In the circuit shown, the BJT has a \(\beta\) of \(100\). The base-emitter junction voltage is \(0.65\) V. The quiescent collector current, \(I_C\), is _______ mA.
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Reduce the base bias divider to its Thevenin voltage and resistance, then apply the base-emitter KVL loop with \(I_E=(\beta+1)I_B\) to find \(I_B\), and multiply by \(\beta\) to get \(I_C\).
Step 1: Identify the bias network as a Thevenin-equivalent base circuit.
The base of the transistor is tied to the midpoint of a voltage divider made of two \(5\) k\(\Omega\) resistors between the \(5\) V rail and ground. Replace this divider by its Thevenin equivalent as seen from the base.
\[
V_{TH} = 5\text{ V} \times \frac{R_{B2}}{R_{B1}+R_{B2}} = 5 \times \frac{5}{5+5} = 2.5 \text{ V}
\]
\[
R_{TH} = R_{B1} \parallel R_{B2} = \frac{5 \times 5}{5+5} = 2.5 \text{ k}\Omega
\]
Step 2: Write the base-emitter loop equation.
Walking from the \(2.5\) V Thevenin source, through \(R_{TH}\) and the base, across the base-emitter junction, and through the \(5\) k\(\Omega\) emitter resistor to ground gives
\[
V_{TH} = I_B R_{TH} + V_{BE} + I_E R_E
\]
where \(I_E = (\beta+1) I_B\) because the emitter current carries both the base and collector currents.
Step 3: Solve for the base current.
Substitute the known values, with \(R_E = 5\) k\(\Omega\), \(V_{BE}=0.65\) V, and \(\beta = 100\):
\[
2.5 = I_B(2.5\text{ k}\Omega) + 0.65 + I_B(101)(5\text{ k}\Omega)
\]
\[
2.5 - 0.65 = I_B(2500 + 505000)
\]
\[
1.85 = I_B(507500)
\]
\[
I_B = \frac{1.85}{507500} = 3.645\ \mu\text{A}
\]
Step 4: Compute the collector current.
Using \(I_C = \beta I_B\):
\[
I_C = 100 \times 3.645\ \mu\text{A} = 364.5\ \mu\text{A} \approx 0.365 \text{ mA}
\]
Step 5: Confirm the transistor is in the active region.
The collector voltage is \(V_C = 5 - I_C R_C = 5 - (0.365\text{ mA})(5\text{ k}\Omega) \approx 3.18\) V, and the emitter voltage is \(V_E = I_E R_E \approx (1.01)(0.365\text{ mA})(5\text{ k}\Omega) \approx 1.84\) V, so \(V_{CE} \approx 1.34\) V, well above saturation. The active-region assumption used in Step 2 is valid.
Final Answer:
The quiescent collector current is \(I_C \approx 0.365\) mA.
\[ \boxed{I_C \approx 0.365 \text{ mA}} \]
Find dV/dt