where

*k*=Boltzmann’s constant ≈1.381×10^{−23} J/K

*T*=absolute temperature in K=°C+273.16

*B*=bandwidth in Hz

*R*=resistance in Ω.

For a typical internal temperature of 40°C (313 K), with a bandwidth of 20 kHz, this is more conveniently expressed as:

${v}_{\mathrm{n}}=1.86\times {10}^{-8}\sqrt{R}$

Using this equation, we find that a perfect 100 kΩ resistor generates 5.9 μV of thermal noise. In this instance, the resistor’s thermal noise has been greatly exceeded by its excess noise. To find the total noise of the resistor, we must add the individual noise *powers*, which, if we remember that

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