How can I find the minimum analog input voltage for an oscilloscope? I've checked the datasheet for the particular one I am using(Agilent DSO-X 2002A) and only the maximum input voltage is listed.
Add a comment
|
3 Answers
\$\begingroup\$
\$\endgroup\$
2
There's no minimum signal amplitude, because too small a signal won't damage the scope.
There are specs indicating how small a signal can be measured:
With 1 mV/division and 8 divisions on the screen, the maximum scale is 8 mV, full scale. But that is just a zoom-in on the 4-mV/division waveform. With 8-bit resolution, that means the minimum detectable signal is about 128 uV. And of course there will also be noise obscuring the signal. With multi-trace averaging, you might be able to detect a slightly smaller signal.
Also, remember that your choice of probes will affect what signal actually reaches the scope's input electronics. Using a 10:1 probe will increase the minimum detectable amplitude by a factor of 10.
answered Sep 9, 2017 at 15:32
-
1\$\begingroup\$ I would not count on 32 uV resolution as it is +/-8 divisions using 8 bits @ 4mV/div while 1mV/div is zoomed. and 12 bit resolution only with averaging.Thus 16 div*4mV/div =256 thus a resolution of 64 uV then SNR is not specified, nor BW or f. \$\endgroup\$– D.A.S.Commented Sep 9, 2017 at 15:48
-
\$\begingroup\$ As a practical limit, with my MSO-X 2002A (same as OP but with digital enabled), when nothing is connected, set to 1:1 probe, and no averaging is done there is a noise signal of roughly 1mVpp. \$\endgroup\$ Commented Sep 11, 2017 at 20:34
\$\begingroup\$
\$\endgroup\$
It's in the datasheet 15 of 24.
1 mV/div and 2 mV/div is a magnification of 4 mV/div setting. For vertical accuracy calculations, use full scale of 32 mV for 1 mV/div and 2 mV/div sensitivity setting
\$\begingroup\$
\$\endgroup\$
I'm not sure I understand the question. The maximum input is a protection for the front end. If you put in more than that it could damage the oscilloscope. However, there is no way to damage the scope by putting in too little signal.
But, you may be asking what is the minimum signal you can see on the scope? That depends on your probing. But, for the Keysight DSOX2002A the minimum hardware voltage setting is 4 mV/division. In normal acquisition modes, you can then divide the total voltage setting on the screen (4 mV x 8 divisions = 32 mV). And, you get 8 bits of resolution in normal acquisition mode, so 2E8 = 256. So, you could theoretically measure signal in increments of 125 uV (32 mV / 256 levels).
But, in other acquisition modes you can get higher resolution.
But this doesn't take into consideration the noise floor of the oscilloscope (what do you measure with nothing connected?) and it doesn't take into consideration triggering. you'll need an edge at least .6 divisions tall (ish) for the scope to trigger and for you to see a stable waveform on the screen.
I hope that's helpful info!
answered Sep 11, 2017 at 19:24