Worked example: voltage-divider bias — Q-point and β-stability

Semiconductor Devices · BJT Biasing · Example

A voltage-divider bias has VCC=12VV_{CC} = 12\,\text{V}, R1=47kΩR_1 = 47\,\text{k}\Omega, R2=10kΩR_2 = 10\,\text{k}\Omega, RC=1kΩR_C = 1\,\text{k}\Omega, RE=220ΩR_E = 220\,\Omega, β=150\beta = 150, VBE=0.7VV_{BE} = 0.7\,\text{V}. Find the Q-point (VBV_B, IEI_E, ICI_C, VCEV_{CE}). Then drop β\beta to 75 and quantify how much the Q-point shifts.

  1. Thevenize the input divider.
    VTH=VCCR2R1+R2=1210572.105VV_{TH} = V_{CC} \cdot \frac{R_2}{R_1 + R_2} = 12 \cdot \frac{10}{57} \approx 2.105\,\text{V}
    RTH=R1R2=4710578.246kΩR_{TH} = R_1 \| R_2 = \frac{47 \cdot 10}{57} \approx 8.246\,\text{k}\Omega
  2. Solve the base-emitter loop. KVL through the Thevenin source, the base–emitter junction, and RER_E:
    IE=VTHVBERE+RTH/(β+1)I_E = \frac{V_{TH} - V_{BE}}{R_E + R_{TH} / (\beta + 1)}
    IE=2.1050.7220+8246/151=1.405220+54.65.12mAI_E = \frac{2.105 - 0.7}{220 + 8246/151} = \frac{1.405}{220 + 54.6} \approx 5.12\,\text{mA}
  3. ICI_C and VBV_B.
    IC=ββ+1IE5.08mAI_C = \frac{\beta}{\beta + 1} \cdot I_E \approx 5.08\,\text{mA}
    VB=VTHIBRTH2.1050.0348.2461.82VV_B = V_{TH} - I_B \cdot R_{TH} \approx 2.105 - 0.034 \cdot 8.246 \approx 1.82\,\text{V}
    (Stiff divider — small base-loading correction.)
  4. Output loop for VCEV_{CE}.
    VCE=VCCICRCIEREV_{CE} = V_{CC} - I_C \cdot R_C - I_E \cdot R_E
    VCE125.081.135.79VV_{CE} \approx 12 - 5.08 - 1.13 \approx 5.79\,\text{V}
    Mid-load-line: the load line runs from VCC=12VV_{CC} = 12\,\text{V} to VCC/(RC+RE)9.84mAV_{CC}/(R_C+R_E) \approx 9.84\,\text{mA}, so VCEQVCC/2V_{CEQ} \approx V_{CC}/2 is exactly where you want it.
  5. Now sweep β\beta from 150 to 75.
    IE(β=75)=1.405220+8246/76=1.405328.54.28mAI_E(\beta=75) = \frac{1.405}{220 + 8246/76} = \frac{1.405}{328.5} \approx 4.28\,\text{mA}
    IC(β=75)4.22mAI_C(\beta=75) \approx 4.22\,\text{mA}
    VCE(β=75)124.220.946.84VV_{CE}(\beta=75) \approx 12 - 4.22 - 0.94 \approx 6.84\,\text{V}
  6. Quantify the drift. Halving β\beta moved the Q-point from IC5.08mAI_C \approx 5.08\,\text{mA} to IC4.22mAI_C \approx 4.22\,\text{mA} — about a 17 % change.
    Compare to fixed bias. If we picked an RBR_B giving the same nominal IC5mAI_C \approx 5\,\text{mA} at β=150\beta = 150, halving β\beta halves ICI_C — a 50 % change. Voltage-divider bias is roughly 3× more β-stable here, all thanks to RER_E.
For voltage-divider bias the Q-point sits at IC(VTHVBE)/REI_C \approx (V_{TH} - V_{BE})/R_E as long as the divider is stiff (RTH(β+1)RER_{TH} \ll (\beta+1)\cdot R_E) and β\beta is reasonably large. The full formula IE=(VTHVBE)/(RE+RTH/(β+1))I_E = (V_{TH} - V_{BE})/(R_E + R_{TH}/(\beta+1)) shows exactly how much β leaks in.

Try the same circuit live — drag the β\beta slider in the Simulate stage and watch the green VDB Q-point barely move while the rose fixed-bias Q-point slides up and down its load line.