A voltage-divider bias has VCC=12V, R1=47kΩ, R2=10kΩ, RC=1kΩ, RE=220Ω, β=150, VBE=0.7V. Find the Q-point (VB, IE, IC, VCE). Then drop β to 75 and quantify how much the Q-point shifts.
- Thevenize the input divider.
VTH=VCC⋅R1+R2R2=12⋅5710≈2.105V RTH=R1∥R2=5747⋅10≈8.246kΩ - Solve the base-emitter loop. KVL through the Thevenin source, the base–emitter junction, and RE:
IE=RE+RTH/(β+1)VTH−VBE IE=220+8246/1512.105−0.7=220+54.61.405≈5.12mA - IC and VB.
IC=β+1β⋅IE≈5.08mA VB=VTH−IB⋅RTH≈2.105−0.034⋅8.246≈1.82V (Stiff divider — small base-loading correction.) - Output loop for VCE.
VCE=VCC−IC⋅RC−IE⋅RE VCE≈12−5.08−1.13≈5.79V Mid-load-line: the load line runs from VCC=12V to VCC/(RC+RE)≈9.84mA, so VCEQ≈VCC/2 is exactly where you want it. - Now sweep β from 150 to 75.
IE(β=75)=220+8246/761.405=328.51.405≈4.28mA IC(β=75)≈4.22mA VCE(β=75)≈12−4.22−0.94≈6.84V - Quantify the drift. Halving β moved the Q-point from IC≈5.08mA to IC≈4.22mA — about a 17 % change.
Compare to fixed bias. If we picked an RB giving the same nominal IC≈5mA at β=150, halving β halves IC — a 50 % change. Voltage-divider bias is roughly 3× more β-stable here, all thanks to RE.
For voltage-divider bias the Q-point sits at
IC≈(VTH−VBE)/RE as long as the divider is stiff (
RTH≪(β+1)⋅RE) and
β is reasonably large. The full formula
IE=(VTH−VBE)/(RE+RTH/(β+1)) shows exactly how much β leaks in.