Measuring small currents over a small sense resistor

Question

I have been having some difficulty recently when trying to measure very low currents when using a small sense resistor.

I am trying to measure in the range of uA (1 - 150uA) with a supply voltage of 3.3V. Now, usually this is not too much of an issue, but the problem I am getting here is that this particular project uses a WiFi module that peaks at around 600mA.

I am needing to use a small current sense resistor (0.1R) because I just can not afford the voltage drop over higher values at peak current consumption. I tried a few different current sense amplifiers but found them to be not quite accurate enough at these small levels. I eventually settled for a chopper stabilised amplifier (TLC2652) because even though it wasnt a specific current sense amplifier, it offered my <1uV offset which I thought was great. I set it up as a differential amplifier with a gain of 100 however, it still struggled with accuracy when getting tot he very low current range. Annoyingly, it works perfectly when using a 1R current sense resistor, but as previously mentioned, I cannot afford this voltage drop. With the 0.1R resistor, it is fine untill it gets below round about 20ish uA then it gets a bit noisy.

Does anyone have any suggestions on a good workaround for this to get it to work?

Here is the way I have it set up:

^{simulate this circuit – Schematic created using CircuitLab}

I will say that it is not highly critical for it to be supremely accurate, this is just part of a circuit to do functionality checks, so I am fine with it being a little noisy in these very low currents, however, it does not stop my curiosity and my desire to find a way to solve it!

I hope I have worded my question alright, please let me know if I have missed any information or if there is something I have not explained correctly.

Also, as a side note, I must state I am still learning electronics so apologies if I have done something incorrectly or done something that some more experienced people may find silly!

Answer 1

Try this

^{simulate this circuit – Schematic created using CircuitLab}

How much noise to expect?

1uVrms, 48dB SNR

And now, with time and frequency combined, in SWE; here is 3Volt Peak signal at 50Hz, with 0dB RMS noise of bandwidth 250Hertz, into 20Hz LowPass filter:

The frequency response of the 20Hz LPF is

$$ 1 / sqrt(1 + (Freq/20Hz)^2)$$

At DC, gain is unity.

At 20Hz, gain is 1/sqrt(1 + 1) = 1/1.414 = 0.707 == -3dB

At 200Hz, gain is nearly 0.1

At 2,000Hz gain is nearly 0.01

EDIT>>> In the first schematic, the 100Kohm and 1uF create a 0.1 second TAU or TimeConstant. For accurate measurements, you need to WAIT after the current changes; 1Tau provides 1 Neper of accuracy (9dB or 1.5bits); 2tau provides 2 Nepers of accuracy because the RC has had twice as long to settle (18dB, 3 bits, or approximately 16% error); 5 TAU or 0.5 seconds achieves 45 dB of accuracy, nearly 0.5% accuracy.

Thus the bandwidth of the system, here set by that 100Kohm and 1uF, determines how LONG YOU MUST WAIT after a big step change to accurately measure that big step change. Small changes only need proportionately less time, before you can accurately measure the value. Of course, small changes may be almost buried in the noise.

Answer 2

I have been having some difficulty recently when trying to measure very low currents.

Use a I-V converter instead.

Quite simple.

edit:

I am trying to measure in the range of uA (1 - 150uA) ... that peaks at around 600mA.

that's a range of 600k:1.

I would say that it is (way) beyond even what 99.99% of expert analog engineers can do. any one coming out with a solution here would be an instant multi-millionaire.

HP happens to have a patent on that front - less than the range you are looking for - that may serve as a starting point for you.

Answer 3

When the problem is problematic, why not change the problem?

^{simulate this circuit – Schematic created using CircuitLab}

This is a simple opamp-based voltage regulator, with an extra current-sense. Compensation and stability is on you, but it's not that hard. Pick a FET which can pass the current...

R3 (100R here) is your sense resistor, so 10µA gives you 1mV on the resistor. Plug your diff-amp input across R3. Choose a proper value for R3. You can use a switch with several value if you want.

D1, of course, solves your dynamic range problem, by limiting the voltage across R3 to one diode drop, so you can use a large resistor AND have your high current, too.

Say you want to measure 150µA... no problem, set R3 to 2k2 kOhm. 150µA gives you 330mV. Easy to mesure. And you set your current sense amp gain to make sure that it does not clip when voltage on R3 reaches one diode drop and it begins conducting.

Because if your current sense amp clips, then its input stage will have a thermal tail, delaying settling by potentially milliseconds. Which is why none of the circuits proposed here will work. You don't use an opamp to measure microvolts on a sense resistor, and trust the results if it occasionnally clips.

Answer 4

Ignoring the schematic error showing the opamp output connected to the positive power rail.....

The opamp noise is about 30 nano volts per root hertz so, if you have a 1 MHz measurement bandwidth, the rms noise at the input will be about 30 uV rms. You have a gain of 100 so this becomes 3 mV of noise at the output.

Now, these are just ballpark figures for you to muse over. If the bandwidth of your measuring circuit is only 10 kHz then the noise will be about 300 uV rms.

This and the LM317 are the likely culprits.

Another problem is the tolerances on the resistors. With such a high gain differential circuit and 1% resistors, the common mode rejection will be pretty bad and noise from the LM317 power rail will be amplified. There are chips that will do this job for you. Linear Technology make "high side" current monitors that will be much more suitable than a diff opamp circuit.