Q:1. In 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

Soln: n-type is obtained by doping the Ge or Si with pentavalent atoms. In n-type semiconductor, electrons are majority carriers and holes are minority carriers, hence option (c) is True.

Q:2. Which of the statements given in Q.1 is true for p-type semiconductors.

Soln: p-type semiconductor is obtained by doping Ge or Si with trivalent atoms. In p-type semiconductor holes are majority carriers and electrons are minority carriers. Hence option (d) is correct.

Q:3. 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

Soln: The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given three elements. Hence option (c) is correct.

Q:4. 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 the above.

Soln:

In an unbiased p-n junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration. Thus option (c) is correct.

Q:5. When a forward bias is applied to p-n junctions, It

(a) raises the potential barrier

(b) reduces the majority carrier current to zero

(c) lower the potential barrier

(d) none of the above.

Soln: When a forward bias is applied across the p-n junction, the applied voltage opposes the barrier voltage. Due to it, the potential barrier across the junction is lowered. Hence option (c) is correct.

Q:6. For a transistor action, which of the following statements are correct:

(a) collector current is equal to the sum of base current and emitter current.

(b) The input resistance depends upon the current Ic in the transistor.

(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.

Soln: (b) and (c); for a transistor,$\large \beta = \frac{I_C}{I_B} $

or, $\large I_B = \frac{I_C}{\beta} $

$\large R_i = \frac{V_i}{I_B} = \frac{V_i}{I_C} \times \beta $

$\large R_i \propto \frac{1}{I_c} $

Therefore, R_{i} depends upon collector current I_{C}. For a transistor action, the emitter junction is forward biased and collector junction is reverse biased.

Q:7. For transistor amplifier, the voltage gain (a) remains constant for all frequencies

(a) is high at high and low frequencies and constant in the middle frequency range

(b) is low at high and low frequencies and constant at mid frequencies

(c) None of the above.

Soln: (c) the voltage gain is low at high and low frequencies and constant and high at mid-frequencies.

Q:8. In half wave rectifier, what is the output frequency of the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.

Soln: The output ripple frequency is 50 Hz for half wave rectifier and 100 Hz for full-wave rectifier.

Q: 9. For a common emitter amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2V. 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.

Soln: Here, R_{0} = 2000 Ω ; V_{0} = 2V, β_{ac} = 100; V_{i} = ? I_{b} = ?; R_{i} = 1000 Ω

$\large A_v = \frac{V_o}{V_i} = \beta_{ac} \frac{R_o}{R_i} $

$\large V_i = \frac{V_o}{\beta_{ac} \frac{R_o}{R_i}}$

$\large = \frac{2}{100 \times (2000/1000)} = 0.01 V$

$\large I_b = \frac{V_i}{R_i}= \frac{0.01}{1000} = 10 \mu A $

Q:10. 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 0.01V, calculate the output a.c. signal.

Soln: Total voltage gain $\large A_v = \frac{\Delta V_o}{\Delta V_i} = A_{V_1} \times A_{V_2} $

$\large \Delta V_o = \Delta V_i \times A_{V_1} \times A_{V_2} $

$\large = 0.01 \times 10 \times 20 $

= 2 V

Q: 11. A p-n junction is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 600 nm ?

Soln: Energy , $\large E = \frac{h c}{\lambda } $

$\large E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-10} \times 1.6 \times 10^{-19} } eV $

= 2.06 eV < 2.8 eV

As E < E_{g} , so p-n junction cannot detect the radiation of given wavelength.

Q:12. The number of silicon atoms per m^{3} is 5 × 10^{28} . This is doped simultaneously with 5 × 10^{22} atoms per m^{3} of Arsenic and 5 × 10^{20} per m^{3} atoms of Indium. Calculate the number of electrons and holes. Given that n_{i} = 1.5 × 10^{16} m^{-3}. Is the material n-type or p-type ?

Soln: n_{e} = 5 × 10^{22} – 5 × 10^{20} = (5 – 0.05) × 10^{22} = 4.95 × 10^{22} m^{-3}

$\large n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} $

= 4.54 × 10^{9} m^{-3}

As , n_{e} > n_{h}, so the material is n-type semiconductor.

Q:13. 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 300 K? Assume that the temperature dependence intrinsic concentration n_{i} is given by

$\large n_i = n_o exp (\frac{-E_g}{2 k_B T}) $ , where n_{0} is a constant and k_{B} = 8.62 × 10^{-5} eV/K.

Sol: $\large \frac{E_g}{2 k} (\frac{1}{T_1} – \frac{1}{T_2}) $

$\large = \frac{1.2}{2 \times 8.62 \times 10^{-5}} (\frac{1}{300} – \frac{1}{600}) $

= 11.6

$\large \frac{n_{600}}{n_{300}} = e^{\frac{E_g}{2k}} (\frac{1}{T_1} – \frac{1}{T_2}) $

$\large \frac{n_{600}}{n_{300}} = e^{11.6} = (2.718)^{11.6}$

= 1.089 x 10^{5}

$\large \frac{\sigma_{600}}{\sigma_{300}} = \frac{n_{600}}{n_{300}} = 1.1 \times 10^5 $