$ \begin{aligned} &s_{21}=1.9 \angle 81^{\circ} \ &S_{22}=0.5 \angle-60^{\circ} \ &F_{\text {min }}=1.6 d B_1 R_n=20 \Omega, \Gamma_{O P T}=0.62 \angle 100^{\circ} \end{aligned} $ Assuming the FET to be unilateral, design an amplifier for maximum possible gain and noise figure not more than $2 \mathrm{~dB}$.

Solution:

stability checks:-

$ \begin{aligned} \\ |\Delta| &=\left|s_{11} s_{22}-s_{12} s_{21}\right| \\\\ &=\left|0.6 L-60^{\circ} \times 0.5 L-60^{\circ}-0.05 L-26^{\circ} \times 1.9 L 81^{\circ}\right|=0.394 \\\\ |\Delta| &\lt1 \\\\ K &=\frac{1-\left|s_{11}\right|^2-\left|s_{22}\right|^2+|\Delta|^2}{2\left|s_{12} . s_{21}\right|} \\ \end{aligned}\\ $

As it is assumed to be unilateral $S_{12}=0 \therefore k=\infty$ $\therefore$ As $|\Delta|\lt1$ \& $K\gt1$ we can consider the transistor to be unconditionally stable \& can be used for amplifier design.

Manimum Gain calculation under unilateral assumption:

$ G_{\sin \alpha x}=\frac{1}{1-|\sin |^2}=\frac{1}{1-0.6^2}=1.5625=1.938 d B \\ $

$ G \operatorname{lmax}=\frac{1}{1-\left|s_{22}\right|^2}=\frac{1}{1-0.5^2}=1.33=1.249 \mathrm{~dB} \\ $

$ G_0=\left|s_{21}\right|^2=1 . g^2=3.61=5.575 \mathrm{~dB} \\ $

Therefore maximum unilateral gain of FET amplifier,

$ G_{\operatorname{romax}}=G_{\sin a x}+G_0+G_{\operatorname{mom} a x}\\ $

$ =1.938 d B+5.575 d B+1.249 d B\\ $

$ \therefore G_{\text {Tumax }}=8.76 \mathrm{~dB} \\ $

plotting of constaut gain circles.

$ \begin{aligned} \\ &g_s=\frac{G_s}{G_{s m a x}} \text { as for max gain } G_s=G_s \text { max } \Rightarrow g_s=1 \\\\ &g_L=\frac{G_L}{G_{\text {Lmax }}} \text { as for max gain } G_L=G_{L \text { max }} \Rightarrow g_L=1 \\\\ &C_g s=\frac{g_s \times s_{11}^*}{1-|s 11|^2 \times\left(1-g_s\right)}=g_s \times s_{11}^*=s_{11}^*=0.6 L 60^{\circ} \\\\ &R_{g s}=\frac{\left(\sqrt{1-g_s}\right)\left(1-\left|s_{11}\right|^2\right)}{1-\left|s_{11}\right|^2 \times(1-g s)}=0\\ \end{aligned}\\ $

$ \begin{aligned}\\ &\text { simila1ly } \\\\ &\operatorname{cg}_L=\frac{g_L \times s_{22}{ }^*}{1-\left|s_{22}\right|^2\left(1-g_L\right)}=g_L \times s_{22}{ }^*=s_{22}{ }^7=0.5 L_{60}{ }^{\circ} \\\\ &R_L=\frac{\left(\sqrt{1-g_i}\right)\left(1-\left|s_{22}\right|^2\right)}{1-\left|s_{22}\right|^2\left(1-g_L\right)}=0 \end{aligned}\\ $

We note that, for maximum gains, the source $\&$ load gain circles will reduce to just points.

plotting of constant Noise Figure circle

Maximum noise figule $=F=2 \mathrm{~dB}=1.584$

$ N=\frac{F-F_{\min }}{4\left(R_n / z_0\right)}\left|1+\Gamma_{O P T}\right|^2 \quad \text { with } F_{\min }=1.6 \mathrm{~dB}=1.445\\ $

$ \begin{aligned} \\ &\therefore N=\frac{(1.584-1.445)}{4 \times(20 \mid 50)}|1+0.62 \angle 100|^2=\frac{0.1625}{1.6}=0.1015 \\\\ &\therefore \text { CNF }_N=\frac{\Gamma_{0 P T}}{N+1}=\frac{0.62 L 100}{0.1015+1}=0.56 L 100^{\circ} \\ \end{aligned}\\ $

$ \begin{aligned} \\ &R N F=\frac{\sqrt{N^2+N\left(1-1 \Gamma_{O P T} I^2\right)}}{N+1} \\\\ &\therefore R_{N F}=\frac{\sqrt{0.1015^2+0.1015\left(1-0.62^2\right)}}{0.1015+1}=\frac{0.2697}{1.1015}=0.244 \end{aligned}\\ $

If we take source gain Gs $=1.7 d B=1.479 \quad$ gs $=0.946$

$ \therefore \quad c_{g s}=0.58 \mathrm{~L} 60^{\circ}, \quad R g s=0.15 \mathrm{~g} \\ $

we get constant gain circle intersecting with constant noise figure circle. The point ' 1 ' is interedition point used for design of matching circuit. Therefore maximum possible gain value obtainable

$ G_{T V}=1.7 d B\left(a_5\right)+5.575\left(a_0\right)+1.249 \\ $ (a_l)

$ \therefore G_{T U}=8.524 \mathrm{~dB} \\ $

Source point (l') matching circuit design.

location of single parallel stub $d=0.266 \lambda$ length of stub $\ell=0.125 \lambda$

load point (CgL) matching circuit design

location of single parallels stub $d=0.251 \lambda$ length of stub $l=0.11 \lambda$