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

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

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 Ohmâ€™s law.

*V *= *IR*

*I *=*V/R *= 15/5 = 3 *A*

EXAMPLE 9. 2

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

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 Ohmâ€™s law

*I *= *V/R *= 3 - (-7) / 1Ã— 103 /103 = 10 =10-2 *A *= 10*mA*

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

Voltage across AB is VZ = 9V

Voltage drop across R = 15 - 9 = 6V

Therefore current through the resistor R,

*I *= 6 / 1Ã—103 =6 *mA*

Voltage across the load resistor = *VAB *= 9*V*

Current through load resistor

*IL *= VAB/RL = 9 / 2Ã—103 = 4.5 *mA*

The current through the Zener diode,

*IZ *= *I *âˆ’*IL *=6 *mA*âˆ’ 4.5*mA *=1.5 *mA*

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 *h *= 6.6 Ã— 10-34 Js).

Solution

*Eg *= *hc **/ Î»*

Therefore

Î»= *hc / Eg* = 6.6Ã—10âˆ’34 Ã—3Ã—108 / 1.875Ã—1.6Ã—10âˆ’19

= 660 *nm*

The wavelength 660 nm corresponds to red colour light.

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

Î±= *IC**/IE*

*IC *=Î± *IE *=0.95Ã—1=0.95 *mA*

*IE *= *IB *+ *IC*

âˆ´ *IB *= *IC *âˆ’*IE *=1âˆ’0.95=0.05 *mA*

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 *V*, *IC *= 1.5 *Î¼*A

When *VCE *is changed to 10 *V*, *IC *= 1.4 *Î¼*A

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 *V*, *VBE *= 0 *V *and *VCE *= 0 *V*. What are the values of *IB , IC , Î²*?

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

*Î² *= 120

Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 Î¼H. The frequency band is from 500 kHz to 1500 kHz.

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?

EXAMPLE 9. 11

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

The output at the 1st AND gate: *A*

The output at the 2nd AND gate: *B *

The output at the OR gate: *Y *= *A*. * *+ . *B*

__De Morganâ€™s Theorem - Numerical Problems Questions
with Answers, Solution__

Simplify the Boolean identity

AC + ABC = AC

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

Step 2: AC . 1 = AC [AND law â€“ 2]

Therefore, AC + ABC = AC

*Circuit Description*

Thus the given statement is proved.

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 6.4Ã—106 m.

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

*d *= *d*1 +*d*2

= âˆš(2*Rh*Â) + âˆš(2*Rh*2)

= âˆš2âˆš*R *(âˆš*h*1 +* *âˆš*h*2)

= âˆš[2 Ã— 6.4 Ã—106] Ã— ( âˆš40 + âˆš30 )

=16 Ã—102 âˆš5 Ã—(6.32 + 5.48)

=42217*m*= 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 D2 is forward biased, so it will pass the current.

The given circuit becomes

Effective resistance Reff = R1 + R3 = 4Î©

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

2. Four silicon diodes and a 10 Î© resistor are connected as shown in figure below. Each diode has a resistance of 1Î©. Find the current flows through the 18Î© 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

= 1Î© + 10Î© + 1Î© = 12Î©

Here silicon diodes are used

âˆ´ Barrier potential for Si is 0.7 V

Net potential (Vnet) = 3 â€“ 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 Î² = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation. [Ans: 56 ÂµA]

Solution;

VCE = 0.2V

Rc = 1*k*Î©

Î² =50

Vcc = 3V

I B = ?

Ic = [ VCC â€“ VCE ] / RC

= (3-0.2) / 103

= 2.8 Ã— 10-3 A

= 2.8 *m*A

Î²= IC/IB

âˆ´ IB = IC / Î² = 2.8 Ã— 10-3 / 50 = 0.056Ã— 10-3

= 56 Ã— 10-6 A

IB=56Î¼A

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

[Ans: 5.33 mA]

Given data:

Î± =0.99

VBE = 0.7V

IC=?

Solution:

Tranistor is in saturation region.

âˆ´ 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

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

2 IC + 12 IB = 11.2 â€¦â€¦â€¦â€¦â€¦(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 â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Solve the equation (1) and (2)

IB = 0.3125 mA

IC = 3.725 mA â‰ˆ 3.73 mA

5. In the circuit shown in the figure, the BJT has a current gain (Î²) of 50. For an emitter â€“ base voltage VEB = 600 mV, calculate the emitter â€“ collector voltage VEC (in volts). [Ans: 2 V]

Given data:

Î² = 50

VE=3V

VEB=60 *m*V

RB=60KÎ©

RE=500Î©

Solution:

VB=VEâˆ’VEB

VB = 3âˆ’0.6 = 2.4V

IB = VB/RB = 2.4 / 60Ã—103 = 40Î¼A

IC = Î²IB = 50Ã—40Î¼A = 2Ã—10-3 A = 2*m*A

VC = RCIC = 500Ã—2Ã—10-3 = 1V

VEC = VE-VC = 3-1 = 2V

VEC = 2V

6. Determine the current flowing through 3Î© and 4Î© 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.

âˆ´ current through 3Î© is = 0

The diode D1 is in forward biased, and it is an ideal diode. So the given circuit becomes as

The current through 4Î© is

âˆ´ *I = V/R* = 12/6 =2*A*

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 + á¾¹B = 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 / 3Ã—103 = 5Ã—10-3 A

IL = 5 mA

Current through RS is

I = VRS/RS = 10/500 = 20Ã—10-3A

I = 20mA

Current through Zener diode is

IZ=I-IL

= (20-5) Ã— 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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