First-order RC and RL response
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| natural response/ˈnætʃərəl rɪˈspɒns/ | phản ứng tự nhiên |
| step response/step rɪˈspɒns/ | phản ứng bậc |
| time constant/taɪm ˈkɒnstənt/ | hằng số thời gian |
| flyback diode/ˈflaɪbæk ˈdaɪəʊd/ | diode giải phóng năng lượng dư |
One formula for every first-order circuit
- Every circuit with one storage element obeys $v(t) = V_{F} + (V_0 - V_{F})e^{-t/τ}$: the natural response 自然响应 decays, the step response 阶跃响应 slides between two fixed ends.
- $V_0$ comes from stored energy (voltage continuity for C, current continuity for L); $V_{F}$ from the new circuit with the element as its steady-state open (C) or short (L); $τ = RC$ or
L/Ruses the resistance seen with sources zeroed.
{exercise:uni-ee17-formula}
A capacitor starts at 1 V and the new circuit would hold it at 5 V, with τ = 0.1 s. What is v at t = 0.1 s?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
v = 5 − 4e^{−1} = 5 − 1.47 = 3.53 V. The gap (4 V) has closed to 37% of itself.
The time constant is the clock
- The time constant 时间常数 $τ$ is where the remaining gap has closed to 37%; after $5τ$ the circuit is within 1% and treated as settled.
- $22 kΩ \times 470 nF \approx 10.3 ms$, so settling takes about 52 ms. The exponent's argument is dimensionless because $τ$ carries the seconds.
$τ = RC = 1 s$: discharging from 12 V gives $v = 12e^{-t}$, so at 2 s it is $12 \times 0.135 \approx 1.62 V$. Charging from 0 to 9 V gives $9(1 - e^{-t})$ — 5.7 V at t = 1 s.
{exercise:uni-ee17-tau}
What does settling after 5τ mean numerically?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
e^{−5} ≈ 0.0067, so the remaining difference from the final value is under 1% — close enough to call finished.
Switching problems: three numbers, one line
- Find $V_0$ from the old circuit, $V_{F}$ from the new one, $τ$ from the new one with sources zeroed — then write the answer and check both endpoints.
- Reset the clock at the switch: after it, a 100 µF capacitor discharging through 20 kΩ has $τ = 2 s$ and $v = 9e^{-t/2}$.
{exercise:uni-ee17-switching}
A 9 V-charged 100 µF capacitor is switched to discharge through 20 kΩ. Which response is correct?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
τ = RC = 20 000 × 100µF = 2 s after the switch, and discharge decays from the initial 9 V — the charging τ no longer applies.
Inductor switching has consequences
- Stopping inductor current abruptly makes
v = L di/dtenormous: 0.1 H losing 0.6 A in 1 ms generates 60 V across the opening switch. - A flyback diode 续流二极管 keeps the current circulating, so the voltage spike collapses to about a diode drop plus IR.
{exercise:uni-ee17-flyback}
A 0.1 H relay coil carrying 0.6 A is interrupted in 1 ms. What voltage appears if nothing limits it?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
v = L di/dt = 0.1 × 600 = 60 V — the arc across the opening contact, tamed by a flyback diode.