WBCHSE Class 12 Physics Semiconductor Electronics Long Answer Questions

WBCHSE Class 12 Physics Semiconductor Electronics Long Questions And Answers

Question 1. When a semiconductor is irradiated by light of a definite wavelength, its resistance decreases. Explain why.
Answer:

• In semiconductors, the greater the number produced, the greater its electrical conductivity and hence lesser its resistance.
• The energy of photons of light of definite wavelength is fixed.
• If the energy of photon incident oh a semiconductor is equal to the difference in the energy level between the conduction band and valence band,
• An electron transits from the valence band to the conduction band by generating a hole in the die valence band.
• Due to an increase in the electrical conductivity of a semiconductor resistance decreases

Question 2. What happens if the amount of reverse bias in a p-n junction diode is gradually increased?
Answer:

• If the reverse bias is increased gradually, the majority of carrier electrons of the TL -region and the majority of carrier holes of the p -region move away from the junction of the diode gradually.
• As a result, the thickness of the depletion layer on the two sides of the junction goes on increasing.
• Hence, although the applied potential difference is increased, the current being only due to minority carriers, is negligible.
• If the reverse bias voltage across the z.p-n junction diode is very high, the minority charge carriers get accelerated.
• Due to their high speed, the car. knock out electrons from the covalent bonds and in turn produce a large reverse current.
• This phenomenon is termed a breakdown

WBBSE Class 12 Semiconductor Electronics Long Answer Questions

Question 3. Ax absolute zero temperature, the conductivities of the current p-n junction both the insulators and the semiconductors are zero. At any higher temperature, the semiconductor may have some conductivity but for insulator is zero. FT plain why
Answer:

• At absolute zero temperature, there is no existence of charge carriers in semiconductors or insulators.
• So, their electrical conductivities are zero.
• At a temperature higher than absolute zero, some electron-hole pairs are generated due to thermal energy, and they act as charge carriers in semiconductors.
• Some electrons or holes of the valence band, in p-type and p-type semiconductors respectively, can cross the narrow band gap and reach the conduction band.
• Hence, semiconductors possess some electrical conductivities.
• But at that temperature, no charge carriers are generated inside insulators and hence their thermal conductivities remain zero.

Question 4. You are given two bars of the same resistance; one of them is a conductor and the other is a semiconductor. How will you distinguish them experimentally?
Answer:

• At first, both the rods are connected separately with the same source of electricity at room temperature, and currents passing through them are measured.
• After that, the temperature of both the rods is increased.
• Then they are again connected with the same source of electricity and currents through them are measured.
• The bar through which the flow of current increases due to temperature rise is a semiconductor.

Question 5. In a transistor, the base is made very thin. Explain the reason. reverse
Answer:

• The number density of majority carriers in the base region is low as it is lightly doped.
• When the emitter-base junction is forward-biased, the majority of charge carriers crossing the emitter-base junction reach the base and result in electron-hole recombination in the base region.
• Since it is thin and lightly doped, only a small amount of electron-hole recombination takes place.
• The rest of the majority of charge carriers cross the barrier and reach the collector under the influence of the reverse bias and produce an appreciable current in the collector circuit.

Long Answer Questions on Transistors and Their Applications

Question 6. Why is a transistor called a temperature-sensitive device?
Answer:

• A transistor has free electrons and holes as charge tiers.
• When the emitter-base junction is forward-biased and the collector-base junction is reverse-biased, the charge carriers carry current through the transistor in the external circuit.
• When temperature increases, theÿelectrons gain sufficient energy to break the covalent bonds and thus produce a large amount of current
• .If the transistor is continued to operate at this temperature a strong current would flow through it.
• As a result, the transistor would get heated excessively and ultimately break down.
• Thus the operations of the transistor are restricted by its temperature

Question 7. In a transistor, the forward bias is always small compared to the reverse bias. Why?
Answer:

• In a transistor, the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
• If the forward bias is made large, the velocity of the majority of carriers entering the collector region through the base region would be very large.
• In addition to this, the number of these carriers is also very large.
• So, the heat produced would be large enough to break up the covalent bonds resulting in permanent damage of the i transistor.
• But the reverse bias applied to the collector is larger than this forward bias, this would restrict the majority of carriers coming from the emitter from reaching the collector with such high speed and thus saving the device from damage

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Question 8. Explain why the input resistance of a transistor is low while the output resistance is high.
Answer:

• While using a transistor the emitter-base junction is always forward biased and the collector-base junction is biased.
• For this reason, a small change in emitter voltage produces a large emitter current.
• This means a small variation of the input voltage produces a large variation in the emitter rent. This implies that the input resistance of the transistor is small.
• On the other hand, the collector being reverse-biased, collects all the charge carriers that enter through the base.; So a very large change in the collector voltage produces a very small change in the collector current.
• This implies that the output resistance of the transistor is very high

Question 9. The gain of a common-emitter amplifier is given by A_v= ~g_mR_L. Does it mean that if we keep on increasing- R_L indefinitely, the gain of the amplifier will also Increase? Explain your answer
Answer:

No, the gain of the amplifier will not increase indefinitely on increasing R_L.

We know that, V_CE = V_CC– I_CR_L

Where, V_CE → Collector-emitter voltage

V_CC → Output reverse battery voltage,

I_C → Current through output circuit.

Thus, when R_L is increased V_CE decreases and ultimately becomes less than the base voltage when both the junctions get forward biased and the collector current I_C gets saturated. Once Ic is saturated there will be no increase in the gain.

Practice Long Answer Questions on Logic Gates

Question 10. Under normal use of transistors, the emitter is forward-biased while the collector is reverse-biased. Can either of these biases be changed? Explain.
Answer:

1. If the emitter is reversed biased, no majority charge carrier would reach the base. So, if both the emitter and collector are reverse-biased, there would be no current through the transistor.
2. When both the emitter and collector are forward-biased, the majority of carriers would enter the base from both of these. So two separate circuits, i.e., one emitter-base and the other collector-base circuit will be formed and the purpose of the transistor will not be served.
3. If the emitter is reverse biased and the collector is forward biased, then the transistor will perform in the opposite direction, i.e., the emitter will become the collector and the collector will become the emitter. The base current will increase and the collector current

Question 11.

1. Draw the output characteristic curves of a n – p -n  transistor in a CE configuration and find the output resistance from it.
2. Explain, with a circuit diagram, the action of a transistor as a switch.

Answer: 

For an n-p-n transistor in CE configuration, the graphs of output current I_C concerning output voltage V_CE when the base current I_B is kept fixed at different values. These are called output characteristic curves of the n-p-n transistor. If two points A and B are taken on the same graph, the change in collector-emitter voltage = Δ V_CE and the change in collector current = ΔI_c for these two points.

∴ Output Resistance  = \(\frac{\Delta V_{C E}}{\Delta I_C}\)

Question 12. Derive an expression for the voltage gain of the amplifier and hence show that the output voltage is in the opposite phase with the input voltage
Answer:

Let us consider the input voltage V_i = 0.

ApplyingKirchhoff’s lawto the outputloop we get,

V_CC = V_CE+I_CR_L

In theinput loop, V_BB = V_BE+I_BR_B

When V_i isnot zero, we get,

V_BE+ V_i = V_BE+ I_BR_B +ΔI_B( R_B+r_i)

The change in V_BE can be related to the input resistance r_i and the change in I_B. Hence,

V_i = ΔI_B (R_B+ r_i) = RΔI_B

The change in I_B causes a change in Ic. We define a parameter /?

β = \(\frac{\Delta I_C}{\Delta I_B}=\frac{I_C}{I_B}\) …………………..  (2)

Again, the change in the I_C due to a change in I_B causes a change in V_CE and the voltage drop across the resistor RL because V_CC is fixed.

These changes can be given by equation (1) as,

ΔV_CC = ΔV_CE  + R_LΔI_C = 0

Or, ΔV_CE =  -R ΔI_C

The change in V_CE is the output voltage V₀. We get

V0 = ΔV_CE =  – βR_L ΔI_B

The voltage gain of the amplifier is

A_V = \(\frac{V_o}{V_i}=\frac{\Delta V_{C E}}{R \Delta I_B}=-\frac{\beta R_L}{R}\)

The negative sign represents that the output voltage is in the opposite phase to the input voltage.

Conceptual Long Answer Questions on Intrinsic and Extrinsic Semiconductors

Question 13. Draw V – I characteristics of a p- n junction

1. Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
2. Why does the reverse current show a sudden increase at the critical voltage?

Answer:

1. Up to a certain reverse bias potential, the small current p-n junction is due only to the saturated current produced by the minority carriers. It is independent of the applied potential.

2. Either,

1. When the reverse bias potential is fairly high, the high velocity attained by the minority carriers becomes sufficient to break the crystal bonds and produce a large number of electron-hole pairs. For this increase in the number of charge carriers (avalanche breakdown), the reverse current shows a sudden increase.
2. When the dopings, of the p-n junction are sufficiently high, even a comparatively small reverse potential can produce a high electric field across the junction. As a result, a large number of charge carriers become free (Zener breakdown). So, the reverse current shows a sudden increase.

The semiconductor device used in this case is called a Zener diode.

Question 14. Show how these characteristics can be used to determine output resistance.

The collector current (I_C) of a transistor is a function of the base current (JB) and the collector-emitter voltage (V). The output resistance of a transistor is r₀ = \(\frac{\Delta V_{C E}}{\Delta I_C}\)  when IB is kept constant On the linear graph in the. the active region of the output characteristics, two convenient points are chosen by a line representing a fixed I_B. The coordinates of the points, (V_CE, I_C) and (V’_CE, I’_C) Then r₀ is calculated from the relation \(r_o=\frac{v_{C E}^{\prime}-v_{C E}}{I_C^{\prime}-I_C}\)

Question 15. For a CE transistor amplifier, the audio signal voltage across the collector 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 1kΩ
Answer:

Voltage amplification factor,

A_v = \(\frac{V_o}{V_i}=\beta \frac{R_C}{R_B}\)

∴ V_i = \(\frac{V_o R_B}{\beta R_C}\)

Now, V₀ = 2 V, R_B = 1 kΩ = 1000 Ω

β = 100, R_C = 2kΩ = 2000 Ω

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

V = \(=\frac{V_l}{R_B}=\frac{0.01}{1000}\) = 10^-5 A = 10 μA

Question 16. The 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 V, calculate the output AC signal.
Answer:

Netvoltage gain, A = A₁ × A₂

∴ A = 10 × 20 = 200

Again A = \(\frac{V_o}{V_i}\)

Question 17. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

The energy of the photon of wavelength λ Is

E = hc/ λ

∴ E = \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9}}\)

= 0.206 eV

The band gap of the semiconductor = 2.8 eV

Since the band gap energy is more than, the energy of the light photon,it cannot be detected

Important Definitions and Explanations in Semiconductor Electronics

Question 18. In an intrinsic semiconductor, the energy gap E_g is 1.2 eV. j 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 by n_i = \(n_0 \exp \left(-\frac{E_g}{2 k_B T}\right)\) where n_i is a constant
Answer:

Let K and K be the conductivity of the material at 600  and 300 K respectively.

⇒ \(\frac{K_1}{K_2}=\frac{n_1}{n_2}=\frac{n_0 e^{-\frac{E_g}{2 k_B T_1}}}{n_0 e^{-\frac{E_g}{2 k_B T_2}}}=e^{\frac{E_g}{2 k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}\)

Here, \(\frac{E_g}{2 k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=\frac{1.2}{2 \times 8.6 \times 10^{-5}}\left(\frac{1}{300}-\frac{1}{600}\right)\)

= 11.279

∴ \(\frac{K_1}{K_2}=e^{11.6279}\)

= 1.12 × 10⁵

Question 19. The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m3 of arsenic and 5 × 10²⁰ atoms per m³ of indium. Calculate the number of electrons and holes. Given that n_C

= 1.5 × 10¹⁶m^-3. Is the material n-type or p-type?

n_e = 5 × 10²²  – 5  × 10²⁰

= 4.95 × 10²² m^-3

The number density of hole

n_h = \(\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}\)

= 4.55 ×  10⁹ m^-3

Since n_e>>n_h, the material is of n-type.

Examples of Applications of Semiconductors in Technology

Question 20. For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2kΩ is 2 V. Given the current amplification factor of the transistors 100, find the input signal voltage and base current, if the base resistance 1kΩ.

Given R_C = 2kΩ ,V_CC =  2V , β = 100 , R_β = 1 kΩ

The current amplification factor β= \(\frac{I_C}{I_B}\)

I_C = \(\frac{V_{C C}}{R_C}=\frac{2}{2 \times 10^3}\)

= 1mA

I_B= \(\frac{I_C}{\beta}=\frac{1}{100}\)

= 0.01 mA

= 10 μA

∴  \(\frac{V_o}{V_i}=\frac{\beta R_C}{R_i}\)

Or, \(\frac{2}{V_i}=100 \times \frac{2000}{1000}\)

Or, V_i = 0.01 V