Measuring small currents over a small sense resistor
Try this







simulate this circuit &ndash; 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.