Determination of the show-stopper values of feedback elements

Exercise

The figure below shows a concept of a negative feedback voltage amplifier with resistive feedback. The amplifier is driven from a voltage source with a source resistance \(R_s=600 \Omega\). The voltage gain of this amplifier should be 10, while its noise figure should be better than 2dB.

Give the maximum value of the feedback resistors.

Concept of a negative feedback voltage amplifier.

Solution

The noise contribution of the feedback resistors is found by considering their parallel connection in series with the signal source. The spectral density \(S_{v_{tot}}\) of the total source-referred noise can then be written as:

\[S_{v_{tot}}=4kT\left( R_s+\frac{R_1R_2}{R_1+R_2} \right).\]

The noise figure \(NF\) is defined as the ratio of the total source-referred noise and the noise of the source itself.

\[NF = \frac{4kTB\left( R_s+\frac{R_1R_2}{R_1+R_2} \right)}{4kTBR_s},\]

where \(B\) is the noise bandwidth of interest. This result can be written as:

\[NF=1+\frac{R_1R_2}{R_s \left( R_1+R_2\right)}.\]

We can express \(R_1\) in the voltage gain of the amplifier and in \(R_2\). The voltage gain \(A_v\) of this amplifier can be found from network analysis:

\[A_v = \frac{R_1+R_2}{R_2},\]

from which we obtain and expression for \(R_1\):

\[R_1 = R_2\left( A_v-1\right).\]

Substitution of this value in the expression for the noise figure yields:

\[NF=1+\frac{R_2}{R_s}\frac{A_v-1}{A_v}.\]

From which we obtain:

\[R_2 = R_s\left( NF-1 \right)\frac{A_v}{A_v-1}.\]

We have:

\[NF=10^{\frac{F}{10}},\]

where \(F\) is the noise figure in dB. After substitution of the expression for \(NF\) in the expression for \(R_2\), we obtain:

\[R_2 \leq R_s\left( 10^{\frac{F}{10}}-1 \right)\frac{A_v}{A_v-1} = 600\left( 10^{\frac{2}{10}}-1 \right)\frac{10}{10-1} = 390\Omega.\]

Hence, the maximum values for the feedback resistors are: \(R_2=390\Omega\) and \(R_1=390\left(10-1 \right)=3.51k\Omega\).