Question:

Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.

Answer:

Step 1: Calculate the output voltage of the first amplifier. Answer: 0.01 V x 10 = 0.1 V

Step 2: Calculate the output voltage of the second amplifier. Answer: 0.1 V x 20 = 2 V

Step 3: Calculate the total output voltage of the two amplifiers connected in series. Answer: 2 V

Question:

The number of silicon atoms per m3 is 5×10^28. This is doped simultaneously with 5×1022 atoms per m^3 of arsenic and 5×10^20 per m3 atoms of indium. Calculate the number of electrons and holes. Given that ni​=1.5×10^16 /m^3. Is the material n-type or p-type?

Answer:

1. Calculate the total number of atoms per m3: 5×10^28 + 5×10^22 + 5×10^20 = 5.005×10^28 atoms per m3

2. Calculate the number of electrons and holes: Number of electrons = 5.005×10^28 × 1.5×10^16 /m^3 = 7.5075×10^44 electrons Number of holes = 5.005×10^28 × 1.5×10^16 /m3 = 7.5075×10^44 holes

3. Determine the type of material: Since the number of electrons is greater than the number of holes, the material is n-type.

Question:

In an unbiased p-n junction, holes diffuse from the p-region to n-region because A Free electrons in the n-region attract them. B They move across the junction by the potential difference. C Hole concentration in p-region is more as compared to n-region. D All of the above

Answer:

Answer: D All of the above

Question:

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Answer:

Step 1: Convert the wavelength into electron volts.

6000 nm = 600 eV

Step 2: Compare the band gap of the semiconductor to the wavelength in electron volts.

2.8 eV < 600 eV

Step 3: Determine if the photodiode can detect the wavelength.

Yes, the photodiode can detect the wavelength.

Question:

In an intrinsic semiconductor the energy gap Eg​ is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni​ is given byni​=no​exp(−Eg​​/2kB​T) Where n0​ is a constant.

Answer:

1. First, calculate the intrinsic carrier concentration ni at 600 K and 300 K using the equation given: ni at 600 K = noexp(-1.2eV/2kB600K) ni at 300 K = noexp(-1.2eV/2kB300K)

2. Next, calculate the conductivity at 600 K and 300 K using the equation conductivity = e*(electron mobility)ni: Conductivity at 600 K = e(electron mobility)ni at 600 K Conductivity at 300 K = e(electron mobility)*ni at 300 K

3. Finally, calculate the ratio between conductivity at 600 K and that at 300 K by dividing the conductivity at 600 K by the conductivity at 300 K: Ratio between conductivity at 600 K and that at 300 K = Conductivity at 600 K/Conductivity at 300 K

Question:

In an n-type silicon, which of the following statement is true: A Electrons are majority carriers and trivalent atoms are the dopants. B Electrons are minority carriers and pentavalent atoms are the dopants. C Holes are minority carriers and pentavalent atoms are the dopants. D Holes are majority carriers and trivalent atoms are the dopants.

Answer:

Answer: D. Holes are majority carriers and trivalent atoms are the dopants.

Question:

For transistor action, which of the following statements are correct: This question has multiple correct options A Base, emitter and collector regions should have similar size and doping concentrations B The base region must be very thin and lightly doped C The emitter junction is forward biased and collector junction is reverse biased D Both the emitter junction as well as the collector junction are forward biased

Answer:

A - Base, emitter and collector regions should have similar size and doping concentrations B - The base region must be very thin and lightly doped C - The emitter junction is forward biased and collector junction is reverse biased D - Both the emitter junction as well as the collector junction are forward biased

All of the above statements are correct.

Question:

In a p-n junction diode, the current I can be expressed as I=I0​exp(eV​/2KB​T−1), where I0​ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kB​ is the Boltzmann constant (8.6×10^−5eV/K) and T is the absolute temperature. If for a given diode Io​=5×10^−12A and T=300 K, then (a) What will be the forward current at a forward voltage of 0.6 V?(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

Answer:

(a) The forward current at a forward voltage of 0.6 V can be calculated as: I = I0exp(eV/2KBT−1) I = 5 x 10^−12 exp(0.6/2(8.6 x 10^−5)(300)) I = 3.2 x 10^−9 A

(b) The increase in current if the voltage across the diode is increased to 0.7 V can be calculated as: I = I0exp(eV/2KBT−1) I = 5 x 10^−12 exp(0.7/2(8.6 x 10^−5)(300)) I = 4.2 x 10^−9 A

Increase in current = 4.2 x 10^−9 - 3.2 x 10^−9 = 1 x 10^−9 A

(c) The dynamic resistance can be calculated as: R = (V2 - V1)/(I2 - I1) R = (0.7 - 0.6)/(4.2 x 10^−9 - 3.2 x 10^−9) R = 1/1 x 10^−9 R = 10^9 Ω

(d) The current if reverse bias voltage changes from 1 V to 2 V can be calculated as: I = I0exp(eV/2KBT−1) I = 5 x 10^−12 exp(-2/2(8.6 x 10^−5)(300)) I = 1.2 x 10^−14 A

Question:

Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg​)C​,(Eg​)Si​ and (Eg​)Ge​.Which of the following statements is true? A (Eg​)Si​<(Eg​)Ge​<(Eg​)C​ B (Eg​)C​<(Eg​)Ge​>(Eg​)Si​ C (Eg​)C​>(Eg​)Si​>(Eg​)Ge​ D (Eg​)C​=(Eg​)Si​=(Eg​)Ge

Answer:

Answer: A (Eg​)Si​<(Eg​)Ge​<(Eg​)C​

Question:

For a transistor amplifier, the voltage gain A Remains constant for all frequencies. B Is high at high and low frequencies and constant in the middle frequency range. C Is low at high and low frequencies and constant at mid frequencies. D None of the above.

Answer:

Answer: D None of the Above.

Question:

When a forward bias is applied to a p-n junction, it A raises the potential barrier. B reduces the majority carrier current to zero. C lowers the potential barrier. D None of the above.

Answer:

Answer: C lowers the potential barrier.

Question:

In half-wave rectification, what is the output frequency if the input frequency is 50 Hz What is the output frequency of a full-wave rectifier for the same input frequency.

Answer:

Answer: In half-wave rectification, the output frequency is 50 Hz. In full-wave rectification, the output frequency is 100 Hz for the same input frequency of 50 Hz.

Question:

For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.

Answer:

1. Calculate the base current: Ib = (2V/2kΩ) x 100 = 100μA

2. Calculate the input signal voltage: Vin = Ib x 1kΩ = 100μA x 1kΩ = 0.1V

Question:

Which of the statements given is true for p-type semiconductors (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

Answer:

Answer: (d) Holes are majority carriers and trivalent atoms are the dopants.