Noise Figure of Resistor

Consider an RF source with output resistance of R_s driving a resistive load R_L as shown in Figure 1.

Figure 1. Noise figure computation of resistive load >> Norton equivalent representation >> Equivalent circuit for noise figure calculation

RF source is directly driving the load, input and output nodes are the same. Therefore the power gain =1
Noise factor is now given by the ratio of noise power at the output to the noise power due to source.

(1)   \begin{eqnarray*}F &=& { i_{n,R_s}^2 + i_{n,R_L}^2 \over i_{n,R_s}^2} \\ &=& {{4kT \over R_s}+{4kT \over R_s} \over {4kT \over R_s}} = 1 + {R_s \over R_L}\end{eqnarray*}

Alternatively, consider the input as current source and taking output as voltage across R_L. The Norton equivalent of the circuit is shown in Figure 1.

Power gain from input to output is

    \[G=({v_o \over i_s})^2 = (R_s || R_L)^2\]

Noise factor is given by,

(2)   \begin{eqnarray*} F &=& 1 + {N_A \over G N_i} = 1 + { 4kT (R_s || R_L) \over (R_s || R_L)^2 . {4kT\over R_s}} \\ &=& 1 + {R_s \over R_L} \end{eqnarray*}

If the load impedance(R_L) is power matched for source resistance(R_s), then NF= 20 \log(2) = 6dB. Its conversion gain is -6dB.
To minimize noise factor R_L should me as high as possible, but R_L=R_s for maximum power transfer from source to load. Therefore a tradeoff comes into picture between noise figure and maximum power transfer.

For example a 6dB RF attenuator or pad has noise figure of 6dB and conversion gain of -6dB. If a signal enters into a attenuator or pad, then the signal is attenuated by 6dB while the noise floor remains constant. Therefore the signal to noise ratio through the pad is degraded by 6dB.

The Noise figure of a passive device is same as that of the conversion gain(in dB sense).

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