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Chapter 6: Problem 32

An N-P-N transistor is connected in common emitter configuration in which collector supply is \(9 \mathrm{~V}\) and the voltage drop across the load resistance of \(1000 \Omega\) connected in the collector circuit is \(1 \mathrm{~V}\). If current amplification factor is (25/26), If the internal resistance of the transistor is \(200 \Omega\), then which of the following options is \((1) \mathrm{V}_{\mathrm{CE}}=8 \mathrm{~V}\) (2) collector current is \(1.0 \mathrm{~mA}\) (3) voltage gain \(\frac{50}{23}\), and power gain is \(4.6\) (4) emitter current is \(2.04 \mathrm{~mA}\)

Short Answer

Expert verified
The correct options are: (1) V_{CE} = 8V, (3) voltage gain is \frac{50}{23}, power gain is 4.6, and (4) emitter current is 2.04 mA.

Step by step solution

01

Calculate Collector Current

Determine the collector current using Ohm's law. The voltage drop across the load resistance is given as 1V and the resistance is 1000 Ω. Using Ohm's law, the collector current (I_C) can be calculated as follows:\[I_C = \frac{V_{R_L}}{R_L} = \frac{1V}{1000 \, \Omega} = 1 \, mA\].
02

Calculate Collector-Emitter Voltage

Calculate the voltage between the collector and the emitter (V_{CE}). The supply voltage is 9V, and the voltage drop across the load is 1V. Hence:\[V_{CE} = V_{CC} - V_{RC} = 9V - 1V = 8V\].
03

Calculate Emitter Current

Use the current amplification factor \beta = \frac{25}{26} and the collector current I_C = 1mA to find the base current I_B. The relationship is:\[\beta = \frac{I_C}{I_B} \implies I_B = \frac{I_C}{\beta} = \frac{1mA}{25/26} \approx 1.04 mA\].The emitter current I_E is then:\[I_E = I_C + I_B = 1mA + 1.04mA = 2.04mA\].
04

Calculate Voltage Gain

The voltage gain (A_V) can be calculated using the formula:\[A_V = \beta \times \frac{R_L}{r_{in}}\], where \beta = \frac{25}{26}, R_L = 1000 \, \Omega, and r_in = 200 \, k\Omega. Thus,\[A_V = \frac{25}{26} \times \frac{1000 \, \Omega}{200 \, \Omega} = \frac{25}{26} \times 5 = \frac{125}{26} = \frac{50}{23}\].
05

Calculate Power Gain

The power gain (P_G) is the product of voltage gain (A_V) and current gain (\beta):\[P_G = A_V \times \beta = \frac{50}{23} \times \frac{25}{26} = 4.6\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's law
Ohm's law is a fundamental principle in electronics that describes the relationship between voltage, current, and resistance in an electrical circuit. It is expressed as:
$$ V = IR $$
where:
  • **V** is the voltage across the resistor (in volts).
  • **I** is the current through the resistor (in amperes).
  • **R** is the resistance (in ohms).
To find the collector current ( I_C), we use Ohm's law. The voltage drop across the load resistor is given as 1V, and the resistance as 1000 \Omega. Applying Ohm's law, we get: $$ I_C = \frac{V_{R_L}}{R_L} = \frac{1V}{1000 \Omega} = 1mA $$
Current amplification factor
The current amplification factor, often designated by the symbol \beta, is a measure of how much the input current is amplified in a transistor. It is defined as the ratio of the collector current (**I_C**) to the base current (**I_B**): $$ \beta = \frac{I_C}{I_B} $$ In our example, \beta is given as 25/26. To find the base current (**I_B**), we rearrange the formula:
**I_B = \frac{I_C}{\beta}** Given **I_C = 1mA**, let's calculate **I_B**: $$ I_B = \frac{1mA}{\frac{25}{26}} \approx 1.04mA $$ Now, the emitter current (**I_E**) is the sum of the collector and base currents:
$$ I_E = I_C + I_B = 1mA + 1.04mA = 2.04mA $$
Voltage gain
Voltage gain (A_V) in a transistor amplifier is defined as the ratio of the change in output voltage ( V_{out} ) to the change in input voltage ( V_{in} ). It can be expressed as: $$ A_V = \frac{\beta \times R_L}{r_{in}} $$ Here:
  • \beta = 25/26
  • R_L = 1000 \Omega (load resistance)
  • r_{in} = 200 \Omega (internal resistance of the transistor)
Let's calculate the voltage gain: $$ A_V = \frac{25}{26} \times \frac{1000 \Omega}{200 \Omega} = \frac{25}{26} \times 5 = \frac{125}{26} = \frac{50}{23} $$
Power gain
Power gain (P_G) in a transistor amplifier combines both voltage gain (A_V) and current gain ( \beta ). It represents the overall amplification of power and is calculated as the product of voltage gain and current gain: $$ P_G = A_V \times \beta $$
With our calculated values: $$ A_V = \frac{50}{23} $$ $$ \beta = \frac{25}{26} $$
The power gain is: $$ P_G = A_V \times \beta = \frac{50}{23} \times \frac{25}{26} = 4.6 $$
Collector-emitter voltage
The collector-emitter voltage (V_{CE}) is the voltage difference between the collector and the emitter of a transistor. It's crucial for analyzing the transistor's operation, as it influences its working region. To calculate V_{CE}, we use the following relationship: $$ V_{CE} = V_{CC} - V_{RC} $$
In our example:
  • V_{CC} = 9V (supply voltage)
  • V_{RC} = 1V (voltage drop across the load resistor)
Plugging in the values: $$ V_{CE} = 9V - 1V = 8V $$ This helps us verify the correct operation and any discrepancies in our circuit.

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