given snr and maximum frequency, what is the lowest analog current measurement possible

All about making the smallest-possible current measurements is spelled out in extreme detail in The Art of Electronics, 3rd edition, in Chapters 5 and 8. Besides the detailed discussion, there are multiple formulas, graphs, and tables, taking up 220 pages for the two chapters. Get the book, study the topic, learn how to ask and answer the question. I seriously doubt that you can instead get a useful two or five paragraph answer online, let alone understand it.

Is the signal a sine wave? If so, there is no lower limit; it just takes longer to measure smaller signals.

As others mention, it's not the frequency that matters, but the spread. If the signal has truly zero bandwidth, as an ideal sine wave does, you can integrate forever with perfect timing, and remove any noise.

Doing it with finite everything (nonzero bandwidth, finite time, finite nonzero impedance), and SNR is derived from the bandwidth, Johnson-Nyquist noise of the source resistance, and noise factor of the receiver.

Tim

By definition 1 A = 1 C/s = 6.2E18 electrons/second. This sets the theoretical minimum slightly below 1 aA. Best electrometers are a few order of magnitude worse

That would suggest that the signal is AC, so with a sufficient long integration time and small bandwidth ultimately the quantization of the electron charge will limit the performance.

Thanks snipped-for-privacy@downunder.com for taking time to review and reply to my pos t...

I am assuming that with typical digital signal with no anticipated signal p attern (see my reply to Tim's response for some more detail about the signa l being measured) can still be measured of the order of nA (which is a few order of magnitudes lower than fA) using an accompanying circuit. I would l ike to confirm this is theoretically possible before thinking about buildin g the circuitry.

Do you th> On Tue, 19 Feb 2019 16:45:56 -0800 (PST), >

nA, pA or fA) if I know the SNR I can get and the frequency of the signal.

As usual you don't give anough info on what you're trying to achieve. Plot the current waveform, or just measure average current over a 1 minute period - the solutions are quite different.

NT

A voltage source through a resistor will have much less noise, than the shot noise limit, so in principle you can measure less than one electron worth of charge. (you need gain) Bipolar transistors are limited by shot noise in the base, I'm not sure what limits FET noise.

George H.

We've been through this before, but the fact that charge is quantized doesn't imply that current is, too.

Jeroen Belleman