Physics 4 Marks Question

Mutual conductance = g_m = 2.0milli mho = 2 × 10^-3mho

Amplification factor $\mu=30$

Load Resistance = RL = ?

We know,

$\text{A}=\frac{\mu}{1+\frac{\text{r}_\text{p}}{\text{R}_\text{L}}}$ where A = Voltage amplification factor

$\Rightarrow\text{A}=\frac{\text{r}_\text{g}\times\text{g}_\text{m}}{1+\frac{\text{r}_\text{p}}{\text{R}_\text{L}}}$ where $\mu=\text{r}_\text{p}\times\text{g}_\text{m}$

$\Rightarrow30=\frac{20\times10^{3}\times2\times10^{-3}}{1+\frac{20000}{\text{R}_\text{L}}}$

$\Rightarrow3=\frac{4\text{R}_\text{L}}{\text{R}_\text{L}+20000}$

$\Rightarrow3\text{R}_\text{L}+60000=4\text{R}_\text{L}$

$\Rightarrow\text{R}_\text{L}=6000\Omega=60\text{K}\Omega$

$\delta\text{V}_\text{p}=48\text{V},\ \delta\text{I}_\text{p}=?$

$\delta\text{I}_\text{p}=\frac{\Big(\frac{\delta\text{V}_\text{p}}{\text{r}_\text{p}}\Big)}{\text{V}_\text{g}}=\text{constant}$

So, $\delta\text{I}_\text{p}=\frac{48}{8000}=0.006\text{A}=6\text{mA}$

Now, Vp is constant.

$\delta\text{I}_\text{p}=6\text{mA}=0.006\text{A}$

$\text{g}_\text{m}=0.0025\Omega$

$\delta\text{V}_\text{g}=\frac{\Big(\frac{\delta\text{I}_\text{p}}{\text{g}_\text{m}}\Big)}{\text{V}_\text{p}}=\text{constant}$

$=\frac{0.006}{00025}=2.4\text{V}$

Consider the two points on Vg = -6 line

$\text{r}_\text{p}=\frac{(240-160)\text{V}}{(13-3)\times10^{-3}\text{A}}=\frac{80}{10}\times10^{3}\Omega=8\text{K}\Omega$

$\text{g}_\text{m}=\Big(\frac{\delta\text{I}_\text{p}}{\delta\text{V}_\text{g}}\Big)\text{v}_\text{p}=\text{constant}$

Considering the points on 200V line,

$\text{g}_\text{m}=\Big(\frac{\delta\text{I}_\text{p}}{\delta\text{V}_\text{g}}\Big)\text{v}_\text{p}=\text{constant}$

Considering the points on 200V line,

$\text{g}_\text{m}=\frac{(13-3)\times10^{-3}}{\big[(-4)+(-8)\big]}\text{A}$

$=\frac{10\times10^{-3}}{4}=2.5\text{milli}\Omega$

$\mu=\text{r}_\text{p}\times\text{gm}$

$\mu=8\times10^{3}\Omega\times2.5\times10^{-3}\Omega^{-1}$

$\mu=8\times1.5=20$

Diff. the equation,

$\text{di}_\text{p}=\text{K}\frac{3}{2}\Big(\text{V}_\text{g}+\frac{\text{V}_\text{p}}{\mu}\Big)^{\frac{1}{2}}\text{dV}_\text{g}$

$\frac{\text{di}_\text{P}}{\text{dV}_\text{g}}=\frac{3}{2}\text{K}\Big(\text{V}_\text{g}+\frac{\text{V}_0}{\mu}\Big)^{\frac{1}{2}}$

$\text{g}_\text{m}=\frac{3}{2}\text{K}\Big(\frac{\text{V}_\text{g}+\text{V}_\text{p}}{\mu}\Big)^{\frac{1}{2}}\ ....(\text{ii})$

From (i) $\text{i}_\text{p}=\text{k}'(\text{gm})^{3}$

$\text{g}_\text{m}\propto3\sqrt{\text{i}_\text{p}}$

$\Rightarrow{\text{di}_\text{p}}=\text{C}\frac{3}{2}\text{V}_\text{p}^{(\frac{3}{2})-1\text{dv}_\text{p}}$

$\Rightarrow\frac{{\text{di}_\text{p}}}{\text{dv}_\text{p}}=\frac{3}{2}\text{CV}_\text{p}^{\frac{1}2}\ ....(\text{ii})$

Dividing (ii) and (i)

$\frac{\text{i}}{\text{i}_\text{p}}\frac{{\text{di}_\text{p}}}{\text{dv}_\text{p}}=\frac{\frac{3}{2}\text{CV}_\text{p}^{\frac{1}2}}{\text{CVp}^{\frac{3}2}}$

$\Rightarrow\frac{\text{i}}{\text{i}_\text{p}}\frac{{\text{di}_\text{p}}}{\text{dv}_\text{p}}=\frac{3}{2\text{V}}$

$\Rightarrow\frac{{\text{dv}_\text{p}}}{\text{di}_\text{p}}=\frac{2\text{V}}{3\text{i}_\text{p}}$

$\Rightarrow\text{R}=\frac{2\text{V}}{3\text{i}_\text{p}}=\frac{2\times60}{3\times10\times10^{-3}}$

$\Rightarrow\text{R}=4\times10^3=4\text{k}\Omega$

$\delta\text{I}_\text{p}=?$

$\delta\text{V}_\text{p}=220-220=20\text{V}$

$\delta\text{I}_\text{p}=\frac{\Big(\frac{\delta\text{V}_\text{p}}{\text{r}_\text{p}}\Big)}{\text{V}_\text{g}}=\text{constant}$

$=\frac{20}{10^4}=0.002\text{A}=2\text{mA}$

$\text{F}=\text{qE}=1.6\times10^{-9}\times5\times10^{3}$

$\text{a}=\frac{\text{qE}}{\text{m}}=\frac{1.6\times5\times10^{-16}}{9.1\times10^{-31}}$

1. S = Distance travelled

$=\frac{1}{2}\text{at}^2=439.56\text{m}=440\text{m}$

2. $\text{d}=1\text{mm}=1\times10^{-3}\text{m}$

$1\times10^{-3}=\frac{1}{2}\times\frac{1.6\times5}{9.1}10^5\times\text{t}^2$

$\text{t}^2=\frac{9.1}{0.8\times5}\times10^{-18}$

$\text{t}=1.508\times10^{-9}\sec$

$\text{t}=1.5\text{ns.}$ 

$\phi=4.5\text{ev}$

$\sigma=6\times10^{-8}\omega/\text{m}^2-\text{k}^4$

$\text{S}=2\times10^{-5}\text{m}^2$

$\text{K}=1.38\times10^{23}\text{J/K}$

$\text{H}=24\omega'$

The cathode acts as a black body,i.e. emissivity = 1

$\therefore\text{ E}=\sigma\text{AT}^4\ (\text{A is area})$

$\text{T}^4=\frac{\text{E}}{\sigma\text{A}}=\frac{24}{6\times10^{-8}\times2\times10^{-5}}=2\times10^{13}\text{K}=20\times10^{12}\text{K}$

$\text{T}=2.1147\times10^{3}=2114.7\text{K}$

Now, $\text{i}=\text{AST}^2\text{e}^{\frac{-\phi}{\text{KT}}}$

$=6\times10^5\times2\times10^{-5}\times(2114.7)^2\times\text{e}^{\frac{-45\times1.6\times10^{-19}}{1.38\times\text{T}\times10^{-23}}}$

$=1.03456\times10^{-3}\text{A}=1\text{mA}$

$\text{g}_\text{m}=2\text{milli mho}=2\times10^{-3}\text{mho}$

$\mu=\text{r}_\text{p}\times\text{g}_\text{m}=2\times10^{3}\times2\times10^{-3}=4$ Amplification factor is 4.

‘e’ is same in all cases.

We know,

$\text{i}=\text{AST}^2\text{e}^{\frac{-\phi}{\text{RT}}}\ \phi=4.52\text{eV},\ \text{K}=1.38\times10^{-23}\text{J/k}$

$\text{n(1000)}=\text{As}\times(1000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times1000}$

$1.7396\times10^{-17}$

1. $\text{T}=300\text{K}$

$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}\times(300)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times300}}{\text{AS}\times1.7396\times10^{-17}}=7.05\times10^{-55}$

2. $\text{T}=2000\text{K}$

$$$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}(2000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23\times2000}}}{\text{AS}\times1.7396\times10^{-17}}=9.59\times10^{11}$

3. $\text{T}=3000\text{K}$

$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}\times(3000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times3000}}{\text{AS}\times1.7396\times10^{-17}}=1.340\times10^{16}$

1. $\text{i}_\text{p}=41(\text{V}_\text{p}+7\text{V}_\text{g})^{1.41}$

$41(250-140)^{1.41}=41\times(110)^{1.41}$

$=30984\mu\text{A}=30\text{mA}$

2. $\text{i}_\text{p}=41(\text{V}_\text{p}+7\text{V}_\text{g})^{1.41}$

Differentiating,

$\text{di}_\text{p}=41\times1.41\times(\text{V}_\text{p}+7\text{V}_\text{g})^{0.41}\times(\text{dV}_\text{p}+7\text{dV}_\text{g})$

Now, $\text{r}_\text{p}=\frac{\text{dV}_\text{p}}{\text{di}_\text{p}}\text{V}_\text{g}=\text{constant}$

or $\frac{\text{dV}_\text{p}}{\text{di}_\text{p}}=\frac{1\times10^{6}}{41\times1.41\times110^{0.41}}$

$=10^{6}\times2.510\times10^{-3}$

$2.5\times10^{3}\Omega=2.5\text{K}\Omega$

3. From above,

$\text{dI}_\text{p}=41\times1.41\times6.87\times7\text{d V}\text{g}$

$\text{g}_\text{m}=\frac{\text{dI}_\text{p}}{\text{dV}_\text{g}}=41\times1.41\times6.87\times7\mu\Omega$

$=2780\mu\Omega=2.78\Omega$

4. Amplification factor,

$\mu=\text{r}_\text{p}\times\text{g}_\text{m}$

$=2.5\times10^{3}\times2.78\times10^{-3}=6.95=7$

$\mu=20,\ \text{V}_\text{p}$

$\text{V}_\text{g}=-7.5\text{V},\ \text{I}_\text{p}=10\text{mA}$

1. $\text{g}_\text{m}=\Big(\frac{\delta\text{I}_\text{p}}{\delta\text{V}_\text{g}}\Big)\text{V}_\text{p}=\text{constant}$

$\delta\text{V}_\text{g}=\frac{\delta\text{I}_\text{p}}{\text{g}_\text{m}}=\frac{15\times10^{3}-10\times10^{-3}}{\frac{\mu}{\text{r}_\text{p}}}$

$\delta\text{V}_\text{g}=\frac{5\times10^{-3}}{\frac{20}{10\times10^{3}}}=\frac{5}{2}=2.5$

$\text{r}'\text{g}=+2.5-7.5=-5\text{V}$

2. $\text{r}_\text{p}=\Big(\frac{\delta\text{V}_\text{p}}{\delta\text{I}_\text{p}}\Big)\text{V}_\text{g}=\text{constant}$

$10^{4}=\frac{\delta\text{V}_\text{p}}{(15\times10^{-3}-10\times10^{-3})}$

$\delta\text{V}_\text{p}=10^4\times5\times10^{-3}=50\text{v}$

$\text{V}'_\text{p}-\text{V}_\text{p}=50$

$\text{V}'_\text{p}=-50+\text{V}_\text{p}=200\text{V}$

$\text{i}'=\text{AST}^{12}\text{e}^{\frac{-\phi}{\text{KT}}}$

$\frac{\text{i}}{\text{i}'}=\frac{\text{T}^2}{\text{T}^{12}}\frac{\text{e}^{\frac{-\phi}{\text{KT}}}}{\text{e}^{\frac{-\phi}{\text{KT}}}}$

$\Rightarrow\frac{\text{i}}{\text{i}'}=\Big(\frac{\text{T}}{\text{T}'}\Big)^2\text{e}^{\frac{-\phi}{\text{KT}+\phi\text{KT}'}}=\Big(\frac{\text{T}}{\text{T}'}\Big)^2\text{e}^{\phi\text{KT}'-\phi/\text{KT}}$

$\Rightarrow\frac{\text{i}}{\text{i}'}=\Big(\frac{2000}{2010}\Big)^2\text{e}^{\frac{4.5\times1.6\times10^{-19}}{1.38\times10^{-23}}}\Big(\frac{1}{2010}-\frac{1}{2000}\Big)=0.8690$

$\Rightarrow\frac{\text{i}}{\text{i}'}=\frac{1}{0.8699}=1.1495=1.14$

$\text{V}_\text{p}=36\text{V},\ \text{V}_\text{p}=49\text{V},\ \text{P}=\text{I}_\text{p}\text{V}_\text{p}$

$\text{I}_\text{p}=\frac{\text{P}}{\text{V}_\text{p}}=\frac{1}{36}$

$\text{I}_\text{p}\propto(\text{V}_\text{p})^{3/2}$

$\text{I}'_\text{p}\propto(\text{V}'_\text{p})^{3/2}$

$\Rightarrow\frac{\text{I}_\text{p}}{\text{I}'_\text{p}}=\frac{(\text{V}_\text{p})^{3/2}}{\text{V}'_\text{p}}$

$\Rightarrow\frac{\frac{1}{36}}{\text{I}'_\text{p}}=\Big(\frac{36}{49}\Big)^{3/2}$

$\Rightarrow\frac{1}{36\text{I}'_\text{p}}=\frac{36}{49}\times\frac{6}{7}$

$\Rightarrow\text{I}'_\text{p}=0.4411$

$\Rightarrow\text{P}'=\text{V}'_\text{p}\text{I}'_\text{p}$

$\Rightarrow\text{P}'=49\times0.4411$

$\Rightarrow\text{P}=2.1613\text{W}=2.2\text{W}$

$\text{i}_1=\text{i},\ \text{i}_2=100\text{mA}$

$\text{A}_1=60\times10^4,\ \text{A}_2=3\times10^4$

$\text{S}_1=\text{S},\ \text{S}_2=\text{S}$

$\text{T}_1=2000,\ \text{T}_2=2000$

$\phi_1=4.5\text{eV},\ \phi_2 =2.6\text{eV}$

$\text{i}=(60\times10^{4})(\text{S})\times(2000)^2\frac{-45\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$

$100=(3\times10^{4})(\text{S})\times(2000)^2\frac{-2.6\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$

Dividing the equation,

$\Rightarrow\frac{\text{i}}{100}=\text{e}^{\big[\frac{-4.5\times1.6\times10}{1.38\times2}\big(\frac{-2.6\times1.6\times10}{1.38\times20}\big)\big]}$

$\Rightarrow\frac{\text{i}}{100}=20\times\text{e}^{-11.014}$

$\Rightarrow\frac{\text{i}}{100}=20\times0.000016$

$\Rightarrow\text{i}=20\times0.0016=0.0329\text{mA}=33\mu\text{A}$

$\text{G}_\text{m}=2\text{g}_\text{m}$

$\text{V}_\text{s}=\text{P}_\text{s}\text{d}_\text{s}$

$\text{V}_\text{L}=\text{P}_\text{L}\text{d}_\text{L}$

$\Rightarrow\frac{\text{V}_\text{s}}{\text{V}_\text{L}}=\frac{\text{P}_\text{s}}{\text{P}_\text{L}}\times\frac{\text{d}_\text{s}}{\text{d}_\text{L}}$

$\Rightarrow\frac{100}{100}=\frac{10}{20}\times\frac{1\text{mm}}{\text{x}}$

$\Rightarrow\text{x}=\frac{1\text{mm}}{2}=0.5\text{mm}$