Question

The Johnson-Nyquist noise in a resistor in a circuit of 1Mhz bandwidth is 70 nV / √hz. What is the magnitude of the noise in the resistor in μV? (Type in a one-decimal number)

Answer 1

Given that,

Noise bandwidth $\Delta f= 1\ Mhz$

Noise voltage = 70n V/√hz

We need to calculate the magnitude of the noise in the resistor

Using formula of noise voltage

$\dfrac{V_{rms}}{\sqrt{\Delta f}}=70 n\dfrac{V}{\sqrt{hz}}$

$V_{rms}=70 n\times \dfrac{V}{\sqrt{hz}}\times\Delta f$

Where, $\Delta f$ = noise bandwidth

Put the value into the formula

$V_{rms}=70\times10^{-9}\times\sqrt{10^{6}}$

$V_{rms}=70\times10^{-6}\ V$

$V_{rms}=70\ \mu V$

Hence, The magnitude of the noise in the resistor is 70 μV.