Exercise Numerical Problems

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 R_eff = R₁ + R₃ = 4Ω

Current through R₁ = V/R_eff = 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 R_eff

= 1Ω + 10Ω + 1Ω = 12Ω

Here silicon diodes are used

∴ Barrier potential for Si is 0.7 V

Net potential (V_net) = 3 – 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 β = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation. [Ans: 56 µA]

Solution;

V_CE = 0.2V

R_c = 1kΩ

β =50

Vcc = 3V

I _B = ?

Ic = [ V_CC – V_CE ] / R_C

= (3-0.2) / 10³

= 2.8 × 10^-3 A

= 2.8 mA

β= I_C/I_B

∴ I_B = I_C / β = 2.8 × 10^-3 / 50 = 0.056× 10^-3

= 56 × 10^-6 A

I_B=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

V_BE = 0.7V

I_C=?

Solution:

Tranistor is in saturation region.

∴ 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

∴ 1(I_C + I_B) + 10 I_B + 0.8 + 1 (I_C + I_B) = 12

2 I_C + 12 I_B = 11.2         ……………(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    ……………….(2)

Solve the equation (1) and (2)

I_B = 0.3125 mA

I_C = 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:

 Î² = 50

V_E=3V

V_EB=60 mV

R_B=60KΩ

R_E=500Ω

Solution:

V_B=V_E−V_EB

V_B = 3−0.6 = 2.4V

I_B = V_B/R_B = 2.4 / 60×10³ = 40μA

I_C = βI_B = 50×40μA = 2×10^-3 A = 2mA

V_C = R_CI_C = 500×2×10^-3 = 1V

V_EC = V_E-V_C = 3-1 = 2V

V_EC = 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 D₁ is 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 +  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 = V₀/R_L = 15 / 3×10³ = 5×10^-3 A

I_L = 5 mA

Current through R_S is

I = V_RS/R_S = 10/500 = 20×10^-3A

I = 20mA

Current through Zener diode is

I_Z=I-I_L

= (20-5) × 10^-3

I_z = 15mA

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

Solution: