Semiconductor Devices and Logic Gates

At absolute zero temperature, an intrinsic semiconductor behave as an insulator.

$ \begin{aligned} y & =\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A B \\ & =\text { Output of AND gate } \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} I_C & =0.98 I_E=0.98\left(I_C+I_B\right) \\ I_C & =0.98 I_C+0.98 I_B \\ \Rightarrow 0.02 I_C & =0.98 I_B \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{I_B}{I_C}=\frac{0.02}{0.98}=\frac{1}{49} \\ & \therefore I_B: I_C=1: 49 \end{aligned} $

$ \text { When } A=0, B=1 \text { and } C=1 \text {; } $

When $A=0, B=1$, output of ' $\mathrm{OR}^{\prime}$ gate is 1 and of NOT gate is $y_1=0$.

When $y_1=0$ and $C=1$ fed to AND gate, output of AND gate is zero.

So, output $y_2$ of NOT gate is 1

$ \therefore\left(y_1, y_2\right)=(0,1) $

A p-n junction diode is in forward biased if $p$-side is connected with more positive voltage than $n$-side.

$ \begin{aligned} & \text { Power gain }=\text { Current gain }\text { × Voltage gain } \\ & =\beta \times\left(\beta \frac{V_{\text {out }}}{V_{\text {in }}}\right) \\ & =\beta^2 \frac{V_C}{V_B} \\ & =100^2 \times \frac{3 \times 10^3}{2 \times 10^3} \\ & =10000 \times \frac{3}{2}=15000 \end{aligned} $

In a transistor size of collector is largest, size of emitter is moderate and size of base is smallest.

$\therefore$ Collector $>$ Emitter $>$ Base

$ \Rightarrow \quad Z>X>Y . $

$ \text { Truth table of above combination is; } $

Output $Y$ resembles output of an NAND gate

The question tells us that the voltage gain divided by the current amplification factor ($ \beta $) equals 4.

We know that the formula for voltage gain in a common emitter amplifier is $ A_V = \frac{\beta \times R_C}{R_B} $, where $ R_C $ is collector resistance and $ R_B $ is base resistance.

So if we divide $ A_V $ by $ \beta $, we get:
$ \frac{A_V}{\beta} = \frac{\frac{\beta \times R_C}{R_B}}{\beta} $

This simplifies to:
$ \frac{R_C}{R_B} $

The question gives $ \frac{A_V}{\beta} = 4 $, so:
$ \frac{R_C}{R_B} = 4 $

This means the ratio of collector resistance to base resistance is 4:1.

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

$ \text { } \begin{aligned} A_V & =300, \beta=60 \\ A_V & =\beta \cdot \frac{R_C}{R_B} \\ 300 & =60 \times \frac{5 \times 10^3}{R_B} \\ R_B & =10^3 \Omega=1 \mathrm{~K} \Omega \end{aligned} $

$ \begin{aligned} & y_1=\overline{\overline{A+B}}=A+B \\ \therefore \quad & y=\overline{y_1}=\overline{A+B}=\text { Output of NOR gate. } \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} Output, \,\,& Y=\overline{\bar{A}} \cdot \overline{\bar{B}} \\ = & \bar{A}+\overline{\bar{B}}=A+B \\ = & \text { Output of OR gate } \end{aligned} \end{aligned} $

The correct answer is:

✅ Option A: Zener diode

Explanation:

A Zener diode is specifically designed to operate in the reverse breakdown region and maintain a nearly constant voltage across it, even when the current or load varies. Therefore, it is commonly used for voltage regulation in electronic circuits.

In a transistor, base region is lightly doped, emitter region is highly doped and collector region is moderately doped.

A transistor works as an amplifier when emitter-base junction is in forward biased and base-collector junction is in reverse biased.

$ \text { Truth table for given input is } $

$ \begin{array}{cccccccc} \hline \mathrm{A} & \mathrm{~B} & \mathrm{C} & x_1 & x_2 & y_1 & y_2 & y_3 \\ \hline 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 \\ \hline \end{array} $

$ V_Z=20 \mathrm{~V} $

$ \begin{aligned} I & =\frac{V-V_Z}{R_s}=\frac{40-20}{2 \times 10^3} \\ & =\frac{20}{2 \times 10^3}=10^{-2} \mathrm{~A}=10 \mathrm{~mA} \\ I_L & =\frac{V_Z}{R_L}=\frac{20}{5 \times 10^3} \\ & =4 \times 10^{-3} \mathrm{~A}=4 \mathrm{~mA} \\ \therefore I_Z & =I-I_L=10-4=6 \mathrm{~mA} \end{aligned} $

$ \begin{aligned} &\text { Total voltage gain }\\ &\begin{aligned} A & =A_1 \times A_2=8 \times 12.5=100.0 \\ A & =100 \\ \therefore \text { Since, } A & =\frac{V_{\text {out }}}{V_{\text {in }}} \\ \Rightarrow \quad V_{\text {out }} & =A \times V_{\text {in }}=100 \times 200 \times 10^{-6} \\ & =0.02 \mathrm{~V}=20 \times 10^{-3} \mathrm{~V} \\ & =20 \mathrm{mV} \end{aligned} \end{aligned} $

For reverse biasing of an ideal diode, the potential of $n$-side should be higher than potential of $p$-side.

$ \begin{aligned} &\text { Output, }\\ &\begin{aligned} Y & =\overline{\bar{A}+\bar{B}}=\overline{\bar{A}}+\overline{\bar{B}} \\ & =A B \\ & =\text { Output of AND gate } \end{aligned} \end{aligned} $

We know that, two inputs $A$ and $B$ are given then,

Boolean expression for NOR gate $=\overline{A+B}$

Boolean expression for AND gate $=A \cdot B$

So, Boolean expression for $Y$ (NAND

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

So, output $Y$ will always be 1 .

$ \text { Truth table: } $

$ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array} $

Given:

Collector side resistance: $ R_C $

Base side resistance: $ R_B $

Current amplification factor: $ \beta $

We know that the voltage gain of a transistor amplifier can be calculated as:

$ \text{Voltage gain} = \text{Current gain} \times \frac{\text{Output resistance}}{\text{Input resistance}} $

In a common emitter configuration, the voltage gain $ A_V $ is derived as follows:

$ A_V = \frac{\Delta I_C}{\Delta I_B} \times \frac{R_C}{R_B} = \frac{\beta \cdot R_C}{R_B} $

This equation shows that the voltage gain is the product of the current gain and the ratio of the collector resistance to the base resistance.

The output voltage in a common emitter transistor configuration can be determined using the formula for voltage gain, $ A_V $. The equation for voltage gain in this configuration is given by:

$ A_V = -\beta \left(\frac{R_C}{R_{\text{in}}}\right) $

Here, the current gain ($\beta$) is 80, the resistance on the collector side ($R_C$) is $5 \, \text{k}\Omega$, and the resistance on the base side ($R_{\text{in}}$) is $1 \, \text{k}\Omega$. Plugging in the values:

$ A_V = -80 \left(\frac{5000}{1000}\right) = -80 \times 5 = -400 $

The negative sign indicates an inversion in the phase between the input and output signals.

Now, to find the output voltage ($V_{\text{out}}$), multiply the voltage gain by the input voltage ($V_{\text{in}} = 2 \, \text{mV} = 2 \times 10^{-3} \, \text{V}$):

$ V_{\text{out}} = A_V \times V_{\text{in}} $

Calculating the output voltage:

$ V_{\text{out}} = -400 \times 2 \times 10^{-3} = -0.8 \, \text{V} $

Therefore, the output voltage is $-0.8 \, \text{V}$, indicating the output is inverted with respect to the input voltage.

The combination of logic gate is as follows.

For $Y_1=\overline{A \cdot B \cdot B}=\overline{A \cdot B}$

For $Y_2=\overline{B+C+C}=\overline{B+C}$

For $A=0, B=1, C=1$

$ Y_1=1, Y_2=0 $

Given:

Initial electron and hole concentration in pure silicon at 300 K:

$ n_h = n_e = 1.5 \times 10^{16} \, \text{m}^{-3} $

After doping, the new hole concentration becomes:

$ n_h' = 3 \times 10^{22} \, \text{m}^{-3} $

Using the mass action law, the relationship between electron and hole concentrations is:

$ n_e n_h = n_e' n_h' $

We need to find the new electron concentration $ n_e' $. Rewriting the equation for $ n_e' $:

$ n_e' = \frac{(n_e)^2}{n_h'} $

Substitute the known values:

$ n_e' = \frac{(1.5 \times 10^{16})^2}{3 \times 10^{22}} $

Calculating the above expression:

$ n_e' = 7.5 \times 10^9 \, \text{m}^{-3} $

Thus, the electron concentration in the silicon after doping is $ 7.5 \times 10^9 \, \text{m}^{-3} $.

Given:

Collector current, $ I_C = 10 \, \text{mA} $

$ 95\% $ of the electrons emitted reach the collector

First, let's determine the emitter current $ I_E $:

$ I_C = 95\% \, \text{of} \, I_E $

So,

$ I_C = \frac{95}{100} \times I_E $

Rearranging for $ I_E $, we get:

$ I_E = \frac{100 \times I_C}{95} $

Substituting the known value of $ I_C $:

$ I_E = \frac{100 \times 10}{95} = 10.53 \, \text{mA} $

Now, we use the relationship between emitter current, collector current, and base current:

$ I_B = I_E - I_C $

Substitute the known values:

$ I_B = 10.53 - 10 $

Therefore, the base current is:

$ I_B = 0.53 \, \text{mA} $

To emit visible light, an LED’s semiconductor must have a band gap energy that falls within the range corresponding to visible photon energies. Here's why:

• The energy of the photon emitted when an electron recombines with a hole is approximately equal to the band gap energy, given by

$ E = \frac{hc}{\lambda} $

where

  - $h$ is Planck's constant,

  - $c$ is the speed of light, and

  - $\lambda$ is the wavelength of the emitted light.

• Visible light covers roughly from 400 nm (violet) to 700 nm (red), which translates to energies approximately between 3.1 eV and 1.8 eV. A semiconductor with a band gap energy lower than 1.8 eV would emit light in the infrared region, not visible.

Given the options:

  - Option A: 0.6 eV

  - Option B: 1.2 eV

  - Option C: 1.8 eV

  - Option D: 0.9 eV

The minimum band gap for a visible LED is therefore about $1.8 \text{ eV}$.

Thus, the correct answer is: Option C.

To find the voltage gain of a common emitter amplifier, we have the following information:

AC current gain ($\beta$) = 40

Input resistance ($R_i$) = $2 \, \mathrm{k}\Omega$

Load resistance ($R_L$) = $10 \, \mathrm{k}\Omega$

The voltage gain ($A_v$) is calculated using the formula:

$ A_v = \beta \cdot \frac{R_L}{R_i} $

Substituting the given values into the formula:

$ \begin{aligned} A_v &= \frac{40 \times 10 \times 10^3}{2 \times 10^3} \\ A_v &= 40 \times 5 \\ A_v &= 200 \end{aligned} $

Thus, the voltage gain is 200.

Given:

Voltage gain, $ A_V = 150 $

Current gain, $ A_i = 50 $

Base resistance, $ R_B = 850 \, \Omega $

To find the resistance in the collector circuit, use the relationship for voltage gain in a common emitter transistor configuration:

$ A_V = A_i \cdot \frac{R_C}{R_B} $

Rearrange to solve for $ R_C $:

$ R_C = \frac{A_V}{A_i} \times R_B $

Substitute the given values:

$ R_C = \frac{150}{50} \times 850 $

$ R_C = 3 \times 850 $

$ R_C = 2550 \, \Omega $

Thus, the resistance in the collector circuit is $ 2550 \, \Omega $.

Let's analyze the energy gap ($E_g$) values typically associated with different materials:

Conductors (Metals):

Metals have overlapping conduction and valence bands, meaning they don’t exhibit a significant band gap.

Semiconductors:

Semiconductors usually have a moderate energy gap, commonly less than about 3 eV. For example, silicon has an energy gap of around 1.1 eV.

Both $n$-type and $p$-type semiconductors are doped forms of intrinsic semiconductors and share similar energy gap values with the pure material.

Insulators:

Insulators typically have a large energy gap, often greater than 3 eV. A 5.4 eV energy gap is well within the range for insulators.

Since the given energy gap is $5.4 \text{ eV}$, the substance falls into the insulator category.

Thus, the correct answer is:

Option A: insulator.

A p-n junction diode is utilized as a rectifier. This is because the diode allows current to flow in only one direction when it is forward-biased and prevents current flow when it is reverse-biased. This unidirectional conduction property makes the diode suitable for converting alternating current (AC) to direct current (DC), which is the primary function of a rectifier.

The answer is Option B: CE - configuration.

Here's why:

In a common emitter (CE) amplifier, the output is taken from the collector while the input is applied to the base.

The CE configuration is known for inverting the input signal, which results in a phase shift of $180^\circ$.

Other configurations like common base (CB) and common collector (CC) do not exhibit this 180° phase inversion.

Thus, when you observe that the output signal is phase-shifted by $180^\circ$, it indicates that the transistor is set up in a common emitter configuration.

For an intrinsic (pure) semiconductor, where the number of electrons is equal to the number of holes ( $n_e=n_h$ ), the intrinsic carrier concentration ( $n_i$ ) in thermal equilibrium is given by,

$ n_i^2=n_e \times n_h \Rightarrow n_i=\sqrt{n_e \times n_h} $

Given, input, $A=1$

$ \begin{aligned} & B=1 \\ & C=1 \end{aligned} $

Here, $B$ and $C$ are inputs for NAND gate.

So, output

$ \begin{aligned} & Y_1=\overline{B \cdot C} \\ & Y_1=\overline{1 \cdot 1} \Rightarrow Y_1=0 \end{aligned} $

Now, $\bar{A}$ and $Y_1$ are inputs for NOR gate.

So, output,

$ \begin{aligned} & Y_2=\overline{\bar{A}+Y_1} \\ & Y_2=\overline{0+0} \Rightarrow Y_2=1 \end{aligned} $

Therefore, $Y_1=0, Y_2=1$

At 0K , semiconductor behaves as an insulator, hence, there is no free electrons in the semiconductor at 0 K .

At room temperature or above room temperature, some covalent bonds are broken, hence some free electrons are found at these temperatures. On increasing temperature, more number of covalent bonds are broken, hence number of free electrons increases with increase of temperature of semiconductor. Conductivity of semiconductor is less than that in conductor, hence number of free electrons in semiconductor is less than that of conductor. Therefore, statements (A), (C), (D) are correct but (B) is false.

The given logic circuit diagram is shown as:

Output of the logic circuit is given as

$\begin{aligned} Y & =(\bar{A}+B)(\bar{A}+\bar{B})(A+\bar{B}) \\ & =(\bar{A} \bar{A}+\bar{A} \bar{B}+B \bar{A}+B \bar{B})(A+\bar{B}) \quad(\because B \bar{B}=0) \\ & =(\bar{A}+\bar{A} \bar{B}+B \bar{A}+0)(A+\bar{B}) \end{aligned}$

$\begin{aligned} &\begin{aligned} & =[\bar{A}(1+\bar{B})+B \bar{A}](A+\bar{B}) \\ & =(\bar{A}+B \bar{A})(A+\bar{B}) \quad (\because 1+\bar{B}=1)\\ & =\bar{A}(1+B)(A+\bar{B}) \\ & =\bar{A} \cdot 1(A+\bar{B}) \quad (\because 1+B=1)\\ & =\bar{A} A+\bar{A} \bar{B} \\ & =0+\bar{A} \bar{B} \\ & =\bar{A} \bar{B} \end{aligned}\\ \end{aligned}$

The current can be written as $ I = I_0 \sin \omega t $, where $ I_0 $ is the highest value (peak current), and $ \omega $ is the angular frequency.

To find the average value of the current ($ I_{\text{av}} $), we need to find the average over half the cycle (because the current repeats every half cycle):

$ I_{\text{av}} = \frac{1}{T/2} \int_0^{T/2} I\, dt $ where $ T $ is the time for one full cycle.

Plug in the current equation:

$ I_{\text{av}} = \frac{2}{T} \int_0^{T/2} I_0 \sin \omega t\, dt $

Take out constants and solve the integral:

$ I_{\text{av}} = \frac{2 I_0}{T} \left[ \frac{-\cos \omega t}{\omega} \right]_0^{T/2} $

Put in the values for $ t = 0 $ and $ t = T/2 $:

$ I_{\text{av}} = \frac{2 I_0}{T \omega} \left( -\cos(\omega \frac{T}{2}) + \cos(0) \right) $

Since $ \omega = \frac{2\pi}{T} $, then $ \omega \frac{T}{2} = \pi $:

$ I_{\text{av}} = \frac{2 I_0}{T \omega} [ -\cos(\pi) + \cos(0) ] $

We know $ \cos(\pi) = -1 $ and $ \cos(0) = 1 $:

$ I_{\text{av}} = \frac{2 I_0}{T \omega} [ -(-1) + 1 ] = \frac{2 I_0}{T \omega} [ 1 + 1 ] = \frac{2 I_0}{T \omega} \times 2 $

Now replace $ \omega $ with $ 2\pi / T $:

$ I_{\text{av}} = \frac{2 I_0}{T \times \frac{2\pi}{T}} \times 2 = \frac{2 I_0}{2\pi} \times 2 = \frac{2 I_0}{\pi} $

So, the average current is: $ I_{\text{av}} = \frac{2 I_0}{\pi} $

Given, change in base emitter voltage,

$\Delta V_{B E}=0.04 \mathrm{~V}$

Change in base current, $\Delta I_B=20 \mu \mathrm{A}$

Change in collector current, $\Delta I_C=2 \mathrm{~mA}$

$\because$ Input resistance,

$\begin{aligned} R_{\text {in }} & =\frac{\Delta V_{B E}}{\Delta I_B} \\ & =\frac{0.04}{20 \times 10^{-6}}=2 \times 10^3 \Omega \\ & =2 \mathrm{k} \Omega \end{aligned}$

$\text { current gain, } \begin{aligned} \beta & =\frac{\Delta I_C}{\Delta I_B} \\ & =\frac{2 \times 10^{-3}}{20 \times 10^{-6}}=10^2=100 \end{aligned}$

Given,

Here, X = A + B

$\therefore$ It is an output of OR gate.

Given, collector voltage, $V_c=8V$

Voltage drop across resistor, $R=800\Omega$

and $V=0.5$ V

Current gain, $\alpha=0.96$

Let, collector, emitter and base current be $I_C, I_E$ and $I_B$, respectively.

Since, $\alpha=\frac{I_C}{I_E}$

and $I_E =I_B+I_C$

$\because \alpha =\frac{I_C}{I_B+I_C}=\frac{1}{\frac{I_B}{I_C}+1}$

$\Rightarrow \frac{I_B}{I_C} =\frac{1}{\alpha}-1$

$=\frac{100}{96}-1=\frac{4}{96}=0.042$

$\begin{aligned} \Rightarrow \quad I_B & =0.042 I_C=0.042 \times \frac{0.5}{800}\left(\because I_C=\frac{V}{R}\right) \\ & =\frac{42 \times 10^{-3} \times 5 \times 10^{-1}}{800} \\ & =2.625 \times 10^{-5} \mathrm{~A} \end{aligned}$