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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 70

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 70
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The logic which produces LOW output when one of the input is HIGH and produces HIGH output only when all its inputs are LOW is called _.
    Solution

  • Question 2
    1 / -0
    In the resistance box, a double wired coil is fixed as shown in the figure so that :-

    Solution

  • Question 3
    1 / -0
    For a diode connected in parallel with a resistor, which is the most likely current (I) -voltage (V) characteristics?

    Solution
    When the value of V is +ve, the diode is forward biased. Thus, the current flowing through the diode is exponential and the current in the resistor varies linearly. Thus, the V−I characteristic is an exponential curve when $$V>0$$
    When the value of V is -ve, the diode is reverse biased. Thus, no current flows through it and the current in the resistor varies linearly with voltage (Ohm's law). Thus, the V−I characteristic is a straight line when $$V<0$$
    So, the correct answer is option (A).

  • Question 4
    1 / -0
    The output of the given logic circuit is 

    Solution

  • Question 5
    1 / -0
    In figure the approximate value of $$V_o$$ across the diode is 

  • Question 6
    1 / -0
    If in a p-n junction diode, a square input signal of $$10V$$ is applied as shown. Then, the output signal across $${R}_{L}$$ will be-

    Solution

  • Question 7
    1 / -0
    Shown in the figure is a combination of logic gates. The output values at $$P$$ and $$Q$$ are correctly represented by which of the following?

    Solution

  • Question 8
    1 / -0
    Which one is in forward bias :
    Solution
    In forward bias of a diode the $$P$$-end must be connected to the positive terminal of the battery and the $$N$$-end is connected to the negative terminal of the battery.
    So, option (B) is correct.

  • Question 9
    1 / -0
    Which of the following P-N junction diode is reverse biased
    Solution
    Current only flows from a higher potential to a lower potential. Also, the ends of the circuit that are grounded remain at zero potential.
    In cases (A), (C) & (D), the diodes are connected in the forward bias.
    In case (B), the $${\text{p - }}$$side is connected to the lower potential, and $${\text{n - }}$$side is connected to the higher potential, suggesting that the diode is in reverse bias.

  • Question 10
    1 / -0

    The resistivity of a pure semiconductor is  $$0.5 \Omega\ m$$. If the electron and hole mobility be $$0.39$$ $$m^2/V-s$$ and $$0.19$$ $$m^2/V-s$$ respectively then calculate the intrinsic carrier concentration.

    Solution
    The resistivity of the semiconductor, $$\rho  = 0.5{\text{ }}\Omega m$$
    The charge of electron, $${q_e} = 1.6 \times {10^{ - 19}}{\text{ C}}$$
    The electron mobility, $${\mu _e} = 0.39{\text{ }}{{\text{m}}^2}{{\text{V}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$$
    The hole mobility, $${\mu _h} = 0.19{\text{ }}{{\text{m}}^2}{{\text{V}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$$
    Now,
    $$\frac{1}{\mu } = {q_e}\left( {{n_i}{\mu _e} + {n_i}{\mu _h}} \right)$$
    $${n_i} = \frac{1}{{\mu .{q_e}\left( {{\mu _e} + {\mu _h}} \right)}}$$
    $${n_i} = \frac{1}{{\left( {0.5{\text{ }}\Omega m} \right).\left( {1.6 \times {{10}^{ - 19}}{\text{ C}}} \right)\left( {\left( {0.39 + 0.19} \right){{\text{m}}^2}{{\text{V}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}} \right)}}$$
    $${n_i} = 4.64 \times {10^{18}}{\text{ }}{{\text{m}}^{ - 3}}$$
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