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?

enter image description here

Is this calculation correct:

There are 10 vertical squares so,

2mV × 10 = 20mV

Min dynamic voltage = 20mV/2^8 = 78.125mV ??

  • 2
    \$\begingroup\$ Your assumption is correct if you change that 78 mV to 78 uV, but only if the 20 mV full scale of the display is the same as the full scale of the ADC. My point: the display might show 20 mV but that is no guarantee that the full range of the ADC is also 20 mV. It might be 100 mV, then the resolution drops with a factor 100mV/20mV = 5. \$\endgroup\$ Oct 1, 2019 at 13:50
  • \$\begingroup\$ Also, "Min dynamic voltage" isn't the correct terminology, we generally call this number "resolution". \$\endgroup\$ Oct 1, 2019 at 13:55
  • \$\begingroup\$ Depending upon the bandwidth and implementation the noise level of the scope may be that much on the most sensitive ranges so the quantization may be covered up by the noise. I have often seen 50-200uV p-p noise at the input of a scope. \$\endgroup\$ Oct 1, 2019 at 15:59
  • \$\begingroup\$ @KevinWhite If I short the inputs of a scope channel and measure the noise floor, do you mean that is the limit of the resolution? \$\endgroup\$
    – floppy380
    Oct 1, 2019 at 16:06
  • \$\begingroup\$ @HelpMee - pretty much. In an analog system the ration between the maximum signal and noise floor is of course the signal to noise ratio. \$\endgroup\$ Oct 1, 2019 at 22:00

1 Answer 1


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


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