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 $R_C=10 \mathrm{~V}$
$$\begin{aligned} I_C & =\frac{\text { Voltage across } R_C}{R_C}=\frac{10}{3 \times 10^3} \\ & =3.33 \times 10^{-3} \mathrm{~A}=3.33 \mathrm{~mA} \\ \beta & =\frac{I_C}{I_B}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}} \\ & =1.33 \times 10^2=133 \end{aligned}$$
Draw the output signals $C_1$ and $C_2$ in the given combination of gates.
First draw the truth table of $C_1$ and $C_2$.
Consider the circuit arrangement shown in figure for studying input and output characteristics of $n-p-n$ transistor in CE configuration. Select the values of $R_B$ and $R_C$ for a transistor whose $V_{\mathrm{BE}}=0.7 \mathrm{~V}$, so that the transistor is operating at point $Q$ as shown in the characteristics (see figure).
Given that the input impedance of the transistor is very small and $V_{C C}=V_{B B}=16 \mathrm{~V}$, also find the voltage gain and power gain of circuit making appropriate assumptions.
$$\begin{aligned} &\begin{aligned} \text{Given,}\quad V_{B E} & =0.7 \mathrm{~V}, V_{\mathrm{CC}}=V_{\mathrm{BB}}=16 \mathrm{~V} \\ V_{\mathrm{CE}} & =8 \mathrm{~V} \text { (from graph) }\\ I_C & =4 \mathrm{~mA}=4 \times 10^{-3} \mathrm{~A} \\ I_B & =30 \propto \mathrm{~A}=30 \times 10^{-6} \mathrm{~A} \end{aligned}\\ \end{aligned}$$
$$\begin{aligned} &\text { For the output characteristic at } \theta \text {, }\\ &\begin{aligned} V_{C C} & =I_C R_C+V_{C E} \\ R_C & =\frac{V_{C C}-V_{C E}}{I_C}=\frac{16-8}{4 \times 10^{-3}}=\frac{8 \times 1000}{4}=2 \mathrm{k} \Omega \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Using the relation, }\\ &\begin{aligned} V_{\mathrm{BB}} & =I_B R_B+V_{\mathrm{BE}} \\ R_B & =\frac{V_{\mathrm{BB}}-V_{\mathrm{BE}}}{I_B}=\frac{16-0.7}{30 \times 10^{-6}} \\ & =510 \times 10^3 \Omega=510 \mathrm{k} \Omega \\ \beta & =\frac{I_C}{I_B}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133 \\ \text { Voltage gain } & =\beta \frac{R_C}{R_B}=\frac{133 \times 2 \times 10^3}{510 \times 10^3}=0.52 \\ \text { Power gain } & =\beta \times \text { Voltage gain }=133 \times 0.52=69 \end{aligned} \end{aligned}$$