NCERT Solutions For Class 12 Physics Chapter 14 Semiconductor Electronic Material Devices And Simple Circuits - 2025-26

```html 1. In an n-type silicon, which of the following statements is true:
Electrons are majority carriers and trivalent atoms are the dopants.
Electrons are minority carriers and pentavalent atoms are the dopants.
Holes are minority carriers and pentavalent atoms are the dopants.
Holes are majority carriers and trivalent atoms are the dopants.
Ans: Electrons are the majority carriers and holes are the minority carriers in an n-type silicon atom. When pentavalent atoms like Phosphorus are added to Silicon, an n-type semiconductor forms.

2. In a p-type silicon, which of the following statements is true:
Electrons are majority carriers and trivalent atoms are the dopants.
Electrons are minority carriers and pentavalent atoms are the dopants.
Holes are minority carriers and pentavalent atoms are the dopants.
Holes are majority carriers and trivalent atoms are the dopants.
Ans: Electrons are the minority carriers while holes are the majority carriers in a p-type silicon atom. If trivalent atoms like Aluminum are doped into Silicon, a p-type semiconductor is made.

3. Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to ${{({{E}_{g}})}_{C}}$, ${{({{E}_{g}})}_{Si}}$ and ${{({{E}_{g}})}_{Ge}}$. Which of the following statements is true?
\[{{({{E}_{g}})}_{Si}}<{{({{E}_{g}})}_{Ge}}<{{({{E}_{g}})}_{C}}\]
\[{{({{E}_{g}})}_{C}}<{{({{E}_{g}})}_{Ge}}<{{({{E}_{g}})}_{Si}}\]
\[{{({{E}_{g}})}_{C}}>{{({{E}_{g}})}_{Si}}>{{({{E}_{g}})}_{Ge}}\]
\[{{({{E}_{g}})}_{C}}={{({{E}_{g}})}_{Si}}={{({{E}_{g}})}_{Ge}}\]
Ans: The energy band gap of Carbon is the largest, and Germanium’s is the smallest.
\[ {{({{E}_{g}})}_{C}}>{{({{E}_{g}})}_{Si}}>{{({{E}_{g}})}_{Ge}} \]

4. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
Free electrons in the n-region attract them.
They move across the junction by the potential difference.
Hole concentration in p-region is more as compared to n-region.
All the above
Ans: The concentration of holes is higher in the p-type region than in the n-type region, so holes diffuse from the p-type region to the n-type region. This happens because particles move from higher to lower concentration naturally.

5. When a forward bias is applied to a p-n junction, it
Raises the potential barrier.
Reduce the majority carrier current to zero.
Lower the potential barrier.
None of the above.
Ans: When a forward bias is applied to a p-n junction, the barrier is opposed by the voltage. This lowers the potential barrier, so the barrier gets smaller.

6. For transistor action, which of the following statements are correct?
Base, emitter and collector regions should have similar size and doping concentrations.
The base region must be very thin and lightly doped.
The emitter junction is forward biased and collector junction is reverse biased.
Both the emitter junction as well as the collector junction are forward biased.
Ans: (b) and (c) are correct. The base must be thin and lightly doped, and for proper transistor action, the emitter junction is forward biased while the collector junction is reverse biased.

7. For a transistor amplifier, the voltage gain
Remains constant for all frequencies.
Is high at high and low frequencies and constant in the middle frequency range.
Is low at high and low frequencies and constant at mid frequencies.
None of the above.
Ans: The voltage gain of a transistor amplifier is constant at mid frequencies and lower at both low and high frequencies.

8. In a half-wave rectifier, find the output frequency if the input frequency is \[~\mathbf{50Hz}\]. Also, find the output frequency of a full-wave rectifier for the same input frequency?
Ans: - For a half-wave rectifier, output frequency is the same as the input frequency, so it's 50 Hz. - For a full-wave rectifier, output frequency is twice the input: 2 × 50 Hz = 100 Hz.
So, for 50 Hz input: half-wave = 50 Hz, full-wave = 100 Hz.

9. For a CE-transistor amplifier, the audio signal voltage across the collector resistance of \[\mathbf{2}k\Omega \] is \[\mathbf{2}V\]. Suppose the current amplification factor of the transistor is $100$, find the input signal voltage and base current, if the base resistance is \[1k\Omega \].
Ans: - The input signal voltage is 0.01 V and base current is 10 μA for this amplifier. - Step-by-step:

• Given: Collector resistance Rc = 2 kΩ = 2000 Ω, Audio signal voltage across Rc = 2 V, β = 100, Base resistance Rb = 1 kΩ = 1000 Ω.
• Voltage amplification: \(\frac{V}{V_i} = β \frac{R_c}{R_B}\)
• Substituting: \(\frac{2}{V_i} = 100 \times \frac{2000}{1000} = 200\), so \(V_i = \frac{2}{200} = 0.01 V\).
• Base current: \(R_B = \frac{V_i}{V_B}\). So \(V_B = \frac{V_i}{R_B} = \frac{0.01}{1000} = 0.00001 = 10 μA\).

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 $0.01$ volt, calculate the output ac signal.
Ans: - The output AC signal for the amplifiers is 2 V. - Steps:

• Total voltage gain: 10 × 20 = 200.
• \(V = \frac{V_o}{V_i}\) so \(V_o = 200 × 0.01 = 2\) V.

11. A p-n photodiode is fabricated from a semiconductor with a band gap of \[\mathbf{2}.\mathbf{8}\text{ }\mathbf{eV}\]. Can it detect a wavelength of \[\mathbf{6000}\text{ }\mathbf{nm}\]?
Ans: No, the photodiode cannot detect a wavelength of 6000 nm because the photon energy is not enough to cross the band gap.

• Band gap = 2.8 eV.
• Wavelength, λ = 6000 nm = \(6 × 10^{-6}\) m.
• Energy of photon, \(E = \frac{hc}{λ} = \frac{6.626 × 10^{-34} × 3 × 10^8}{6 × 10^{-6}} = 3.313 × 10^{-20}\) J = 0.207 eV.
• This energy is much less than 2.8 eV, so this photodiode cannot detect such a long wavelength.

12. The number of silicon atoms per ${{m}^{3}}$ is $5\times {{10}^{28}}$. This is doped simultaneously with $5\times {{10}^{22}}$ atoms per ${{m}^{3}}$ of Arsenic and $5\times {{10}^{20}}$ per ${{m}^{3}}$ atoms of indium. Calculate the number of electrons and holes. Given that${{n}_{i}}=1.5\times {{10}^{16}}{{m}^{-3}}$. Is the material n-type or p-type?
Ans: The material is n-type. There are about \(4.99 × 10^{22}\) electrons per \({m}^3\) and about \(4.51 × 10^{9}\) holes per \({m}^3\).

• Electrons: \(n_e = n_{As} - n_i = 5×10^{22} - 1.5×10^{16} ≈ 4.99×10^{22}\).
• Holes: \(n_h = \frac{n_i^2}{n_e} = \frac{(1.5×10^{16})^2}{4.99×10^{22}} ≈ 4.51×10^9\).
• Since electrons are much more than holes, it's n-type.

13. In an intrinsic semiconductor the energy gap ${{E}_{g}}$ is \[1.2\text{ }eV\]. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at \[600K\] and that at $300K$? Assume that the temperature dependence of intrinsic carrier concentration ${{n}_{i}}$ is given by ${{n}_{i}}={{n}_{o}}\exp \left( -\frac{{{E}_{g}}}{2{{k}_{B}}T} \right)$.
Ans: The conductivity at 600 K is about \(1.09 × 10^5\) times that at 300 K.

• Carrier concentration ratio \(\frac{n_{i2}}{n_{i1}}\) depends on temperature.
• \(\frac{n_{i2}}{n_{i1}} = \exp\left(\frac{E_g}{2k_B}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right)\)
• Using numbers, ratio = \(\exp(11.6) = 1.09 × 10^5\).
• This means the higher temperature has much higher conductivity.

14. In a p-n junction diode, the current $I$ can be expressed as $I={{I}_{0}}\exp \left( \left( \frac{eV}{{{k}_{B}}T} \right)-1 \right)$ where ${{I}_{0}}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative bias, and $I$ is the current through the diode, ${{k}_{B}}$ is the Boltzmann constant $(8.62\times {{10}^{-5}}eV/K)$ and $T$ is the absolute temperature. If for a given diode ${{I}_{0}}=5\times {{10}^{-12}}A$and\[\mathbf{T}\text{ }=\mathbf{300}\text{ }\mathbf{K}\], then
(a) What will be the forward current at a forward voltage of \[\mathbf{0}.\mathbf{6}\text{ }\mathbf{V}\]?
Ans: The forward current is 0.0256 A.

(b) What will be the increase in the current if the voltage across the diode is increased to \[\mathbf{0}.\mathbf{7}\text{ }\mathbf{V}\]?
Ans: The current increases by 1.23 A when the voltage is increased from 0.6 V to 0.7 V.

(c) What is dynamic resistance?
Ans: Dynamic resistance is the ratio of change in voltage to change in current (\(\frac{\Delta V}{\Delta I}\)). Here, it is calculated as \(0.0336/\Omega\).

(d) What will be the current if reverse bias voltage changes from \[\mathbf{1V}\] to \[2\mathbf{V}\]?
Ans: The current stays almost the same (equal to \(I_0\)) for both reverse bias voltages, making the dynamic resistance very large (almost infinite).

15. You are given the two circuits as shown in figure. Show that current acts as OR gate
Ans: The output of the circuit is \(Y = A + B\), so the circuit acts as an OR gate (it gives output 1 when A or B is 1).

(b) acts as AND gate
Ans: The circuit output is \(Y = A \cdot B\), so the circuit functions as an AND gate (output is 1 only when both A and B are 1).

16. Write the truth table for a NAND Gate connected as given in figure. Hence identify the exact logic operation carried out by this circuit.
Ans: The circuit’s output is \(Y = \overline{A}\), which is NOT operation. The truth table is:

A	Y
0	1
1	0

The operation is a NOT gate.

17. You are given two circuits as shown in figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits.
Ans: In the first circuit, the output is \(Y=AB\), so it acts as an AND gate.

(b)
Ans: In the second circuit, the output is \(Y=A+B\), so it acts as an OR gate.

18. Write the truth table for circuit given in figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
Ans: The truth table matches that of an OR gate.

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

So, this circuit is an OR gate.

19. Write the truth table for the circuits given in the figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
a)
Ans: Only input A is used. Output is \(Y = \overline{A}\) (NOT gate).

A	Y
0	1
1	0

b)
Ans: When both A and B inputs are used, output is \(Y=AB\) (AND gate).

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

So, this one is an AND gate.

Key Concepts Covered in Semiconductor Electronic Material Devices And Simple Circuits Class 12 NCERT Solutions

• Learn about intrinsic and extrinsic semiconductors, and how doping changes their properties.
• Understand how p-n junctions, diodes, and transistors work in simple electronic circuits.
• Practice solving Class 12 Physics Chapter 14 question answer formats, including numericals and logic gate tables.
• See how biasing affects current flow and device characteristics in semiconductor electronics.
• Recognize basic logic gates and their truth tables as used in digital circuits.

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