Question

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.

• A. \(0.365\)
• B. \(0.435\)
• C. \(0.625\)
• D. \(0.862\)

Hint:

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 \(I_C\).

Answer 1

The Correct Option is A

Instead of first reducing the base divider to a Thevenin source, this method writes the current balance at the base node directly, together with the base-emitter loop, and solves the two equations as a pair.

1. Set up the base node current balance: Working in mA and k$\Omega$ so that $V = IR$ holds without unit conversion, the current arriving at the base node from the $5$ V rail through the top $5$ k$\Omega$ resistor splits into the current flowing down the bottom $5$ k$\Omega$ resistor to ground and the base current $I_B$ entering the transistor: $$\frac{5 - V_B}{5} = \frac{V_B}{5} + I_B$$
2. Set up the base-emitter loop: The base voltage also equals the base-emitter drop plus the voltage built up across the $5$ k$\Omega$ emitter resistor by the emitter current, where $I_E = (\beta+1)I_B = 101 I_B$: $$V_B = 0.65 + 5(101 I_B)$$
3. Combine the two equations: Clearing fractions in the first equation by multiplying through by $5$: $$5 - V_B = V_B + 5I_B \implies V_B = 2.5 - 2.5 I_B$$ Substituting into the base-emitter loop equation: $$2.5 - 2.5 I_B = 0.65 + 505 I_B$$ $$2.5 - 0.65 = 505 I_B + 2.5 I_B$$ $$1.85 = 507.5 I_B$$ $$I_B = \frac{1.85}{507.5} = 0.003645 \text{ mA} = 3.645\ \mu\text{A}$$
4. Find the collector current and check the answer: $$I_C = \beta I_B = 100 \times 0.003645 = 0.3645 \text{ mA}$$ As a check, $V_B = 2.5 - 2.5(0.003645) \approx 2.491$ V, and separately $0.65 + 505(0.003645) \approx 2.491$ V, so both equations agree with each other.

The quiescent collector current works out to about $0.365$ mA once rounded to three significant figures, which matches option (A).

Let's summarize:

• Writing the node and loop equations directly, without naming a Thevenin step, gives the same result as reducing the divider first.
• Both methods need the factor $(\beta+1)$ on the emitter resistor, since the emitter carries base current plus collector current.

So the quiescent collector current is about 0.365 mA, option (A).