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