"Feedback Network Design"

Feedback Network Design

Noise figure Budget:

I can't use the entire noise budget for the feedback network; so instead I want the feedback network to have a fraction of the budget:

\begin{equation} \alpha_{NF fb}=0.3 \end{equation}

Design Equation formulation:

I have three design variables $R_a$, $R_b$ and $R_c$ so I will use these three design equations:

\begin{equation} B=\frac{0.5}{A_{y}} \end{equation} \begin{equation} D=\frac{0.5}{A_{y} Z_{i}} \end{equation} \begin{equation} NF=10^{0.1 NF \alpha_{NF fb}} \end{equation}

I can now use the three circuit analysis equations:

\begin{equation} B=\frac{R_{a} R_{b}}{R_{a} + R_{b} + R_{c}} \end{equation} \begin{equation} D=\frac{R_{b} + R_{c}}{R_{a} + R_{b} + R_{c}} \end{equation} \begin{equation} NF=\frac{0.5 R_{a} \left(R_{b} - R_{s}\right)^{2}}{R_{s} \left(R_{a} + R_{b} + R_{c}\right)^{2}} + \frac{R_{b} \left(R_{a} + R_{c} + R_{s}\right)^{2}}{R_{s} \left(R_{a} + R_{b} + R_{c}\right)^{2}} + \frac{0.5 R_{c} \left(R_{b} - R_{s}\right)^{2}}{R_{s} \left(R_{a} + R_{b} + R_{c}\right)^{2}} + 1 \end{equation}

Solutions:

Setting these system of equations equal to each other and solving I get the following design equations for $R_a$, $R_b$ and $R_c$:

\begin{equation} R_{a}=\frac{0.5 \left(A_{y} Z_{i} - 1\right) \left(A_{y} Z_{i} + 1\right)}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)} + \frac{\left(A_{y} Z_{i} - 1\right) \left(10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 0.5 \cdot 10^{0.1 NF \alpha_{NF fb}} + 0.25 A_{y}^{2} Z_{i}^{2} - 0.5 A_{y} Z_{i} + 0.25\right)^{0.5}}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)} \end{equation} \begin{equation} R_{b}=\frac{Z_{i}}{2 A_{y} Z_{i} - 1} \end{equation} \begin{equation} R_{c}=\frac{- Z_{i} + \frac{0.5 \left(A_{y} Z_{i} - 1\right) \left(A_{y} Z_{i} + 1\right)}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)} + \frac{\left(A_{y} Z_{i} - 1\right) \left(10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 0.5 \cdot 10^{0.1 NF \alpha_{NF fb}} + 0.25 A_{y}^{2} Z_{i}^{2} - 0.5 A_{y} Z_{i} + 0.25\right)^{0.5}}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)}}{2 A_{y} Z_{i} - 1} \end{equation}

Numerical Solutions:

\begin{equation} R_{a}=1082.0 \end{equation} \begin{equation} R_{b}=13.04 \end{equation} \begin{equation} R_{c}=34.01 \end{equation}

I can also find the standard deviation of the resistors using $\Delta = \sqrt{ | \frac{\partial R}{\partial Z_i}|^2\Delta_{Z_i}^2+|\frac{\partial R}{\partial A_y}|^2\Delta_{A_y}^2}$

\begin{equation} \Delta_{R a}=47.0 \end{equation} \begin{equation} \Delta_{R b}=1.361 \end{equation} \begin{equation} \Delta_{R c}=5.23 \end{equation}

Updated specifications

budget specification

Table budget specification
symboldescriptionvalueunits
$\alpha_{NF fb}$Fraction of Noise Figure Budget for Feedback Network$0.3$$\mathrm{1}$

design specification

Table design specification
symboldescriptionvalueunits
$\Delta_{R a}$Required Standard Deviation of Feedback Resistor $R_a$$47.0$$\mathrm{\Omega}$
$\Delta_{R b}$Required Standard Deviation of Feedback Resistor $R_b$$1.361$$\mathrm{\Omega}$
$\Delta_{R c}$Required Standard Deviation of Feedback Resistor $R_c$$5.23$$\mathrm{\Omega}$
$R_{a}$Feedback Resistor $R_a$$1082.0$$\mathrm{\Omega}$
$R_{b}$Feedback Resistor $R_b$$13.04$$\mathrm{\Omega}$
$R_{c}$Feedback Resistor $R_c$$34.01$$\mathrm{\Omega}$

designequations specification

Table designequations specification
symboldescriptionvalueunits
$B_{eq}$Typical Voltage -> Current Gain$\frac{0.5}{A_{y}}$$\mathrm{\frac{V}{A}}$
$D_{eq}$Typical Current -> Current Gain$\frac{0.5}{A_{y} Z_{i}}$$\mathrm{1}$
$R_{aeq}$Feedback Resistor $R_a$$\frac{0.5 \left(A_{y} Z_{i} - 1\right) \left(A_{y} Z_{i} + 1\right)}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)} + \frac{\left(A_{y} Z_{i} - 1\right) \left(10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 0.5 \cdot 10^{0.1 NF \alpha_{NF fb}} + 0.25 A_{y}^{2} Z_{i}^{2} - 0.5 A_{y} Z_{i} + 0.25\right)^{0.5}}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)}$$\mathrm{\Omega}$
$R_{beq}$Feedback Resistor $R_b$$\frac{Z_{i}}{2 A_{y} Z_{i} - 1}$$\mathrm{\Omega}$
$R_{ceq}$Feedback Resistor $R_c$$\frac{- Z_{i} + \frac{0.5 \left(A_{y} Z_{i} - 1\right) \left(A_{y} Z_{i} + 1\right)}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)} + \frac{\left(A_{y} Z_{i} - 1\right) \left(10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 0.5 \cdot 10^{0.1 NF \alpha_{NF fb}} + 0.25 A_{y}^{2} Z_{i}^{2} - 0.5 A_{y} Z_{i} + 0.25\right)^{0.5}}{A_{y} \left(2 \cdot 10^{0.1 NF \alpha_{NF fb}} A_{y} Z_{i} - 10^{0.1 NF \alpha_{NF fb}} - 2 A_{y} Z_{i}\right)}}{2 A_{y} Z_{i} - 1}$$\mathrm{\Omega}$

Go to index

SLiCAP: Symbolic Linear Circuit Analysis Program, Version 2.0.1 © 2009-2023 SLiCAP development team

For documentation, examples, support, updates and courses please visit: analog-electronics.tudelft.nl

Last project update: 2023-11-25 20:52:48