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Fixed bias: IB, IC, VCE (20 V)

Short answer

IB = (20−0.7)/200k = 96.5 µA. Active assumption gives VCE < 0, so transistor is saturated: IC ≈ 20/80k = 0.25 mA, VCE ≈ 0.2 V.

For the given circuit with β = 75, determine IB, IC and VCE. [Figure: +20 V supply; RB = 200 kΩ from supply to base; RC = 80 kΩ from supply to collector; emitter grounded.]

Given: VCC = 20 V, RB = 200 kΩ, RC = 80 kΩ, β = 75, emitter grounded. Assume VBE = 0.7 V for silicon.

Base current

IB = (VCC − VBE)/RB = (20 − 0.7)/200 k = 96.5 µA

Collector current (if active)

IC = β IB = 75 × 96.5 µA = 7.24 mA

Check VCE

VCE = VCC − IC RC = 20 − 7.24 mA × 80 kΩ = 20 − 579 = −559 V

A negative VCE is impossible, so the transistor is not active; it is in saturation. The largest collector current possible is

IC(sat) = (VCC − VCE(sat))/RC ≈ 20/80 k = 0.25 mA

(more exactly (20 − 0.2)/80 k = 0.2475 mA).

Results

  • IB = 96.5 µA
  • IC ≈ 0.25 mA (saturation value, less than βIB)
  • VCE ≈ 0.2 V (VCE(sat), approximately 0 V)

Since IC < β IB, the transistor works as a closed switch. (If RC were 80 Ω instead of 80 kΩ the transistor would be active, with IC = 7.24 mA and VCE = 20 − 0.58 = 19.4 V.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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