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.
$$\begin{aligned} &\begin{aligned} \text{Given,}\quad V_{B E} & =0.7 \mathrm{~V}, V_{\mathrm{CC}}=V_{\mathrm{BB}}=16 \mathrm{~V} \\ V_{\mathrm{CE}} & =8 \mathrm{~V} \text { (from graph) }\\ I_C & =4 \mathrm{~mA}=4 \times 10^{-3} \mathrm{~A} \\ I_B & =30 \propto \mathrm{~A}=30 \times 10^{-6} \mathrm{~A} \end{aligned}\\ \end{aligned}$$
$$\begin{aligned} &\text { For the output characteristic at } \theta \text {, }\\ &\begin{aligned} V_{C C} & =I_C R_C+V_{C E} \\ R_C & =\frac{V_{C C}-V_{C E}}{I_C}=\frac{16-8}{4 \times 10^{-3}}=\frac{8 \times 1000}{4}=2 \mathrm{k} \Omega \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Using the relation, }\\ &\begin{aligned} V_{\mathrm{BB}} & =I_B R_B+V_{\mathrm{BE}} \\ R_B & =\frac{V_{\mathrm{BB}}-V_{\mathrm{BE}}}{I_B}=\frac{16-0.7}{30 \times 10^{-6}} \\ & =510 \times 10^3 \Omega=510 \mathrm{k} \Omega \\ \beta & =\frac{I_C}{I_B}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133 \\ \text { Voltage gain } & =\beta \frac{R_C}{R_B}=\frac{133 \times 2 \times 10^3}{510 \times 10^3}=0.52 \\ \text { Power gain } & =\beta \times \text { Voltage gain }=133 \times 0.52=69 \end{aligned} \end{aligned}$$
Assuming the ideal diode, draw the output waveform for the circuit given in fig. (a), explain the waveform.
When the input voltage is equal to or less than 5 V , diode will be revers biased. It will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.
If input voltage is more than +5 V , diode will be conducting as if forward biased offering low resistance in comparison to $R$. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across $R$.
When input voltage is negative, there will be opposition to 5 V battery in $p-n$ junction input voltage becomes more than $-$5 V , the diode will be reverse biased. It will offer high resistance in comparison to resistance $R$ in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals.
The output waveform is shown here in the fig. (b)
Suppose a $n$-type wafer is created by doping Si crystal having $5 \times 10^{28}$ atoms $/ \mathrm{m}^3$ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ' $p$ ' region in this wafer. Considering $n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}$, (i) Calculate the densities of the charge carriers in the $n$ and $p$ regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
When As is implanted in Si crystal, $n$ - type wafer is created. The number of majority carriers electrons due to doping of As is
$$\begin{aligned} n_e & =N_D=\frac{1}{10^6} \times 5 \times 10^{28} \\ & =5 \times 10^{22} / \mathrm{m}^3 \end{aligned}$$
Number of minority carriers (holes) in $n$-type wafer is
$$\begin{aligned} n_h & =\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}} \\ & =0.45 \times 10^{10} / \mathrm{m}^3 \end{aligned}$$
When $B$ is implanted in Si crystal, p-type wafer is created with number of holes,
$$n_h=N_A=\frac{200}{10^6} \times\left(5 \times 10^{28}\right)=1 \times 10^{25} / \mathrm{m}^3$$
Minority carriers (electrons) created in p-type wafer is
$$\begin{aligned} n_e & =\frac{n_i^2}{n_h}=\frac{\left(1.5 \times 10^{16}\right)^2}{1 \times 10^{25}} \\ & =2.25 \times 10^{27} / \mathrm{m}^3 \end{aligned}$$
When $p-n$ junction is reverse biased, the minority carrier holes of $n$-region wafer ( $n_h=0.45 \times 10^{10} / \mathrm{m}^3$ ) would contribute more to the reverse saturation current than minority carrier electrons $\left(n_e=2.25 \times 10^7 / \mathrm{m}^3\right)$ of pregion wafer.