Q. Draw the voltage current characteristic of a zener diode
Q. In the following circuits which one of the two diodes is forward biased and which is reverse biased

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Sol. (i) Reverse biased (ii) Forward biased.
Q. In half wave rectifier, 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.
[NCERT]
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Sol. The output ripple frequency is 50 Hz for half wave rectifier and 100 Hz for full wave rectifier
Q. Which one of the transistors p-n-p and n-p-n is more useful and why?
[NCERT]
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Sol. n-p-n is more useful because in it the current carriers are mainly electrons which are more speedy carriers than holes (the current carriers in p-n-p)
Q. Distinguish between N-type and P-type semiconductors on the basis of energy band diagrams
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Sol. N-type semiconductor : Electrons are majority charge carriers in conduction band and holes are minority charge carriers in valance band. Donor energy level lies just below the conduction band. P-type semiconductor : Electrons are minority charge carriers in conduction band and holes are majority charge carriers in valence band. Impurity (acceptor) energy level lies just above the valence band.

Q. Draw a circuit diagram to show how a photo diode is biased. Draw it’s characteristic curves for two different illumination intensities
Q. What do the acronyms ‘LASER’ and ‘LED’ stand for? Name the factor which determines (i) Frequency, and (ii) Intensity of light emitted by LED.
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Sol. LASER stands for light amplification by stimulated emission of radiation. LED stands for light emitting diode (i) The frequency of light emitted by an LED depends on the band gap of the semiconductor used in LED. (ii) The intensity of light depends on the doping level of the semiconductor used.
Q. Explain how the width of depletion layer in a p-n junction diode changes when the junction is (i) forward biased (ii) reverse biased.
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Sol. When a p-n junction is forward biased, the majority charge carries flow towards the junction, the width of the depletion layer decreases, as shown on next page. (iii) Semiconductors : In semiconsuctors, a finite but small band gap $\left(E_{g}<3 e V\right)$ exists between valence band and conduction band. Because of the small band gap, at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the conduction band. For Ge band gap is 0.7 eV and for silicon it is about 1.1 eV

Q. Explain how an intrinsic semiconductor can be converted into (i) N-type and (ii) P-type semiconductor. Give one example of each and their energy band diagrams.
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Sol. (i) N-Type Semiconductor : Suppose a pure semiconductor say silicon in which each atom has four valence electrons is doped with pentavalent impurity say Arsenic (As). The four of the five valence electrons of impurity atom will form covalent bonds by sharing the electrons with the adjoining four atoms of silicon, while the fifth electron is very loosely bound with the parent impurity atom and is comparatively free to move. Thus each impurity atom donate one free electron to the crystal, that is why impurity is called donor impurity. On giving one electron, the donor atom becomes positively charged. However, the crystal remains electrically neutral as a whole. The semiconductor so formed is called N-type semi conductor. In N-type semiconductors, electrons are majority charge carriers and holes are minority charge carriers.


Q. What is zener diode? how is it symbolically represented? With the help of a circuit diagram, explain the use of zener diode as a voltage stabilizer
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Sol. Zener diode : The specially designed junction diodes, which can operate in the reverse breakdown voltage region continuously without getting damaged, are called zener diodes.


Q. Explain briefly, with the help of circuit diagram, how V-I characteristics of a p-n junction diode are obtained in (i) Forward bias, and (ii) Reverse bias. Draw the shape of the curves obtained
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Sol. Figures below shows a forward biased p-n junction and its voltage-current graph


Q. Explain (i) Forward biasing (ii) Reverse biasing of a P-N junction diode. With the help of a circuit diagram, explain the use of this device as a half-wave rectifier
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Sol. Forward biasing : A P-N junction is said to be forward biased if the positive terminal of the external battery is connected to the P-side and the negative terminal to the n-side of P-N junction. In forward biasing, potential barrier height is reduced and width of depletion layer decreases. In forward biasing the resistance of P-N junction is low to the flow of current.




Q. A transistor is a temperature sensitive device. Explain
[NCERT]
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Sol. In a transistor, free electrons and holes are the charge carriers and are responsible for the current through the transistor as well as in the external circuit. If temperature rises, more covalent bonds are broken in the semi-conducting material of the transistor giving rise to additional free electrons and holes. The results in larger current in the transistor and in the external circuit. The effect may be cumulative resulting in excessive heat and ultimately in permanent damage of transistor structure.
Q. Explain with the help of a labelled circuit diagram the use of a transistor as an oscillator
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Sol. Circuit diagram

Q. Explain with the help of a labelled circuit diagram, the use of n-p-n transistor as an amplifier in common emitter configuration. Explain how the input and output voltages are out of phase by $180^{\circ}$ for a common emitter transistor amplifier
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Sol.

Q. Draw a circuit diagram to obtain the characteristics of a NPN transistor in common emitter configuration. Describe how you will obtain input and output characteristics. Give shape of the curves. Define the term (i) Output resistance and (ii) Current amplification factor
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Sol. Circuit diagram


Q. A photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect on wavelength of 6000 nm ? Justify
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Sol. $\lambda=6000 n m=6 \times 10^{-6} \mathrm{m}$ $\therefore E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-6}}=3.3 \times 10^{-20} J$ $=\frac{3.3 \times 10^{-20}}{1.6 \times 10^{-19}}=0.206 e V$ As the energy of the photon is less than $E_{g}(=2.8 e V)$ of the semiconductor, so a wavelength of 6000 Å cannot be detected.
Q. A semiconductor has equal electron and hole concentration of $6 \times 10^{8} / \mathrm{m}^{3}$. On doping with certain impurity, electron concentration increases to $9 \times 10^{12} / m^{3}$ (i) Identify the new semiconductor obtained after doping
(ii) Calculate the new hole concentration.
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Sol. $n_{i}=6 \times 10^{8} / m^{3}, n_{e}=9 \times 10^{12} / m^{3}$ As $n_{e} n_{h}=n_{i}^{2}$ $\therefore n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left(6 \times 10^{8}\right)^{2}}{9 \times 10^{12}}=4 \times 10^{4} / m^{3}$ As the new electron concentration is greater than the new hole concentration, so the new semiconductor is of n-type
Q. The input resistance of a transistor is 1000 $\Omega$. On changing its base current by $10 \mu \mathrm{A}$, the collector current increases by 2 mA. If a load resistance of 5 $K \Omega$is used in the circuit, calculate.
(i) The current gain (ii) Voltage gain of the amplifier
Q. Suppose a pure Si crystal has $5 \times 10^{28}$ atoms $m^{-3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_{1}=1.5 \times 10^{16} \mathrm{m}^{-3}$
[NCERT]
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Sol. Note that thermally generated electrons $\left(n_{i} \approx 10^{16} m^{-3}\right)$ are negligibly small as compared to those produced by doping. Therefore $n_{e}=N_{D}$

Q. 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 V, calculate the output ac signal
[NCERT]
Q. For a common emitter transistor amplifier, the audio signal voltage across the collector resistance of $2 k \Omega$is 2V. If the current amplification factor of the transistor is 100 calculate (i) Input signal voltage, (ii) Base current and (iii) power gain. Given that the value of the base resistance is 1 $k \Omega$
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Sol. $R_{C}=2 k \Omega=2000 \Omega, R_{B}=1 k \Omega=1000 \Omega, \beta=100, V_{0}=2 V$

Q. An n-p-n transistor in a common-emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of a 8 V battery is connected to the collector through a load resistance
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Sol. For the given combination of gates, the truth table is

Q. The output of an OR gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table.
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Sol. Logic circuit of the combination is shown below

Q. Identify the logic gates marked X, Y in the following figure. Write down the output at Z, when A=1, B=1 and A = 0, B = 1

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Sol. X is NAND gate and Y is OR gate

Q. In the figure below, circuit symbol of a logic gate and two input waveforms A and B are shown.
(i) Name the logic gate (ii) Write its truth table
(iii) Give the output waveform

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Sol. (i) The logic gate is AND gate. (ii) Truth table of AND gate


Q. Give the logic symbol for an OR gate. Draw the output wave form for input wave forms A and B for this gate

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Sol. Logic symbol of OR gate is given below

Q. Two signals A and B shown in the given figure are used as two inputs of a NAND gate. Draw its output wave form. Give the logic symbol of NAND gate.
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Sol. The output waveform of NAND gate is shown below

Q. In the figure, circuit symbol of logic gate and input wave forms are given. Name the logic gate. Write its truth table and give the output wave form.

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Sol. The logic gate is NOR gate. The output waveform is shown below.


Q. Show that a bubbled AND gate is equivalent to a NOR gates. Hence prove the identity
[NCERT]
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Sol. A bubbled AND gate is shown in figure (A) which is the combination of two NOT gates and one AND gate as shown in figure (B).

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