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BJT ICQ and VCEQ with RB 100k, RC 3k

Short answer

IB = (3-0.7)/100kΩ = 23μA. ICQ = β IB = 2.3mA. VCEQ = VCC - ICQ RC = 10 - 6.9 = 3.1V. The transistor is in active region.

Determine the dc collector current ICQ and collector emitter voltage VCEQ for the following circuit. Given β = 100. [Figure: NPN transistor with base resistor RB = 100 kΩ from source VBB = 3 V to base; collector resistor RC = 3 kΩ from a 10 V supply to collector; emitter grounded]

Use the base-emitter loop to find IB, then IC = βIB, then the collector loop for VCE. Assume the active region (to be checked) and VBE = 0.7 V.

Given

VBB = 3 V, RB = 100 kΩ, RC = 3 kΩ, VCC = 10 V, β = 100.

Base loop

IB = (VBB - VBE)/RB = (3 - 0.7)/100 kΩ = 23 µA

Collector current

ICQ = β IB = 100 × 23 µA = 2.3 mA

Collector-emitter voltage

VCEQ = VCC - ICQ RC = 10 - (2.3 mA)(3 kΩ) = 10 - 6.9 = 3.1 V

Check

VCEQ = 3.1 V > VCE(sat) ≈ 0.2 V, so the transistor is in the active region and the assumption is valid.

Answer: ICQ = 2.3 mA, VCEQ = 3.1 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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