Johnson-Nyquist or thermal noise

What do you need to know to understand this topic?

What is Johnson-Nyquist or thermal noise?

This type of noise is generated by the thermal agitation of charge carriers, such as electrons.
Agitation is caused by heating associated with resistive loads. Then, thermal noise is always associated with some resistance value. If the power spectral density (PSD) of thermal noise is $\overline{v_n^2}$, then $$\overline{v_n^2} = 4k_BTR$$ where $k_B$ is the Boltzmann's constant (Joule/┬║Kelvin), $T$ is the resistance's absolute temperature (Kelvin) and $R$ is the resistance value ($\Omega$). The higher the temperature or resistance, the higher the thermal noise. The PSD does not depend on the frequency because thermal noise is white (equal power density for all frequencies).
Instead, the thermal noise can be defined as a current $\overline{i_n^2}$: $$\overline{i_n^2} = \frac{4k_BT}{R}$$

In order to find out how much noise power the resistor would generate in some circuit, multiply the PSD by the equivalent noise bandwidth of the system ($\Delta f$): $$ v_n^2 = \overline{v_n^2}\Delta f = 4k_BTR \Delta f$$

Thermal noise in a RC circuit

It is interesting that the thermal noise in a RC circuit does not depend on the value of the resistance. That is because a larger resistance creates more noise, but it also filters it more, so both effects cancel each other.
For a RC circuit, the bandwidth is $\omega_{3dB} = 1/RC$. Since it is a first-order low-pass filter, we know that the equivalent noise bandwidth is $$\Delta f = \frac{\pi}{2}f_{3dB} = \frac{\pi\omega_{3dB}}{2*2\pi} = \frac{\omega_{3dB}}{4} = \frac{1}{4RC}.$$ Then the total noise power in the RC circuit becomes $$ v_n^2 = \overline{v_n^2}\Delta f = 4k_BTR \Delta f = \frac{k_BT}{C}. $$ If instead of voltage we look at the charge variance at the capacitor, we have $$Q_n^2 = C^2 v_n^2 = C^2 \frac{k_BT}{C} = k_BTC.$$ For this reason, thermal noise in a capacitor is also known as kTC noise.