Question

The current amplification factor of a transistor in common emitter configuration is 80. If the emitter current is 2.43 mA, then the base current is

• A. 15 $\mu$A
• B. 1.5 $\mu$A
• C. 3 $\mu$A
• D. 30 $\mu$A

Hint:

For a transistor, always remember the core current relationship ($I_E=I_C+I_B$) and the current gain for the common emitter configuration ($\beta=I_C/I_B$). This allows you to quickly derive the relation $I_E = (\beta+1)I_B$, which is the key to solving this type of problem.

Answer 1

The Correct Option is D

Step 1: State the relations for a Common Emitter (CE) configuration.
In a transistor, the emitter current ($I_E$), collector current ($I_C$), and base current ($I_B$) are related by: \[ I_E = I_C + I_B. \] The current amplification factor in the CE configuration is $\beta$, which is the ratio of the collector current to the base current: \[ \beta = \frac{I_C}{I_B}. \]

Step 2: Relate the emitter current to the base current using $\beta$.
From the definition of $\beta$, we have $I_C = \beta I_B$. Substitute this into the emitter current relation: \[ I_E = (\beta I_B) + I_B = (\beta+1)I_B. \] This gives the base current as: \[ I_B = \frac{I_E}{\beta+1}. \]

Step 3: Substitute the given values.
We are given the current amplification factor $\beta = 80$. The emitter current is $I_E = 2.43 \text{ mA} = 2.43 \times 10^{-3} \text{ A}$. \[ I_B = \frac{2.43 \text{ mA}}{80 + 1} = \frac{2.43 \text{ mA}}{81}. \]

Step 4: Calculate the numerical value of the base current.
The fraction $\frac{2.43}{81}$ can be calculated: \[ \frac{2.43}{81} = \frac{243}{8100}. \] Since $81 \times 3 = 243$, we have: \[ I_B = \frac{3}{100} \text{ mA} = 0.03 \text{ mA}. \]

Step 5: Convert the result to $\mu$A.
Since $1 \text{ mA} = 1000 \mu\text{A}$: \[ I_B = 0.03 \times 1000 \mu\text{A} = 30 \mu\text{A}. \] \[ \boxed{I_B = 30 \mu\text{A}}. \]