Semiconductor Devices and Logic Gates

Current amplification (CE)

$ \begin{aligned} & =\beta=\frac{\Delta I_C}{\Delta I_B} \\ \Rightarrow \Delta I_B & =\frac{\Delta I_C}{\beta} \Rightarrow \Delta I_B=\frac{2.43}{80} \mathrm{~mA} \\ & =0.030 \mathrm{~mA}=30 \times 10^{-6} \mathrm{~A} \end{aligned} $

or $30 \mu \mathrm{~A}$

Negative feedback counteract the changes of output signal.

As frequent charges are depressed, there is low noise and distortion in output.

At absolute zero temperature (0 Kelvin), there is no thermal energy available.

1. Semiconductors and Energy Bands: In a semiconductor, electrons reside in the valence band at lower energy levels. For conduction to occur, electrons need to jump from the valence band to the conduction band, leaving behind holes in the valence band. The energy required for this jump is called the band gap energy.

2. Effect of Temperature:

   • At typical room temperatures, some electrons gain enough thermal energy to cross the band gap and move into the conduction band, contributing to electrical conductivity.

   • As temperature decreases, the number of thermally excited electrons decreases.

   • At absolute zero temperature (0 K), there is no thermal energy. Therefore, no electrons can gain enough energy to jump from the valence band to the conduction band.

3. Result:

   • The valence band remains completely filled.

   • The conduction band remains completely empty.

   • A material with a completely filled valence band and an empty conduction band, separated by a band gap, behaves as an insulator. There are no free charge carriers available for conduction.

Therefore, at absolute zero temperature, a semiconductor behaves like an insulator.

The final answer is $\boxed{D}$

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

$ \begin{aligned} y_1 & =\overline{A B}=\overline{1.0} \\ & =\overline{0}=1 \\ y_2 & =\overline{B+C}=\overline{0+1}=\overline{1}=0 \\ y_3 & =y_1 \cdot y_2=1.0=0 \end{aligned} $

From the graph

In region I, output voltage is high as $V_0$ when the input voltage $V_i$ is low. i.e., transistor is off. This is cutoff region.

In region II, output voltage decreases as input voltage increases. This is the

situation when transistor operates as an amplifier.

Thus, in region II, transistor is in active mode.

In region III, output voltage is low when input voltage is high i.e., maximum current flow through the collector. Thus, transistor is in saturation region.

NAND and NOR gate are knows as universal gate because any logic gate can be implemented using these two gates.

To use transistor as a switch in a common emitter configuration, the transistor operates in cutoff and saturation region.

In cutoff region, the transistor is "off" and acts like an open switch, allowing no current to flow between the collector and emitter.

In saturation region, the transistor is "on" and acts like a closed switch (short circuit) between the collector and emitter, allowing maximum current to flow.

In active region, transistor works as an amplifier.

$ \begin{aligned} \lambda & =\frac{h c}{E} \\ & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \\ & =6.53 \times 10^{-7} \mathrm{~m} \\ & \simeq 653 \mathrm{~nm} \end{aligned} $

The visible light spectrum ranges approximately 400 nm (violet) to 700 nm (red). A wavelength of 653 nm falls with in the red part of the spectrum. Therefore, the colour of the light emitted by the LED is red.

Given:

• Number of electrons entering emitter per second (or per given time)

  $ N_E = 10^{10} $ electrons in $ t = 0.4 \, \mu s $

• $ 5\% $ of electrons are lost in the base.

So, only $ 95\% $ of electrons reach the collector.

Step 1: Find the emitter current

Total charge entering emitter in $ 0.4 \, \mu s $:

$ Q_E = N_E \times e = 10^{10} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-9} \, C $

Emitter current:

$ I_E = \frac{Q_E}{t} = \frac{1.6 \times 10^{-9}}{0.4 \times 10^{-6}} = 4.0 \, mA $

Step 2: Find the collector current

Since $ 5\% $ are lost, $ 95\% $ reach collector:

$ I_C = 0.95 \, I_E = 0.95 \times 4.0 \, mA = 3.8 \, mA $

✅ Final Answer:

$ \boxed{I_C = 3.8 \, mA} $

Correct option: D (3.8 mA)

According to conservation of energy

$ \begin{aligned} & \frac{1}{2} m_e v_i^2=\frac{1}{2} m_e v_f^2+e V_b \\ \Rightarrow \quad & \frac{1}{2} m_e v_f^2=\frac{1}{2} m_e v_i^2-e V_b \\ \Rightarrow \quad & v_f^2=v_i^2=\frac{2 e V_b}{m_e} \\ & =\left(5 \times 10^5\right)^2-\frac{2 \times 1.6 \times 10^{-19} \times 0.4}{9 \times 10^{-31}} \\ & =9 \times 10^{10} \\ \Rightarrow & v_f=\sqrt{9 \times 10^{10}}=3 \times 10^5 \mathrm{~m} / \mathrm{s} \end{aligned} $

Power gain $=$ Current $\times$ Voltage gain

$ \Rightarrow 1800=\text { Current gain }(\beta) \times 60 $

$ \begin{array}{ll} \Rightarrow & \beta=\frac{1800}{60} \Rightarrow \beta=30 \\ \Rightarrow & \frac{\Delta I_C}{\Delta I_B}=30 \Rightarrow \frac{\Delta I_B-\Delta I_B}{\Delta I_B}=30 \\ \Rightarrow & \frac{\Delta I_E}{\Delta I_B}-1=30 \\ \Rightarrow & \frac{0.62}{\Delta I_B}=31 \\ \Rightarrow & \Delta I_B=\frac{0.62}{31}=0.02 \mathrm{~mA} \\ \therefore & \Delta I_C=\Delta I_E-\Delta I_B=0.62-0.02 \\ & =0.60 \mathrm{~mA} \end{array} $

$ \begin{aligned} & y_1=\overline{1+1}=\overline{1}=0 \\ & y_2=\overline{1+1}=\overline{1}=0 \\ & y_3=y_1 \cdot y_2=0.0=0 \end{aligned} $

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} $

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).

Transistor Transistors are commonly used as amplifiers to increase the amplitude of a signal.

(b) Diode Diode are used in rectifier circuits to convert

alternating current $ (\mathrm{AC}) $ to direct current (DC).

(c) Zener diode Zener diodes are used in voltage regulator circuits to maintain a constant voltage.

(d) Capacitor Capacitors are used in filter circuits to smooth out fluctuations in voltage by filtering out noise.

So, the correct matching is Transistor : Amplifier Diode : Rectified Zener diode : Voltage regulator Capacitor : Filter circuit

From left side, the first gate is OR gate and next one is AND gate. So, the outpur of OR gate is $ A+B $ and the output of AND gate $ Y=(A+B) \cdot C $. We know that, for AND gate, the output will be 1 if both inputs are 1 . So,

$ A+B=1 \text { and } C=1 $

For $ A+B=1 $, either of the inputs must be one.

Therefore, from the given option (c) is the correct answer.

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

An $n$-type semiconductor, after being doped with donor atoms, has electrons as the majority charge carriers. However, the semiconductor remains electrically neutral overall.

Here's why:

Initially, the semiconductor material is intrinsically neutral, with an equal number of protons (positive charge) and electrons (negative charge).

Doping introduces donor atoms, which have more electrons than necessary to form bonds with the host atoms. These extra electrons become free carriers.

Despite the increased number of free electrons, the donor atoms themselves are neutral because they provide one more electron in their outer shell but don’t remove the existing positive charges in the nucleus.

Thus, even though there is an excess of free electrons compared to holes, the $n$-type semiconductor as a whole is charge neutral because the additional electrons are matched by the additional positive charge in donor atom's nucleus.

Therefore, the answer is:

Option C: neutral

In the output voltage versus input voltage graph of a transistor, the region where a transistor can be used as an amplifier is the active region (Option A).

In the active region, the transistor acts as a linear amplifier. The base-emitter junction is forward-biased, and the collector-base junction is reverse-biased. This configuration allows the transistor to amplify small input signals into larger output signals while maintaining proportionality and phase characteristics.

Explanation:

Active Region:

Base-Emitter Junction: Forward-Biased

Collector-Base Junction: Reverse-Biased

Used for Amplification

Proportional relationship between input and output voltage

Cut-off Region (Option B):

Both junctions are reverse-biased.

The transistor is off, and no current flows between collector and emitter.

Saturation Region (Option C):

Both junctions are forward-biased.

The transistor is fully on, allowing maximum current to flow between collector and emitter.

Not suitable for amplification as it works like a closed switch.

Passive Region (Option D):

This term is not typically used in the context of transistor operation regions.

Therefore, the active region is crucial for the amplification function of transistors.

Given,

Voltage gain $ =160=\frac{V_{\text {out }}}{V_{\text {in }}} $

$ I_{C}=\text { ? } $

Base current, $ I_{B}=100 \mu \mathrm{~A} $

Load resistance, $ R_{L}=4 \mathrm{k} \Omega $

Input resistance, $ R_{I}=1 \mathrm{k} \Omega $

Voltage gain $ =\beta \frac{R_{L}}{R_{I}} $

where $ \beta $ is current gain

$ \begin{aligned} \beta & =\frac{I_{C}}{I_{B}} \\\\ 160 & =\frac{I_{C}}{I_{B}} \cdot \frac{R_{L}}{R_{I}} \\\\ 160 & =\frac{I_{C}}{100 \mu \mathrm{~A}} \cdot \frac{4 \mathrm{k} \Omega}{1 \mathrm{k} \Omega} \\\\ I_{C} & =4 \times 10^{-3} \mathrm{~A} \\\\ I_{C} & =4 \mathrm{~mA} \end{aligned} $

The presence of the capacitor across the output terminal of rectifier eliminates the pulsation and fluctuation in the output voltage. The capacitor hold charge during the charging phase and discharging slowly during the discharging phase, maintaing relative constant voltage across load.

Given, $V_z=30 \mathrm{~V}, V_{\max }=100 \mathrm{~V}$

Voltage drop across zener branch and its parallel branch will be the same i.e.,

Current in $6 \mathrm{k} \Omega$ resistor,

$ i_1=\frac{30 \mathrm{~V}}{6 \mathrm{k} \Omega}=5 \mathrm{~mA} $

Current through rest of resistor $=\frac{V-V_z}{R_z}$

$ \begin{aligned} & =\frac{70 \mathrm{~V}}{7 \mathrm{k} \Omega}=10 \mathrm{~mA} \\\\ \therefore I_{\text {mener }} & =10 \mathrm{~mA}-5 \mathrm{~mA}=5 \mathrm{~mA} \end{aligned} $

From diagram

$ \begin{aligned} & Y_1=\overline{A \cdot(\overline{A+B})} \\\\ & Y_2=\overline{A+B} \end{aligned} $

Truth table,

In an intrinsic semiconductor, the concentration of electrons ($n_i$) and holes ($p_i$) are equal and given by:

$ n_i = p_i = 6 \times 10^{15} \, \mathrm{m}^{-3}. $

When the semiconductor is doped with an impurity, the electron concentration increases to $4 \times 10^{22} \, \mathrm{m}^{-3}$, which is denoted by $n$. The product of the electron concentration ($n$) and the hole concentration ($p$) in a semiconductor at thermal equilibrium is given by:

$ n \cdot p = n_i^2. $

To find the hole concentration ($p$) in the doped semiconductor, use the above relationship:

Calculate $n_i^2$:

$ n_i^2 = (6 \times 10^{15} \, \mathrm{m}^{-3})^2 = 36 \times 10^{30} \, \mathrm{m}^{-6}. $

Using the relation $n \cdot p = n_i^2$, solve for $p$:

$ p = \frac{n_i^2}{n} = \frac{36 \times 10^{30} \, \mathrm{m}^{-6}}{4 \times 10^{22} \, \mathrm{m}^{-3}}. $

Calculate $p$:

$ p = \frac{36}{4} \times 10^{30 - 22} \, \mathrm{m}^{-3} = 9 \times 10^8 \, \mathrm{m}^{-3}. $

Therefore, the concentration of holes in the doped semiconductor is $9 \times 10^8 \, \mathrm{m}^{-3}$, which corresponds to Option C.

Given, inputs are

$ A=1 \Rightarrow B=1 $

NOR GATE

When inputs $A=1$ and $B=1$, then NOR gate will gives output of $X_1=0$

AND GATE

When inputs $A=1$ and $B=0$, then AND gate will gives output of $Y_1=0$.

OR GATE

When inputs $A=0$ and $B=1$, then gate wil gives output of $Y_2=1$

Thus, value of $Y_1=0$ and $Y_2=1$.

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} $.