An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is `100 Omega` and the output load resistance is `10 kOmega`.The common emitter current gain `beta` is :

Options

A. `10^2`
B. `60`
C. `6xx10^2`
D. `10^4`

Answer

To find the common emitter current gain (β) of the npn transistor operating as a common emitter amplifier, we will follow these steps:

### Step 1: Convert Power Gain from Decibels to Linear Scale
The power gain (AP) in decibels is given by the formula:
\[ AP_{dB} = 10 \log_{10}(AP) \]
Given that \(AP_{dB} = 60 \, dB\), we can rearrange the formula to find \(AP\):
\[ 60 = 10 \log_{10}(AP) \]
Dividing both sides by 10:
\[ 6 = \log_{10}(AP) \]
Now, converting from logarithmic to linear form:
\[ AP = 10^6 \]

### Step 2: Use the Power Gain Formula
The power gain can also be expressed in terms of voltage gain (AV) and current gain (β):
\[ AP = AV \cdot \beta \]
We also know that the voltage gain (AV) for a common emitter amplifier is given by:
\[ AV = \beta \cdot \frac{R_{output}}{R_{input}} \]

### Step 3: Substitute Voltage Gain into Power Gain Formula
Substituting the expression for AV into the power gain formula:
\[ AP = \beta \cdot \left(\beta \cdot \frac{R_{output}}{R_{input}}\right) \]
This simplifies to:
\[ AP = \beta^2 \cdot \frac{R_{output}}{R_{input}} \]

### Step 4: Rearrange to Solve for β
Rearranging the equation to solve for β:
\[ \beta^2 = AP \cdot \frac{R_{input}}{R_{output}} \]
Taking the square root gives:
\[ \beta = \sqrt{AP \cdot \frac{R_{input}}{R_{output}}} \]

### Step 5: Substitute the Known Values
We have:
- \(AP = 10^6\)
- \(R_{input} = 100 \, \Omega\)
- \(R_{output} = 10 \, k\Omega = 10^4 \, \Omega\)

Substituting these values:
\[ \beta = \sqrt{10^6 \cdot \frac{100}{10^4}} \]
Calculating the fraction:
\[ \frac{100}{10^4} = \frac{100}{10000} = 0.01 \]
Thus:
\[ \beta = \sqrt{10^6 \cdot 0.01} = \sqrt{10^4} = 10^2 = 100 \]

### Final Answer
The common emitter current gain \( \beta \) is:
\[ \beta = 100 \]

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