oscilloscope
In general Vertical sensitivity of the oscilloscopes are in 2mV to 10V/div range.
Neglecting the noise contributions, if the scope has an 8-bit resolution and if I set the vertical sensitivity to 2mV as shown below how can I calculate the minimum voltage the scope can measure or show on screen?
Is this calculation correct:
There are 10 vertical squares so,
2mV × 10 = 20mV
Min dynamic voltage = 20mV/2^8 = 78.125mV ??
If you have a single frequency spectral point at say 2kHz and it is -55 dBu then it has a voltage level of 0.001377449 Vrms (see this for the conversion). If you also had another spectral point at 2.1kHz and it was -60 dBu (0.000774597 Vrms), then you'd calculate the RMS value of these two points by this: -
RMS Value of A and B together = \$\sqrt{A^2 + B^2}\$
You'd get a value of 0.001580305. That's two points.
Now extend this method to all the points within the spectral band of interest. If it means summing 10,000 points then taking the sq root of this sum, then that's what you have to do.
For the sake of discussion, let's assume that the MCU absorbs around 1mA (I suspect much less, actually). This means 2mV across the sensing resistor, which is pretty low. I think you are just measuring common mode noise, and your DC component is swamped by that.
You could try to activate an averaging acquisition mode, averaging over several waveforms. You should see some improvement.
You could also try to deactivate all possible sources of interference in the room (if you have LED lighting fixtures their power supply could contribute to the radiated and conducted noise, for example).
Try also to avoid using the ground clip of your scope probe: that acts as an antenna! Locate a ground point near your resistor and connect to it using this kind of probe accessory:
EEVBlog #441 - "How to track down common mode noise" gives additional hints and I think it's relevant.
Anyway to measure such tiny currents you may need some additional amplification. Dave Jones, from EEVBlog, has designed his μCurrent amplifier just for this kind of things.
Best Answer
You're basically correct. This is known as the scope's "resolution." For 99.9% digital oscilloscopes, the ADC only looks at the height of the screen and does not go off screen. For this reason, they also don't make measurements on signal data above or below the screen.
So, you can take the "full scale" of the screen - the volts-per-division times the number of vertical divisions - and divide by the number of "q-levels." Q-levels, or quantization levels, are the number of different levels an oscilloscope's ADC can measure. So, a scope with an 8-bit ADC will have 2^8 q-levels.
Now, just because the resolution means it's precise, but it doesn't mean it's accurate. Oscilloscopes have a "DC vertical accuracy" specification that tells you how accurate those levels actually are. That spec is a cocktail of other accuracy specs. And, to make things more complicated, the actual accuracy may vary depending on the V/div setting.
My team put together a great whitepaper on measuring oscilloscope vertical noise here: http://literature.cdn.keysight.com/litweb/pdf/5989-3020EN.pdf
And a video here: https://www.youtube.com/watch?v=fZNp5h3JjrY
I hope this is helpful!
Source: I work for Keysight with oscilloscopes