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Chapter: 12th Physics : UNIT 10a : Semiconductor Electronics

Example Solved Numerical Problems

Physics : Semiconductor Electronics: Book Back, Exercise, Example Numerical Question with Answers, Solution: Example Solved Numerical Problems with Answers, Solution

Junction diode - Numerical Problems Questions with Answers, Solution

EXAMPLE 9. 1

An ideal diode and a 5 Ω resistor are connected in series with a 15 V power supply as shown in figure below. Calculate the current that flows through the diode.


Solution

The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage. Therefore, current that flows through the diode can be calculated using Ohm’s law.

IR

=V/R = 15/5 = 3 A

 

EXAMPLE 9. 2

Consider an ideal junction diode. Find the value of current flowing through AB is


Solution

The barrier potential of the diode is neglected as it is an ideal diode.

The value of current flowing through AB can be obtained by using Ohm’s law

V/R = 3 - (-7) / 1× 103 /10= 10 =10-2 = 10mA


Zener diode - Numerical Problems Questions with Answers, Solution


EXAMPLE: 9.3

Find the current through the Zener diode when the load resistance is 1 KΩ. Use diode approximation.


Solution

Voltage across AB is VZ = 9V

Voltage drop across R = 15 - 9 = 6V

Therefore current through the resistor R,

= 6 / 1×10=6 mA

Voltage across the load resistor = VAB = 9V

Current through load resistor

IL = VAB/RL = 9 / 2×10= 4.5 mA

The current through the Zener diode,

IZ IL =6 mA− 4.5mA =1.5 mA


Optoelectronic devices using Diodes - Numerical Problems Questions with Answers, Solution


EXAMPLE 9.4

Determine the wavelength of light emitted from LED which is made up of GaAsP semiconductor whose forbidden energy gap is 1.875 eV. Mention the colour of the light emitted (Take = 6.6 × 10-34 Js).

Solution

Eg hc / λ

Therefore

λ= hc / Eg  = 6.6×10−34 ×3×108  / 1.875×1.6×10−19

= 660 nm

The wavelength 660 nm corresponds to red colour light.


Transistor action in the common base mode - Numerical Problems Questions with Answers, Solution


EXAMPLE 9. 5

In a transistor connected in the common base configuration, α=0.95 , IE =1 mA . Calculate the values of IC and IB .

Solution

α= IC/IE

IC =α IE =0.95×1=0.95 mA

IE IB IC

 IB IC IE =1−0.95=0.05 mA


Transistor in Common Emitter Mode - Numerical Problems Questions with Answers, Solution


EXAMPLE 9. 6

The output characteristics of a transistor connected in common emitter mode is shown in the figure. Determine the value of IC when VCE = 15 V. Also determine the value of IC when VCE is changed to 10 V.


When VCE = 15 VIC = 1.5 μA

When VCE is changed to 10 V,  IC = 1.4 μA

The collector current is independent of the collector- emitter voltage in the active region.

 

EXAMPLE 9.7

In the circuit shown in the figure, the input voltage Vi is 20 VVBE = 0 and VCE = 0 V. What are the values of IB , IC , β?


Bipolar Junction Transistor [BJT] - Numerical Problems Questions with Answers, Solution


EXAMPLE: 9.8

The current gain of a common emitter transistor circuit shown in figure is 120. Draw the dc load line and mark the Q point on it. (VBE to be ignored).


Solution

β = 120


Transistor as an oscillator - Numerical Problems Questions with Answers, Solution


EXAMPLE 9. 9

Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 μH. The frequency band is from 500 kHz to 1500 kHz.



Logic gates - Numerical Problems Questions with Answers, Solution



EXAMPLE 9. 10

What is the output Y in the following circuit, when all the three inputs A, B, and C are first 0 and then 1?


Solution


 

EXAMPLE 9. 11

In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B.


Solution

The output at the 1st AND gate: A

The output at the 2nd AND gate: 

The output at the OR gate: A B


De Morgan’s Theorem - Numerical Problems Questions with Answers, Solution


EXAMPLE: 9 . 12

Simplify the Boolean identity

AC + ABC = AC

Solution

Step 1: AC (1 + B) = AC.1 [OR law-2]

Step 2: AC . 1 = AC [AND law – 2]

Therefore, AC + ABC = AC

Circuit Description


Thus the given statement is proved.


Space Wave Propagation - Numerical Problems Questions with Answers, Solution


EXAMPLE 10.1

A transmitting antenna has a height of 40 m and the height of the receiving antenna is 30 m. What is the maximum distance between them for line-of-sight communication? The radius of the earth is 6.4×106 m.


Solution:

The total distance between the transmitting and receiving antennas will be the sum of the individual distances of coverage.

d1 +d2

= √(2Rh­) + √(2Rh2)

= √2√(√h1 + h2)

= √[2 × 6.4 ×106] × ( √40 + √30 )

=16 ×102 √5 ×(6.32 + 5.48)

=42217m= 42.217 km


Numerical Problems


1. The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R1 [Ans: 2.5 A]


Solution:

Barrier potential for ideal diode is zero. The diode D1 is reverse biased, so it will block the current and diode Dis forward biased, so it will pass the current.

The given circuit becomes


Effective resistance Reff = R1 + R3 = 4Ω

Current through R1 = V/Reff = 10/4 = 2.5A


2. Four silicon diodes and a 10 Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 18Ω resistor. [Ans: 0.13 A]


Solution:

In the given circuit D2 & D3 are in forward bias so they conduct current while D1 &D4 are in reverse bias so they do not conduct current. So the equivalents circuit will be


the effective resistance is Reff

= 1Ω + 10Ω + 1Ω = 12Ω

Here silicon diodes are used

 Barrier potential for Si is 0.7 V

Net potential (Vnet) = 3 – 0.7 - 0.7

Vnet = 1.6 V

Current (I) = Vnet / R eff = 1.6 / 12 = 0.133A


3. Assuming VCEsat = 0.2 V and β = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation. [Ans: 56 µA]


Solution;

VCE = 0.2V

Rc = 1kΩ

β =50

Vcc = 3V

I B = ?


Ic = [ VCC – VCE ] / RC

= (3-0.2) / 103

= 2.8 × 10-3 A

= 2.8 mA

β= IC/IB

 IB = IC / β = 2.8 × 10-3 / 50 = 0.056× 10-3

= 56 × 10-6 A

IB=56μA


4. A transistor having α =0.99 and VBE = 0.7V, is given in the circuit. Find the value of the collector current. 

[Ans: 5.33 mA]


Given data:

α =0.99

VBE = 0.7V

IC=?

Solution:


Tranistor is in saturation region.

 Ic = Ic(sat)

IC and IB are independent

VCE(sat) = 0.2 V, VBE(sat) = 0.8 V for silicon transistor (ie, standard value)

Apply KVR across B-E loop:

V1k + V10k + VBE sat + V1k = VCC

 1(IC + IB) + 10 IB + 0.8 + 1 (IC + IB) = 12

2 IC + 12 IB = 11.2         ……………(1)

Apply KVR across C-E loop:

V1k + V1k + VCE sat + V1k = VCC

l(IC + IB) + 1 IC + 0.2 + I(IC + IB) = 12

3IC + 2IB = 11.8    ……………….(2)

Solve the equation (1) and (2)

IB = 0.3125 mA

IC = 3.725 mA ≈ 3.73 mA


5. In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter – base voltage VEB = 600 mV, calculate the emitter – collector voltage VEC (in volts). [Ans: 2 V]


Given data:

 β = 50

VE=3V

VEB=60 mV

RB=60KΩ

RE=500Ω

Solution:

VB=VE−VEB

VB = 3−0.6 = 2.4V

I= VB/RB = 2.4 / 60×103 = 40μA

I= βIB = 50×40μA = 2×10-3 A = 2mA

VC = RCIC = 500×2×10-3 = 1V

VEC = VE-V= 3-1 = 2V

VEC = 2V


6. Determine the current flowing through 3Ω and 4Ω resistors of the circuit given below. Assume that diodes D1 and D2 are ideal diodes.


Solution:

The diode D2 is in reverse biased. So does not conduct current.

∴ current through 3Ω is = 0

The diode Dis in forward biased, and it is an ideal diode. So the given circuit becomes as


The current through 4Ω is

∴ I = V/R = 12/6 =2A

 

7. Prove the following Boolean expressions using the laws and theorems of Boolean algebra.

i) (A + B) (A + ) = A

ii) A  ( +B) = AB

iii) (A + B) (A + C) = A + BC

Solution:


 

8. Verify the given Boolean equation A +  ᾹB = A + B using truth table.

Solution:


Hence, verified

 

9. In the given figure of a voltage regulator, a Zener diode of breakdown voltage 15V is employed. Determine the current through the load resistance, the total current and the current through the diode. Use diode approximation.


Solution:

Voltage across RL(VO) = Vz = 15V

 Voltage across RS(VRS) = 25 -15 = 10V

current through RL is

IL = V0/RL = 15 / 3×103 = 5×10-3 A

I= 5 mA

Current through RS is

I = VRS/RS = 10/500 = 20×10-3A

I = 20mA

Current through Zener diode is

IZ=I-IL

= (20-5) × 10-3

Iz = 15mA

 

10. Write down Boolean equation for the output Y of the given circuit and give its truth table.


Solution:


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