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Voltage divider bias, DC load line, Q point (15 V)

Short answer

By Thevenin: IB = 27.7 µA, IC = 2.08 mA. Q-point: VCEQ = 8.72 V, ICQ = 2.08 mA. DC load line joins IC(sat) = 5 mA (VCE = 0) and VCE = 15 V (IC = 0).

Draw the DC load line and determine the Q point of the voltage divider biased transistor circuit having VCC = 15 V, RC = 1 kΩ, R1 = 10 kΩ, R2 = 5 kΩ, RE = 2 kΩ [printed as Rc = 2 kΩ] and β = 75.

Given: VCC = 15 V, RC = 1 kΩ, R1 = 10 kΩ, R2 = 5 kΩ, RE = 2 kΩ, β = 75. Assumption: the figure label "Rc = 2 kΩ" is read as RE = 2 kΩ, with RC = 1 kΩ. VBE = 0.7 V.

           +15 V
         |       |
       [R1]     [RC]
       10k       1k
         |       |C
         +---B--|  NPN
         |      |E
       [R2]     |
        5k    [RE] 2k
         |       |
        GND     GND

Thevenin equivalent of the base circuit

  • VTH = 15 × 5/15 = 5 V
  • RTH = (10 × 5)/15 = 3.33 kΩ

Base current

IB = (VTH − VBE)/(RTH + (β + 1) RE) = (5 − 0.7)/(3.33 k + 76 × 2 k) = 4.3/155.33 k = 27.68 µA

Currents

  • IC = β IB = 75 × 27.68 µA = 2.076 mA
  • IE = 76 × 27.68 µA = 2.104 mA

VCE

VCE = VCC − IC RC − IE RE = 15 − 2.076 × 1 − 2.104 × 2 = 15 − 2.076 − 4.208 = 8.72 V

DC load line

  • Saturation: IC(sat) = VCC/(RC + RE) = 15/3 k = 5 mA (at VCE = 0)
  • Cut-off: VCE = 15 V (at IC = 0)
 IC (mA)
 5    |\
      | \
 2.08 |  *Q
      |   \
      +----\------ VCE (V)
         8.7  15

Answer: Q point: ICQ = 2.08 mA, VCEQ = 8.72 V (IB = 27.7 µA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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