1mA and needs resolutions of nearly 1pA.

I know about the TIA topology and that you can measure currents of 1pA and below with proper component selection and PCB layout but I think they have very limited range, like 1pA to 10pA. I want resolution, hence, the question is what resolutions can you expect from TIAs for input currents of around 1uA to 1mA? And what other current sensing topologies are there?

Below is the schematic of the TIA I currently have and the layout of the PCB.

[RoGeorge]

Since the circuit is analog, the main limitation in resolution is caused by noise and by time (by time because of how many measurements you can afford to average, so to filter out some of the noise).

[moffy]

Bob Pease, semiconductor designer for National Semiconductor, designed some CMOS opamps that had sub pico amp bias currents. The LMC660 data sheet: https://www.ti.com/lit/ds/symlink/lmc660.pdf
has a section on page 11 about how they measured the input bias current using integration. Integration provides a way to measure very low currents.

[Someone]

mA to pA? 180dB dynamic range..... thats off into imaginary territory. Look at what the absolute best current measurement instruments offer. What magic do you have to do it better?

[Kleinstein]

The resolution is limited by the noise and drift. With a large resistor in the feedback the resistor noise can be a major part.  Here it can definitely help if a voltage of more than 5 V is available for the TIA output. So more like +-10 or 12 V at the TIA output and maybe a +-15 V supply. This is especially the case when using a more drifty JFET input OP-amp, that have a hard time to resolve down to the low µV range. The input to the TIA may want a series resistor / inductor as the TIA may not be stable with a capacitive signal source.  The soure impedance can also be important for the noise - a low impedance signal source will cause extra noise.

1 µA with 1 pA resolution is around 20 bits - so quite some resolution, but still not impossible.
The speed / bandwidth of the measurement may not be very high as more BW also means more noise. Because of parasitic capacitance the measurement tends to be not very fast anyway.

For the 1 mA range one can run into another problem, with the heat from 1 mA*10 V, that can already start to cause measurable self heating of the resistor.  For 1 pA required resolution one is in a range where a low bias zero drift OP amp like LTC2054 or max4238 can still be an option (with a 2nd amplifier for more voltage). Not yet a nead to look for the more electrometer OP-amps with there larger drift. That gets an issue in the fA range.

For lowest leakage air wiring may be an option and easier than a super clean PCB.

[David Hess]

As the previous posters have pointed out, dynamic range is limited by noise and drift.  The input current noise, input voltage noise, and resistor noise are all important.  Drift may be considered low frequency noise.

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

[moffy]

As the previous posters have pointed out, dynamic range is limited by noise and drift.  The input current noise, input voltage noise, and resistor noise are all important.  Drift may be considered low frequency noise.

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

The transistor Vbe junction on a number of small signal transistors e.g. 2N3904, is much more like the ideal diode than diodes are, and of course the LM194 had excellent log conformity from 3na to 300ua because of its low Rb, but drift is an issue. Good idea though.

[David Hess]

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

The transistor Vbe junction on a number of small signal transistors e.g. 2N3904, is much more like the ideal diode than diodes are, and of course the LM194 had excellent log conformity from 3na to 300ua because of its low Rb, but drift is an issue. Good idea though.

The transistor could be used as a high performance diode, but the translinear configuration uses it as a transistor and has even better log conformance at lower currents than a diode.  Accurate operation down to picoamps then becomes possible.

[moffy]

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

The transistor Vbe junction on a number of small signal transistors e.g. 2N3904, is much more like the ideal diode than diodes are, and of course the LM194 had excellent log conformity from 3na to 300ua because of its low Rb, but drift is an issue. Good idea though.

The transistor could be used as a high performance diode, but the translinear configuration uses it as a transistor and has even better log conformance at lower currents than a diode.  Accurate operation down to picoamps then becomes possible.

That is very interesting, do you have any references for that? I would be very interested.

P.S. Correct me if I am wrong, but are you talking about connecting the base and collector together, so that the transistor still operates as a transistor?

[moffy]

I ran a simulation on a 2N3904(base and collector connected) in LTSpice fed by an exponential current source and got a quite linear Vbe from 2pa to 10ma, not sure how accurate it is but an interesting result anyway.

[Marco]

do you have any references for that?

Not with transistors, but with LEDs which also tend to be really clean junctions.

https://www.semanticscholar.org/paper/Logarithmic-current-electrometer-using-light-diodes-Acharya-Aggarwal/ca5510a8c52a8bb979eaef2437991430f069a782

[moffy]

do you have any references for that?

Not with transistors, but with LEDs which also tend to be really clean junctions.

https://www.semanticscholar.org/paper/Logarithmic-current-electrometer-using-light-diodes-Acharya-Aggarwal/ca5510a8c52a8bb979eaef2437991430f069a782

That is a really interesting article. The great thing about LEDs (as long as they are blacked out) is their higher bandgap and therefore lower leakage.

[David Hess]

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

The transistor Vbe junction on a number of small signal transistors e.g. 2N3904, is much more like the ideal diode than diodes are, and of course the LM194 had excellent log conformity from 3na to 300ua because of its low Rb, but drift is an issue. Good idea though.

The transistor could be used as a high performance diode, but the translinear configuration uses it as a transistor and has even better log conformance at lower currents than a diode.  Accurate operation down to picoamps then becomes possible.

That is very interesting, do you have any references for that? I would be very interested.

My workstation with all of my references is in another state at the moment.   Bob Pease has written articles about it.  Linear Technology published an example of a very wide dynamic range photodiode amplifier, like 160dB, but I could not find it online.

P.S. Correct me if I am wrong, but are you talking about connecting the base and collector together, so that the transistor still operates as a transistor?

No, the transistor is connected with the base grounded, or offset to trim error, and the emitter driven by the operational amplifier output, with the collector tied to the inverting input.  The transistor then operates with a zero base-to-collector voltage.  There are plenty of examples of this available online.

If the base and collector are tied together, then the transistor is operating as a diode.

[RoGeorge]

...connecting the base and collector together, so that the transistor still operates as a transistor?

I think the referred topology is different, the transistor should be connected like in the logarithmic amplifier seen at page 2 of 5 here https://www.arsdcollege.ac.in/wp-content/uploads/2020/05/logamp.pdf, with the CE in the negative feedback loop of an opamp, and B to GND.

In this one at page 4 of 6 is mentioned the 120dB dynamic range instead of only 40-60dB, when the diode is replaced with a transistor with Base to GND:  https://www.analog.com/media/en/training-seminars/tutorials/MT-077.pdf

[moffy]

Much greater dynamic range can be had by replacing the feedback resistor with a diode junction, or better a bipolar transistor in the translinear (?) configuration, to take the logarithm of the input current.

The transistor Vbe junction on a number of small signal transistors e.g. 2N3904, is much more like the ideal diode than diodes are, and of course the LM194 had excellent log conformity from 3na to 300ua because of its low Rb, but drift is an issue. Good idea though.

The transistor could be used as a high performance diode, but the translinear configuration uses it as a transistor and has even better log conformance at lower currents than a diode.  Accurate operation down to picoamps then becomes possible.

That is very interesting, do you have any references for that? I would be very interested.

My workstation with all of my references is in another state at the moment.   Bob Pease has written articles about it.  Linear Technology published an example of a very wide dynamic range photodiode amplifier, like 160dB, but I could not find it online.

P.S. Correct me if I am wrong, but are you talking about connecting the base and collector together, so that the transistor still operates as a transistor?

No, the transistor is connected with the base grounded, or offset to trim error, and the emitter driven by the operational amplifier output, with the collector tied to the inverting input.  The transistor then operates with a zero base-to-collector voltage.  There are plenty of examples of this available online.

If the base and collector are tied together, then the transistor is operating as a diode.

Yeah I saw some examples in the LM194 datasheet where the base was grounded, collector to inverting input and emitter connected to the output of the op amp.
With the base and collector tied together, it still works as a transistor with a Vbe bias voltage across collector and emitter, with a similar Beta of the LM194 example. Iin = Ib + Ic, Ic = Ib*Beta and Ie = Ic + Ib. When running the LTSpice sim the Beta was significant. Running it this way also reduces the non perfect response due to Rb at higher currents because only a small portion of the current flows through the base contact. Hadn't really thought about it before, kind of surprising. It seems simple but it is actually a little bit complex. To run it just as a diode the collector should be left open circuit.

[moffy]

...connecting the base and collector together, so that the transistor still operates as a transistor?

I think the referred topology is different, the transistor should be connected like in the logarithmic amplifier seen at page 2 of 5 here https://www.arsdcollege.ac.in/wp-content/uploads/2020/05/logamp.pdf, with the CE in the negative feedback loop of an opamp, and B to GND.

In this one at page 4 of 6 is mentioned the 120dB dynamic range instead of only 40-60dB, when the diode is replaced with a transistor with Base to GND:  https://www.analog.com/media/en/training-seminars/tutorials/MT-077.pdf

Thanks for the references, I was under the mistaken impression that Ie = Is.exp(Vbe/Vt) when it is really Ic = Is.exp(Vbe/Vt) hence that is why the grounded base, removing the base current from the input current.

[magic]

You were under the right impression.
Ie is the exponential BE junction current and Ic depends on beta which depends on phase of the moon, among other factors.

[MasterT]

1 pA x 100 kOhm = 0.1 uV
opa2192 0.1-10 Hz noise is 1.3 uVp-p

[Marco]

One of the citations of the led paper did exactly what OP wanted BTW, using a second led with fixed current for temperature compensation.

[moffy]

You were under the right impression.
Ie is the exponential BE junction current and Ic depends on beta which depends on phase of the moon, among other factors.

I thought so too, but the Ebers-Moll model uses these equations: Ic = Is.(exp(Vbe/vt) - 1) and Ib = (Is/Bf).(exp(Vbe/vt) - 1)
So from these we can conclude that Ic is exponentially related to Vbe, no Beta to worry about, but Ib has the 1/Bf factor. I find it counter intuitive, but that's nothing new.

Reference: https://www.chu.berkeley.edu/wp-content/uploads/2020/01/Chenming-Hu_ch8-2.pdf page 305

[mawyatt]

The Vbe inferred in these equations is the Vbe at the actual Base Emitter Junction, not the Base Emitter terminals which include parasitic resistances. This intrinsic active junction Vbe is not even the Vbe on a chip transistor Base and Emitter because of the doping, contact and metallization resistances to get to the actual active junction where this Vbe is referenced. Then of course the junction is not an ideal "Step Junction" and is distributed over some distance with some additional finite resistances.

If you include the base and emitter contact & parasitic resistances, and the junction distribution resistance, then you discover that the external Vbe (Vbe') is far removed from the active intrinsic junction Vbe. This is influenced by these resistances which allow Beta to come into play by means of the emitter resistances which alter the active emitter side of the junction as Re*Ie = Re(Ic + Ib) = Re*Ic(1+ 1/Beta), where Re is the sum of all the emitter resistances. Likewise the base junction side is influnced by the total base resistances as Rb(Ib) = Rb*Ic/Beta.

Taking these into account and one arrives at;

Vbe = Vbe' -Re(Ic(1+ 1/Beta)) -Ic*Rb/Beta, where Vbe is the active intrinsic Vbe

And Vbe' is the extrinsic Base Emitter Terminal Voltage, and Vbe is the active junction Vbe.

Now introducing back into the classic equation:

Ic = Is{exp(Vbe' -Re(Ic(1+ 1/Beta)) -Ic*Rb/Beta)/vt -1}, which of course is transcendental.

Also note this does not include the distributed junction resistive effects which further complicates things!

If the resistances are very low, as is Ic, then this reduces to,

Ic = Is{exp(Vbe'/vt) -1}

Please check the math as this was just derived in my head and likely an error introduced.

Edit: Yep found one, equation above lacked a close ")" bracket!!

Anyway, the point is the terminal Vbe' is quite removed and different from the actual active junction Vbe, and this is how the Beta enters and influences the collector current.

Most modern bipolar models include some of the parasitic resistances and produce reasonable results under simulation.

Note the above assumes the transistor is not Saturated nor in Quasi-Saturation.

Best

[mawyatt]

You were under the right impression.
Ie is the exponential BE junction current and Ic depends on beta which depends on phase of the moon, among other factors.

I thought so too, but the Ebers-Moll model uses these equations: Ic = Is.(exp(Vbe/vt) - 1) and Ib = (Is/Bf).(exp(Vbe/vt) - 1)
So from these we can conclude that Ic is exponentially related to Vbe, no Beta to worry about, but Ib has the 1/Bf factor. I find it counter intuitive, but that's nothing new.

Reference: https://www.chu.berkeley.edu/wp-content/uploads/2020/01/Chenming-Hu_ch8-2.pdf page 305

Think the above should read Ib = (Ic/Bf) and not Ib = (Is/Bf).

Sorry didn't see the additional term, it looked like the "." was the sentence end and not the multiplicand, my bad!!

BTW the Ebers-Moll model is one of the more elementary bipolar models, more modern models are VBIC, MEXTRAM and HiCUM which encompass more of the non-ideal bipolar behavior.

Best,

[magic]

I thought so too, but the Ebers-Moll model uses these equations: Ic = Is.(exp(Vbe/vt) - 1) and Ib = (Is/Bf).(exp(Vbe/vt) - 1)
So from these we can conclude that Ic is exponentially related to Vbe, no Beta to worry about, but Ib has the 1/Bf factor. I find it counter intuitive, but that's nothing new.

Reference: https://www.chu.berkeley.edu/wp-content/uploads/2020/01/Chenming-Hu_ch8-2.pdf page 305

I have also seen versions of this equation with I_E instead.

But maybe there is something to it, because the plot on page 301 implies that β falloff at low currents is due to Ib decreasing disproportionately slowly.
Unfortunately, the supposed explanation in section 4.9.1 is missing from this excerpt.

Interesting, I will have to look it up.


BTW, I'm not entirely convinced that any of that is relevant to the OP. I think he imagined measuring mA scale currents with pA resolution, which is a completely different kind of dynamic range problem.

[moffy]

BTW, I'm not entirely convinced that any of that is relevant to the OP. I think he imagined measuring mA scale currents with pA resolution, which is a completely different kind of dynamic range problem.

I agree, it is at best tangential to the OP, but I have learned quite a bit. Who knows when that knowledge might come in handy.