# Noise figure minimized proof

Why is the noise figure of an amplifier minimized when the source impedance is equal to: $$Rs=\frac{E_n}{I_n},$$ En,In are respectively the noise voltage and noise current of the amplifier Noise figure is given by : $$F=1+\frac{E_n^2}{4kTBR_s}+\frac{I_n^2R_s}{4kTB}$$ k is Boltzmann’s constant ,T is the temperature in degrees Kelvin, B is the bandwidth of the system in Hz.

• Note that En/In = noise resistance. – Brian Drummond Jan 18 '18 at 23:12
• I think you should add a definition of what the noise figure is to your question. (Then we can look at some ratio's.) – George Herold Jan 19 '18 at 0:26

$E_n$ and $I_n$ are independent. If the source impedance is "high" then $E_n$ has little effect because $I_n$ is dominant and $I_n$ contributes most of the net noise signal. If the source impedance is low then $E_n$ is the dominant effect. Somewhere half way both noise sources contribute equally.
However, due to the way non-coherent noise sources add there is always going to be a resistance that produces the smallest net contribution of noise. They add as a sum of squares such that if the effect from each is a value of ten then the net effect is $\sqrt{10^2+10^2}$ = 14.14.
If one was doubled and the other halved (due to the source resistance not being optimized) we would get $\sqrt{5^2+20^2}$ = 20.62.