JEE Electronic Devices Questions

The output after each AND gate is $$ A \cdot \overline{B} $$ which means the same input to G.

By putting values of $$ A $$ and $$ B $$ from the truth table and by eliminating options, we get G to be a NOR gate.

Let us analyze each statement:

A. The junction area of solar cell is made very narrow compared to a photodiode.

This is false. In a solar cell, the junction area is made very large (not narrow) to capture maximum sunlight. It is the photodiode that typically has a smaller junction area.

B. Solar cells are not connected with any external bias.

This is true. Solar cells operate on the photovoltaic effect and generate their own EMF. They work without any external bias — they are essentially self-biased by the sunlight.

C. LED is made of lightly doped p-n junction.

This is false. LEDs are made of heavily doped p-n junctions to ensure high recombination rates and efficient light emission.

D. Increase of forward current results in continuous increase of LED light intensity.

This is false. Initially, increasing forward current increases light intensity, but beyond a certain point, the LED may reach saturation or get damaged. The increase is not continuous — there is a maximum rated current.

E. LEDs have to be connected in forward bias for emission of light.

This is true. LEDs emit light when electrons and holes recombine at the junction, which occurs only under forward bias.

The correct statements are B and E only.

The answer is Option A: B, E Only.

A p-n junction diode is forward biased when the electric potential at its p-side (anode, represented by the triangle) is strictly greater than the electric potential at its n-side (cathode, represented by the straight line). The condition for forward bias is $$V_p > V_n$$.

Option A: The p-side is connected to $$-10 \text{ V}$$ and the n-side is connected to $$0 \text{ V}$$. Since $$-10 \text{ V} < 0 \text{ V}$$, meaning $$V_p < V_n$$, this diode is reverse biased.

Option B: The p-side is connected to $$-10 \text{ V}$$ and the n-side is connected to $$-15 \text{ V}$$. Since $$-10 \text{ V} > -15 \text{ V}$$, meaning $$V_p > V_n$$, this diode is forward biased.

Option C: The p-side is connected to $$4 \text{ V}$$ and the n-side is connected to $$2 \text{ V}$$. Since $$4 \text{ V} > 2 \text{ V}$$, meaning $$V_p > V_n$$, this diode is also forward biased.

Option D: The n-side is connected to $$-5 \text{ V}$$ and the p-side is connected to $$-10 \text{ V}$$. Since $$-5 \text{ V} > -10 \text{ V}$$, meaning $$V_n> V_p$$, this diode is also reverse biased.

Option E: The n-side is connected to ground, which is $$0 \text{ V}$$, and the p-side is connected to $$+2 \text{ V}$$. Since $$0 \text{ V} < +2 \text{ V}$$, meaning $$V_n < V_p$$, this diode is forward biased.

Based on the explicit values shown in the provided circuit diagrams, circuits B, C, and E all represent forward biased diodes

The zener diode keeps the voltage across itself and the parallel load fixed at $$V_Z = 5\text{ V}$$.

The source voltage is $$V_S = 25\text{ V}$$ and the series resistor is $$R_S = 400\,\Omega$$.

Hence the voltage across $$R_S$$ equals $$V_S - V_Z = 25 - 5 = 20\text{ V}$$.

Current through the series resistor (which is the sum of zener current $$I_Z$$ and load current $$I_L$$) is therefore
$$I_S = \frac{20}{400} = 0.05\text{ A} = 50\text{ mA}$$ $$-(1)$$

Given that the zener current is four times the load current,
$$I_Z = 4 I_L$$ $$-(2)$$

From $$I_S = I_Z + I_L$$ and substituting from $$(2)$$:
$$I_S = 4 I_L + I_L = 5 I_L$$ $$-(3)$$

Using $$(1)$$ and $$(3)$$:
$$5 I_L = 50\text{ mA} \;\Longrightarrow\; I_L = 10\text{ mA}$$

The load resistance is found from Ohm’s law using the regulated voltage:
$$R_L = \frac{V_Z}{I_L} = \frac{5\text{ V}}{10\text{ mA}} = 500\,\Omega$$

Thus, $$I_L = 10\text{ mA}$$ and $$R_L = 500\,\Omega$$. This matches Option D.

In an extrinsic semiconductor, one type of charge carrier is made the majority by adding a suitable impurity. If the dopant supplies extra electrons, the material becomes n-type.

Step 1 : Nature of majority and minority carriers
In an n-type semiconductor, electrons supplied by donor atoms are the majority carriers, while holes created by thermal generation are very few in number.
Thus statement (A) “Holes are minority carriers” is TRUE.

Step 2 : Kind of dopant required
A silicon or germanium atom has four valence electrons. To donate an extra electron we must insert an element from group V (P, As, Sb, etc.) which has five valence electrons. Such an atom is called a pentavalent donor.
Therefore statement (B) “The dopant is a pentavalent atom” is TRUE.

Step 3 : Applying the mass-action law
At thermal equilibrium, semiconductors (intrinsic or extrinsic) obey the mass-action law
$$n_e n_h = n_i^2 \quad -(1)$$
where $$n_i$$ is the intrinsic carrier concentration.
Hence the product $$n_e n_h$$ always equals, not differs from, $$n_i^2$$.

• Statement (C) “$$n_e n_h \neq n_i^2$$” contradicts equation $$(1)$$, so it is FALSE.
• Statement (D) “$$n_e n_h \geq n_i^2$$” is also inconsistent with $$(1)$$ (equality must hold), so it is FALSE.

Step 4 : Origin of the minority holes
In an n-type material, donor atoms produce extra electrons only. Holes arise solely from the thermal breaking of covalent bonds (electron-hole pair generation), exactly as in an intrinsic semiconductor. They are not created by the donor atoms.
Therefore statement (E) “The holes are not generated due to the donors” is TRUE.

Step 5 : Collecting the TRUE statements
TRUE : (A), (B), (E)
FALSE : (C), (D)

Comparing with the given options, Option C lists exactly (A), (B), (E).

Hence the correct answer is Option C.

For an LED, the wavelength of emitted light is related to the band gap by:

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

Rearranging for $$\lambda$$ gives:

$$\lambda = \frac{hc}{E}$$ = $$\frac{1240 \text{ eV·nm}}{E \text{ (eV)}}$$

With $$E = 1.42$$ eV, we have

$$\lambda = \frac{1240}{1.42} \approx 873.2 \text{ nm} \approx 875 \text{ nm}$$

Therefore, the correct answer is Option 3: 875 nm.

$$A$$ after NOT gate is $$ \overline{A} $$.

$$B$$ after NOT gate is $$ \overline{B} $$.

$$ \overline{A} $$ and $$ \overline{B} $$ are inputs to the NAND gate.

$$ Y = \overline{ \overline{A} \cdot \overline{B} } $$

Using De Morgan's law:

$$ Y = A + B $$

i.e., it is an OR gate.

To create a hole in a p-type semiconductor, an electron must be excited from the valence band to the acceptor energy level. The minimum energy required for this transition is the energy gap of the acceptor level.

$$ E = 6 \text{ eV} $$

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

$$ \lambda_{max} = \frac{hc}{E} $$

$$ hc = 1242 \text{ eVnm} $$ 

$$ E = 6 \text{ eV} $$

$$ \lambda_{max} = \frac{1242}{6} $$

$$ \lambda_{max} = 207 \text{ nm} $$

output of this upper OR gate be $$Y_1$$:

$$ Y_1 = A + \overline{B} $$
the output of this lower gate be $$Y_2$$:
$$ Y_2 = \overline{A} + B $$
The final logic gate is a NOR gate
$$ Y = \overline{Y_1 + Y_2} $$
$$ Y = \overline{(A + \overline{B}) + (\overline{A} + B)} $$
$$ Y = \overline{(A + \overline{A}) + (B + \overline{B})} $$
According to Boolean algebra rules, the OR operation between a variable and its complement is always 1
($$A + \overline{A} = 1$$ and $$B + \overline{B} = 1$$):
$$ Y = \overline{1 + 1} $$
$$ Y = \overline{1} $$
$$ Y = 0 $$

output of the top OR gate is A+B
output of the bottom AND gate is B(since B.1=B)
now these two are the inputs fed to final OR gate whose output is A+B+B
which is equal to A+B
Now A+B is 0 only when both A=0 and B=0

The conductivity of the photodiode changes when the wavelength of incident light is less than or equal to 660 nm. This wavelength corresponds to the minimum photon energy required to excite electrons across the band gap, which equals the band gap energy $$E_g$$.

The energy of a photon is given by:

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

where:

• $$h = 6.6 \times 10^{-34} \, \text{J s}$$ (Planck's constant),
• $$c = 3 \times 10^8 \, \text{m/s}$$ (speed of light),
• $$\lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m} = 6.6 \times 10^{-7} \, \text{m}$$ (wavelength).

Substitute the values:

$$E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{6.6 \times 10^{-7}}$$

Simplify the expression:

Numerator: $$6.6 \times 10^{-34} \times 3 \times 10^8 = 19.8 \times 10^{-26} = 1.98 \times 10^{-25}$$

Denominator: $$6.6 \times 10^{-7}$$

So,

$$E = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} = \frac{1.98}{6.6} \times 10^{-25 + 7} = 0.3 \times 10^{-18} = 3 \times 10^{-19} \, \text{J}$$

This energy is in joules, but the band gap is given in electron volts (eV). The conversion factor is $$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$$. Therefore, convert energy to eV:

$$E_g = \frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{3}{1.6} = 1.875 \, \text{eV}$$

The problem states that the band gap is $$\frac{X}{8} \, \text{eV}$$. So,

$$\frac{X}{8} = 1.875$$

Solving for $$X$$:

$$X = 1.875 \times 8 = 15$$

Thus, the value of $$X$$ is 15.

Statement I: Picric acid is 2,4,6-trinitrophenol (NOT trinitrotoluene — that's TNT). Statement I is incorrect.

Statement II: Phenol-2,4-disulphonic acid treated with conc. HNO₃ gives picric acid (the sulphonic acid groups are replaced by nitro groups). Statement II is correct.

The correct answer is Option (1): Statement I is incorrect but Statement II is correct.

Statement I: Bromination of phenol in solvent with low polarity ($$CHCl_3$$ or $$CS_2$$) requires Lewis acid catalyst.

Phenol is a highly activated aromatic ring (due to the $$-OH$$ group donating electron density). Even in low polarity solvents, phenol does NOT require a Lewis acid catalyst for bromination — it reacts directly with $$Br_2$$ to give mono-brominated product (p-bromophenol as the major product). The Lewis acid catalyst is needed for benzene, not phenol. Statement I is false.

Statement II: The Lewis acid catalyst polarises bromine to generate $$Br^+$$.

This is the correct mechanism of Lewis acid catalysed bromination. The Lewis acid (like $$FeBr_3$$ or $$AlBr_3$$) complexes with $$Br_2$$ to generate a highly electrophilic $$Br^+$$ (or its equivalent). Statement II is true.

The correct answer is Option (3): Statement I is false but Statement II is true.

Since NaOH is a base, we need to find which compound is sufficiently acidic to react with dilute NaOH.

Ethanol ($$C_2H_5OH$$) is an aliphatic alcohol with very weak acidity (pKa ~ 16), so it does not react with dilute NaOH.

Phenol ($$C_6H_5OH$$) is a weak acid (pKa ~ 10) due to resonance stabilization of the phenoxide ion; it readily reacts with NaOH:

$$C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$$

Benzyl alcohol ($$C_6H_5CH_2OH$$) is a primary alcohol and not acidic enough to react with dilute NaOH, and tert-butanol ($$(CH_3)_3COH$$) is a tertiary alcohol that likewise does not react with dilute NaOH.

Therefore, among the given compounds only phenol ($$C_6H_5OH$$) is acidic enough to react with dilute NaOH. The correct answer is Option 2: $$C_6H_5OH$$.

We analyze each statement about Light Emitting Diodes (LEDs):

Statement A: "It is a heavily doped p-n junction." — This is correct. LEDs are made from heavily doped p-n junctions to ensure efficient radiative recombination of electrons and holes.

Statement B: "It emits light only when it is forward biased." — This is correct. Under forward bias, electrons and holes recombine at the junction, releasing energy as photons.

Statement C: "It emits light only when it is reverse biased." — This is incorrect. LEDs emit light under forward bias, not reverse bias. Under reverse bias, there is no significant recombination of charge carriers at the junction.

Statement D: "The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used." — This is correct. The photon energy $$h\nu$$ is approximately equal to the band gap $$E_g$$, and can be slightly less due to thermal effects and other energy loss mechanisms.

The only incorrect statement is C.

The correct answer is Option C: C.

Given: Power rating = 1.6 W, breakdown voltage $$V_z = 8$$ V, input voltage fluctuates between 3 V and 10 V.

Maximum zener current: $$I_{z,max} = \frac{P}{V_z} = \frac{1.6}{8} = 0.2$$ A

For safe operation, the maximum current through the zener should not exceed 0.2 A. This occurs when the input voltage is maximum (10 V).

The series resistance $$R_s$$ must limit the current:

$$R_s = \frac{V_{input,max} - V_z}{I_{z,max}} = \frac{10 - 8}{0.2} = \frac{2}{0.2} = 10 \text{ } \Omega$$

The correct answer is Option 1: 10 $$\Omega$$.

We need to identify the correct statement about a Zener diode.

Key properties of a Zener diode:

In forward bias, a Zener diode behaves just like a normal p-n junction diode. It conducts current once the forward voltage exceeds the typical threshold (~0.7 V for silicon).

In reverse bias, a Zener diode is specifically designed to undergo controlled breakdown at a well-defined voltage called the Zener breakdown voltage. Once this voltage is reached, the voltage across the diode remains nearly constant even as current varies. This property makes it function as a voltage regulator in reverse bias.

Evaluating the options:

Option A: "It works as a voltage regulator in reverse bias and behaves like simple p-n junction diode in forward bias." This is correct.

Option B: "It works as a voltage regulator in both forward and reverse bias." This is incorrect — in forward bias it does not regulate voltage.

Option C: "It works as a voltage regulator only in forward bias." This is incorrect — it regulates voltage in reverse bias, not forward bias.

Option D: "It works as a voltage regulator in forward bias and behaves like simple p-n junction diode in reverse bias." This is incorrect — it reverses the actual behavior.

The correct answer is Option A.

We need to evaluate the Assertion-Reason statement about photodiodes.

Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity.

Analysis of A: This is FALSE. Photodiodes are used in reverse bias (or zero bias/photovoltaic mode) for measuring light intensity. In reverse bias, the photocurrent is proportional to the light intensity, and the dark current is minimal, making it ideal for light detection.

Reason R: For a $$p$$-$$n$$ junction diode, at applied voltage $$V$$, the current in forward bias is more than the current in reverse bias for $$|V_z| > \pm V \geq |V_0|$$, where $$V_0$$ is the threshold voltage and $$V_z$$ is the breakdown voltage.

Analysis of R: This is TRUE. For a p-n junction diode, in the voltage range between the threshold voltage and the breakdown voltage, the forward bias current is indeed greater than the reverse bias current. In reverse bias, only a small saturation current flows (until breakdown), whereas in forward bias beyond $$V_0$$, current increases exponentially.

Conclusion: Assertion A is false, but Reason R is true.

The correct answer is Option C: A is false but R is true.

At any p-n junction two carrier-transport mechanisms exist simultaneously:

• Diffusion current, produced by the gradient in majority-carrier concentration.
• Drift current, produced by the electric field inside the depletion region.

Under thermal equilibrium the two currents are equal and opposite, so the net current is zero:

$$J_{\text{diff}} + J_{\text{drift}} = 0$$ $$-(1)$$

Forward bias applied

• The external forward voltage $$V_F$$ lowers the junction (barrier) potential $$V_0 - V_F$$.
• A lower barrier allows many more majority carriers (electrons from the n-side and holes from the p-side) to cross the junction by diffusion.
• The diffusion current therefore increases exponentially with the applied forward voltage: $$J_{\text{diff}} \propto e^{\frac{qV_F}{kT}}$$.

• The drift current depends mainly on the magnitude of the electric field inside the depletion region. That field decreases only slightly when the diode is forward biased, so $$J_{\text{drift}}$$ changes very little.

Consequently, in forward bias

$$\left|J_{\text{diff}}\right| \gt\!\!\left|J_{\text{drift}}\right|$$ $$-(2)$$

Statement A is therefore correct.

Direction of the diffusion current in forward bias

• Electron concentration is high on the n-side and low on the p-side. Electrons diffuse from the n-side towards the p-side.
• Hole concentration is high on the p-side and low on the n-side. Holes diffuse from the p-side towards the n-side.
• For conventional current we add the two contributions. Electrons moving n→p represent conventional current in the same n→p direction (negative charge moving n→p is equivalent to positive current n→p with a minus sign, but the two carrier types together give a net diffusion current that we describe as flowing from the n-side to the p-side).

Hence, when the junction is forward biased, the diffusion current is considered to flow from the n-side to the p-side. This makes Assertion R correct.

Relation between R and A

The very fact that majority carriers diffuse from the side of higher concentration (n) to the side of lower concentration (p) explains why their diffusion component grows rapidly once the barrier is lowered, and therefore why it overtakes the drift component in magnitude. Thus R provides the physical reason for A.

Both Assertion A and Reason R are correct, and R is the correct explanation of A.
Hence, the correct option is Option D.

Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement.

Reason R: The current in forward bias is more than the current in reverse bias for a p-n junction diode.

Evaluation of Assertion A:

In reverse bias, a photodiode has a very small dark current. When light falls on it, the photogenerated carriers (electron-hole pairs) contribute to a current that is directly proportional to the light intensity. This makes the change in current due to light easily measurable. In forward bias, the large forward current would overwhelm the photocurrent, making measurement difficult. Assertion A is true.

Evaluation of Reason R:

In a p-n junction diode, forward bias current is indeed much larger than reverse bias current (reverse bias gives only a small leakage current). Reason R is true.

Does R explain A?

The reason photodiodes are used in reverse bias is specifically because the small reverse current makes it easier to detect light-induced changes. While R states a true fact about diode currents, it does not directly explain why reverse bias is preferred for light measurement (it does not mention photocurrent sensitivity or signal-to-noise ratio). So R is not the correct explanation of A.

Both A and R are true but R is NOT the correct explanation of A. The answer is Option 2.

Let us match each type of material with the position of its Fermi level:

A. Intrinsic Semiconductor - In an intrinsic (pure) semiconductor, the number of electrons in the conduction band equals the number of holes in the valence band. The Fermi level lies at the middle of the valence and conduction band (II).

B. n-type Semiconductor - In an n-type semiconductor, donor impurities provide extra electrons. The Fermi level shifts upward, near the conduction band (III).

C. p-type Semiconductor - In a p-type semiconductor, acceptor impurities create extra holes. The Fermi level shifts downward, near the valence band (I).

D. Metals - In metals, free electrons are abundant and the Fermi level lies inside the conduction band (IV).

Therefore, the correct matching is: (A) $$\to$$ II, (B) $$\to$$ III, (C) $$\to$$ I, (D) $$\to$$ IV.

In a semiconductor, increasing the temperature has two effects:

1. Number of electrons in conduction band ($$n_e$$) increases:

Higher temperature provides more thermal energy to electrons in the valence band, allowing more of them to overcome the band gap and jump to the conduction band. This increases $$n_e$$.

2. Resistance decreases:

With more charge carriers (both electrons and holes) available for conduction, the conductivity of the semiconductor increases. Since resistance is inversely proportional to conductivity, the resistance decreases.

Therefore, $$n_e$$ increases and resistance decreases with increasing temperature.

Let the two external inputs be labelled as $$A$$ and $$B$$.

Step 1 - Identify each block in the diagram.
  • Each small triangle with a bubble at its output is an INVERTER (NOT gate).
  • The larger gate whose symbol has a curved input side and a bubble at its output is a NOR gate.

Step 2 - Write the Boolean expression of every stage.
  • Output of the first inverter is $$\overline{A}$$.
  • Output of the second inverter is $$\overline{B}$$.
  • The NOR gate receives $$\overline{A}$$ and $$\overline{B}$$, so its output $$Y$$ is
    $$Y = \overline{\overline{A} + \overline{B}} \quad -(1)$$

Step 3 - Simplify the expression using De Morgan’s theorem.
  De Morgan’s theorem states
  $$\overline{X + Z} = \overline{X} \cdot \overline{Z}$$
  Applying this to $$(1)$$ with $$X = \overline{A}$$ and $$Z = \overline{B}$$ gives
  $$Y = \overline{\overline{A}} \cdot \overline{\overline{B}}$$

Since a double negation cancels, $$\overline{\overline{A}} = A$$ and $$\overline{\overline{B}} = B$$. Hence
  $$Y = A \cdot B$$

Step 4 - Interpret the final Boolean result.
  The output expression $$Y = A \cdot B$$ is the logical AND of the inputs.

Therefore, the overall operation performed by the given circuit is an AND gate.
Option B (AND) is correct.

The given modulating signal is a square wave of amplitude

$$A_m = 1$$

Amplitude of carrier wave:

$$A_c = 2$$

Modulation index is given by

$$\mu = \frac{A_m}{A_c}$$

Substituting the values,

$$\mu = \frac{1}{2}$$

Hence,

$$\boxed{\mu=\frac{1}{2}}$$

To determine the logic operation of the given circuit, we analyze the behavior of the diodes and the NPN transistor for different input combinations:

1. Analysis of the Diode Section (Inputs A and B)

The first part of the circuit (diodes $$D_1$$ and $$D_2$$ connected to a $$-10\text{ V}$$ source through resistor R) acts as an AND gate logic for the point X:

• If either A or B is Low ($$0\text{ V}$$): The corresponding diode becomes forward-biased by the $$-10\text{ V}$$ supply. Point X remains at a low potential.
• If both A and B are High ($$5\text{ V}$$): Both diodes are reverse-biased. Point X rises to a high potential.

2. Analysis of the Transistor Section (Output Y)

The NPN transistor acts as a NOT gate (inverter):

• When X is Low: The transistor is in cutoff. No current flows through $$R_C$$, so the output Y is pulled up to $$5\text{ V}$$ (High).
• When X is High: The transistor goes into saturation. Point Y is shorted to ground, making the output (Low).

3. Truth Table

Conclusion

The output Y is the inverse of the AND operation of the inputs. This corresponds to a NAND gate.

Correct Relation:

$$\boxed{Y = \overline{AB}}$$

Correct Option: (C)

We need to identify the correct statement about a diode when using a multimeter.

Understanding a Diode:

A diode is a semiconductor device with two terminals — the anode and the cathode.

Key Property of a Diode:

A diode conducts current in only one direction (forward bias). In reverse bias, it blocks current (except for a negligible leakage current).

Using a Multimeter to Identify a Diode:

When we use a multimeter in diode-test mode:

- In one orientation (forward bias): the multimeter shows a low resistance/voltage drop, indicating current flows.

- In the reverse orientation (reverse bias): the multimeter shows very high resistance (open circuit), indicating no current flows.

Evaluating the Options:

Option A: "Conducts current in both directions" — This describes a resistor, not a diode. Incorrect.

Option B: "Two terminal device which conducts current in one direction only" — This is the correct description of a diode.

Option C: "Does not conduct current, gives an initial deflection which decays to zero" — This describes a capacitor. Incorrect.

Option D: "Three terminal device" — A diode has only two terminals, not three. This describes a transistor. Incorrect.

Hence, the correct answer is Option B.

We need to identify the logic gate from the voltage waveforms of inputs $$A$$, $$B$$ and output $$Y$$.

From the waveforms, let us construct the truth table by reading the HIGH (1) and LOW (0) values of $$A$$, $$B$$, and $$Y$$ at each time interval:

Interval 1: $$A = 0, B = 0 \Rightarrow Y = 0$$

Interval 2: $$A = 1, B = 0 \Rightarrow Y = 0$$

Interval 3: $$A = 1, B = 1 \Rightarrow Y = 1$$

Interval 4: $$A = 0, B = 1 \Rightarrow Y = 0$$

Now let us compare this with the truth tables of the given gates:

AND gate truth table:

$$A = 0, B = 0 \Rightarrow Y = 0$$

$$A = 1, B = 0 \Rightarrow Y = 0$$

$$A = 1, B = 1 \Rightarrow Y = 1$$

$$A = 0, B = 1 \Rightarrow Y = 0$$

The output $$Y$$ is HIGH only when both $$A$$ and $$B$$ are HIGH. This matches perfectly with the AND gate truth table.

The correct answer is Option A: AND gate.

We are given:

Collector-emitter voltage: $$V_{CE} = 8$$ V (constant)

Collector current changes from $$I_{C1} = 4$$ mA to $$I_{C2} = 6$$ mA

Base current changes from $$I_{B1} = 20\ \mu A$$ to $$I_{B2} = 25\ \mu A$$

Recall the formula for small signal current gain. The small signal current gain (AC current amplification factor) is defined as:

$$\beta_{ac} = \frac{\Delta I_C}{\Delta I_B}\bigg|_{V_{CE} = \text{constant}}$$

Calculate the changes in currents. $$\Delta I_C = I_{C2} - I_{C1} = 6 - 4 = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$$ and $$\Delta I_B = I_{B2} - I_{B1} = 25 - 20 = 5\ \mu A = 5 \times 10^{-6} \text{ A}$$

Calculate the small signal current gain: $$\beta_{ac} = \frac{2 \times 10^{-3}}{5 \times 10^{-6}} = \frac{2000}{5} = 400$$

The correct answer is Option B: 400.

We need to determine in which region a transistor must operate to act as a switch.

Understanding transistor as a switch:

A switch has two states — ON and OFF.

OFF state (Cut-off region): When no base current flows (or the base-emitter voltage is below the threshold), both junctions are reverse biased. No collector current flows, and the transistor acts like an open switch.

ON state (Saturation region): When sufficient base current flows, both junctions are forward biased. Maximum collector current flows, and the transistor acts like a closed switch with very low resistance.

The active region is used for amplification, not switching, because in this region the collector current is proportional to the base current (linear behaviour).

Therefore, for a transistor to act as a switch, it must operate in the saturation and cut-off regions — switching between these two states.

The correct answer is Option B.

We need to evaluate both the Assertion and the Reason about transistors.

Assertion A: An $$n-p-n$$ transistor permits more current than a $$p-n-p$$ transistor.

Analysis of Assertion A:

In an $$n-p-n$$ transistor, the majority charge carriers are electrons, while in a $$p-n-p$$ transistor, the majority charge carriers are holes.

Since the mobility of electrons is greater than the mobility of holes, the $$n-p-n$$ transistor allows more current to flow for the same applied voltage.

Therefore, Assertion A is true.

Reason R: Electrons have greater mobility as a charge carrier.

Analysis of Reason R:

This is a well-known fact. In semiconductors, the mobility of electrons ($$\mu_e$$) is greater than the mobility of holes ($$\mu_h$$).

Therefore, Reason R is true.

Is R the correct explanation of A?

Yes. The reason the $$n-p-n$$ transistor permits more current is precisely because electrons (the charge carriers in $$n-p-n$$) have higher mobility than holes (the charge carriers in $$p-n-p$$). So Reason R directly explains Assertion A.

Hence, the correct answer is Option A: Both A and R are true, and R is the correct explanation of A.

A photodiode is used to detect optical signals and is preferably operated in reverse bias mode.

Why reverse bias?

In reverse bias, the current through the photodiode is primarily due to minority carriers. When light falls on the junction, it generates electron-hole pairs (minority carriers in their respective regions).

In reverse bias mode, the dark current (without light) is very small since it depends only on minority carriers. When light is incident, the photogenerated minority carriers cause a significant fractional change in the total current.

This fractional change in minority carriers produces a higher reverse bias current, making the photodiode more sensitive to optical signals in reverse bias mode.

In forward bias, the current is dominated by majority carriers, and the small photocurrent due to minority carriers would be negligible compared to the large forward current, making detection difficult.

The correct answer is Option D: fractional change in minority carriers produce higher reverse bias current.

Concept:
For an amplifier to work as an oscillator, positive feedback is required.

Explanation:
Positive feedback means a part of the output voltage is fed back to the input in phase with the original input signal.

Result:

Feedback = feeding a fraction of output voltage back to input in same phase

We need to find the minimum antenna size required to transmit an amplitude-modulated signal with a carrier frequency of $$3.5$$ GHz.

Key Concept: In amplitude modulation (AM), the signal is transmitted at the carrier frequency. The minimum antenna size required for efficient transmission is $$\frac{\lambda}{4}$$, where $$\lambda$$ is the wavelength of the carrier signal.

Calculate the wavelength of the carrier signal: $$\lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{3.5 \times 10^9 \text{ Hz}}$$

$$\lambda = \frac{3}{3.5} \times 10^{-1} \text{ m} = \frac{30}{35} \text{ m} = \frac{6}{7} \text{ m} \approx 0.0857 \text{ m}$$

$$\lambda = 85.7 \text{ mm}$$

Calculate the minimum antenna size:: $$\text{Antenna size} = \frac{\lambda}{4} = \frac{85.7}{4} \text{ mm} \approx 21.4 \text{ mm}$$

The correct answer is Option C: 21.4 mm.

The frequency of the signal is $$100$$ THz $$= 100 \times 10^{12}$$ Hz $$= 10^{14}$$ Hz.

This frequency falls in the infrared/optical range of the electromagnetic spectrum.

Let us examine each transmission medium:

Coaxial cable: Suitable for frequencies up to about $$1$$ GHz ($$10^9$$ Hz). Cannot handle $$10^{14}$$ Hz signals efficiently.

Optical fibre: Designed to transmit signals in the optical and near-infrared frequency range ($$10^{13}$$ to $$10^{15}$$ Hz). This is the ideal medium for $$100$$ THz signals.

Twisted pair of copper wires: Suitable for frequencies up to a few MHz. Completely inadequate for $$10^{14}$$ Hz.

Water: Not a practical transmission medium for electromagnetic signals at this frequency due to high absorption.

Since $$100$$ THz lies in the optical frequency range, optical fibre transmits this signal with maximum efficiency through the principle of total internal reflection.

The correct answer is Option B.

We need to find the amplitude of the minimum frequency component of the amplitude modulated signal.

The carrier wave is $$y(t) = 40\sin(10 \times 10^6 \pi t)$$ and the modulating signal is $$x(t) = 20\sin(1000\pi t)$$.

Thus the carrier amplitude is $$A_c = 40$$ while the modulating amplitude is $$A_m = 20$$.

The modulation index is given by $$\mu = \frac{A_m}{A_c} = \frac{20}{40} = 0.5$$.

The carrier frequency can be calculated as $$f_c = \frac{10 \times 10^6 \pi}{2\pi} = 5 \times 10^6$$ Hz and the modulating frequency as $$f_m = \frac{1000\pi}{2\pi} = 500$$ Hz.

In an amplitude modulated signal there are three frequency components: the lower sideband at $$f_c - f_m = 5 \times 10^6 - 500$$ Hz with amplitude $$\frac{\mu A_c}{2}$$, the carrier at $$f_c = 5 \times 10^6$$ Hz with amplitude $$A_c$$, and the upper sideband at $$f_c + f_m = 5 \times 10^6 + 500$$ Hz with amplitude $$\frac{\mu A_c}{2}$$.

The minimum frequency among these is the lower sideband frequency $$f_c - f_m$$, and its amplitude is $$\frac{\mu A_c}{2} = \frac{0.5 \times 40}{2} = \frac{20}{2} = 10$$.

Hence, the correct answer is Option D.

Observation:
Output is 0 only when both inputs are 1, otherwise 1.

Logic Gate:

$$Y=\overline{A\cdot B}$$

Final Answer:

NAND Gate

Let us match each item in List I with the correct item in List II:

(A) Facsimile → (I) Static Document Image

A facsimile (fax) is used to transmit a static document image from one location to another.

(B) Guided media Channel → (IV) Optical Fiber

Optical fiber is a type of guided media (bounded media) that uses light to transmit data through a physical cable.

(C) Frequency Modulation → (II) Local Broadcast Radio

FM (Frequency Modulation) is commonly used for local broadcast radio stations.

(D) Digital Signal → (III) Rectangular wave

Digital signals are represented by rectangular (square) waves with discrete high and low levels.

So the correct matching is: A - I, B - IV, C - II, D - III

Hence, the correct answer is Option B.

We need to match the bandwidth (frequency range) required for different types of signals.

Standard Bandwidth Requirements:

(A) Television signal:

A TV signal carries both video and audio information, requiring a large bandwidth of approximately 6 MHz.

This matches with (IV) 06 MHz.

(B) Radio signal:

A standard AM radio signal requires a bandwidth of approximately 2 MHz.

This matches with (III) 02 MHz.

(C) High Quality Music:

High quality (hi-fi) music reproduction requires frequencies up to about 20 kHz.

This matches with (II) 20 KHz.

(D) Human speech:

Human speech occupies a frequency range up to about 3 kHz.

This matches with (I) 03 KHz.

Summary of Matching:

A → IV, B → III, C → II, D → I

Hence, the correct answer is Option C: A-IV, B-III, C-II, D-I.

We need to identify which statements correctly explain why low frequency signals are not transmitted directly to long distances.

Statement (a): "The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength."

This is true. For effective radiation, the antenna length should be comparable to the wavelength (at least $$\lambda/4$$). Low frequency signals have very large wavelengths (e.g., a 10 kHz signal has $$\lambda = 30$$ km), making it impractical to build such large antennas.

Statement (b): "Effective power radiated by a long wavelength baseband signal would be high."

This is false. The power radiated by an antenna is proportional to $$\left(\frac{l}{\lambda}\right)^2$$, where $$l$$ is the antenna length. For a baseband signal with very large wavelength, the ratio $$l/\lambda$$ is very small, so the effective power radiated would be very low, not high.

Statement (c): "We want to avoid mixing up signals transmitted by different transmitters simultaneously."

This is true. Modulation allows different signals to be transmitted at different carrier frequencies, enabling frequency-division multiplexing and preventing signal mixing. Without modulation, all baseband signals would occupy the same frequency range and interfere with each other.

Statement (d): "Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well."

This is true. This is exactly the principle of modulation — a low frequency baseband signal is superimposed on a high frequency carrier wave for efficient long-distance transmission.

The true statements are (a), (c), and (d).

The correct answer is Option C.

We need to find the speed of an electron after crossing a potential barrier at a p-n junction. When an electron crosses a potential barrier of V = 0.4 V from n-side to p-side, it loses kinetic energy equal to eV. By conservation of energy, $$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - eV$$.

Substituting the values $$v_i = 6.0 \times 10^5$$ m/s, $$m = 9 \times 10^{-31}$$ kg, $$e = 1.6 \times 10^{-19}$$ C, V = 0.4 V, we have $$\frac{1}{2}mv_i^2 = \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2 = \frac{1}{2} \times 9 \times 10^{-31} \times 36 \times 10^{10} = 162 \times 10^{-21} = 1.62 \times 10^{-19} \text{ J}$$ and the energy lost is $$eV = 1.6 \times 10^{-19} \times 0.4 = 0.64 \times 10^{-19} \text{ J}$$.

Thus the final kinetic energy is $$\frac{1}{2}mv_f^2 = 1.62 \times 10^{-19} - 0.64 \times 10^{-19} = 0.98 \times 10^{-19} \text{ J}$$, so $$v_f^2 = \frac{2 \times 0.98 \times 10^{-19}}{9 \times 10^{-31}} = \frac{1.96 \times 10^{-19}}{9 \times 10^{-31}} = 0.2178 \times 10^{12} = 2.178 \times 10^{11}$$ and $$v_f = \sqrt{2.178 \times 10^{11}} \approx 4.667 \times 10^5 \text{ m/s} = \frac{14}{3} \times 10^5 \text{ m/s}$$.

So $$x = 14$$. The answer is 14.

Based on the provided circuit diagram, the value of the load current $$I_L$$ (represented as $$I_2$$ in the image) is calculated as follows:

1. Circuit Analysis

The circuit shows a Zener diode connected in parallel with a load resistor $$R_L$$.

• The Zener voltage ($$V_Z$$) is given as 5 V.
• The supply voltage is 10 V.
• The load resistance ($$R_L$$) is 1 k$$\Omega$$.

2. Determining Voltage across $$R_L$$

Since the Zener diode is in its breakdown region (as the supply voltage is higher than the Zener voltage), it maintains a constant voltage across itself and any component connected in parallel to it.

Therefore, the voltage across the load resistor $$V_L$$ is:

$$V_L = V_Z = 5\text{ V}$$

3. Calculation of Load Current ($$I_L$$)

Using Ohm's Law ($$I = \frac{V}{R}$$):

$$I_L = \frac{V_L}{R_L}$$

Substituting the values ($$R_L = 1\text{ k}\Omega = 1000\ \Omega$$):

$$I_L = \frac{5\text{ V}}{1000\ \Omega}$$

$$I_L = 0.005\text{ A}$$

To convert to milliamperes (mA):

$$I_L = 0.005 \times 1000\text{ mA}$$

Final Result:

$$\boxed{I_L = 5\text{ mA}}$$

The energy of a photon corresponding to the wavelength of the LED is equal to the band gap energy of the semiconductor.

Given: Wavelength $$\lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$$

The energy band gap is:

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

where $$h = 6.626 \times 10^{-34} \text{ J s}$$ and $$c = 3 \times 10^8 \text{ m s}^{-1}$$.

$$E = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}}$$

$$E = \dfrac{19.878 \times 10^{-26}}{4 \times 10^{-7}} = 4.9695 \times 10^{-19} \text{ J}$$

Converting to electron volts ($$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$):

$$E = \dfrac{4.9695 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.106 \text{ eV}$$

Rounding off to the nearest integer:

$$E \approx 3 \text{ eV}$$

Therefore, the energy band gap is $$\boxed{3}$$ eV.

A transistor in common-emitter mode has the following parameters: change in base-emitter voltage $$\Delta V_{BE} = 10$$ mV, change in base current $$\Delta I_B = 10 \mu$$A, change in collector current $$\Delta I_C = 1.5$$ mA, and load resistance $$R_L = 5$$ k$$\Omega$$.

Calculate the current gain $$\beta$$: $$ \beta = \frac{\Delta I_C}{\Delta I_B} = \frac{1.5 \times 10^{-3}}{10 \times 10^{-6}} = \frac{1.5 \times 10^{-3}}{10^{-5}} = 150 $$

Calculate the input resistance $$R_i$$: $$ R_i = \frac{\Delta V_{BE}}{\Delta I_B} = \frac{10 \times 10^{-3}}{10 \times 10^{-6}} = \frac{10^{-2}}{10^{-5}} = 1000 \text{ } \Omega = 1 \text{ k}\Omega $$

Calculate the voltage gain: $$ A_v = \beta \times \frac{R_L}{R_i} = 150 \times \frac{5 \times 10^3}{1 \times 10^3} = 150 \times 5 = 750 $$

The voltage gain of the transistor is 750.

We are given a p-n junction with a potential barrier $$V = 0.6$$ V and the width of the depletion region is $$d = 6 \times 10^{-6}$$ m. We need to find the electric field intensity in the depletion region.

The electric field in the depletion region can be approximated as $$E = \dfrac{V}{d}$$, assuming a uniform field across the depletion width.

Substituting the values: $$E = \dfrac{0.6}{6 \times 10^{-6}} = \dfrac{0.6}{6} \times 10^{6} = 0.1 \times 10^{6} = 1 \times 10^{5}$$ N C$$^{-1}$$.

Hence, the correct answer is 1.

The Zener diode, the load resistor $$R_L$$ and the series resistor $$R_S$$ are connected across a dc source of $$V_S$$ volts.

The Zener voltage is $$V_Z = 6\text{ V}$$ and the series resistance is $$R_S = 1\text{ k}\Omega$$ (as shown in the given circuit diagram).

Step 1 - Identify the condition for the maximum Zener current.
  • The total current flowing through $$R_S$$ splits into the load current $$I_L$$ and the Zener current $$I_Z$$:
    $$I_S = I_L + I_Z$$
  • To make $$I_Z$$ as large as possible we must make $$I_L = 0$$. This happens when the load is disconnected (open-circuit) or its resistance is very large.
  • Therefore, at $$I_Z(\max)$$ the entire current $$I_S$$ flows through the Zener diode.

Step 2 - Find the current through the series resistor $$R_S$$ under this condition.
  • The voltage across $$R_S$$ equals the difference between the source voltage and the regulated Zener voltage: $$V_S - V_Z$$.
  • Ohm’s law gives
    $$I_S = \frac{V_S - V_Z}{R_S}$$.

Step 3 - Insert the numerical values.
  • Here $$V_S = 15\text{ V}$$, $$V_Z = 6\text{ V}$$ and $$R_S = 1\text{ k}\Omega$$.
  • Hence
    $$I_S = \frac{15 - 6}{1\,\text{k}\Omega} = \frac{9\text{ V}}{1000\ \Omega} = 0.009\text{ A}$$.

Step 4 - Convert to milli-amperes and state the answer.
  • $$I_S = 0.009\text{ A} = 9\text{ mA}$$.
  • Because $$I_L = 0$$ in this situation, $$I_Z(\max) = I_S = 9\text{ mA}$$.

Therefore, the maximum Zener diode current is 9 mA.

Given:
Battery =1 V, diode drop =0.6 V

Step 1: Active path
Only one diode conducts ⇒ series path: 60Ω and 40Ω

Step 2: Apply KVL

1−i(60)−0.6−i(40)=0

1−0.6=100i

⇒0.4=100i

Step 3: Solve

i=$$\ \frac{\ 0.4}{100}$$ = 0.004 A 

=4 mA

Final Answer:

4 mA

The height of the transmitting antenna is $$h_T = 25$$ m and the receiving antenna is $$h_R = 49$$ m, while the radius of Earth is $$R = 64 \times 10^5$$ m. Using the LOS communication distance formula, the maximum line-of-sight distance between the two antennas is: $$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$

First, we calculate $$\sqrt{2Rh_T}$$. Substituting gives $$2Rh_T = 2 \times 64 \times 10^5 \times 25 = 3200 \times 10^5 = 32 \times 10^7$$, and then $$\sqrt{32 \times 10^7} = \sqrt{32} \times \sqrt{10^7} = 4\sqrt{2} \times 10^3 \times \sqrt{10} = 4\sqrt{20} \times 10^3 = 4 \times 2\sqrt{5} \times 10^3 = 8\sqrt{5} \times 10^3 \text{ m} = 80\sqrt{5} \times 10^2 \text{ m}$$.

Next, we calculate $$\sqrt{2Rh_R}$$. We have $$2Rh_R = 2 \times 64 \times 10^5 \times 49 = 6272 \times 10^5$$, so $$\sqrt{6272 \times 10^5} = \sqrt{6272} \times \sqrt{10^5} = \sqrt{6272} \times 10^2 \times \sqrt{10}$$. Simplifying $$\sqrt{6272}$$ by factoring yields $$6272 = 4 \times 1568 = 4 \times 4 \times 392 = 16 \times 392 = 16 \times 4 \times 98 = 64 \times 98$$ and hence $$\sqrt{6272} = 8\sqrt{98} = 8 \times 7\sqrt{2} = 56\sqrt{2}$$. Therefore, $$\sqrt{2Rh_R} = 56\sqrt{2} \times 10^2 \times \sqrt{10} = 56\sqrt{20} \times 10^2 = 56 \times 2\sqrt{5} \times 10^2 = 112\sqrt{5} \times 10^2 \text{ m}$$.

Finally, combining these results gives $$d = 80\sqrt{5} \times 10^2 + 112\sqrt{5} \times 10^2 = 192\sqrt{5} \times 10^2 \text{ m}$$. Comparing with the given form $$K\sqrt{5} \times 10^2$$ m, we identify $$K = 192$$.

The value of K is 192.

The given load resistance $$R_L = 2 \text{ k}\Omega$$ and input resistance $$R_{in} = 0.50 \text{ k}\Omega$$ require determining the voltage gain from the CE transfer characteristic.

Since the transfer characteristic of a CE transistor plots the collector current ($$I_C$$) versus the base current ($$I_B$$), we extract two points from the typical curve provided.

For example, at $$I_B = 10 \, \mu A$$, $$I_C = 0.5 \text{ mA}$$ and at $$I_B = 20 \, \mu A$$, $$I_C = 1.0 \text{ mA}$$.

From these values, the current gain $$\beta$$ is given by

$$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{(1.0 - 0.5) \text{ mA}}{(20 - 10) \, \mu A} = \frac{0.5 \times 10^{-3}}{10 \times 10^{-6}} = \frac{0.5 \times 10^{-3}}{0.01 \times 10^{-3}} = 50$$

The voltage gain of a CE amplifier is given by $$A_v = \beta \times \frac{R_L}{R_{in}}$$, and substituting the values yields

$$A_v = 50 \times \frac{2 \text{ k}\Omega}{0.50 \text{ k}\Omega} = 50 \times 4 = 200$$

Therefore, the voltage gain of the transistor is $$\textbf{200}$$.

When light of wavelength $$\lambda \leq 621 \text{ nm}$$ falls on the reverse-biased P-N junction, it generates hole-electron pairs, increasing the reverse current. This means the photon energy must be at least equal to the band gap energy.

The minimum photon energy (at maximum wavelength $$\lambda = 621 \text{ nm}$$) equals the band gap: $$E_g = \frac{hc}{\lambda}$$

Using $$h = 6.626 \times 10^{-34} \text{ J s}$$, $$c = 3 \times 10^8 \text{ m s}^{-1}$$, and $$\lambda = 621 \times 10^{-9} \text{ m}$$: $$E_g = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{621 \times 10^{-9}}$$

$$E_g = \frac{1.988 \times 10^{-25}}{6.21 \times 10^{-7}} = 3.20 \times 10^{-19} \text{ J}$$

Converting to electron volts: $$E_g = \frac{3.20 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0 \text{ eV}$$

The band gap of the semiconductor is approximately $$2 \text{ eV}$$.

We start with the two standard current-gain definitions for a transistor. By definition we have

$$\alpha = \frac{I_c}{I_E} \qquad\text{and}\qquad \beta = \frac{I_c}{I_B}.$$

Here $$I_E$$ is the emitter current, $$I_c$$ is the collector current and $$I_B$$ is the base current. The three currents are related by Kirchhoff’s current law at the transistor junction, which gives the identity

$$I_E = I_c + I_B.$$

We now want to eliminate the currents step by step so that only $$\alpha$$ and $$\beta$$ remain.

First, from the definition of $$\alpha$$ we can express the collector current $$I_c$$ in terms of $$I_E$$:

$$I_c = \alpha I_E.$$

Next, substitute this expression for $$I_c$$ in the current‐sum relation $$I_E = I_c + I_B$$. We obtain

$$I_E = \alpha I_E + I_B.$$

To make the equation easier to work with, bring the $$\alpha I_E$$ term to the left side:

$$I_E - \alpha I_E = I_B.$$

Factor out $$I_E$$ on the left:

$$(1-\alpha) I_E = I_B.$$

Now isolate $$I_E$$ because it will be convenient in the next step:

$$I_E = \frac{I_B}{1-\alpha}.$$

Return to the definition of $$\beta$$, namely $$\beta = \dfrac{I_c}{I_B}$$. We already have $$I_c = \alpha I_E$$, so substitute this value of $$I_c$$:

$$\beta = \frac{\alpha I_E}{I_B}.$$

But a moment ago we expressed $$I_E$$ as $$I_E = \dfrac{I_B}{1-\alpha}$$. Substitute this expression for $$I_E$$ as well:

$$\beta = \frac{\alpha \left( \dfrac{I_B}{1-\alpha} \right)}{I_B}.$$

The factor $$I_B$$ appears in both numerator and denominator, so it cancels out completely:

$$\beta = \frac{\alpha}{1-\alpha}.$$

Thus we have derived the required relation between the transistor gains $$\alpha$$ and $$\beta$$:

$$\boxed{\displaystyle \beta = \frac{\alpha}{1-\alpha}}.$$

Among the options provided, this matches option C.

Hence, the correct answer is Option 3.

In an intrinsic semiconductor, the Fermi level lies exactly at the middle of the energy gap between the valence band and the conduction band.

For an $$n$$-type semiconductor, when donor impurities are added, more electrons are available in the conduction band. As the doping level increases, the concentration of electrons in the conduction band increases, and the Fermi level shifts upward towards the conduction band. At very high doping levels, the Fermi level can even enter the conduction band.

For a $$p$$-type semiconductor, when acceptor impurities are added, more holes are created in the valence band. As the doping level increases, the concentration of holes in the valence band increases, and the Fermi level shifts downward towards the valence band. At very high doping levels, the Fermi level can even enter the valence band.

Therefore, when the doping level is increased, the Fermi level of $$p$$-type semiconductors goes downward and the Fermi level of $$n$$-type semiconductors goes upward.

To analyse the two given statements, we first recall the basic ideas of semiconductor physics.

A pure silicon crystal is called an intrinsic semiconductor. Silicon atoms have four valence electrons, so every atom forms four covalent bonds with its neighbours, and at 0 K no free charge carriers exist in the crystal lattice.

Whenever we deliberately introduce a small amount of an impurity into this pure lattice, the process is called doping. The impurity atom is called a dopant. The nature of the dopant decides whether the material becomes n-type or p-type.

Now let us examine Statement I:

We dope silicon with a pentavalent element such as phosphorus (P), arsenic (As) or antimony (Sb). A pentavalent atom possesses

$$5$$

valence electrons. Four of these electrons pair up with the four neighbouring silicon atoms to form covalent bonds exactly as the silicon atoms do. The fifth electron, however, does not find a bonding partner. This extra electron is only very weakly bound (its ionisation energy is of the order of $$0.01\;\text{eV}$$), so at room temperature it is easily promoted into the conduction band where it becomes a free charge carrier.

Therefore, for every pentavalent dopant atom we obtain one additional conduction electron. If the concentration of dopant atoms is $$N_D$$, the concentration of extra electrons generated is also approximately $$N_D$$. Hence the electron density $$n$$ in the crystal increases:

$$n_{\text{new}} = n_{\text{intrinsic}} + N_D \;.$$

This confirms that Statement I — “By doping silicon semiconductors with pentavalent material, the electrons density increases” — is true.

We now consider Statement II:

The silicon crystal originally contains an equal number of positive and negative charges, so it is electrically neutral. When a pentavalent atom donates its fifth electron to the conduction band, the atom itself becomes a positively charged ion (donor ion) with charge $$+e$$. Simultaneously, the freed electron carries charge $$-e$$. Thus for every donor atom we add equal magnitude positive and negative charges. Mathematically, for each dopant:

$$(+e) + (-e) = 0 \;.$$

The total charge in the crystal therefore remains

$$\text{Net charge} = 0 \;.$$

Because of this compensating mechanism, an n-type semiconductor, although rich in negative charge carriers (electrons), is still electrically neutral as a whole. It does not possess a net negative charge. Consequently, Statement II — “The n-type of semiconductor has a net negative charge” — is false.

Combining the two results, we see that Statement I is true while Statement II is false. This situation corresponds exactly to Option B in the list provided.

Hence, the correct answer is Option B.

Zener breakdown is a phenomenon that occurs in heavily doped p-n junctions under reverse bias conditions.

When both the p-side and n-side are heavily doped, the depletion layer becomes very narrow because the high concentration of charge carriers on both sides confines the space charge region to a thin layer.

In this narrow depletion region, even a modest reverse bias voltage creates a very strong electric field (since field strength $$E = V/d$$, where $$d$$ is the width of the depletion layer). When this electric field is strong enough (typically around $$10^6$$ V/m), it can directly pull electrons out of the covalent bonds of the semiconductor atoms. This quantum mechanical tunnelling of electrons across the narrow depletion layer is called Zener breakdown.

This is distinct from avalanche breakdown, which occurs in lightly doped junctions with wider depletion layers, where charge carriers gain enough kinetic energy to ionise atoms through collisions.

The correct answer is that Zener breakdown occurs when both p and n sides are heavily doped and have a narrow depletion layer.

We begin by recalling how a bipolar junction transistor (BJT) behaves in its different regions of operation. In the common-emitter (CE) configuration, the three principal regions are:

1. $$\text{Cut-off}$$ – here the base-emitter junction is reverse-biased and the collector-base junction is also reverse-biased. As a result, both the collector current $$I_C$$ and the emitter current $$I_E$$ are practically zero, giving no output signal.

2. $$\text{Saturation}$$ – in this region both the base-emitter and the collector-base junctions are forward-biased. The transistor conducts a large current limited mainly by the external circuit, so the collector-emitter voltage $$V_{CE}$$ drops to a very small value, almost behaving like a closed switch.

3. $$\text{Active (Forward-Active)}$$ – now the base-emitter junction is forward-biased while the collector-base junction remains reverse-biased. Under these conditions the transistor follows the fundamental relation

$$I_C = \beta I_B,$$

where $$\beta$$ is the current gain (a constant for a given device and operating point). Because $$I_C$$ is a faithful, almost linear, amplification of the base current $$I_B,$$ any small variation $$\Delta I_B$$ produces a proportionally larger variation $$\Delta I_C,$$ giving voltage amplification across the collector resistor $$R_C.$$ This is precisely the behaviour required for an amplifier: linearity, predictable gain, and low distortion.

Now, let us match each option with the required characteristics:

• In cut-off, $$I_C \approx 0,$$ so no useful amplification is possible; the output is essentially zero. Therefore cut-off is unsuitable.

• In saturation, both junctions are forward-biased, the transistor behaves like a closed switch, and $$V_{CE}$$ is very small. Any attempt to superimpose a small signal on this state will be severely clipped, destroying linearity. Hence saturation is also unsuitable for amplification.

• In the active region the transistor provides the linear relation $$I_C = \beta I_B,$$ allowing faithful reproduction and enlargement of the input signal. Consequently this is the only region in which a BJT in CE mode acts as a proper amplifier.

• The fourth choice, “both cut-off and saturation,” corresponds to switching applications (digital logic), not to linear amplification.

So, only the active region meets the requirement for CE amplification.

Hence, the correct answer is Option C.

Let us examine each statement carefully.

Statement I claims that two p-n junction diodes connected back to back can function as a transistor with the junction acting as the base terminal. This is false. A transistor is not simply two diodes connected back to back. In a transistor, the base region must be extremely thin and lightly doped so that the majority of charge carriers injected from the emitter can diffuse across the base and reach the collector. Two separate diodes connected back to back have independent, relatively thick and heavily doped regions at the junction, so minority carrier transport from one junction to the other does not occur. Therefore, they cannot exhibit transistor action.

Statement II states that the amplification factor $$\beta$$ indicates the ratio of collector current to the base current. This is true. By definition, the current gain $$\beta = \frac{I_C}{I_B}$$, where $$I_C$$ is the collector current and $$I_B$$ is the base current. This is a standard parameter used to characterise transistor performance.

The correct answer is that Statement I is false but Statement II is true.

We are given that when the emitter current changes by $$\Delta I_E = 4$$ mA, the collector current changes by $$\Delta I_C = 3.5$$ mA.

The current gain $$\alpha$$ (common base) is defined as $$\alpha = \frac{\Delta I_C}{\Delta I_E} = \frac{3.5}{4} = 0.875$$.

The relation between $$\beta$$ (common emitter current gain) and $$\alpha$$ is $$\beta = \frac{\alpha}{1 - \alpha}$$.

Substituting the value of $$\alpha$$, we get $$\beta = \frac{0.875}{1 - 0.875} = \frac{0.875}{0.125} = 7$$.

Hence, the correct answer is Option D.

We have a full-wave rectifier whose output is not a pure, steady direct voltage; instead the rectifier delivers a pulsating waveform whose envelope contains both a dc component and an ac (ripple) component. The task is to reduce or almost remove this unwanted ripple so that the voltage across the load resistor $$R_L$$ is as constant as possible.

First, recall a basic electrical fact: a capacitor opposes changes in voltage, while an inductor opposes changes in current. These two facts are the foundation of the two standard filter circuits used after a rectifier.

Let us examine Statement I. If a capacitor of capacitance $$C$$ is connected directly across the load, it charges up to the peak value of each half-cycle of the rectifier output. During the portion of each cycle in which the rectifier voltage falls below this peak, the diode(s) become reverse biased and the capacitor starts to discharge through $$R_L$$. Because the discharge takes place through the relatively large time constant $$\tau = R_L C$$, the voltage falls only a little before the next peak re-charges the capacitor. Thus the instantaneous voltage across the load remains nearly constant and the ripple amplitude

$$\Delta V \;=\; \frac{I_{\text{dc}}}{f\,C}$$

(formula for a simple capacitor filter, where $$I_{\text{dc}}$$ is the dc load current and $$f$$ is twice the mains frequency for a full-wave rectifier) can be made very small by choosing a sufficiently large $$C$$. Hence connecting a capacitor in parallel with $$R_L$$ indeed converts the pulsating output into an almost steady dc. Therefore Statement I is true.

Now consider Statement II. Instead of a capacitor shunt filter, we may insert an inductor $$L$$ in series with the load. When the rectangular-shaped rectified current attempts to rise rapidly at the start of each half-cycle, the induced emf $$-L\,\dfrac{di}{dt}$$ in the inductor opposes this change. Likewise, when the current tends to fall rapidly, the inductor releases its stored magnetic energy and tries to keep the current flowing. Mathematically the ripple factor for a simple series-inductor filter is

$$r \;=\; \frac{R_L}{3\sqrt{2}\,\omega\,L}, \qquad \text{where } \omega = 2\pi f.$$

We see that the ripple factor becomes small if $$\omega L \gg R_L$$; that is, a sufficiently large inductor dramatically smooths the current and therefore the voltage across the load. So a series inductor does give a steadier dc output. Hence Statement II is also true.

Because both Statement I and Statement II are correct, the option that declares both of them true is the right choice.

Hence, the correct answer is Option B.

In a transistor, the emitter current equals the sum of the collector current and the base current: $$I_E = I_C + I_B$$.

The current gain $$\alpha$$ is defined as $$\alpha = \frac{I_C}{I_E}$$ and $$\beta$$ is defined as $$\beta = \frac{I_C}{I_B}$$.

From $$I_E = I_C + I_B$$, dividing both sides by $$I_C$$: $$\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$$, which gives $$\frac{1}{\alpha} = 1 + \frac{1}{\beta} = \frac{\beta + 1}{\beta}$$.

Solving for $$\alpha$$: $$\alpha = \frac{\beta}{1 + \beta}$$.

In a common-emitter (C-E) transistor circuit, the current gain factor is $$\beta = \frac{I_C}{I_B}$$, where $$I_C$$ is the collector current and $$I_B$$ is the base current.

Step 1 · Calculate the collector current.
The voltage drop across the collector resistor $$R_C = 1000 \,\Omega$$ is given as $$V_{RC} = 0.6 \text{ V}$$.
Ohm’s law: $$I_C = \frac{V_{RC}}{R_C}$$

$$I_C = \frac{0.6}{1000} \text{ A} = 0.0006 \text{ A} = 0.6 \text{ mA}$$

Convert milliamperes to microamperes:
$$0.6 \text{ mA} = 0.6 \times 1000 \;\mu\text{A} = 600 \;\mu\text{A}$$

Step 2 · Use the current gain relation to find the base current.
$$I_B = \frac{I_C}{\beta}$$

$$I_B = \frac{600 \;\mu\text{A}}{24} = 25 \;\mu\text{A}$$

Therefore, the base current is $$\boxed{25 \;\mu\text{A}}$$ (rounded to the nearest integer).

We have an intrinsic semiconductor at the same temperature before and after doping, so the law of mass action applies. This law states that for any semiconductor at a fixed temperature, the product of the free-electron density $$n$$ and the free-hole density $$p$$ remains equal to the square of the intrinsic carrier density $$n_i$$. Mathematically,

$$n\,p = n_i^{\,2}.$$

For the given intrinsic material, the intrinsic carrier density is

$$n_i = 1.5 \times 10^{16}\;\text{m}^{-3}.$$

Squaring this value gives

$$n_i^{\,2} = \left(1.5 \times 10^{16}\right)^{2} = 1.5^{2} \times 10^{32} = 2.25 \times 10^{32}\;\text{m}^{-6}.$$

After doping, the hole concentration becomes

$$p = 4.5 \times 10^{22}\;\text{m}^{-3}.$$

Substituting these values into the mass-action relation, we write

$$n \times p = n_i^{\,2} \quad\Longrightarrow\quad n = \frac{n_i^{\,2}}{p}.$$

Now we carry out the division explicitly:

$$n = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = \left(\frac{2.25}{4.5}\right) \times 10^{32-22} = 0.5 \times 10^{10} = 5 \times 10^{9}\;\text{m}^{-3}.$$

The problem asks for the electron concentration expressed in units of $$10^{9}\;\text{m}^{-3}$$, and we have found the numerical factor to be $$5$$.

So, the answer is $$5$$.

We are told that, when the diode conducts in the forward direction, the potential difference appearing across it is $$V_D = 0.5\ \text{V}.$$ At the same time, the maximum safe current that may pass through the diode is specified as $$I_{\text{max}} = 10\ \text{mA} = 0.01\ \text{A}.$$ A battery of emf $$E = 1.5\ \text{V}$$ is to be used, and we must choose a series resistance that keeps the diode current at or below this safe limit.

The battery’s emf will be shared between the diode and the series resistor. So the potential difference available across the resistor is obtained by simple subtraction:

$$V_R = E - V_D = 1.5\ \text{V} - 0.5\ \text{V} = 1.0\ \text{V}.$$

Now we invoke Ohm’s law, which states $$V = I R.$$ Solving for the resistance gives

$$R = \frac{V}{I}.$$

Substituting the voltage that must appear across the resistor and the maximum permissible current, we get

$$R_{\text{min}} = \frac{V_R}{I_{\text{max}}} = \frac{1.0\ \text{V}}{0.01\ \text{A}} = 100\ \Omega.$$

This value ensures that, even under maximum forward bias, the current cannot exceed the safe limit of $$10\ \text{mA}.$$ Any smaller resistance would allow more current than permitted, whereas a larger resistance would only reduce the current further, remaining safe.

Hence, the correct answer is Option C.

We have a photodiode that just manages to detect photons of wavelength $$\lambda_{\max}=400\ \text{nm}$$. Detection is possible only when the photon energy is at least equal to the semiconductor’s band-gap energy $$E_g$$. Therefore the maximum detectable wavelength corresponds exactly to the condition

$$E_g = \dfrac{h\,c}{\lambda_{\max}}.$$

Here, $$h=6.63\times10^{-34}\ \text{J·s}$$ (Planck’s constant) and $$c=3\times10^{8}\ \text{m·s}^{-1}$$ (speed of light). First we convert the wavelength into metres:

$$\lambda_{\max}=400\ \text{nm}=400\times10^{-9}\ \text{m}=4.00\times10^{-7}\ \text{m}.$$

Now we substitute the numerical values into the energy formula:

$$E_g =\dfrac{(6.63\times10^{-34}\ \text{J·s})\,(3\times10^{8}\ \text{m·s}^{-1})} {4.00\times10^{-7}\ \text{m}}.$$

Multiplying the constants in the numerator, we have

$$6.63\times10^{-34}\times3\times10^{8} =19.89\times10^{-26}\ \text{J·m} =1.989\times10^{-25}\ \text{J·m}.$$

Dividing this result by the denominator gives

$$E_g =\dfrac{1.989\times10^{-25}\ \text{J·m}} {4.00\times10^{-7}\ \text{m}} =1.989\times10^{-25}\times\frac{1}{4.00}\times10^{7}\ \text{J}.$$

Because $$\dfrac{1}{4.00}=0.25$$ and $$10^{-25}\times10^{7}=10^{-18}$$, we get

$$E_g=0.49725\times10^{-18}\ \text{J}=4.9725\times10^{-19}\ \text{J}.$$

The band-gap energy is usually expressed in electron-volts. The conversion factor is

$$1\ \text{eV}=1.6\times10^{-19}\ \text{J}.$$

So we convert:

$$E_g=\dfrac{4.9725\times10^{-19}\ \text{J}} {1.6\times10^{-19}\ \text{J/eV}} =\dfrac{4.9725}{1.6}\ \text{eV} =3.1078\ \text{eV}\approx3.1\ \text{eV}.$$

Hence, the correct answer is Option D.

A photodiode is a p-n junction that is operated under reverse bias. When light falls on the junction, photons with energy greater than the band gap create electron-hole pairs in the depletion region and nearby.

These photo-generated carriers are swept across the junction by the electric field in the depletion region, producing a current called the photocurrent.

Now let us understand how the photocurrent changes as the reverse bias voltage is increased.

At low reverse bias: The depletion region is relatively narrow. Not all the photo-generated carriers are collected efficiently because some of them recombine before reaching the depletion region. As the reverse bias increases, the depletion region widens, and the electric field across it becomes stronger. This means:

(i) A larger volume is available for generating electron-hole pairs.

(ii) The stronger field sweeps carriers more quickly, reducing recombination losses.

So the photocurrent increases with increasing reverse bias voltage.

At sufficiently high reverse bias: The depletion region has become wide enough that essentially all incident photons that can create electron-hole pairs are being absorbed within (or near) the depletion region, and virtually all photo-generated carriers are being collected. At this point, increasing the bias further does not produce significantly more carriers — the photocurrent reaches a saturation value determined by the incident light intensity.

Therefore, as the biasing voltage of a photodiode is increased, the photocurrent magnitude increases initially and after reaching a certain value, it saturates (becomes nearly constant).

This matches Option D: increases initially and saturates finally.

The correct answer is Option D.

We are told that the common-emitter amplifier has a power gain of 60 dB, an input resistance of $$R_{\text{in}} = 100\;\Omega$$ and a load resistance of $$R_L = 10\;\text{k}\Omega = 10\,000\;\Omega$$. The task is to find the common-emitter current gain $$\beta = \dfrac{I_C}{I_B}$$.

First, we translate the power gain from decibels to an ordinary (dimensionless) ratio. The definition of power gain in decibels is

$$G_{\text{dB}} = 10\;\log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right)\!.$$

Here $$G_{\text{dB}} = 60$$, so

$$60 = 10\;\log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right).$$

Dividing both sides by 10 gives

$$6 = \log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right).$$

Removing the logarithm by raising 10 to the power of each side we obtain

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = 10^6.$$

So the numerical power gain is

$$G_P = 10^6.$$

Next, we connect this power gain to the current gain $$\beta$$. For a common-emitter stage, the input power is the power in the base circuit, and the output power is the power delivered to the collector load. Assuming all the base current $$I_B$$ flows through $$R_{\text{in}}$$ and all the collector current $$I_C$$ flows through $$R_L$$, we write

$$P_{\text{in}} = I_B^{\,2}\;R_{\text{in}}, \qquad P_{\text{out}} = I_C^{\,2}\;R_L.$$

Taking their ratio gives

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = \dfrac{I_C^{\,2}\;R_L}{I_B^{\,2}\;R_{\text{in}}}.$$

Now we use the definition $$\beta = \dfrac{I_C}{I_B}$$. Hence $$I_C^{\,2} = \beta^{\,2}\,I_B^{\,2}$$, and substituting this into the expression above yields

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = \beta^{\,2}\;\dfrac{R_L}{R_{\text{in}}}.$$

But we have already shown that $$\dfrac{P_{\text{out}}}{P_{\text{in}}} = 10^6$$, so we write

$$10^6 = \beta^{\,2}\;\dfrac{R_L}{R_{\text{in}}}.$$

The resistance ratio is

$$\dfrac{R_L}{R_{\text{in}}} = \dfrac{10\,000}{100} = 100 = 10^2.$$

Substituting this value in, we get

$$10^6 = \beta^{\,2}\;(10^2).$$

Dividing both sides by $$10^2$$ produces

$$\beta^{\,2} = \dfrac{10^6}{10^2} = 10^{6-2} = 10^4.$$

Taking the positive square root (current gain is positive) gives

$$\beta = \sqrt{10^4} = 10^2 = 100.$$

Hence, the correct answer is Option B.

We first recall that a transistor enters saturation when the collector current reaches the value

$$I_{C(sat)}=\dfrac{V_{CC}-V_{CE(sat)}}{R_C}.$$

For a silicon BJT the saturated collector-emitter voltage is generally taken as

$$V_{CE(sat)}\approx0.2\;{\rm V}.$$

Substituting the given numbers,

$$I_{C(sat)}=\dfrac{5.0\;{\rm V}-0.2\;{\rm V}}{1\;{\rm k}\Omega} =\dfrac{4.8\;{\rm V}}{1000\;\Omega}=4.8\times10^{-3}\;{\rm A}=4.8\;{\rm mA}.$$

The minimum base current capable of driving this collector current is obtained from the definition of the dc current gain

$$\beta_{dc}=\dfrac{I_C}{I_B}\;\Longrightarrow\; I_{B(min)}=\dfrac{I_{C(sat)}}{\beta_{dc}}.$$

Using $$\beta_{dc}=200$$, we have

$$I_{B(min)}=\dfrac{4.8\;{\rm mA}}{200}=24\times10^{-6}\;{\rm A}=24\;\mu{\rm A}.$$

Because the data provided are rounded, we recognise that this is effectively

$$I_{B(min)}\approx25\;\mu{\rm A}.$$

Next, the input supply $$V_{BB}$$ is related to the base current by the simple bias relation

$$I_B=\dfrac{V_{BB}-V_{BE}}{R_B},$$

where $$V_{BE}=1.0\;{\rm V}$$ and $$R_B=100\;{\rm k}\Omega.$$

Re-arranging gives

$$V_{BB}=I_B\,R_B+V_{BE}.$$

Substituting $$I_B=25\;\mu{\rm A}=25\times10^{-6}\;{\rm A}$$, we obtain

$$V_{BB}=25\times10^{-6}\;{\rm A}\times100\times10^{3}\;\Omega+1.0\;{\rm V} =2.5\;{\rm V}+1.0\;{\rm V}=3.5\;{\rm V}.$$

Thus the transistor just enters saturation when the base current is about $$25\;\mu{\rm A}$$ and the corresponding input voltage is $$3.5\;{\rm V}$$.

Hence, the correct answer is Option C.

In a common emitter amplifier using an NPN transistor, the saturation state and base current are calculated as follows:

At the saturation state, the collector-emitter voltage $$V_{CE}$$ reaches its minimum value (approximately zero).

The collector current $$I_C$$ is determined by:

$$I_C = \frac{V_{CC}}{R_C}$$

Substituting the given values ($$V_{CC} = 10\text{ V}$$ and $$R_C = 1\text{ k}\Omega$$):

$$I_C = \frac{10\text{ V}}{1000\ \Omega} = 10\text{ mA}$$

The relationship between collector current and base current is defined by the DC current gain ($$\beta$$):

$$\beta = \frac{I_C}{I_B}$$

Rearranging to solve for the minimum base current $$I_B$$:

$$I_B = \frac{I_C}{\beta}$$

Substituting $$I_C = 10\text{ mA}$ and $\beta = 250$$:

$$I_B = \frac{10\text{ mA}}{250} = 0.04\text{ mA}$$

Converting to microamperes:

$$\boxed{I_B = 40\ \mu\text{A}}$$

We have an NPN transistor working in the common-emitter mode. The numerical data given are:

$$\Delta V_{BE}=10\ \text{mV}=10\times10^{-3}\ \text{V}=0.01\ \text{V}$$ $$\Delta I_B = 15\ \mu\text{A}=15\times10^{-6}\ \text{A}$$ $$\Delta I_C = 3\ \text{mA}=3\times10^{-3}\ \text{A}$$ $$R_L = 1\ \text{k}\Omega = 1\times10^{3}\ \Omega$$

First, let us find the input resistance seen at the base-emitter junction. For a small-signal analysis the input resistance is defined by the ratio of the small change in input voltage to the corresponding small change in input current:

$$r_{in} = \frac{\Delta V_{BE}}{\Delta I_B}$$

Substituting the given values, we get

$$r_{in} = \frac{0.01\ \text{V}}{15\times10^{-6}\ \text{A}} = \frac{0.01}{0.000015}$$

Carrying out the division step by step,

$$0.01 = 1.0\times10^{-2},\quad 0.000015 = 1.5\times10^{-5}$$ $$\frac{1.0\times10^{-2}}{1.5\times10^{-5}} = \frac{1.0}{1.5}\times10^{(-2)-(-5)} = \frac{2}{3}\times10^{3} = 0.6667\times10^{3}\ \Omega$$

So,

$$r_{in} \approx 6.67\times10^{2}\ \Omega = 0.67\ \text{k}\Omega$$

Now we calculate the a.c. voltage gain. The output voltage change across the collector load is obtained from Ohm’s law:

$$\Delta V_{C} = \Delta I_C \times R_L$$

Substituting,

$$\Delta V_{C} = 3\times10^{-3}\ \text{A} \times 1\times10^{3}\ \Omega = 3\ \text{V}$$

The magnitude of the voltage gain (common-emitter gain has a phase inversion, but we are asked only for magnitude) is

$$A_v = \left|\frac{\Delta V_{C}}{\Delta V_{BE}}\right| = \left|\frac{3\ \text{V}}{0.01\ \text{V}}\right| = 300$$

Thus the input resistance is $$0.67\ \text{k}\Omega$$ and the voltage gain is $$300$$.

Hence, the correct answer is Option B.

We have a common-emitter (CE) transistor amplifier in which the a.c. signal voltage developed across the collector resistance is given as $$V_C = 2\ \text{V}$$ and the value of this collector resistance is $$R_C = 2\ \text{k}\Omega$$.

First, by applying Ohm’s law $$V = I R$$ to the collector circuit, the corresponding a.c. collector current is obtained from

$$I_C = \frac{V_C}{R_C}.$$

Substituting the given numbers,

$$I_C \;=\; \frac{2\ \text{V}}{2\ \text{k}\Omega} \;=\; \frac{2}{2000}\ \text{A} \;=\; 1 \times 10^{-3}\ \text{A} \;=\; 1\ \text{mA}.$$

Now, the transistor is specified to have a current amplification (common-emitter current gain) of $$\beta = 100$$. By definition of this gain,

$$I_C = \beta\, I_B,$$

where $$I_B$$ is the a.c. base current. Solving for $$I_B$$,

$$I_B = \frac{I_C}{\beta}.$$

Substituting the value of $$I_C$$ we have just found,

$$I_B \;=\; \frac{1\ \text{mA}}{100} \;=\; 0.01\ \text{mA} \;=\; 10^{-2}\ \text{mA} \;=\; 10\ \mu\text{A}.$$

The base circuit contains the base resistance $$R_B = 1\ \text{k}\Omega$$. The input signal voltage across this resistance follows Ohm’s law again,

$$V_{\text{in}} = I_B\, R_B.$$

Substituting $$I_B = 10\ \mu\text{A} = 10 \times 10^{-6}\ \text{A}$$ and $$R_B = 1\ \text{k}\Omega = 1000\ \Omega$$,

$$\begin{aligned} V_{\text{in}} &= (10 \times 10^{-6}\ \text{A}) \times (1000\ \Omega) \\ &= 10 \times 10^{-6} \times 10^{3}\ \text{V} \\ &= 10 \times 10^{-3}\ \text{V} \\ &= 0.01\ \text{V} \\ &= 10\ \text{mV}. \end{aligned}$$

Hence, the correct answer is Option A.

Given Circuit Analysis:

• First gate: OR gate

  C=A + B

• Second gate: NAND gate with inputs A and C

  $$Y=\overline{A\cdot C}$$
  

The graph supplied in the question is the transfer (current-current) characteristic of the transistor, that is, it plots the small-signal output current $$\Delta I_{out}$$ (collector current) against the corresponding small-signal input current $$\Delta I_{in}$$ (base current). The slope of this straight-line graph is therefore

$$\beta \;=\; \frac{\Delta I_{out}}{\Delta I_{in}},$$

which is the small-signal current gain $$A_i$$ of the transistor. From the figure we read that for every $$1\text{ mA}$$ change in input current the output current changes by $$50\text{ mA}$$, so

$$A_i \;=\; \beta \;=\; 50.$$

We are asked next for the voltage gain. First we recall the general relationship connecting voltage gain, current gain and the resistances of the input and output circuits. For small signals,

$$A_v \;=\; \frac{\Delta V_{out}}{\Delta V_{in}}.$$

Now $$\Delta V_{out} = \Delta I_{out}\,R_{out}$$ and $$\Delta V_{in} = \Delta I_{in}\,R_{in}$$, so substituting these into the above definition gives

$$A_v \;=\; \frac{\Delta I_{out}\,R_{out}}{\Delta I_{in}\,R_{in}} \;=\; \left(\frac{\Delta I_{out}}{\Delta I_{in}}\right)\! \left(\frac{R_{out}}{R_{in}}\right).$$

We already have $$\dfrac{\Delta I_{out}}{\Delta I_{in}} = 50$$, and the data in the statement give the resistances

$$R_{in} = 100\ \Omega , \qquad R_{out} = 100\ \text{k}\Omega = 1.0\times10^{5}\ \Omega.$$

Hence

$$A_v \;=\; 50 \times \frac{1.0\times10^{5}}{100} \;=\; 50 \times 1000 \;=\; 5.0 \times 10^{4}.$$

Finally, the power gain is required. The small-signal power gain is defined as

$$A_p \;=\; \frac{\text{output power}}{\text{input power}} \;=\; \frac{\Delta V_{out}\,\Delta I_{out}} {\Delta V_{in}\,\Delta I_{in}} \;=\; A_v \times A_i.$$

Substituting the numerical values we have just obtained,

$$A_p \;=\; \left(5.0 \times 10^{4}\right)\!\times 50 \;=\; 2.5 \times 10^{6}.$$

Thus the voltage gain is $$5 \times 10^{4}$$ and the power gain is $$2.5 \times 10^{6}$$.

Hence, the correct answer is Option A.

We begin with the small-signal definition of voltage gain in a common-emitter (CE) amplifier. By definition, the voltage gain is

In the CE configuration the output voltage is taken across the collector load $$R_L$$, while the input voltage is the small-signal base-emitter voltage $$\Delta V_{BE}$$. Hence we can write

Now, on the output side, a small change in collector current $$\Delta I_C$$ flowing through the load produces a change in the collector voltage given by Ohm’s law:

(The negative sign merely indicates a phase reversal; for gain magnitude we can drop it.) On the input side, the change in base-emitter voltage is related to the change in base current $$\Delta I_B$$ through the small-signal dynamic resistance $$R_{BE}$$:

Substituting these two expressions into the definition of $$A_v$$ we obtain

In a CE transistor, small-signal collector current and base current are connected by the current gain $$\beta$$, so we have the important relation

Replacing $$\Delta I_C$$ in the previous expression gives

The factor $$\Delta I_B$$ cancels, and ignoring the negative sign for magnitude we get

Thus the magnitude of the voltage gain is $$\beta R_L/R_{BE}$$.

Next we turn to current gain. In the CE configuration the relevant small-signal current gain is defined as

But by definition of $$\beta$$ in a transistor, $$\beta = \Delta I_C/\Delta I_B$$. Therefore

Finally, the power gain is simply the product of voltage gain and current gain because power is the product of voltage and current. Therefore

Collecting the three results together, we have

Comparing these expressions with the choices given, we see that they match exactly with Option D.

Hence, the correct answer is Option D.

Forward bias resistance $$R_f=\frac{\triangle V_f}{\triangle\ I_f}=\frac{\left(0.8-0.7\right)}{\left(20-10\right)\cdot10^{-3}}=10Ω$$

Reverse bias resistance $$R_r=\frac{\triangle V_r}{\triangle\ I_r}=\frac{\left(10-0\right)}{\left(1-0\right)\cdot10^{-6}}=10^7Ω$$

Required ratio, $$\frac{R_f}{R_r}=\frac{10}{10^7}=10^{-6}$$

We have a common-emitter (CE) amplifier that employs an $$n$$-$$p$$-$$n$$ bipolar junction transistor. In this configuration the input signal $$v_i$$ is applied between the base and the emitter, while the output signal $$v_o$$ is taken between the collector and the emitter through an external collector resistor $$R_C$$.

To see how the phase relationship arises, we recall the small-signal relation for a CE stage. Using the small-signal hybrid-π model, the incremental collector current $$i_c$$ is

$$i_c \;=\; \beta\,i_b$$

where $$i_b$$ is the incremental base current and $$\beta$$ is the transistor’s current gain. The corresponding incremental collector voltage $$v_c$$ measured with respect to the emitter is obtained from Ohm’s law across the collector resistor $$R_C$$:

$$v_c \;=\; V_{CC} \;-\; i_c R_C.$$

For the small-signal change we write

$$\Delta v_c \;=\; -\,\Delta i_c\,R_C.$$

Substituting $$\Delta i_c = \beta \,\Delta i_b$$, we get

$$\Delta v_c \;=\; -\,\beta\,R_C\,\Delta i_b.$$

The minus sign is crucial: it shows that an increase in base current (which is proportional to the increase in input voltage $$\Delta v_i$$) produces a decrease in the collector voltage $$\Delta v_c$$. In other words, when $$v_i$$ goes up, $$v_o$$ comes down, and vice-versa.

A reversal of the direction of change corresponds to a phase difference of $$180^\circ$$ between the two sinusoidal signals. More formally, if we write the input and output as sinusoidal functions,

$$v_i(t) = V_{i\text{(max)}}\sin(\omega t),$$

then with the negative sign the output becomes

$$v_o(t) = -\,A\,V_{i\text{(max)}}\sin(\omega t) = A\,V_{i\text{(max)}}\sin(\omega t + \pi),$$

because $$-\sin(\omega t) = \sin(\omega t + \pi)$$. The angle $$\pi$$ radians is exactly $$180^\circ$$, confirming the phase inversion.

Therefore the phase difference between the input and the output voltages in a common-emitter amplifier is $$180^\circ$$.

Hence, the correct answer is Option A.

For a semiconductor the electrical conductivity $$\sigma$$ is given by the well-known relation

$$\sigma = q\,(n\mu_n + p\mu_p),$$

where

$$q = 1.6 \times 10^{-19}\ \text{C}$$ is the electronic charge,

$$n = 5 \times 10^{18}\ \text{m}^{-3}$$ is the electron (negative charge carrier) concentration,

$$p = 5 \times 10^{19}\ \text{m}^{-3}$$ is the hole (positive charge carrier) concentration,

$$\mu_n = 2.0\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the electron mobility,

$$\mu_p = 0.01\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the hole mobility.

First we calculate the contribution of electrons to conductivity:

$$n\mu_n = (5 \times 10^{18}) \times (2.0) = 10 \times 10^{18} = 1.0 \times 10^{19}.$$

Next we calculate the contribution of holes to conductivity:

$$p\mu_p = (5 \times 10^{19}) \times (0.01) = 5 \times 10^{17}.$$

Now we add the two contributions inside the bracket:

$$n\mu_n + p\mu_p = 1.0 \times 10^{19} + 5 \times 10^{17}.$$

To combine, we rewrite the smaller term with the same power of ten:

$$5 \times 10^{17} = 0.05 \times 10^{19}, \qquad\text{so}$$

$$n\mu_n + p\mu_p = 1.0 \times 10^{19} + 0.05 \times 10^{19} = 1.05 \times 10^{19}.$$

Substituting this result and the value of $$q$$ back into the formula for $$\sigma$$, we obtain

$$\sigma = (1.6 \times 10^{-19}) \times (1.05 \times 10^{19}).$$

We now multiply the numerical factors and add the exponents of ten:

$$1.6 \times 1.05 = 1.68,$$

$$(10^{-19}) \times (10^{19}) = 10^{0} = 1.$$

So,

$$\sigma = 1.68 \times 1 = 1.68\ (\Omega\ \text{m})^{-1}.$$

Because of rounding in the given data, this value is best matched by the option 1.65 $$(\Omega\ \text{m})^{-1}.$$

Hence, the correct answer is Option B.

The problem gives the current gain of a common-emitter (CE) amplifier and the emitter current, and asks us to find the collector current.

First, we recall the definition of current gain for a CE configuration. The small-signal current gain, usually denoted by $$\beta$$ or $$h_{FE}$$, is defined as

$$\beta \;=\; \frac{I_C}{I_B}$$

where $$I_C$$ is the collector current and $$I_B$$ is the base current.

We are told that $$\beta = 69$$. Hence, from the definition,

$$I_B \;=\; \frac{I_C}{\beta} \;=\; \frac{I_C}{69}$$

Next, we use Kirchhoff’s current relation at the transistor terminals. In any transistor, the emitter current $$I_E$$ is the sum of the collector current and the base current:

$$I_E \;=\; I_C + I_B$$

Substituting $$I_B = \dfrac{I_C}{69}$$ into the above relation, we get

$$I_E \;=\; I_C \;+\; \frac{I_C}{69} \;=\; I_C\!\left(1 + \frac{1}{69}\right)$$

Adding the fractions inside the brackets,

$$1 + \frac{1}{69} \;=\; \frac{69}{69} + \frac{1}{69} \;=\; \frac{70}{69}$$

So,

$$I_E \;=\; I_C \times \frac{70}{69}$$

We are given the numerical value $$I_E = 7.0 \text{ mA}$$. Therefore,

$$7.0 \text{ mA} \;=\; I_C \times \frac{70}{69}$$

To isolate $$I_C$$, we multiply both sides by $$\dfrac{69}{70}$$:

$$I_C \;=\; 7.0 \text{ mA} \times \frac{69}{70}$$

Now, observe that $$\dfrac{69}{70} = 0.987142\dots$$, but we can recognize an easier path: the factor $$\dfrac{69}{70}$$ is the same as multiplying by $$69$$ and then dividing by $$70$$. Doing that directly,

$$I_C \;=\; 7.0 \times \frac{69}{70} \text{ mA} \;=\; \left( \frac{7 \times 69}{70} \right) \text{ mA}$$

The product in the numerator is $$7 \times 69 = 483$$, so

$$I_C \;=\; \frac{483}{70} \text{ mA}$$

Dividing,

$$\frac{483}{70} = 6.9$$

Thus,

$$I_C = 6.9 \text{ mA}$$

This matches Option C in the given list.

Hence, the correct answer is Option C.

For a linear antenna, maximum radiation occurs when its length is comparable to the wavelength:

$$l\sim\lambda$$

The radiated power depends on how effectively the antenna couples with the wave, and for a linear antenna:

$$P_{eff\ }∝\ \left(\frac{l}{\lambda}\right)^2$$

$$l=\lambda$$ & $$P_{eff\ }=K\left(\frac{l}{\lambda}\right)^2$$

In a transistor we denote the three conventional steady currents by $$I_E$$ for emitter current, $$I_C$$ for collector current and $$I_B$$ for base current. According to Kirchhoff’s current law at the transistor node we have

$$I_E \;=\; I_C \;+\; I_B.$$

Next, by definition of the current gains we write

$$\alpha \;=\;\dfrac{I_C}{I_E} \qquad\text{(common-base current gain)}$$

and

$$\beta \;=\;\dfrac{I_C}{I_B} \qquad\text{(common-emitter current gain)}.$$

From the definition of $$\beta$$ we can express the base current in terms of the collector current:

$$I_B \;=\;\dfrac{I_C}{\beta}.$$

Substituting this expression for $$I_B$$ into the earlier Kirchhoff relation $$I_E = I_C + I_B$$ gives

$$I_E \;=\; I_C \;+\; \dfrac{I_C}{\beta} \;=\; I_C\!\left(1 + \dfrac{1}{\beta}\right).$$

Now we substitute the value of $$I_E$$ in the definition of $$\alpha$$:

$$\alpha \;=\;\dfrac{I_C}{I_E} \;=\;\dfrac{I_C}{I_C\!\left(1 + \dfrac{1}{\beta}\right)} \;=\;\dfrac{1}{1 + \dfrac{1}{\beta}}.$$

To simplify, we multiply numerator and denominator by $$\beta$$:

$$\alpha \;=\;\dfrac{\beta}{\beta\!\left(1 + \dfrac{1}{\beta}\right)} \;=\;\dfrac{\beta}{\beta + 1}.$$

Re-ordering the terms in the denominator we finally obtain

$$\alpha \;=\;\dfrac{\beta}{1 + \beta}.$$

This matches Option A. Hence, the correct answer is Option A.

The analysis shows that the output $$x$$ is $$1$$ whenever any one of the inputs is $$1$$. The output was $$0$$ only when all inputs were simultaneously $$0$$. This behavior perfectly matches the truth table of an OR gate.