I need to sense the current of a target MCU through various periods of its on/sleep/off state and then need to send this data back to another interface MCU to read out the value. They will be both on the same board, powered by 5 V USB.

However these are the requirements I have:

  1. 5 V USB PSU
  2. Must be able to sense currents in the range of nA to mA with high accuracy/precision. (I figure 1 nA to 500 mA)
  3. Only measure the current of the target MCU and not the interface.
  4. Must output maximum 3.3 V interface MCU

I have looked at available options and current sense amplifiers by Texas Instruments can't sense nA due to higher bias currents. So therefore I feel I would need a precision amplifier.

However I am stuck on how to proceed with this because, I need to somehow have a automatic dynamic current range and I am fairly new to current sensing and am not really sure on all the specifics.

  • 10
    \$\begingroup\$ You are essentially talking about 9 orders of magnitude, or 30bits (noise-free) resolution. This is highly non-trivial. \$\endgroup\$
    – corecode
    Sep 2, 2016 at 8:25
  • \$\begingroup\$ Why? Manually adjustable gain... Like three resistors between ldo and 12V: 1R, 100R and 10k. \$\endgroup\$
    – user76844
    Sep 2, 2016 at 8:28
  • \$\begingroup\$ Gregory this is what i thought, i have seen it done with the uCurrent Gold device. Would there be a way to possible automatically adjust this gain? \$\endgroup\$ Sep 2, 2016 at 8:34
  • 1
    \$\begingroup\$ If you are okay with switching the ranges manually, it gets much simpler. However, this will fail as soon as your target system does the power mode switching by itself. Trying to draw several mA through a 10k sense resistor will droop the supply voltage and will brownout the chip. \$\endgroup\$
    – corecode
    Sep 2, 2016 at 8:38
  • 1
    \$\begingroup\$ What is your definition of high accuracy and precision? Also, what frequencies are you measuring at? You have many more options if you can take the time to make multiple integrated measurements than if you need to accomplish this in real time at 5 megasamples or something high like that. \$\endgroup\$
    – Cort Ammon
    Sep 2, 2016 at 16:06

4 Answers 4



A circuit is presented based on a regulator topology, stable into any capacitive load, which includes a diode in series with the output current. The voltage developed across this diode is nominally the log of the current, which allows a very wide range of current to be measured with a single voltage range. Excellent dynamic stability has been demonstrated in simulation.

At low current, the circuit is noisy and slow (no big suprise). The present results show about +/- 5% rms noise at low currents, for settling times of 10uS for currents of 1uA and above, increasing to 1 second settling time for currents down to 1nA.


I suspect you don't need high accuracy. You only think you do because of the huge range from nA to 500mA. Obviously +/- 1nA at 500mA would require collossal accuracy. I suspect that +/- 10% at 500mA simultaneously with +/- 10% at a few nA and a single range to cover both without switching would be useful.

The initial thought, that I threw down as a suggestion initially, is shown at the bottom of the post for reference.

Unfortunately it has a fatal flaw. While it can indicate 1nA well enough, as the current suddenly increases, the opamp output doesn't initially move, due to both its internal compensation and C1. As a result, the output voltage drops by over 1v (needed to get the current flowing through Q1 and D1) for a moment, which would severely embarrass any MCU being supplied by that line.

The 'solution' is to incorporate the MCU rail decoupling capacitance into the analysis. However, extra C on the MCU line provokes instability, as it is in shunt with the opamp inverting input, and practically uncompensatable over a the wide range we want.

So the next thought was 'this is basically a transimpedance amplifier, albeit with a very non-linear feedback resistor, how are they stable?' A quick search for those brought me to Bob Pease's article (Nat Semi's RAP, Bob Pease - must reading for any analogue designer. If you take nothing else from this answer, dig out and read some of his stuff!)

It was quickly apparent that the assumed capacitance on the op-amp inverting node, although big compared to pF, was very small compared to the 10uF that we might find on a VCC line, and the high speed tweaking assumed a constant feedback resistor, so this topology was a non-starter.

So then I thought, if we are not going to brown-out the MCU when the current changes, it must behave like a regulator. I recalled the tantalum versus ceramic output capacitor issues of LDOs. Architectures that rely on the half ohm ESR of a tantalum to be stable are not stable with ceramics. When the topology is changed to tolerate ceramics' zero ESR, they can tolerate any large value above the specified minimum.

In order to cope with a large output capacitor, it is designed to be the dominant pole, with an output current source turning it into an integrator, keeping the rest of the control chain with less than 45 degrees phase shift. Once that flip has been made, the output capacitor can be any size larger, and the LDO will still be stable. The output capacitor of the regulator provides all the voltage hold-up during a current change event.

Now I searched for LDO app notes. This is the new design as a result. It is broadly similar to the original one in DC concept, but is built round the output capacitor, and uses the tricks employed by LDOs designed for ceramics, to get sufficient stability.

enter image description here


Q2 is the series pass PNP device, configured with R2 to be current output. That particular type is 1 Amp, 200 hfe 150, 50v, 400MHz ft fairly cooking-grade part that was in the LTSpice library. I1 biasses it to a nominal 10mA, to reduce the delta V required when suddenly required to increase current from zero, and to provide a substantial current sink to cope with a sudden reduction in current output.

D1 is our old friend the non-linear element across which the output current develops the log voltage. I've used 1n4148 as it was in the library. It's joined by R1, to define the bottom end of the current range (10mV for 1nA), D3 to catch reverse voltages when the current suddenly decreases, and C2 as it improves stability and output overshoot. Note that if the 1N4148 is replaced by beefier 1n400x types, their higher capacitance will be completely absorbed by C2, so are well enough modelled for stability.

I would have modelled a TL071. I first tried an LTC1150 which had a GBW of 1.5MHz, but struggled to get reasonable stability. I then switched to the LT1022 shown. This is a bit quicker at 8MHz GBW, but there are many much quicker parts around.

The network around it includes R3 to sense 0v, C3 for stability, and R4 to add a zero to C3, as suggested in LDO app notes. With these values, arrived at by hope'n'poke, it's already not bad. I'm sure it could be better with a bit of proper analysis. Rather than using a yet faster unity gain stable amplifier, it should be better to use one that's decompensated.

It certainly looks stable enough for the purpose. Anybody building this circuit to use in anger may find some more unmodelled parasitics that reduce stability, but I would suggest they start with a yet faster amplifier to give themselves some more elbow room.

I2 provides the time dependent current loading for the demo. As you can see from the parameter string, it's slewing 100pA to 100mA with 100nS risetime (so changing current on in one cycle of 10MHz), and back again. The diode D2 provides a convenient way for the simulation to show the log current, and is no part of the target circuit.

When doing simulations, I prefer to have all the 'action' around 0v, so for the rails of -5, 0v and +5v shown here, read 0v, +5v and +10v respectively for the OP's application.

This is the overall transient plot

enter image description here

The initial DC value of the output voltage is 0.5mV for 100pA, and when I go from 1nA, it's about 5mV, so we have sensible discrimination at, and below, the 1nA level.

There is a slight overshoot of the measurement value when the current increases.

The slew hits the diode limits when the current decreases. There is also a 20mS reading tail when switching down from 100mA to 100pA, I don't know how to improve that, perhaps somebody has a suggestion. The tail is still present when switching down to 10nA, but when switching down to 100nA or more, the tail is absent. For this application, I would imagine that's OK.

In the next three plots, we look at the all important output rail voltage stability.

On the increase from 100pA to 100mA

enter image description here

The up-going rail transient is only 12mV, and dead beat. You won't find many commercial LDOs delivering that sort of performance for such a violent current change.

and on the way back down again to 100pA

enter image description here

Without D3 to provide reverse conduction, Vmeas would swing to the -ve rail for a while rather than to -0.6v.

enter image description here

The down-going rail transient is also limited to 12mV. You can see the rate-limited downwards slew which is the result of the I1 current sink.

I'm not going to say it's a proof of principle, but I think it is a very good proof of plausibility. The simulation includes a lot of parasitics, Q2 Miller C, the opamp's compensation, and with performance rivalling an LDO, I think that's a pretty good basis from which to start developing something that can power an MCU, at different currents, reading over a large range.

This shows Vmeas as the output. As indicated in the original post, thermal accuracy will be improved if it is measured with respect to another diode at the same temperature. Vmeas is a low impedance output, so this is very straightforward to do with a simple differential amplifier.

As before, replacing R1 with a lower value resistor will give a more accurate, linear range output, for voltages for which D1 is not conducting.

Noise issues

Now that a stable circuit has been developed, we can start to look at noise. The following graph shows the gain from op-amp input, with a 1nF capacitor fitted at C2. The curves cover 100pA to 100mA. The 100pA and 1nA curves are indistinguishable at bright blue, and very close to the red 10nA curve. 1uA is pink, 1mA is dark blue, the 100mA curve is lowest as purple.

enter image description here

Using LTSpice's .noise simulation, and using .measure to integrate the output noise over a bandwidth of 10mHz to 10MHz, using a 33nF capacitor for C2, resulted in a relatively constant 2mV rms noise for currents 1nA to 100uA, with noise falling as currents increased to around 100uV rms at 100mA.

The penalty of the increased value of C3 was increased settling time following a step reduction in current. The time to within 1mV of the final value was approximately 10mS to 1uA, 60mS to 100nA, 500mS to 10nA, and 900mS to 1nA.

The present op amp, LT1022, claims several 10s of nV at 1kHz. Bob Pease's transimpedance amplifier article referred to earlier suggests that 3nV is feasible with a low current FET input, using discrete low noise FETs as the front end to a composite amplifier. Using such an improved opamp should reduce the noise levels by an order of magnitude.

This is the original suggestion, for reference.


simulate this circuit – Schematic created using CircuitLab

The opamp will servo the current through Q1 and D1 to maintain the output voltage at 5v, so your MCU is always seeing its correct operating voltage.

The voltage you measure between the two diodes is proportional to the log of the ratio of D1 current to D2 current. While you can work with the voltage across D1 alone, it is temperature dependent. This method uses D2 to compensate that dependence.

  • \$\begingroup\$ Take an op-amp with an excellant noise voltage density of 1 nV per sqrt(Hz) and a bandwidth of 10 kHz (to suit measuring pulses of current taken by the MCU). The noise voltage on the output is going to be 30 odd nV RMS and this is above 100 Hz (typically). At 1 Hz it's going to be microvolts of noise so how can you say this circuit works down to 1nA with any respectable degree of accuracy ? Then you have to look at the noise gain of the OP-AMP. The NG will be substantial given the nature of the load (low impedance). I've not downvoted BTW. \$\endgroup\$
    – Andy aka
    Sep 2, 2016 at 11:46
  • 1
    \$\begingroup\$ @Andyaka not sure the point you're trying to make here Andy. It's log reading. Let's say we had 1mV of noise, a generous over estimate, rather than microvolts you worry about. I've just measured a 1N4007, and it's about 100mV per decade of current (317mV at 1uA, 599mV at 1mA, 909mV at 1A), so 1mV of noise is one centi-decade, or about +/- 2.3%, well within my WAG for accuracy of 10%. More to the point is that 300mV at 1uA projects on down to 100nA/200mV, 10nA/100mV and 1nA for zero bias, so something's going to give somewhere at low enough current. Thanks for your contribution. \$\endgroup\$
    – Neil_UK
    Sep 2, 2016 at 13:41
  • 1
    \$\begingroup\$ Edited the answer to include the low voltage/low current regime. \$\endgroup\$
    – Neil_UK
    Sep 2, 2016 at 13:47
  • \$\begingroup\$ That's a neat circuit. Will the leakage current from the bridge affect much of anything? \$\endgroup\$
    – TLW
    Sep 2, 2016 at 15:32
  • \$\begingroup\$ My recent measurement of IN4007 suggested around 1nA leakage at around zero volts, I would guess typical 1A bridges will use similar silicon. \$\endgroup\$
    – Neil_UK
    Sep 2, 2016 at 15:44

Must be able to sense currents in the range of nA to mA with high accuracy/precision. (I figure 1 nA to 500 mA)


I need to sense the current of a target MCU through various periods of it's on/sleep/off state

OK assuming you will place a small value resistor in the power feed, that resistor must not "drop" more than (say) 0.1 volts at 500 mA. If it did drop a significant voltage then you are compromising the measurement and possibly causing the target device to run at too low a voltage.

So, 500 mA and 0.1 volts requires a resistor of value 0.2 ohms. Now, that resistor when fed 1 nA will produce a measurement voltage of 0.2 nV.

Do you see the first problem? There isn't really a cheap and reliable tech that can do this because any op-amp will have noise significantly bigger than what you are trying to measure and, given that you appear to want to make dynamic measurements, your required bandwidth may be tens of kHz and you'll just measure noise!

EDIT - log amp considerations

  1. Assuming a 10 kHz noise bandwidth (about 7 kHz signal bandwidth in order to adequately measure the changes in the target's current), means that an op-amp with 1 nV/sqrt(Hz) voltage noise has the equivalent to 100 nV RMS at the non-inverting input. An op-amp with this low a noise value is a rare beast indeed and comes with a host of other problems that will dog this design.
  2. Using a diode in the feedback loop looks attractive but at round about 100 nA being delivered to the load it will have maybe 300 mV across it. As an impedance, this sets the noise gain of the op-amp circuit. So, 300 mV / 100 nA is a dynamic resistance of 3 Mohms and this resistance value will only increase as supply current falls below 100 nA i.e. things will get worse at lower currents.
  3. That resistance (the dynamic resistance of the diode in the feedback circuit), along with the dynamic impedance of the load, produce noise gain in the op-amp circuit so, if the dynamic impedance of the load is 1 ohm then the noise gain is 3,000,000 (assuming the op-amp could deliver this open-loop).
  4. The op-amp input noise (as mentioned above) is 100nV RMS or (using 6 sigma), 600 nV p-p. Half of this gets discarded due to the diode blocking it thus leaving 300 nV amplified by 3,000,000 and hence potentially producing a peak voltage of 0.9 volts.
  5. This is the "potential" noise voltage that could be seen at the output of the log-amp. However, if the voltage noise rises above 300 mV then the dynamic impedance of the diode falls from 3 Mohm and gain reduces and, the upshot of all this is that the peak noise voltage will probably find a level peaking at about 400 mV maximum. But, up to that point (0 nA to 100 nA) all bets are off trying to get any decent measurement.

If the dynamic impedance of the load is 10 ohms (rather than 1 ohm) then that is a different story but will this be possible given the likelihood of 100 nF caps on the power rails and the possible presence of higher values.

How tricky will it be to find an op-amp with such a low voltage source noise that has really low input noise currents? Also remember that for most op-amps, the noise voltage dramatically rises as the frequency falls below (about) 100 Hz so this is a real problem.

So, to make a log-amp work, bandwidth has to be significantly restricted but does this give the OP the chance to adequately measure dynamic changes of current when (say) the target MCU executes different routines?

  • 1
    \$\begingroup\$ Reading the OP, I think she does see the problem, with comments like 'automatic ranging' and suchlike. Don't you think your rhetorical question is somewhat patronising, especially as you give her no route to a solution. \$\endgroup\$
    – Neil_UK
    Sep 2, 2016 at 13:42
  • \$\begingroup\$ @Neil_UK I don't see a solution (not yours) unless more detail is provided about bandwidth being very limited. If I come over as patronizing then, on this particular occasion, that would be you misreading me. \$\endgroup\$
    – Andy aka
    Sep 2, 2016 at 13:48
  • \$\begingroup\$ I see the Neil's idea as excellent. The OP was charged with monitoring of power state of MCU, but the requirements were not aligned with specifics of the task, which has led to absurd range. In practice, you need fast response time only when MCU is in active mode, and consumes mAmps. When it sleeps, nobody cares how fast it switches in nA range, and therefore the bandwidth can be reduced to zero. What people care in this mode whether the MCU did enter the low power state, and whether there are design/software bugs that do not let MCU to reach its power targets. \$\endgroup\$ Sep 2, 2016 at 17:54
  • \$\begingroup\$ @AliChen if you do not speak on behalf of the OP then I suggest that you do not try and 2nd guess the situation. \$\endgroup\$
    – Andy aka
    Sep 2, 2016 at 18:34
  • \$\begingroup\$ @Andy: I am simply speaking from experience. Consider my comment above as a question to OP, and allow me to discard your suggestion. \$\endgroup\$ Sep 2, 2016 at 18:49

As long as you don't need fast switching of the gain. You could do a TIA opamp circuit with relays being used to switch feedback resistance as you got to the top and bottom end of the ranges. Getting above ~10-30 mA is hard for the typical opamp so the high range needs a bit more thought. Do you need to sense bipolar currents?


Measuring current in such a wide range without significant precision loss requires a current sense circuit with adjustable resistance. Usually it is either a set of resistors with different values paired with FETs or just FET transistors connected in series. This circuit is driven by a feedback loop: when the measured current changes, either resistor values are switched of FET gate voltage adjusted. Agilent implements the latter method in some of their power supplies.


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