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

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 D_{1} is reverse biased,
so it will block the current and diode D_{2 }is forward biased, so it
will pass the current.

The
given circuit becomes

Effective
resistance R_{eff} = R_{1} + R_{3} = 4Î©

Current
through R_{1} = V/R_{eff} = 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 D_{2} & D_{3} are in forward bias so they
conduct current while D_{1} &D_{4} are in reverse bias so
they do not conduct current. So the equivalents circuit will be

the
effective resistance is R_{eff}

=
1Î© + 10Î© + 1Î© = 12Î©

Here
silicon diodes are used

âˆ´ Barrier potential for Si is 0.7
V

Net
potential (V_{net}) = 3 â€“ 0.7 - 0.7

V_{net}
= 1.6 V

Current
(I) = V_{net} / 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;**

V_{CE}
= 0.2V

R_{c}
= 1*k*Î©

Î²
=50

Vcc
= 3V_{ }

I_{
B} = ?

Ic
= [ V_{CC} â€“ V_{CE} ] / R_{C}

=
(3-0.2) / 10^{3}

=
2.8 Ã— 10^{-3} A

=
2.8 *m*A

Î²=
I_{C}/I_{B}

âˆ´ I_{B} = I_{C} /
Î² = 2.8 Ã— 10^{-3} / 50 = 0.056Ã— 10^{-3}

=
56 Ã— 10^{-6 }A

I_{B}=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

V_{BE}
= 0.7V

I_{C}=?

**Solution:**

Tranistor
is in saturation region.

âˆ´ I_{c = }I_{c(sat)}

I_{C}
and I_{B} are independent

V_{CE(sat)}
= 0.2 V, V_{BE(sat)} = 0.8 V for silicon transistor (ie, standard
value)

**Apply KVR across B-E loop:**

V_{1k}
+ V_{10k} + V_{BE sat} + V_{1k} = V_{CC}

âˆ´ 1(I_{C} + I_{B})
+ 10 I_{B} + 0.8 + 1 (I_{C} + I_{B}) = 12

2
I_{C} + 12 I_{B} = 11.2 â€¦â€¦â€¦â€¦â€¦(1)

**Apply KVR across C-E loop:**

V_{1k}
+ V_{1k} + V_{CE sat} + V_{1k} = V_{CC}

l(I_{C}
+ I_{B}) + 1 I_{C} + 0.2 + I(I_{C} + I_{B}) =
12

3I_{C}
+ 2I_{B} = 11.8 â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Solve
the equation (1) and (2)

I_{B}
= 0.3125 mA

I_{C}
= 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

V_{E}=3V

V_{EB}=60
*m*V

R_{B}=60KÎ©

R_{E}=500Î©

**Solution:**

V_{B}=V_{E}âˆ’V_{EB}

V_{B}
= 3âˆ’0.6 = 2.4V

I_{B
}= V_{B}/R_{B} = 2.4 / 60Ã—10^{3} = 40Î¼A

I_{C
}= Î²I_{B} = 50Ã—40Î¼A = 2Ã—10^{-3} A = 2*m*A

V_{C}
= R_{C}I_{C} = 500Ã—2Ã—10^{-3} = 1V

V_{EC
}= V_{E}-V_{C }= 3-1 = 2V

V_{EC}
= 2V

**6. Determine the
current flowing through 3Î© and 4Î© resistors of the circuit given below. Assume
that diodes D _{1} and D_{2} are ideal diodes.**

**Solution:**

The
diode D_{2} is in reverse biased. So does not conduct current.

âˆ´ current through 3Î© is = 0

The
diode D_{1 }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 R_{L}(V_{O}) = Vz = 15V

Voltage across R_{S}(V_{RS}) =
25 -15 = 10V

current
through R_{L} is

I_{L}
= V_{0}/R_{L} = 15 / 3Ã—10^{3} = 5Ã—10^{-3} A

I_{L
}= 5 mA

Current
through R_{S} is

I
= V_{RS}/R_{S} = 10/500 = 20Ã—10^{-3}A

I
= 20mA

Current
through Zener diode is

I_{Z}=I-I_{L}

=
(20-5) Ã—
10^{-3}

I_{z}
= 15mA

** **

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

**Solution:**

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