Explain why elemental semiconductor cannot be used to make visible LEDs.
In elemental semiconductor, the band gap is such that the emission are in infrared region and not in visible region.
Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.
The circuit resemble AND gate. The boolean expression of this circuit is, $V_0=A . B$ i.e., $V_0$ equals A AND B. The truth table of this gate is as given below
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V , what should be the value of $R_s$ for safe operation (see figure)?
Given,
$$\begin{aligned} \text { power } & =1 \mathrm{~W} \\ \text { Zener breakdown } V_z & =5 \mathrm{~V} \\ \text { Minimum voltage } V_{\min } & =3 \mathrm{~V} \\ \text { Maximum voltage } V_{\max } & =7 \mathrm{~V} \\ \text { Current } I_{Z_{\max }} & =\frac{P}{V_Z}=\frac{1}{5}=0.2 \mathrm{~A} \end{aligned}$$
The value of $R_s$ for safe operation $R_s=\frac{V_{\max }-V_z}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}=10 \Omega$
If each diode in figure has a forward bias resistance of $25 \Omega$ and infinite resistance in reverse bias, what will be the values of the currents $I_1, I_2, I_3$ and $I_4$ ?
$$\begin{aligned} \text { Given, } \quad & \text { forward biased resistance }=25 \Omega \\ & \text { Reverse biased resistance }=\infty \end{aligned}$$
As the diode in branch $C D$ is in reverse biased which having resistance infinite,
SO $$I_3=0$$
Resistance in branch $A B=25+125=150 \Omega$ say $R_1$
Resistance in branch $E F=25+125=150 \Omega$ say $R_2$
AB is parallel to EF.
$$\begin{array}{ll} \text { So, } \quad & \text { resultant resistance } \frac{1}{R^{\prime}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150} \\ \Rightarrow & R^{\prime}=75 \Omega \end{array}$$
Total resistance $R=R^{\prime}+25=75+25=100 \Omega$
Current $I_1=\frac{V}{R}=\frac{5}{100}=0.05 \mathrm{~A}$
$$\begin{aligned} &\begin{aligned} I_1 & =I_4+I_2+I_3 \\ \text{So,}\quad I_1 & =I_4+I_2\quad \text { (Here } I_3=0 \text { ) } \end{aligned}\\ \end{aligned}$$
Here, the resistances $R_1$ and $R_2$ is same. i.e.,
$$I_4=I_2$$
$$\begin{array}{l} \therefore \quad I_1=2 I_2 \\ \Rightarrow \quad I_2=\frac{I_1}{2}=\frac{0.05}{2}=0.025 \mathrm{~A} \\ \text { and } \quad I_4=0.025 \mathrm{~A} \\ \text { Thus, } \quad I_1=0.05 \mathrm{~A}, I_2=0.025 \mathrm{~A}, I_3=0 \text { and } I_4=0.025 \mathrm{~A} \end{array}$$
In the circuit shown in figure, when the input voltage of the base resistance is $10 \mathrm{~V}, V_{\mathrm{BE}}$ is zero and $V_{\mathrm{CE}}$ is also zero. Find the values of $I_B, I_c$ and $\beta$.
Given,
$$\begin{aligned} \text { voltage across } R_B & =10 \mathrm{~V} \\ \text { Resistance } R_B & =400 \mathrm{k} \Omega \\ V_{\mathrm{BE}} & =0, V_{\mathrm{CE}}=0 R_{\mathrm{C}}=3 \mathrm{k} \Omega \\ I_B & =\frac{\text { Voltage across } R_B}{R_B} \\ & =\frac{10}{400 \times 10^3}=25 \times 10^{-6} \mathrm{~A}=25 \propto \mathrm{~A} \end{aligned}$$
Voltage across $R_C=10 \mathrm{~V}$
$$\begin{aligned} I_C & =\frac{\text { Voltage across } R_C}{R_C}=\frac{10}{3 \times 10^3} \\ & =3.33 \times 10^{-3} \mathrm{~A}=3.33 \mathrm{~mA} \\ \beta & =\frac{I_C}{I_B}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}} \\ & =1.33 \times 10^2=133 \end{aligned}$$