Exams › JEE Main › Physics › Semiconductor Electronics
68 questions with worked solutions.
Answer: 0.2 mA
I = e^(1000V/T)-1 = 5 mA -> e^(1000V/T) = 6. Then dI/dV = (1000/T)e^(1000V/T) = (1000/300)(6) = 20 mA/V. Error dI = 20 x 0.01 = 0.2 mA.
Answer: 6.2 V
Voltage gain Av = beta*Rc/Ri = 62*5000/500 = 620. Output voltage = Av*Vin = 620*0.01 = 6.2 V.
Answer: 40%
Half-wave rectifier efficiency eta = 0.406/(1 + Rf/RL) = 0.406/(1 + 10/1000) = 0.402, i.e. about 40%.
Answer: 2.37 × 10¹⁹ m−3
The intrinsic conductivity of a semiconductor is determined by the product of the charge-carrier concentration and their mobilities. By using the given mobilities of electrons and holes along with the intrinsic conductivity, we can calculate the charge-carrier concentration, which confirms that option A is correct.
Answer: It is a pnp transistor with R as the base
With the meter's positive (red) lead on P or Q and the negative (black) lead on R, both junctions conduct, so P and Q are p-type (anodes) and R is the common n-type region. Two p regions sharing one n region is a PNP transistor with R as the base.
Answer: 1.8 × 10⁵
The doping introduces a significant number of electrons into the conduction band, calculated by determining the number of antimony atoms in 1 m³ of silicon and considering that half of these contribute to electron concentration. This results in a charge-carrier concentration increase of 1.8 × 10⁵, which reflects the effective contribution of the dopant.
Answer: -48.78
The current gain in a common-emitter amplifier can be calculated using the formula that incorporates the load impedance and the transistor parameters. Given the values, the calculation yields a current gain of approximately -48.78, indicating a significant amplification of the input current with a phase inversion.
Q8. The depletion region thickness of a p–n junction is typically of the order of:
Answer: 10⁻⁶ m
The depletion region of a p-n junction is typically about a micrometre wide, i.e. of the order of 10^-6 m.
Answer: 2/3 G
For a common-emitter amplifier the gain is proportional to the transconductance (gain = gm * R_load). Replacing gm = 0.03 by 0.02 scales the gain by 0.02/0.03 = 2/3, giving (2/3)G.
Q10. To produce visible light, an LED must have a band gap lying in which of the following ranges?
Answer: 1.7 eV to 3.0 eV
Visible light spans roughly 1.8 to 3.1 eV in photon energy, so an LED emitting visible light needs a band gap of about 1.7 to 3.0 eV.
Answer: 10 mV
Voltage gain Av = beta * Rc/Rin = 100 * 2000/1000 = 200. Input signal = Vout/Av = 2/200 = 0.01 V = 10 mV.
Answer: 500
Power gain = current gain^2 * (R_load / R_in) = 10^2 * (100k / 20k) = 100 * 5 = 500.
Answer: 7/5
Current I = n*e*vd*A, so Ie/Ih = (ne/nh)*(vde/vdh). Thus 7/4 = (ne/nh)*(5/4), giving ne/nh = 7/5.
Answer: The resistance of copper falls, while that of germanium rises
As temperature decreases, the resistance of metals like copper typically decreases due to reduced thermal vibrations of atoms, which allows electrons to flow more freely. In contrast, germanium, a semiconductor, experiences an increase in resistance at low temperatures because fewer charge carriers are available for conduction.
Answer: 0.29 mA
Collector current Ic = alpha*Ie = 0.96*7.2 = 6.912 mA. Base current Ib = Ie - Ic = 7.2 - 6.912 = 0.288 ~ 0.29 mA.
Answer: 7
alpha = dIc/dIe = 3.5/4 = 0.875. Then beta = alpha/(1-alpha) = 0.875/0.125 = 7 (equivalently dIc/dIb = 3.5/0.5 = 7). The stored value 0.875 is alpha, not beta.
Answer: 0.2 mA
I = e^(1000V/T) - 1, so at I = 5 mA we have e^(1000V/T) = 6. dI/dV = (1000/T) e^(1000V/T) = (1000/300)(6) = 20 per volt. With dV = 0.01 V, dI = 20 * 0.01 = 0.2 mA.
Answer: In phase
In a common-base amplifier configuration, the output voltage is in phase with the input signal voltage, meaning that both signals reach their maximum and minimum values simultaneously, resulting in no phase shift between them.
Answer: 49
alpha = Ic/Ie = 5.488/5.60 = 0.98. The current amplification factor beta = alpha/(1 - alpha) = 0.98/0.02 = 49.
Answer: Covalent bonding
A solid opaque to visible light whose conductivity increases with temperature is a semiconductor (e.g. Si, Ge), which is held together by covalent bonding.
Answer: 5/4
Current ratio Ie/Ih = (ne/nh)*(ve/vh). So ve/vh = (7/4)/(7/5) = 5/4.
Answer: The concentration of free electrons available for conduction is appreciable in Si and Ge, but very low in C.
Silicon and germanium are semiconductors that can generate free electrons at room temperature, allowing for appreciable conduction, while carbon, in its common forms like graphite or diamond, has a much lower concentration of free electrons, making it less effective for conduction.
Answer: It is a pnp transistor, with R as the base
The behavior described indicates that when the negative lead is connected to R and the positive lead to P or Q, it shows resistance, which is characteristic of a pnp transistor where the base (R) is connected to the negative lead, allowing current to flow from the emitter to the collector.
Answer: 180°
In a common-emitter amplifier the output voltage is 180 degrees out of phase with the input voltage.
Answer: 0.4Ωm
The resistivity of a semiconductor can be calculated using the formula ρ = (1)/(q n μ), where q is the charge of an electron (approximately 1.6 × 10⁻¹⁹ C), n is the electron density, and μ is the mobility. Substituting the given values, we find that the resistivity is approximately 0.4 Ωm, confirming that option C is correct.
Answer: 0.67 kΩ, 300
The input resistance is calculated using the change in base current and the applied signal voltage, resulting in 0.67 kΩ. The voltage gain is determined by the ratio of the change in collector current to the change in base-emitter voltage, yielding a gain of 300.
Answer: α = β/(1−β)
Correct CE relations: alpha=beta/(1+beta) and equivalently 1/alpha=1/beta+1. The relation alpha=beta/(1-beta) is the incorrect one, so it is the required answer (index 1).
Answer: 180°
In a common emitter amplifier, the output voltage is inverted relative to the input voltage, resulting in a phase difference of 180 degrees. This means that when the input signal goes positive, the output signal goes negative, confirming the phase inversion characteristic of this configuration.
Answer: β R_L/R_BE, ΔI_C/ΔI_B, β² R_L/R_BE
The correct option accurately reflects the relationships in a common emitter configuration, where the voltage gain is determined by the ratio of load resistance to small signal resistance, the current gain is the ratio of collector current change to base current change, and the power gain is the square of the current gain multiplied by the voltage gain.
Answer: 0.67 kΩ, 300
The input resistance is calculated using the change in base-emitter voltage and base current, yielding 0.67 kΩ. The voltage gain is determined by the ratio of the change in collector current to the change in base-emitter voltage, resulting in a gain of 300.
Answer: α = β/(1+β)
With alpha = Ic/Ie and beta = Ic/Ib, and Ie = Ib + Ic, we get alpha = beta/(1+beta). Equivalently beta = alpha/(1-alpha).
Answer: 2 eV
The threshold wavelength corresponds to the band gap: Eg = hc/lambda = 1240 (eV.nm)/621 nm ~ 2.0 eV. So the band gap is nearly 2 eV.
Answer: Statement I is false but Statement II is true
Statement I is false: two PN diodes connected back to back do not behave as a transistor (a transistor requires a single thin, lightly doped base shared between the junctions). Statement II is true: the current amplification factor beta = collector current / base current. So I is false but II is true.
Q34. Zener breakdown occurs in a p-n junction having p and n both:
Answer: heavily doped and have narrow depletion layer.
Zener breakdown occurs in a p-n junction that is heavily doped, which creates a narrow depletion layer. This allows for a strong electric field to develop, enabling the tunneling of charge carriers at lower reverse voltages.
Answer: Statement-I is true but Statement-II is false.
Doping silicon with pentavalent materials, like phosphorus, introduces extra electrons, increasing electron density, making Statement-I true. However, n-type semiconductors are electrically neutral overall, as the added electrons balance the positive charge of the silicon lattice, rendering Statement-II false.
Q36. For a transistor in CE mode to be used as an amplifier, it must be operated in:
Answer: The active region only
The active region is where a transistor can amplify signals, as it allows for a linear relationship between input and output, making it suitable for amplification purposes.
Answer: Both Statement I and Statement II are true
Both statements are correct as they describe methods to smooth out the pulsating DC output from a full wave rectifier. A capacitor across the load helps to filter the voltage fluctuations, while an inductor in series can also smooth the current by opposing changes in current flow.
Answer: fractional change in minority carriers produce higher reverse bias current
In reverse bias mode, the photodiode operates by generating a larger current response due to the movement of minority carriers, which are more sensitive to changes in light intensity. This allows for more effective detection of optical signals.
Answer: Both A and R are true, and R is correct explanation of A.
The assertion is correct because n-p-n transistors do allow for higher current flow due to the greater mobility of electrons compared to holes in p-n-p transistors, making the reason a valid explanation for the assertion.
Answer: 14
The speed of the electron decreases as it crosses the potential barrier due to energy conservation, where the kinetic energy lost equals the potential energy gained. By calculating the initial kinetic energy and equating it to the potential energy change, we find that the speed on the p-side is reduced to 4.67 × 10⁵ m/s, leading to x being 14.
Q41. Choose the correct statement about Zener diode
Answer: It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias.
The Zener diode is specifically designed to operate in reverse bias, where it maintains a stable output voltage, making it effective as a voltage regulator. In forward bias, it behaves like a standard diode, allowing current to flow with a small voltage drop.
Answer: Both A and R are true but R is NOT the correct explanation of A
Photodiodes are indeed operated in reverse bias to enhance their sensitivity to light, allowing for more accurate measurements of light intensity. However, the reason provided about forward bias current being greater than reverse bias current does not directly explain why reverse bias is preferred for photodiodes.
Answer: Statement I is correct but statement II is incorrect
Statement I accurately describes the doping process of silicon, where boron creates holes (P-type) and arsenic adds electrons (N-type). However, Statement II is incorrect because a current does not automatically flow in a P-N junction without an external voltage applied; it requires a bias to create a current.
Answer: 875 nm
The wavelength of light emitted by a semiconductor can be calculated using the energy-band gap relation, where the wavelength is inversely proportional to the energy. Given the band gap of GaAs is 1.42 eV, the corresponding wavelength is approximately 875 nm, which falls within the infrared region of the spectrum.
Answer: I_L = 10 mA; R_L = 500 Ω
The correct option is right because if the zener current is four times the load current, and the total current through the resistor is the sum of the load current and the zener current, we can derive that the load current must be 10 mA to satisfy the given conditions, leading to a load resistance of 500 Ω when using Ohm's law.
Answer: B, E Only
Statements B and E are correct because solar cells operate without external bias and LEDs must be connected in forward bias to emit light. The other statements either misrepresent the characteristics of solar cells or LEDs.
Q47. By increasing the temperature, the specific resistance of a conductor and a semiconductor
Answer: increases, decreases
As temperature rises, the specific resistance of a conductor typically increases due to increased lattice vibrations that impede electron flow. In contrast, for semiconductors, higher temperatures provide more energy to electrons, allowing them to move more freely, which decreases their resistance.
Q48. At absolute zero, Si acts as
Answer: insulator
At absolute zero, silicon's thermal energy is minimized, causing its electrons to occupy the lowest energy states and preventing them from conducting electricity, which classifies it as an insulator.
Q49. The energy band gap is maximum in
Answer: insulators
Insulators have a large energy band gap, which prevents electrons from easily moving from the valence band to the conduction band, thus inhibiting electrical conductivity. This characteristic distinguishes them from metals and semiconductors, which have smaller band gaps.
Q50. The part of a transistor which is most heavily doped to produce large number of majority carriers is
Answer: emitter
The emitter is designed to be heavily doped to ensure a high concentration of majority carriers, which facilitates efficient charge injection into the base region of the transistor, enhancing its overall performance.