Supplies: ripple and the zener window
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| ripple/ˈrɪpl/ | gợn sóng |
| peak inverse voltage | The maximum reverse voltage a rectifier diode must survive without breaking down. |
| filter capacitor | filter capacitor: tụ lọc |
| zener diode | zener diode: diode zener |
Supplies: ripple and the zener window
- The bridge conducts both half-cycles at a cost of 1.4 V (two diodes at a time) and every diode must survive one secondary peak as peak inverse voltage. The filter capacitor supplies the load between recharge peaks: ripple $\Delta V = I_{load}/(2fC)$.
- The zener diode holds the voltage while its series resistor absorbs the surplus — a complete design needs one inequality from EACH end of the load range.
12 V RMS secondary, 50 Hz, 100 mA load, 2200 µF: peak $12\sqrt{2} - 1.4 = 15.6$ V, ripple $0.1/(100 \times 2200\mu) = 0.455$ V — the output lives between 15.6 and 15.1 V (ripple 2.9% of peak ✓). Zener side: 12 V to a 5.1 V zener, load 0–10 mA, minimum zener current 2 mA: $R \le (12-5.1)/12\mathrm{m} = 575$ Ω → 560 Ω; no-load $I_Z = 12.3$ mA ($P_Z = 63$ mW), full-load $I_Z = 2.3$ mA ≥ 2 mA ✓ — both extremes verified.

{exercise:uni-ee5x-supply}
A bridge on a 12 V RMS, 50 Hz secondary with 2200 uF feeding 100 mA has ripple
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
dV = I/(2fC) = 0.1/(100 x 2200u) = 0.455 V. Half-wave would double it to 0.91 V.
Each bridge diode must survive a peak inverse voltage of about
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The blocked pair sees one full secondary peak, 12 x sqrt(2) = 17 V - rate the diodes above that.
A 12 V to 5.1 V zener regulator, load 0-10 mA, minimum zener current 2 mA, allows R up to
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
R <= (12-5.1)/(10m + 2m) = 575 ohm from the FULL-load extreme; then check no-load: I_Z = 12.3 mA, P_Z = 63 mW - both ends verified.