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Fixed bias: VBE, IB, IC, VCE (12 V)

Short answer

VBE = 0.7 V. IB = (12−0.7)/200k = 56.5 µA, IC = βIB = 4.24 mA, VCE = 12 − 4.24×2.2k = 2.68 V. Active region.

For the given circuit with β = 75, determine VBE, IB, IC and VCE. [Figure: VCC = +12 V; RB = 200 kΩ from supply to base; RC = 2.2 kΩ from supply to collector; emitter grounded.]

Given: VCC = 12 V, RB = 200 kΩ, RC = 2.2 kΩ, β = 75, emitter grounded. Silicon transistor.

         +12 V
        |      |
      [RB]    [RC]
      200k    2.2k
        |      |C
        +---B--|  NPN
               |E
              GND

VBE

For a silicon transistor in the active region, VBE = 0.7 V.

Base current

IB = (VCC − VBE)/RB = (12 − 0.7)/200 k = 56.5 µA

Collector current

IC = β IB = 75 × 56.5 µA = 4.2375 mA ≈ 4.24 mA

Collector-emitter voltage

VCE = VCC − IC RC = 12 − 4.2375 mA × 2.2 kΩ = 12 − 9.3225 = 2.68 V

VCE > 0.2 V, so the transistor is in the active region (the assumption is valid).

Answer: VBE = 0.7 V, IB = 56.5 µA, IC = 4.24 mA, VCE = 2.68 V

✓ Answer checked by Claude Opus 5.5 (high effort)

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