In a circuit given below the base current IB is
\(\dfrac{V_{BB}-V_{BE}}{R_B+(\beta+1)R_E}\)
To determine the base current \( I_B \) in the given circuit, we start by analyzing the transistor circuit in terms of voltages and currents. The circuit includes \( V_{BB} \) as the biasing voltage, and resistances \( R_B \) and \( R_E \). The transistor has a base-emitter voltage drop \( V_{BE} \), and an emitter current \( I_E \).
Using KVL (Kirchhoff's Voltage Law) in the base-emitter loop of the transistor:
\(V_{BB} = I_B \cdot R_B + V_{BE} + I_E \cdot R_E\)
Now, recall the relation between base current \( I_B \), collector current \( I_C \), and emitter current \( I_E \) in a transistor:
\(I_E = I_B + I_C\)
For a transistor, we also know:
\(I_C = \beta \cdot I_B\)
So,
\(I_E = I_B + \beta \cdot I_B = (\beta + 1) \cdot I_B\)
Substitute \( I_E \) in the KVL equation:
\(V_{BB} = I_B \cdot R_B + V_{BE} + (\beta + 1) \cdot I_B \cdot R_E\)
Rearrange this to solve for \( I_B \):
\(I_B \cdot (R_B + (\beta + 1) \cdot R_E) = V_{BB} - V_{BE}\)
Finally, the base current \( I_B \) is given by:
\(I_B = \dfrac{V_{BB} - V_{BE}}{R_B + (\beta + 1) \cdot R_E}\)
The correct option, as calculated, is:
\(\dfrac{V_{BB} - V_{BE}}{R_B + (\beta + 1)R_E}\)
This demonstrates that the correct choice among the given options is indeed the expression that accounts for both base bias resistance \( R_B \) and the influence of emitter resistance \( R_E \) scaled by \((\beta + 1)\).
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