What is the proper procedure to measure RMS noise with an oscilloscope?

I have 2 standard resistors, 1k and 100k and a 1.5GHz BW, 5GSa/s Keysight DSO. I have a probe which offers 1:1 (1MOhm//100pF) and 10:1 (10MOhm//15pF) ratios.

I use AC coupling, 1:1 probe and "AC RMS - Full Scale" measurement - which 1mV/div and 100ms/div.

I measure 340uVrms for 1k and 940uVrms for 100k (just clipping the resistor between the probe).

However, I cannot reproduce this result using equations. Three attempts:

  1. The resistor is in parallel with the probe which is 1MOhm and 100pF for 1x. Hence the bandwidth will be 1MOhm//R//1pF ~ 1/(2*pi*RC). Since the total integrated noise is given by 4kTRB = 4kTR/(2*piRC) = 2kT/(piC) = 2/pi kT/C the result should be independent of the resistor 5.1363uVrms. This is far off from the numbers above but even worse - the numbers above are different

  2. I assume the bandwidth limitation comes from the probe itself which is 6MHz for 1x. The result would be sqrt(4*kT*100e3*6e6)=100uVrms for 100k and 10uVrms for 1k. Again, both far off.

  3. I assume the bandwidth is limited by the bandwidth spec of the scope which is 1.5 GHz. This gives 157.68uVrms for 1k and 1.5768mVrms for 100k.

Again, not consistent.

How do I measure noise with an oscilloscope?

  • 1
    \$\begingroup\$ some ideas here: electronics.stackexchange.com/questions/280860/… \$\endgroup\$
    – glen_geek
    Commented Dec 15, 2017 at 2:08
  • 1
    \$\begingroup\$ Are you really hoping to measure Johnson noise directly with an oscillocope? How are you accounting for the other sources of noise in the system, which are orders of magnitude higher? Measuring low levels of noise requires careful differential techniques. \$\endgroup\$
    – Dave Tweed
    Commented Dec 15, 2017 at 2:27
  • \$\begingroup\$ Bandwidth of a scope with a probe is equal to PRODUCT of their transfer functions. Say you have a 1 GHz scope and a 1 GHz probe, the resulting bandwidth is just 700 MHz. \$\endgroup\$ Commented Dec 15, 2017 at 5:09
  • \$\begingroup\$ Do you see the noise increasing/decreasing if you short the noisy resistor ? If not: you're not measuring the resistors's noise. Maybe this video will also help: youtube.com/watch?v=Znwp0pK8Tzk In my opinion you would need to amplify the noise (using a low noise amplifier with a known gain and BW) before you can properly measure the noise on a scope. A scope's input impedance is 1 Mohm and not designed for low noise. You would need a scope with 50 ohm inputs. If available: use a spectrum analyzer. \$\endgroup\$ Commented Dec 15, 2017 at 7:18

1 Answer 1


A system with a single node: some R and some C, will have the total integrated noise defined EXACTLY by sqrt (K*T/C).

Thus a 10pF cap, at 290 degree K, produces exactly 20 microVolts RMS, regardlss of the value of the resistor.

Also, a 100pF cap would produce 1/sqrt(10) less noise, or about 6 microVolts RMS, regardless of the value of resistor.

To measure 6uV RMS in 10GHz bandwidth, the system noise density must be lower than 6uS/sqrt(10GHZ) = 6uV/sqrt(10^10) = 6uV/100,000 = 0.06 nanoVolts per rootHertz.

Given a 60 ohm resistor produces 1nanoVolt/rtHz noise density, and we need 0.06nV, or 16X smaller than 1nV, the total front end Rnoise my be 60/(16*16) 60/256 or 0.25 ohms.

Achieving Rnoise of 0.25 ohms will not happen.

Typical Rnoise (the rbb') of bipolars is 10 ohms or 100 ohms.

I cannot speak for typical rbb' for silicon-germanium bipolars.


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