Physics M.C.Q (1 Marks)

$\frac{10}{15}=\frac{10}{5+R_D}$

The diode can conduct and have resistance $R_D=10 \Omega$ because diode have dynamic resistance. In that case bridge will be balanced.

(image)

$B$: In reverse biased $p n$ junction diode, the current measured in $(\mu A)$, is due to minority charge carrier.

$=\bar{A}$

$Y_2 =\overline{B+B}$

$=\bar{B}$

$Y  =\overline{Y_1+Y_2}$

$=\overline{\bar{A}+\bar{B}}$

$=\overline{\bar{A}} \cdot \overline{\bar{B}}$

$=A \cdot B$ is similar to output of $AND$ Gate

According to given truth table, output is independent on value of $A$

$\therefore$ Output $Y=\bar{B}$

Statement $II$ : We use zener diode in reverse biased condition, when reverse biased voltage more than break down voltage than it act as stablizer.

It is $OR$ gate.

Since both are in series, therefore equal potential will drop across the junction.

Depletion region is thin

It is operated in Reverse Bias region

Zener voltage $\left( V _{ z }\right)$ is constant

$\text { and } I _{ E }=\frac{ I _{ C }}{\alpha}$

$I _{ E }=\frac{24 mA }{0.8}=30\,mA$

$\therefore I _{ B }= I _{ E }- I _C$

$=6\,mA \text { (into the base) }$

using De-Morgan Theorem

$C=\overline{A \cdot B+\bar{A} \cdot B}$

$C=\overline{B(A+\bar{A})}=\bar{B}$

Therefore

$OR$ Gate

For semiconductors and Insulators $\alpha$ is $(-)ve$

$f_{\text {in }}=f_{\text {out }}$

$\Rightarrow f_{\text {out }}=60 Hz$

$I=\operatorname{neAV}_{d}=\operatorname{neA}(\mu \mathrm{E})$

$\mu_{\mathrm{e}}>\mu_{\mathrm{h}}=\mathrm{I}_{\mathrm{e}}>\mathrm{I}_{\mathrm{h}}$

for Ge Potential barrier $\mathrm{V}_{0}=0.3\, \mathrm{~V}$

Si Potential barrier $\mathrm{V}_{0}=0.7\, \mathrm{~V}$

$\mathrm{Y}=\mathrm{AB}+\overline{\mathrm{BC}}$

$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}$

$V _{ L }= I _{ C } R _{ C }$

$0.8= I _{ C } \times 800$

$I _{ C }=1 mA$

It is a $NAND$ Gate

$\therefore \quad$ Collector current, $I_{C}=\frac{(20-0)}{4 \times 10^{3}}$

or $I_{C}=5 \times 10^{-3} \mathrm{A}=5\, \mathrm{mA}$

Input voltage, $V_{i}=V_{B E}+I_{B} R_{B}$

or $\quad V_{i}=0+I_{B} R_{B}$     or       $20=I_{B} \times 500 \times 10^{3}$

$\therefore \quad I_{B}=\frac{20}{500 \times 10^{3}}=40\, \mu \mathrm{A}$

$\therefore $        Current gain, $\beta=\frac{I_{C}}{I_{B}}=\frac{5 \times 10^{-3}}{40 \times 10^{-6}}=125$

${\beta=100}$

Voltage gain of the $CE$ amplifier,

$A_{V}=-\beta_{a c}\left(\frac{R_{C}}{R_{B}}\right)=-100\left(\frac{3}{2}\right)=-150$

Power gain, $A_{P}=\beta \times A_{V}=100 \times(-150)$

$=-15000$

Negative sign represents that output voltage is in opposite phase with the input voltage.

$=\overline{ A }+\overline{ B }+\overline{ B }$

$=\overline{ A }+\overline{ B }$

$Y =\overline{ A . B }$

$NAND$ Gate

At output, the truth table corresponds to $NOR$ gate

$I _{ B }=\frac{ I _{ C }}{\beta}$

$I _{ B }=15 \mu A$

$V _{ CC }= I _{ B } R _{ B }+ V _{ BE }$

$V _{ BE }=24-\left(15 \times 10^{-6} \times 220 \times 10^{3}\right)$

$V _{ BE }=20.7 V$

$V _{ BC }= I _{ C } R _{ L }- I _{ B } R _{ B }=7.05-3.3=3.75 V$

Hence, equivalent circuit becomes as shown in the figure.

Current in the circuit $=$ Current flowing through the resistance

$R_{1}=\frac{10}{2+2}=2.5 \mathrm{A}$

$\alpha=100, R_{B}=1 \mathrm{k} \Omega=1000 \Omega, V_{i}=?$

Voltage gain, $A=\beta \frac{R_{C}}{R_{B}}=100 \times \frac{2000}{1000}=200$

$ \text { Also, } A =\frac{V_{0}}{V_{i}} \text { or } V_{i}=\frac{V_{0}}{A}=\frac{4}{200} $

$=\frac{2}{100} \mathrm{V}=20 \mathrm{mV}$

current gain, $\beta=0.96$

Voltage gain $=$ Current gain $\times$ Resistance gain

                         $=0.96 \times \frac{800}{192}=4$

Power gain $=$ [Current gain ] $\times$ [ Voltage gain ]

                  $=0.96 \times 4=3.84$

If $A=B=C=0$ then $Y_{0}=\overline{0}=1$

If $A=B=C=0$ then $Y_{1}=\overline{1}=0$

$\therefore {{I}_{AB}}=\frac{{{V}_{A}}-{{V}_{B}}}{{{R}_{AB}}}=$ $\frac{4\text{V}-(-6\text{V})}{1\text{k}\Omega }=$ $\frac{10}{1000}\text{A}={{10}^{-2}}\text{A}$