BJT biasing: the Q-point and the β shield
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| active region | active region: vùng khuếch đại |
| bias point | bias point: điểm làm việc |
BJT biasing: the Q-point and the β shield
- In the active region $I_C = \beta I_B$ with $V_{BE} \approx 0.7$ V; the bias must land the Q-point (the bias point) firmly active ($V_{CE} \gg 0.2$ V) and survive β varying 2:1 between devices.
- Thévenin the divider, solve the base loop exactly ($I_E = (\beta+1)I_B$), check the region — and let $R_E$ turn any current rise into a $V_{BE}$ reduction: negative feedback inside the bias.
$V_{CC} = 12$ V, $R_1 = 68$ kΩ, $R_2 = 12$ kΩ, $R_C = 2.2$ kΩ, $R_E = 1$ kΩ, $\beta = 100$: $V_B = 1.8$ V, $R_B = 10.2$ kΩ. Exact: $I_B = 1.1/(10.2\mathrm{k} + 101\mathrm{k}) = 9.9$ µA → $I_C = 0.99$ mA, $V_{CE} = 8.8$ V ✓ active. Double β to 200: $I_C = 1.04$ mA — a 5% move. The quick estimate (1.1 mA) was 11% high from divider loading; both error sources now have numbers.

{exercise:uni-ee5x-bias}
The 12 V divider-bias stage (68k/12k, RC 2.2k, RE 1k, beta 100) has IC and VCE of
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
VB = 1.8 V, RB = 10.2 k; exact base loop: IB = 9.9 uA, IC = 0.99 mA, VCE = 8.8 V - active. The 1.1 mA quick estimate is 11% high from divider loading.
Doubling beta from 100 to 200 in that stage moves IC by
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
IC goes 0.99 -> 1.04 mA: the emitter resistor turns the current rise into a VBE reduction - negative feedback inside the bias. With RE = 0 it would double.
The BJT stays in the active region as long as
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Saturation starts at about VCE = 0.2 V (both junctions forward). Checking VCE > 0 is the classic slip - it accepts a saturated stage.