Electronics Numerical Problems - Chapter 18 of FSc/ICS Physics

Electronics Numerical Problems - Chapter 18

Problem 18.1: The current flowing into the base of a transistor is 100 μA. Find its collector current \( I_c \), its emitter current \( I_E \) and the ratio \( \frac{I_c}{I_E} \), if the value of current gain β is 100.

Given Data:

• Base Current \( I_B = 100 \, \mu A = 100 \times 10^{-6} A = 10^{-4} A \)
• Current Gain β = 100

To Determine:

• Collector Current \( I_c = ? \)
• Emitter Current \( I_E = ? \)
• Ratio \( \frac{I_c}{I_E} = ? \)

Calculations:

(a) Collector Current:

\[ I_c = β I_B \]

\[ = 100 \times 100 \times 10^{-6} \]

\[ = 10^{-2} A \]

\[ = 10 \, mA \]

(b) Emitter Current:

\[ I_E = I_c + I_B \]

\[ = 10^{-2} A + 10^{-4} A \]

\[ = 0.0101 A \]

\[ = 10.1 \, mA \]

(c) Ratio:

\[ \frac{I_c}{I_E} = \frac{10 \, mA}{10.1 \, mA} \]

\[ = 0.99 \]

Problem 18.2: Fig. shows a transistor which operates a relay as the switch S is closed. The relay is energized by a current of 10 mA. Calculate the value \( R_S \) which will just make the relay operate. The current gain β of the transistor is 200. When the transistor conducts, its \( V_{BE} \) can be assumed to be 0.6 V.

Circuit Diagram: Transistor operating a relay with \( V_{CC} = 9V \), base resistor \( R_S \), and relay in collector circuit.

Given Data:

• Collector Current \( I_c = 10 \, mA = 10^{-2} A \)
• Current Gain β = 200
• Base-Emitter Voltage \( V_{BE} = 0.6 \, V \)
• Supply Voltage \( V_{CC} = 9 \, V \)

To Determine:

• Base Resistance \( R_S = ? \)

Calculations:

\[ β = \frac{I_c}{I_B} \Rightarrow I_B = \frac{I_c}{β} \]

\[ = \frac{10^{-2}}{200} \]

\[ = 0.5 \times 10^{-4} A \]

\[ = 50 \, \mu A \]

Applying Kirchhoff's Voltage Rule on input circuit:

\[ V_{CC} - I_B R_S - V_{BE} = 0 \]

\[ \Rightarrow I_B R_S = V_{CC} - V_{BE} \]

\[ \Rightarrow R_S = \frac{V_{CC} - V_{BE}}{I_B} \]

\[ = \frac{9 - 0.6}{0.5 \times 10^{-4}} \]

\[ = \frac{8.4}{5 \times 10^{-5}} \]

\[ = 168000 \, Ω \]

\[ = 168 \, kΩ \]

Problem 18.3: In circuit (Fig.), there is negligible potential drop between B and E, if β is 100. Calculate:

1. base current
2. collector current
3. potential drop across \( R_C \)
4. \( V_{CE} \)

Circuit Diagram: Common emitter configuration with \( V_{CC} = 9V \), \( R_B = 800kΩ \), \( R_C = 1kΩ \)

Given Data:

• Supply Voltage \( V_{CC} = 9 \, V \)
• Base-Emitter Voltage \( V_{BE} \approx 0 \)
• Base Resistor \( R_B = 800 \, kΩ = 800000 \, Ω \)
• Collector Resistor \( R_C = 1 \, kΩ = 1000 \, Ω \)
• Current Gain β = 100

To Determine:

• Base Current \( I_B = ? \)
• Collector Current \( I_c = ? \)
• Potential Drop Across \( R_C = ? \)
• Collector-Emitter Voltage \( V_{CE} = ? \)

Calculations:

(i) Base Current:

Applying KVR on Input Circuit:

\[ V_{CC} - I_B R_B - V_{BE} = 0 \]

\[ \Rightarrow I_B R_B = V_{CC} - V_{BE} \]

\[ \Rightarrow I_B = \frac{V_{CC} - V_{BE}}{R_B} \]

\[ = \frac{9 - 0}{800000} \]

\[ = 11.25 \times 10^{-6} A \]

\[ = 11.25 \, \mu A \]

(ii) Collector Current:

\[ I_c = β I_B \]

\[ = 100 \times 11.25 \times 10^{-6} \]

\[ = 11.25 \times 10^{-4} A \]

\[ = 1.125 \, mA \]

(iii) Potential Drop Across \( R_C \):

\[ V_C = I_C R_C \]

\[ = 11.25 \times 10^{-4} \times 1000 \]

\[ = 1.125 \, V \]

(iv) Collector-Emitter Voltage \( V_{CE} \):

Applying KVR on Output Circuit:

\[ V_{CC} - I_C R_C - V_{CE} = 0 \]

\[ \Rightarrow V_{CE} = V_{CC} - I_C R_C \]

\[ = 9 - 1.125 \]

\[ = 7.875 \, V \]

Problem 18.4: Calculate the output of the op-amp circuit shown in Fig.

Circuit Diagram: Op-amp with three input resistors: \( R_1 = 10kΩ \) with 5V input, \( R_2 = 4kΩ \) with -2V input, and feedback resistor \( R_3 = 20kΩ \)

Given Data:

• Input Resistor \( R_1 = 10 \, k\Omega = 10000 \, \Omega \) with input voltage 5V
• Input Resistor \( R_2 = 4 \, k\Omega = 4000 \, \Omega \) with input voltage -2V
• Feedback Resistor \( R_3 = 20 \, k\Omega = 20000 \, \Omega \)

To Determine:

• Output Voltage \( V_o = ? \)

Calculations:

Let \( I_1 \) = Current Through \( R_1 \)

Let \( I_2 \) = Current Through \( R_2 \)

Let \( I_3 \) = Current Through \( R_3 \)

By Kirchhoff's Current Rule at inverting input:

\[ I_1 + I_2 = I_3 \]

\[ \Rightarrow \frac{(5-0)}{10000} + \frac{(-2-0)}{4000} = \frac{(0-V_o)}{20000} \]

\[ \Rightarrow 0.5 \times 10^{-3} - 0.5 \times 10^{-3} = \frac{-V_o}{20000} \]

\[ \Rightarrow \frac{-V_o}{20000} = 0 \]

\[ \Rightarrow V_o = 0 \]

Problem 18.5: Calculate the gain of non-inverting amplifier shown in Fig.

Circuit Diagram: Non-inverting op-amp configuration with input voltage applied to non-inverting terminal, \( R_1 = 10kΩ \), and \( R_2 = 40kΩ \)

Given Data:

• Input Resistor \( R_1 = 10 \, k\Omega = 10000 \, \Omega \)
• Feedback Resistor \( R_2 = 40 \, k\Omega = 40000 \, \Omega \)

To Determine:

• Gain \( G = ? \)

Calculations:

For Non-Inverting Op-Amp:

\[ G = 1 + \frac{R_2}{R_1} \]

\[ = 1 + \frac{40000}{10000} \]

\[ = 1 + 4 \]

\[ = 5 \]