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:

In transistor circuits, the relationship between currents is $I_E = I_C + I_B$. In the Common Emitter (CE) configuration, the current gain is $\beta = I_C/I_B$. These relations are combined to give $I_E = (\beta+1)I_B$, which is often the quickest way to solve for the base current.

Answer 1

The Correct Option is D

Step 1: State the current relationships in a transistor.
The emitter current ($I_E$) is the sum of the base current ($I_B$) and the collector current ($I_C$): \[ I_E = I_C + I_B. \] The current amplification factor in the Common Emitter (CE) configuration is $\beta$, which is defined as: \[ \beta = \frac{I_C}{I_B}. \]

Step 2: Express the emitter current in terms of the base current and $\beta$.
From the definition of $\beta$, we have $I_C = \beta I_B$. Substitute $I_C$ into the emitter current equation: \[ I_E = \beta I_B + I_B = (\beta + 1)I_B. \] The base current is given by: \[ I_B = \frac{I_E}{\beta + 1}. \]

Step 3: Substitute the given numerical values.
We are given $\beta = 80$ and $I_E = 2.43 \text{ mA}$. \[ I_B = \frac{2.43 \text{ mA}}{80 + 1} = \frac{2.43 \text{ mA}}{81}. \]

Step 4: Calculate the base current.
We can simplify the fraction $\frac{2.43}{81}$: \[ \frac{2.43}{81} = \frac{243}{8100} = \frac{3 \times 81}{81 \times 100} = \frac{3}{100} = 0.03. \] \[ I_B = 0.03 \text{ mA}. \]

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