Question

In an n-p-n transistor 200 electrons enter the emitter in $10^{-8}\ \text{second}$. If $1\%$ electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [$e = 1.6 \times 10^{-19}\ \text{C}$]

• A. $2 \times 10^{-10}\ \text{A}$ and $49$
• B. $3.2 \times 10^{-9}\ \text{A}$ and $99$
• C. $1.6 \times 10^{-19}\ \text{A}$ and $90$
• D. $1.7 \times 10^{-11}\ \text{A}$ and $70$

Hint:

You can find the amplification factor $\beta$ using a quick percentage trick without calculating the currents! If base loss is $1\%$, then the alpha current gain is $\alpha = 99\% = 0.99$. Using the conversion formula: $$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$$ This allows you to identify the correct option immediately by matching $\beta = 99$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem describes charge carrier flow inside an n-p-n bipolar junction transistor (BJT). Given the quantity of electrons moving through the emitter per unit time and the percentage of carriers lost to recombination inside the base, we need to calculate the emitter current ($I_E$) and the common-emitter current gain/amplification factor ($\beta$).

Step 2: Key Formula or Approach:
1. Emitter current ($I_E$) is the total charge passing through per unit time: $$I_E = \frac{q}{t} = \frac{N \cdot e}{t}$$ 2. The current distribution equation for a transistor is $I_E = I_B + I_C$. 3. If $1\%$ of electrons are lost to the base, the base current is $I_B = 0.01 I_E$ and the remaining collector current is $I_C = 0.99 I_E$. 4. The current amplification factor ($\beta$) is defined as: $$\beta = \frac{I_C}{I_B}$$

Step 3: Detailed Explanation:
Let's first calculate the emitter current ($I_E$) using $N = 200$, $e = 1.6 \times 10^{-19}\ \text{C}$, and $t = 10^{-8}\ \text{s}$: $$I_E = \frac{200 \times 1.6 \times 10^{-19}}{10^{-8}}$$ $$I_E = \frac{3.2 \times 10^{-17}}{10^{-8}} = 3.2 \times 10^{-9}\ \text{A}$$ Next, determine the ratio between currents. Since $1\%$ of the current forms the base current and $99\%$ reaches the collector: $$I_B = 0.01 I_E$$ $$I_C = 0.99 I_E$$ Now calculate the current amplification factor ($\beta$): $$\beta = \frac{I_C}{I_B} = \frac{0.99 I_E}{0.01 I_E} = \frac{0.99}{0.01} = 99$$ Combining the computed components gives $I_E = 3.2 \times 10^{-9}\ \text{A}$ and $\beta = 99$.

Step 4: Final Answer:
The emitter current and amplification factor match option (B).