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2009-01-04 10:33:14

The , or voltage variance (mean square) per of , is given by

\bar {v_{n}^2} = 4 k_B T R

where kB is in per , T is the resistor's absolute in kelvins, and R is the resistor value in (Ω). Use this equation for quick calculation:


\sqrt{\bar {v_{n}^2}} = 0.13 \sqrt{R} ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.

For example, a 1 kΩ resistor at a temperature of 300 K has


\sqrt{\bar {v_{n}^2}} = \sqrt{4 \cdot 1.38 \cdot 10^{-23}~\mathrm{J}/\mathrm{K} \cdot 300~\mathrm{K} \cdot 1~\mathrm{k}\Omega} = 4.07  ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.

For a given bandwidth, the (RMS) of the voltage, vn, is given by


v_{n}  = \sqrt{\bar {v_{n}^2}}\sqrt{\Delta f } = \sqrt{ 4 k_B T R \Delta f }

where Δf is the bandwidth in hertz over which the noise is measured. For a 1 kΩ resistor at room temperature and a 10 kHz bandwidth, the RMS noise voltage is 400 nV.

The noise generated at the resistor can transfer to the remaining circuit; the maximum noise power transfer happens with when the resistance of the remaining circuit is equal to the noise generating resistance. In this case the noise power transfer to the circuit is given by


P = k_B \,T \Delta f

where P is the thermal noise power in watts. Notice that this is independent of the noise generating resistance

Noise in decibels

In , power is often measured in relative to 1 milliwatt (), assuming a 50 ohm load resistance. With these conventions, thermal noise for a resistor at can be estimated as:


P_\mathrm{dBm} = -174 + 10\ \log(\Delta f)
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