Biasing and the Q-Point
β varies 4:1 for one part number, so any bias scheme that depends on it depends on luck. The fix is a feedback loop in a DC circuit.
Skip to the animationBiasing sets the DC operating point on the load line, and a usable scheme must be independent of β — which voltage-divider bias with an emitter resistor achieves by closing a negative feedback loop that corrects the operating point automatically.
The load line and the Q-point
The supply and collector resistor fix a load line: every DC operating point the stage can have lies on it. The quiescent point is where it sits with no signal, and the signal then swings it along that line.
| Q-point position | What happens to the signal |
|---|---|
| Near cut-off | The negative half clips — the transistor turns off |
| Near saturation | The positive half clips, and quiescent dissipation is high |
| Mid-line | Maximum symmetric swing — the usual small-signal choice |
Why fixed bias fails
A single resistor from the supply to the base sets I_B directly, so I_C = βI_B inherits the full 4:1 spread of β. Replacing the transistor with another of the same part number can move the Q-point to either end of the load line.
Temperature makes it worse: both β and V_BE drift, and V_BE falls about 2 mV per °C, so a warming stage conducts harder at the same drive.
Emitter degeneration
An emitter resistor closes a negative feedback loop in the DC circuit:
- 1I_C rises for any reason — a hotter device, a higher-β part.
- 2The emitter voltage rises with it, since V_E = I_E·R_E.
- 3With the base held at a fixed voltage, V_BE therefore falls.
- 4Lower V_BE reduces I_C, opposing the change that started it.
The correction is automatic and continuous. It is the same negative-feedback idea as a closed-loop controller, applied to a bias point rather than a set-point.
Voltage-divider bias
A divider holds the base at a stiff fixed voltage, provided its own current is roughly ten times I_B. Then I_E ≈ (V_B − 0.7)/R_E — with no β anywhere in it.
- Divider current about ten times I_B: stiff enough to be independent of the base, without wasting supply current.
- V_E about a tenth of V_CC: enough stabilisation without spending too much headroom.
- This one circuit is why discrete amplifier stages are reproducible at all.
The bypass capacitor
The emitter resistor feeds back the signal as well as the bias, dropping the gain to roughly R_C/R_E. A bypass capacitor across it looks like a short at signal frequencies and an open at DC — recovering the gain while keeping the stabilisation.
It sets the stage's low-frequency cutoff, so it is often the largest component on the board. Leaving a small unbypassed portion trades some gain back for better linearity, which is a common deliberate choice.
Thermal runaway
In a power stage the feedback can run the wrong way. V_BE falls about 2 mV per °C, so a warming transistor conducts harder at the same drive, heats further, and conducts harder still — positive feedback with a physical mechanism.
Emitter degeneration, adequate heatsinking, and a bias diode mounted on the same heatsink so that it tracks the transistor's temperature are the standard defences. All three are usually present in an audio output stage.
The numbers you will be asked for
- Load line
V_CE = V_CC − I_C · R_C
- Divider base voltage
V_B = V_CC · R₂ / (R₁ + R₂)
- Emitter current
I_E ≈ (V_B − 0.7) / R_E
no β term
- Stability rule of thumb
I_divider ≈ 10 · I_B
- V_BE temperature drift
dV_BE/dT ≈ −2 mV/°C
Watch it work
Check yourself
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