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Question

A single stage amplifier employing one active device is powered by a 9 V battery which has a current drain of 20 mA. If load voltage is 3 V at 12 mA, then determine η.

The correct answer is

20%

An amplifier's efficiency, often denoted by the Greek letter eta (\(\eta\)), is a critical performance parameter that indicates how effectively the amplifier converts the DC power supplied by its power source into AC power delivered to the load. In simpler terms, it tells us what percentage of the input power is successfully transformed into useful output power.

To determine the efficiency of the given single-stage amplifier, we need to calculate two main components:

  • The total power consumed by the amplifier from the battery (Input Power).
  • The power delivered by the amplifier to the load (Output Power).

Once both input and output powers are known, we can use the efficiency formula.

Input Power Determination for the Amplifier

The input power is the total power drawn from the battery by the amplifier circuit. It is calculated using the battery's voltage and the total current drained from it.

Given values for input:

  • Battery Voltage (\(V_{battery}\)) = 9 V
  • Current Drain (\(I_{drain}\)) = 20 mA

First, convert the current from milliamperes (mA) to amperes (A):

\(I_{drain}\) = 20 mA = \(20 \times 10^{-3}\) A = 0.02 A

The formula for input power (\(P_{in}\)) is:

\[P_{in} = V_{battery} \times I_{drain}\]

Now, substitute the values:

\[P_{in} = 9 \text{ V} \times 0.02 \text{ A}\]

\[P_{in} = 0.18 \text{ W}\]

So, the amplifier draws 0.18 Watts of power from the battery.

Output Power Determination to the Load

The output power is the power delivered by the amplifier to the connected load. It is calculated using the voltage across the load and the current flowing through the load.

Given values for output:

  • Load Voltage (\(V_{load}\)) = 3 V
  • Load Current (\(I_{load}\)) = 12 mA

Convert the load current from milliamperes (mA) to amperes (A):

\(I_{load}\) = 12 mA = \(12 \times 10^{-3}\) A = 0.012 A

The formula for output power (\(P_{out}\)) is:

\[P_{out} = V_{load} \times I_{load}\]

Now, substitute the values:

\[P_{out} = 3 \text{ V} \times 0.012 \text{ A}\]

\[P_{out} = 0.036 \text{ W}\]

Thus, the amplifier delivers 0.036 Watts of power to the load.

Efficiency Calculation of the Amplifier

The efficiency (\(\eta\)) of an amplifier is the ratio of the output power to the input power, usually expressed as a percentage.

The formula for efficiency is:

\[\eta = \frac{P_{out}}{P_{in}} \times 100\%\]

Now, substitute the calculated values for \(P_{out}\) and \(P_{in}\):

\[\eta = \frac{0.036 \text{ W}}{0.18 \text{ W}} \times 100\%\]

\[\eta = 0.2 \times 100\%\]

\[\eta = 20\%\]

Therefore, the efficiency of the single-stage amplifier is 20%.

Parameter Value Unit
Battery Voltage (\(V_{battery}\)) 9 V
Current Drain (\(I_{drain}\)) 20 mA
Load Voltage (\(V_{load}\)) 3 V
Load Current (\(I_{load}\)) 12 mA
Calculated Input Power (\(P_{in}\)) 0.18 W
Calculated Output Power (\(P_{out}\)) 0.036 W
Calculated Efficiency (\(\eta\)) 20 %

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Important Questions from Power Amplifier

  1. _______ is usually used in RF power amplifier and in amateur radio.

  2. The collector current of a class C amplifier is _______.

  3. A tuned Class-C amplifier has a power supply voltage of 12 V. What is the ideal peak-to-peak output voltage?
  4. The type of amplifier which exhibits crossover distortion in its output is:
  5. In push-pull amplifier there occurs cancellation of:

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