0
\$\begingroup\$

enter image description hereI am trying to plot a Noise Spectral Density from 10 kHz to 1 MHz. The Graph shows a cure which at 1 MHz the noise is 76 pV/rootHz. I am unable to understand that unit definition here. Becuase the unit Says that it is per root of Hertz it has 76 pV. Is that true? Should I apply a formula to convert it to frequency. Please let me know. I have attached an image.

Thanks in advance.

\$\endgroup\$
  • \$\begingroup\$ Welcome to EE.SE! The unit itself should be your first clue. Take the root from your frequency and multiply it with your value and you will get the noise figure in pV. \$\endgroup\$ – winny Nov 22 '18 at 9:42
  • \$\begingroup\$ The noise is almost constant density. Ideally you would integrate the area under the curve, but since it's almost constant the noise over 1MHz is about 76.9pV*sqrt(10^6) = 76.9nV RMS. \$\endgroup\$ – Spehro Pefhany Mar 24 '19 at 4:03
1
\$\begingroup\$

A 62 ohm resistor produces 1 nanoVolt of noise in 1 Hertz bandwidth.

A 6.2 ohm resistor produces 1nV/sqrt(10) = 1nV/3.16 = 0.312 nV

or

312 picoVolt noise in 1 Hertz bandwidth.

The noise power increases linearly with bandwidth. As "Andy aka" explained, the noise voltage increases with the squareroot of bandwidth, hence the dimensions you are seeing.

\$\endgroup\$
  • \$\begingroup\$ Than you for the information, My another question is that, is 312 picoVolt noise is for 1Hz bandwidth or for 1Mhz Bandwidth ? If it is only for 1Hz Bandwidth how should I calculate for 1MHz bandwidth? \$\endgroup\$ – dachu darshan Dec 10 '18 at 6:23
0
\$\begingroup\$

If you had a value of watts per Hz would that confuse you?

Probably not so, given that power is proportional to the square of volts, would it be that surprising to find that the voltage spectral density is the square root of the power spectral density? Given also that power spectral density is measured in watts per Hz, then the square root is volts per root Hz.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.