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IC, Q-point, load line, IB = 10 µA

Short answer

IC = 50 ×10 µA = 0.5 mA, VCE = 15 − 0.5×10k = 10 V. DC load line from (0 V, 1.5 mA) to (15 V, 0 mA). Q-point (10 V, 0.5 mA).

Find the value of collector current, Q-point and DC load line for common emitter circuit having VCC = 15 V, RC = 10 kΩ, IB = 10 µA and β = 50.

Given: VCC = 15 V, RC = 10 kΩ, IB = 10 µA, β = 50.

Collector current

IC = β IB = 50 × 10 µA = 0.5 mA

Q point

VCE = VCC − IC RC = 15 − 0.5 mA × 10 kΩ = 15 − 5 = 10 V

Q point = (VCE = 10 V, IC = 0.5 mA)

DC load line

From VCC = IC RC + VCE:

  • Saturation end (VCE = 0): IC(sat) = 15/10 k = 1.5 mA
  • Cut-off end (IC = 0): VCE = 15 V
 IC (mA)
 1.5 |\
     | \
 0.5 |  *Q
     |   \
     +----\------ VCE (V)
         10  15

Answer: IC = 0.5 mA; Q point (10 V, 0.5 mA); the load line joins (0 V, 1.5 mA) and (15 V, 0 mA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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