Semiconductor Devices and Logic Gates

If $V_A>V_B$, both the junction diodes are forward biased and the given circuit diagram becomes a balanced Wheatstone bridge. The equivalent resistance in this case becomes $4 \Omega$.

If $V_A < V_B$, the diodes are reverse biased. In that case $4 \Omega, 5 \Omega$ and $4 \Omega$ are in series. So, equivalent resistance becomes $13 \Omega$.

With rise in temperature, resistivity of semi conductor decreases non-linearly. The temperature $(T) (T)$ dependence of resistivity $(\rho)$ of a semiconductor is best represented by option (c).

Given circuit can be shown as below

Output, $ Y=\overline{(\bar{A})(\bar{B})}=\bar{A}+\bar{B}=A+B $

The truth table will be

The corresponding input and ou put waveforms are shown below.

To find the power gain of the amplifier, we first need to understand the relationship between voltage gain, input impedance, and output impedance.

The power gain (G_P) of an amplifier is given by the product of the voltage gain (G_V) and the current gain (G_I). However, we can derive it directly from the voltage gain and the impedance values.

Given:

Voltage gain (G_V) = 50

Input impedance (R_in) = 100 $\Omega$

Output impedance (R_out) = 400 $\Omega$

We know that:

$ G_P = \frac{V_{out}^2 / R_{out}}{V_{in}^2 / R_{in}} $

Since the voltage gain is:

$ G_V = \frac{V_{out}}{V_{in}} = 50 $

Substituting $ G_V = 50 $ into the power gain formula:

$ G_P = \left(\frac{V_{out}}{V_{in}}\right)^2 \cdot \frac{R_{in}}{R_{out}} = G_V^2 \cdot \frac{R_{in}}{R_{out}} $

So we have:

$ G_P = 50^2 \cdot \frac{100}{400} $

$ G_P = 2500 \cdot 0.25 $

$ G_P = 625 $

Hence, the power gain of the amplifier is 625.

Therefore, the correct option is:

Option C - 625

$ \begin{aligned} \therefore Y & =\overline{\overline{(\bar{A} \cdot \overline{A B}})(\bar{B} \cdot \overline{A B})} \\ & \overline{A \cdot \overline{A B}}+\overline{B \cdot \overline{A B}} \\ & =\bar{A}+A B+\bar{B}+A B \\ & =\overline{A B}+A B=1 \end{aligned}$

For an insulator, E_g > 3 eV, that is why electron transition from valence band to conduction band is not possible. For semiconductor E_g is 0.2 eV to 0.3 eV, while for metals E_g > 5 eV.

${l_{initial}} = 0$ for ${V_i} \le \,+ 1\,V$

This diode conduct only beyond ${V_i} = \, + 1\,V$

${I_{final}} = {5 \over {250}} = 0.02\,A$

So, change in $I = 0.02\,A = 20\,mA$

According to the given figure X is at lower potential w.r.t Y. Hence, both diodes are in reverse biasing, so, equivalent circuit can be redrawn as follows

Equivalent resistance between X and Y

R_eq = 8 + 2 + 6 = 16$\Omega$

Given, V_CC = 8V

Voltage drop, V₀ = $\Delta$I_C R_L = 0.6V

Load Resistor, R_L = 500 $\Omega$

Gain factor, $\alpha$ = 0.96

$\because$ $\beta = {\alpha \over {1 - \alpha }} = {{0.96} \over {1 - 0.96}} = 24$

Now, V₀ = $\Delta$I_C R_L

0.6 = $\Delta$I_C $\times$ 500

$\Delta$I_C = $\frac{0.6}{500}$ = 1.2 $\times$ 10^$-$3 A

Now, $\beta = {{\Delta {I_C}} \over {\Delta {I_B}}}$

$\therefore$ Base current $\Delta {I_B} = {{\Delta {I_C}} \over \beta } = {{1.2 \times {{10}^{ - 3}}} \over {24}} = 5 \times {10^{ - 5}}$

= 50 $\times$ 10^$-$6

$\Delta$I_B = 50 $\mu$A

There is a phase difference of 180$^\circ$ between the signal voltage and output voltage in a common emitter amplifier. It is also known as phase reversal.

Diode D₁ does not conduct because it is connected in reversed biased position.

Hence, only D₂ conducts as it is forward biased.

$\therefore$ Current in circuit is given by

$i = {V \over R}$

$ = {5 \over {30 + 20}} = {5 \over {20}}A \Rightarrow i = 0.1A$

The given combination of gates is shown below

The output of this combination can be determined by Boolean algebra.

$\therefore$ Output, $X = \overline {\overline A \,.\,\overline B } = \overline{\overline A} + \overline{\overline B} $ [Using De Morgan's theorem]

$ \Rightarrow X = A + B$

This is the output of OR gate. Hence, the given combination of gates yield OR gate.

In this case, the diode is forward biased. Potential drop in a germanium diode in forward bias is around 0.3V.

Hence, current through the given circuit,

$I = {{{V_{net}}} \over {{R_{eq}}}} = {{5 - 0.3} \over {2 + 3}} = {{4.7} \over 5}$

I = 0.94A