In a circuit shown below, the base current is
11.3 μA
Recognise the circuit. This is the classic fixed-bias (base-resistor bias) stage: one resistor RB from the supply to the base, one resistor RC from the supply to the collector, emitter grounded. Its defining feature is that the base current is set entirely by the base loop and does not depend on β at all.
Step 1 — write KVL around the base–emitter loop. Starting at VCC, through RB, across the base-emitter junction, to ground:
\(V_{CC}=I_B R_B + V_{BE}\)
Step 2 — solve for IB. A forward-biased silicon base-emitter junction sits at VBE ≈ 0.7 V:
\(I_B=\dfrac{V_{CC}-V_{BE}}{R_B}=\dfrac{12-0.7}{1\times10^{6}}\)
\(I_B=\dfrac{11.3}{10^{6}}=11.3\times10^{-6}\ \text{A}=11.3\ \mu\text{A}\)
Why RC plays no part. The collector resistor sits in a different loop. It fixes VCE once IC is known, but the base current is decided before the transistor's gain enters the picture. Knowing RC = 2 kΩ you could go on to find, for a typical β = 100, that \(I_C=\beta I_B=1.13\ \text{mA}\) and \(V_{CE}=12-(1.13\text{m})(2\text{k})=9.74\ \text{V}\) — comfortably in the active region.
Where the distractors come from. 12.0 μA is what you get by forgetting VBE altogether, i.e. taking 12 V across RB; the 0.7 V drop is exactly what turns 12.0 into 11.3. 0.7 μA is 0.7 V divided by RB, the discarded part rather than the answer. 6.0 μA would follow from halving the supply.
The weakness of this bias scheme. Since IB is fixed, \(I_C=\beta I_B\) drifts in direct proportion to β, which varies widely between devices and with temperature. That is why fixed bias, despite needing only one resistor, is replaced in practice by voltage-divider bias with an emitter resistor, whose stability factor is far smaller.
Hence, the base current is 11.3 μA.
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