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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.

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  • \$\begingroup\$ Note that En/In = noise resistance. \$\endgroup\$ – Brian Drummond Jan 18 '18 at 23:12
  • \$\begingroup\$ I think you should add a definition of what the noise figure is to your question. (Then we can look at some ratio's.) \$\endgroup\$ – George Herold Jan 19 '18 at 0:26
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You can find a detailed derivation in the following paper: Fundamentals of Low-Noise Analog Circuit Design which is available on-line. Look at equations 25 and 29. You could also just take the derivative of your equation with respect to Rs to find the value that minimizes F. Then solve for Rs. You will find that for a minimum F, Rs=En/In.

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This is how I see it:

\$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.

Can you take it from here?

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