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DC load line and Q-point, IB = 15 µA

Short answer

IC = 40 × 15 µA = 0.6 mA, VCE = 10 − 0.6×8k = 5.2 V. DC load line connects IC(sat) = 1.25 mA (VCE=0) and VCE = 10 V (IC=0). Q-point (5.2 V, 0.6 mA).

In BJT circuit if VCC = 10 V and RC = 8 kΩ, draw the dc load line. Determine the Q-point (operating point) for zero input signal if IB = 15 µA and β = 40.

Given: VCC = 10 V, RC = 8 kΩ, IB = 15 µA, β = 40 (common-emitter circuit with emitter grounded).

DC load line

KVL in the collector circuit: VCC = IC RC + VCE

  • Saturation point (VCE = 0): IC(sat) = VCC/RC = 10/8 k = 1.25 mA
  • Cut-off point (IC = 0): VCE = VCC = 10 V

The line joining the points (VCE = 0, IC = 1.25 mA) and (VCE = 10 V, IC = 0) is the dc load line.

Q point

  • IC = β IB = 40 × 15 µA = 0.6 mA
  • VCE = VCC − IC RC = 10 − 0.6 mA × 8 kΩ = 10 − 4.8 = 5.2 V
 IC (mA)
 1.25 |\
      | \
 0.6  |  *Q  (IB = 15 uA)
      |   \
      +----\------ VCE (V)
         5.2  10

Since VCE = 5.2 V is well above VCE(sat) ≈ 0.2 V, the transistor is in the active region.

Answer: Q point = (VCE = 5.2 V, IC = 0.6 mA)

✓ Answer checked by Claude Opus 5.5 (high effort)

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