A transistor is connected in common base configuration, the collector supply is 8V and the voltage drop across a resistor of `800 Omega` in the collector circuit is 0.5V. If the current gain `alpha` is 0.96, then the base current is

Options

A. `50xx10^(-6)A`
B. `450xx10^(-6)A`
C. `80xx10^(-6)A`
D. `26xx10^(-6)A`

Answer

To solve the problem step by step, we will use the information provided and apply the relevant formulas.

### Step 1: Calculate the Collector Current (Ic)
The collector current (Ic) can be calculated using Ohm's law. The voltage drop across the collector resistor (Rc) is given as 0.5V, and the resistance (Rc) is 800 ohms.

\[ I_c = \frac{V_{drop}}{R_c} = \frac{0.5V}{800 \Omega} \]

Calculating this gives:

\[ I_c = \frac{0.5}{800} = 0.000625 \, A = 625 \, \mu A \]

### Step 2: Calculate the Current Gain (β)
The current gain (β) is related to α (the common base current gain) by the formula:

\[ \beta = \frac{\alpha}{1 - \alpha} \]

Given that α = 0.96, we can substitute this value into the equation:

\[ \beta = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04} = 24 \]

### Step 3: Calculate the Base Current (Ib)
The base current (Ib) can be calculated using the relationship between collector current (Ic) and base current (Ib) with the current gain (β):

\[ I_b = \frac{I_c}{\beta} \]

Substituting the values we have:

\[ I_b = \frac{625 \times 10^{-6} \, A}{24} \]

Calculating this gives:

\[ I_b = \frac{625 \times 10^{-6}}{24} \approx 26.04167 \times 10^{-6} \, A \approx 26 \, \mu A \]

### Conclusion
Thus, the base current \( I_b \) is approximately \( 26 \, \mu A \).

### Final Answer
The base current is \( 26 \, \mu A \).

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