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**Very Short Answer Type Questions**

**Q.17. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?Ans.** When pure semiconductor material is mixed with small amounts of certain specific impurities with valency different from that of the parent material, the number of mobile electrons/holes drastically changes. The process of addition of impurity is called doping. The size of the dopant atom should be compatible such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Silicon or Germanium atoms, which are provided by group XIII or group XV elements.

Ans.

In conductors, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for semiconductor the energy gap is moderate.

The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV related to their atomic size. Therefore Sn is a conductor, C is an insulator, and Ge and Si are semiconductors

On the average the potential barrier in P-N junction is ~0.5 V.

**Q.20. Draw the output waveform across the resistor (Figure).Ans.** The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased. So the output is obtained only when positive input is given,so the output waveform is

(i) if dc supply voltage is 10V?

(ii) if dc supply voltage is 5V?

Ans.

According to the problem, voltage gain in X, v

voltage gain in Y; v

voltage gain in Z, v

ΔV

And Total voltage amplification = v

ΔV

= 10 *20 * 30 * 10

(i) If DC supply voltage is 10 V, then output is 6 V, since theoretical gain is equal to practical gain, i.e., output can never be greater than 6 V.

(ii) If DC supply voltage is 5 V, i.e., V

Ans.

(i) ac current gain:

(ii) dc current gain :

(iii) Voltage gain:

(iv) Power gain:

The power gain is very high in CE transistor amplifier. In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violate the law of conservation.

**SHORT ANSWER TYPE QUESTIONS**

**Q.23.**

**(i) Name the type of a diode whose characteristics are shown in Figure (A) and Fig. (B). ****(ii) What does the point P in Figure ****(A) represent? ****(iii) What does the points P and Q in Figure ****(B) represent?****Ans.**

(i) Figure (a) represents the characteristics of Zener diode and curve (b) is of solar cell.

(ii) In figure (a), point P represents Zener breakdown voltage.

(iii) In figure (b), the point Q represents zero voltage and negative current. Which means the light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point Q the battery is short circuited. Hence it represents the short circuit current.

And the point Pin figure (b) represents some open circuit, voltage on solar cell with zero current through solar cell.

It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.**Q.24. Three photo diodes D _{1}, D_{2} and D_{3} are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength ?**

Ans.

According to the problem,

Wavelength of light

energy of the light photon

The incident radiation which is detected by the photodiode D

Ans.

so, I

Basic current

So, R

Now, the current in ammeter is collector current I

I

Ans.

This means that if any one input is high, output will high otherwise low.

The device is shown like this:

So, OR gate gives the desired output

Ans.

(1) It has only one input and only one output.

(2) Boolean expression is Y = Ᾱ and is read as “y equals not A” .

Logical symbol of NOT gate.

(3) Realization of NOT gate: The transistor is so biased that the collector voltage V

The resistors R and R

If the input is high, the transistor current is in saturation and the net voltage at the output Y is 0 (in state 0)

(4) Truth table for Not gate:

Ans.

Specially designed diodes, which give out light radiations when forward biases. LED 's are made of GaAsp, Gap etc.

These are forward biased P-N junctions which emits spontaneous radiation.

In elemental semiconductor, the band gap is such that the emission arc in infrared region and not in visible region.

for Si; E

for Ge; E

**Q.29. Write the truth table for the circuit shown in Figure. Name the gate that the circuit resembles.Ans.**

(1) it has two inputs (A and B) and only one output (Y)

(2) Boolean expression is Y = A.B is read as "Y equals A AND B"

(3) Realization of AND gate

(i) A = 0 , B = 0

The voltage supply through R is forward biasing diodes D_{1} and D_{2} (offers low resistance), the voltage V would drop across R.

The output voltage at Y= the voltage across diode = 0

(ii) A = 0, B = I

D_{1 = }Conducts, D2 = Not Conducts

The out voltage at Y= The voltage across the diode (D_{2}) = 0

(iii) A = 1 , B = 0

D_{1} = Conducts, D_{2} = Not conducts

The out voltage at Y= The voltage across the diode (D_{2}) = 0

(iv) A = 1 , B = 1

None of the diode conducts

The out voltage at Y= Battery voltage = I

(4) Truth table for ‘AND' gate

**Q.30. A Zener of power rating 1 W is to be used as a voltage regulator. If zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7V, what should be the value of Rs for safe operation (Figure)?Ans.**

According to the problem power = 1 W

Zener breakdown voltage, V

Minimum voltage, V

Maximum voltage, V

We know, P = VI

for safe operations R

**Long Answer Type Questions**

**Q.31. If each diode in Figure has a forward bias resistance of 25Ω and infinite resistance in reverse bias, what will be the values of the current I _{1}, I_{2}, I_{3} and I_{4}?**

Ans.

According to the problem, forward biased resistance = 25 Ω and reverse biased resistance = ∞.

As shown in the figure, the diode in branch CD is in reverse biased which having infinite resistance.

So, current in that branch is zero, i.e. I

Resistance in branch AB = 25 + 125 = 150 Ω, say R

Resistance in branch EF - 25 + 125 = 150 Ω, say R

AB is parallel to EF.

So, effective resistance

⇒ R' = 75Ω

Total resistance R of the circuit - R' + 25 = 75 + 25 = 100 U.

According to the kirchoff’s, current law (K.CL),

I

So, I

Here, the resistances R

i.e., I

and I

Therefore, we get, I

Ans.

V

Ans.

**(b)**

**Select the values of R _{B} and R_{C} for a transistor whose V_{BE} = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in Figure (b).** For output characteristic at point Q

Given that the input impedance of the transistor is very small and V_{CC} = V_{BB} = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.

Ans.

V

Applying Kirchhoff’s law in collector-emitter loop

V

RC= 2000Ω

Now applying Kirchhoff’s loop law in base-emitter circuit,

V_{BB}=I_{B}R_{B }+ V_{BE}

Average Voltage A_{V}

(-) sign shows change in phase angle of output is by input voltage.

Power gain = I.V**Q.35. Assuming the ideal diode, draw the output waveform for the circuit given in Figure Explain the waveform.**

In reverse biased when the input voltage is equal to or less than 5 V diode,then it will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.

If input voltage is greater than +5 V, diode is in conducting state, then it will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.

When input voltage is negative, there will be opposition to 5 V battery in p-n junction input voltage becomes more than -5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals.

The output waveform will be like this (as shown below).

(i) Calculate the densities of the charge carriers in the n & p regions.

(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

Ans.

When As (pentavalent) is added to Si the n-type water is created.

So the number of majority carriers in n-type water,

For number of minority carriers n

=0.3 × 1.5 × 10

When boron(trivalent) is implanted in Si crystal, p- type wafer is formed with number of holes,

n

n

Minority carrier in p-type water

=2.25 × 10

When reversed bias is applied on p-n junction then the minority charge carrier moves toward depletion layer i.e., holes n

It is represented by following logic relation

Build this gate using AND, OR and NOT gates.

When

Y

Now Y can be obtained as output from OR gate, where Y

Thus, the logic circuit of this relation is given below.

**Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.****A student performs an experiment in which any two of these three terminals are connected in the circuit shown in (b). (b)The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.The graphs are(i) when A is positive and B is negative(c)(ii) when A is negative and B is positive**

**(d)**(iii) When B is negative and C is positive**(e)**(iv) When B is positive and C is negative**(f)**(v) When A is positive and C is negative**(g)**(vi) When A is negative and C is positive**(h)**From these graphs of current – voltage characteristic shown in Figure (c) to (h), determine the arrangement of components between A, B and C.**Ans. **The V-I characteristics of these graph is discussed in points:

(a) In V-I graph of condition (i), a reverse characteristics is shown in figure (c). Here A is connected to n-side of p-n junction I and B is connected top-side of p-n junction I with a resistance in series.

(b) In V-I graph of condition (ii), a forward characteristics is shown in figure (d), where 0.7 V is the knee voltage of p-n junction I. 1/slope = (1/1000) Ω.

It means A is connected to n-side of p-n junction I and B is connected to p-side of p-n junction I and resistance R is in series of p-n junction I between A and B.

(c) In V-I graph of condition (iii), a forward characteristics is shown in figure (e) , where 0.7 V is the knee voltage. In this case p-side of p-n junction II is connected to C and n-side of p-n junction II to B.

(d) In V-I graphs of conditions (iv), (v), (vi) also concludes the above connection of p-n junctions I and II along with a resistance R.

Thus, the arrangement of p-n I, p-n II and resistance R between A, B and C will be as shown in the figure.

**Q.39. For the transistor circuit shown in Figure, evaluate V _{E}, R_{B}, R_{E} given I_{C} = 1 mA, V_{CE} = 3V, V_{BE} = 0.5 V and V_{CC} = 12 V, β = 100.**

**Ans. **Let us redraw the circuit diagram given here to solve this problem.

As we know the base current is very small. So,

I_{c} ≈ I_{E}

R_{c} = 7.8 kΩ

From the figure, I_{C}(R_{C} + R_{E}) + V_{CE} = 12

(R_{E} + R_{C}) x 1 x 10^{-3} + 3= 1 2

R_{E} + R_{C} = 9 x I0^{3} = 9 k Ω

R_{E} = 9 - 7.8 = 1.2 kΩ

V_{E} = I_{E} x R_{E}

= l x 10^{-3} x 1.2 x 10^{3} = 1.2 V

Voltage V_{B} = V_{E} +V_{BE}= 1. 2+ 0.5 = 1.7 V**Q.40. In the circuit shown in Figure, find the value of RC.**

**Ans.** Let us consider the circuit diagram to solve this problem,

I_{E} = I_{C} + I_{B} and I_{C} = βI_{B} ..(i)

I_{C}R_{C} + V_{CE} + I_{E}R_{E} = V_{CC} ..(ii)

RI_{B} + V_{BE} + I_{E}R_{E} = V_{CC} ..(iii)

∴ I_{E}≈I_{C} = βI_{B}

from (iii)

(R + βR_{E})I_{B} = V_{CC }- V_{BE}^{}

from (ii),^{}^{}^{}

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