BJT Bias Calculator (Q-Point IC and VCE)
Work out the base and collector currents, VCE and operating region of a voltage-divider biased NPN stage, and see how far the Q-point moves when β changes.
Leave R2 empty for fixed bias with a single base resistor, or set RE to 0 for no emitter resistor.
DC load line and Q-point
How it is calculated
The base side is replaced by its Thévenin equivalent, VTH = VCC·R2/(R1+R2) and RTH = R1‖R2. Then IB = (VTH − VBE)/(RTH + (β+1)RE), IC = β·IB and VCE = VCC − ICRC − IERE. If VCE comes out below 0.2 V the transistor is treated as saturated and the current is capped at what the resistors allow.
The defaults are a 2N3904 self-bias circuit from a PSpice simulation (β and VBE are the simulated values). The BJT biasing guide compares these hand calculations with PSpice.
Worked example: the Q-point of a self-bias circuit
- Thévenin: VTH = 6 × 33k ÷ 133k = 1.489 V, RTH = 100k ‖ 33k = 24.8 kΩ
- Base current: IB = (1.489 − 0.66) ÷ (24.8 kΩ + 133 × 1 kΩ) = 5.25 µA
- Collector current IC = 132 × IB = 693 µA, VCE = 4.47 V → forward active
This calculator against PSpice (sweeping RE)
| RE | β | VBE | IC (this calculator) | IC (PSpice) | Diff. | VCE (calc) | VCE (PSpice) |
|---|---|---|---|---|---|---|---|
| 220 Ω | 149 | 0.68 V | 2.08 mA | 2.07 mA | +0.7% | 3.04 V | 3.05 V |
| 470 Ω | 142 | 0.67 V | 1.26 mA | 1.26 mA | +0.3% | 3.89 V | 3.89 V |
| 1 kΩ | 132 | 0.66 V | 693 µA | 700 µA | −1.0% | 4.47 V | 4.46 V |
| 2.2 kΩ | 121 | 0.64 V | 350 µA | 350 µA | +0.1% | 4.80 V | 4.80 V |
A 2N3904 circuit with R1 = 100 kΩ, R2 = 33 kΩ, RC = 1.2 kΩ and VCC = 6 V, using only β and VBE from PSpice. With the right β the hand calculation lands within about 1%. The BJT biasing guide explains it in detail.
Frequently asked questions
What is the Q-point of a transistor?
The quiescent point is the DC collector current and collector-emitter voltage with no signal applied. For an amplifier it should sit near the middle of the DC load line so the output can swing equally both ways without hitting saturation or cut-off.
Why add an emitter resistor to a voltage-divider bias?
Without it the collector current is proportional to β, which can vary two or three times between parts. With an emitter resistor large enough that (β+1)RE is much bigger than R1‖R2, the current is set by (VTH − VBE)/RE and barely depends on β.
How do I know if the transistor is saturated?
If the calculated VCE comes out below about 0.2 V, or the base-collector junction is forward biased (VB above VC), the transistor is saturated and IC is limited by the resistors rather than by β.