About this calculator
A transistor amplifier only works if it is biased into its active region first — sitting at a steady operating point with enough headroom to swing both ways around it. Voltage-divider bias is the standard way to establish that point, because it holds steady even when the transistor's current gain varies.
This calculator takes the six values that define the stage — supply, the two divider resistors, collector and emitter resistors, and β — and returns the complete DC operating point plus the small-signal gain and input impedance.
How it works
The two divider resistors are replaced by their Thévenin equivalent: a source of Vcc × R2/(R1+R2) behind a resistance of R1 ∥ R2. That turns the input network into something you can walk a single loop around.
Going round that loop, the Thévenin voltage has to cover the base-emitter drop plus the voltage developed across the emitter resistor by the emitter current. Since emitter current is (β+1) times base current, solving gives Ib = (Vth − Vbe) / (Rth + (β+1)Re).
Everything else falls out from there. Collector current is β × Ib, the emitter sits at Ie × Re, the base one diode drop above that, and the collector at Vcc − Ic × Rc. The difference between collector and emitter is Vce, which is the number that tells you whether the stage is usable.
The point of the emitter resistor is negative feedback. If temperature or a different transistor pushes the collector current up, the emitter voltage rises, which reduces the base-emitter voltage, which pushes the current back down. That is what makes this topology stable where simpler biasing schemes are not.
For AC behaviour, the transistor presents a small internal emitter resistance re′ ≈ 26 mV / Ie. Gain is the collector resistance divided by the total emitter resistance, so leaving Re unbypassed gives modest, predictable gain, while bypassing it with a capacitor gives much higher but far less stable gain.
Worked example
A general-purpose audio preamp stage on 12 V, using a 2N3904 with β around 150.
- Divider: R1 = 47 kΩ, R2 = 10 kΩ → Vth = 12 × 10/57 = 2.11 V, Rth = 8.25 kΩ
- Ib = (2.11 − 0.7) / (8250 + 151 × 470) = 1.41 / 79220 = 17.8 µA
- Ic = 150 × 17.8 µA = 2.67 mA
- Ve = 2.69 mA × 470 = 1.26 V, Vb = 1.96 V
- Vc = 12 − 2.67 mA × 2.2 kΩ = 6.13 V
- Vce = 6.13 − 1.26 = 4.87 V
- re′ = 26 mV / 2.69 mA = 9.7 Ω → Av = −2200 / 480 = −4.6
The stage sits at 4.87 V across the transistor with 2.67 mA flowing — comfortably in the active region with room to swing several volts either way. Gain is about 4.6×, inverting. Bypassing the emitter resistor would raise that to roughly 227× at the cost of stability.
Practical notes
- Keep the divider current at least 10× the base current. If the divider is too weak, base current loads it and the bias point starts tracking β, which is exactly what this topology exists to avoid.
- A good starting point is the emitter at about 10% of Vcc and Vce at roughly half of what remains, which gives symmetric clipping headroom.
- β is not a design parameter you can rely on. A 2N3904 is specified anywhere from 100 to 300, and it varies with current and temperature. Any design that depends on a specific β is fragile.
- Vbe falls by about 2 mV per °C of junction temperature. Without the emitter resistor to counteract it, that drift alone can drag a stage into saturation as it warms.
- Bypassing Re gives you gain but makes it depend on
re′, which is set by current and therefore temperature. A common compromise is splitting Re into two parts and bypassing only one. - Real input impedance is the divider in parallel with the transistor's own input resistance, and it is usually dominated by the divider. That matters when the preceding stage cannot drive a low impedance.
Frequently asked questions
What is the Q-point of a transistor?
The quiescent operating point — the steady DC collector current and collector-emitter voltage with no signal applied. The signal swings around that point, so it needs to sit far enough from both saturation and cutoff for the swing to fit.
Why use voltage-divider bias instead of a single base resistor?
A single base resistor sets base current directly, so collector current becomes β times a fixed value — and β varies hugely between parts and with temperature. A divider sets base voltage, and the emitter resistor then fixes the current almost independently of β.
What does it mean if Vce is nearly zero?
The transistor is saturated — fully on, with the collector pulled down close to the emitter. It cannot amplify there because increasing the base drive no longer increases collector current. Reduce Rc, reduce Re, or lower the divider ratio.
How do I calculate the gain of a common-emitter amplifier?
Divide the collector resistance by the total AC emitter resistance: Av = −Rc / (re′ + Re), where re′ = 26 mV / Ie. The minus sign means the output is inverted.
What is re prime?
The transistor's intrinsic small-signal emitter resistance, approximately 26 mV divided by the emitter current. It comes from the diode equation, and the 26 mV is the thermal voltage at room temperature. It is not a real resistor you can see, but it limits gain just like one.