To find the value of the base resistor ($R_B$) required to switch the transistor "ON" when the input terminal voltage exceeds 2.5 V, we need to use the following relationships between the base current ($I_B$), collector current ($I_C$), and base-emitter voltage ($V_{BE}$).

Step 1: Transistor Current Relations

The collector current ($I_C$) and base current ($I_B$) are related by the transistor’s current gain ($\beta$):

$I_C = \beta I_B$

Given values:

• $\beta = 200$
• $I_C = 4 \, \text{mA} = 4 \times 10^{-3} \, \text{A}$
• $I_B = 2 \times 10^{-6} \, \text{A}$

Using $I_C = \beta I_B$, we can verify that:

$I_B = \frac{I_C}{\beta} = \frac{4 \times 10^{-3}}{200} = 2 \times 10^{-5} \, \text{A}$

So, the base current needed to achieve the desired collector current is $I_B = 2 \times 10^{-5} \, \text{A}$.

Step 2: Base-Emitter Voltage Assumption

For most standard transistors, the base-emitter voltage $V_{BE}$ is around 0.7 V when the transistor is fully on (saturated).

Step 3: Applying Ohm’s Law

To find the base resistor $R_B$, we can use Ohm's Law. The voltage across $R_B$ is the difference between the input terminal voltage ($V_{in}$) and the base-emitter voltage $V_{BE}$:

$R_B = \frac{V_{in} - V_{BE}}{I_B}$

Given:

• $V_{in} = 2.5 \, \text{V}$
• $V_{BE} = 0.7 \, \text{V}$
• $I_B = 2 \times 10^{-5} \, \text{A}$

Substitute these values into the formula:

$R_B = \frac{2.5 \, \text{V} - 0.7 \, \text{V}}{2 \times 10^{-5} \, \text{A}} = \frac{1.8 \, \text{V}}{2 \times 10^{-5} \, \text{A}} = 90,000 \, \Omega = 90 \, \text{k}\Omega$

Final Answer:

The required base resistor $R_B$ is 90 kΩ.

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Would you like more details on this calculation or further clarifications?

Here are five related questions to explore:

1. How would changing $\beta$ to a lower value affect the required base resistor?
2. What would happen if $V_{in}$ were reduced below 2.5V?
3. Can we assume $V_{BE} = 0.7 \, \text{V}$ for all transistors, or does it vary?
4. What are the implications of using a higher base resistor than calculated?
5. How does saturation affect the transistor’s switching performance?

Tip: Ensure that the base current is sufficient to saturate the transistor to fully switch it on.