First-order RC and RL response
도입됨| English | 의미 |
|---|---|
| natural response/ˈnætʃərəl rɪˈspɒns/ | 자연 응답 |
| step response/step rɪˈspɒns/ | 계단 응답 |
| time constant/taɪm ˈkɒnstənt/ | 시간 상수 |
| flyback diode/ˈflaɪbæk ˈdaɪəʊd/ | 플라이백 다이오드 |
Why does the light fade up?
- Press a button and a light fades up over a second; a sensor reading creeps to its new value over milliseconds. The stored quantity cannot jump, so it slides.
- A circuit with one storage element is first order, and every such circuit has the same shape of answer. Learn the shape once.
One formula for every first-order circuit
- Every circuit with one storage element obeys $v(t) = V_{F} + (V_{0} - V_{F})e^{-t/\tau}$. The natural response 自然响应 decays to zero; the step response 阶跃响应 slides between two fixed values.
- $V_{0}$ comes from stored energy (voltage continuity for C, current continuity for L). $V_{F}$ comes from the new circuit with the element as an open (C) or short (L). $\tau = RC$ or $\dfrac{L}{R}$, with the sources zeroed.
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?
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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 $\tau$ is when the remaining gap has shrunk to 37%. After $5\tau$ the circuit is within 1% and treated as settled.
- $22\,\text{k}\Omega \times 470\,\text{nF} \approx 10.3\,\text{ms}$, so settling takes about 52 ms. The exponent has no unit because $\tau$ carries the seconds.
$\tau = RC = 1\,\text{s}$: discharging from 12 V gives $v = 12e^{-t}$, so at 2 s it is $12 \times 0.135 \approx 1.62\,\text{V}$. Charging from 0 to 9 V gives $9(1 - e^{-t})$: 5.7 V at $t = 1\,\text{s}$.
What does settling after 5τ mean numerically?
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e^{−5} ≈ 0.0067, so the remaining difference from the final value is under 1% — close enough to call finished.
Starting from a charged capacitor
- The formula works between any two values, not just from zero. The capacitor starts where the old circuit left it.
- A capacitor at 2 V is switched, through a resistor, onto a 9 V supply with $\tau = 1\,\text{s}$: $v(t) = 9 + (2 - 9)e^{-t} = 9 - 7e^{-t}$.
At $t = 1\,\text{s}$: $9 - 7 \times 0.368 = 6.42\,\text{V}$. The 7 V gap has closed by 63.2%, exactly as the definition says. Endpoint checks: $v(0) = 2\,\text{V}$, and $v(5) \approx 9\,\text{V}$.
A capacitor at 1 V is switched onto a 5 V supply through a resistor, with τ = 2 s. What is its voltage after 2 s, in volts? Give two decimal places.
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$v = 5 + (1 - 5)e^{-1} = 5 - 4 \times 0.368 \approx 3.53\,\text{V}$. One time constant closes 63% of the 4 V gap.
Switching problems: three numbers, one line
- Find $V_{0}$ from the old circuit, $V_{F}$ from the new one, and $\tau$ from the new one with sources zeroed. Then write the answer and check both endpoints.
- Restart the clock at the switch: afterwards, a 100 µF capacitor discharging through 20 kΩ has $\tau = 2\,\text{s}$ and $v = 9e^{-t/2}$.
A 9 V-charged 100 µF capacitor is switched to discharge through 20 kΩ. Which response is correct?
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τ = RC = 20 000 × 100µF = 2 s after the switch, and discharge decays from the initial 9 V — the charging τ no longer applies.
Put the steps for a switching problem in order.
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The first two values come from different snapshots of the circuit; τ belongs to the new circuit.
The RL mirror
- For an inductor the same shape holds for current, with $\tau = \dfrac{L}{R}$.
- 20 mH in series with 100 Ω switched onto 10 V: $\tau = \dfrac{0.02}{100} = 0.2\,\text{ms}$ and $I_{F} = \dfrac{10}{100} = 0.1\,\text{A}$, so $i(t) = 0.1(1 - e^{-t/\tau})$. After $3\tau = 0.6\,\text{ms}$ it reaches 95%: 0.095 A.
A 50 mH inductor is in series with 25 Ω. What is the time constant, in ms?
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$\tau = \dfrac{L}{R} = \dfrac{0.05}{25} = 0.002\,\text{s} = 2\,\text{ms}$.
Inductor switching has consequences
- Stopping inductor current abruptly makes $v = L\dfrac{di}{dt}$ enormous: 0.1 H losing 0.6 A in 1 ms gives 60 V across the opening switch.
- A flyback diode 续流二极管 keeps the current circulating, so the spike shrinks to about a diode drop plus the coil's own $IR$.
A 0.1 H relay coil carrying 0.6 A is interrupted in 1 ms. What voltage appears if nothing limits it?
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v = L di/dt = 0.1 × 600 = 60 V — the arc across the opening contact, tamed by a flyback diode.
Three common slips
- Mixing the snapshots. $V_{0}$ comes from the old circuit and $V_{F}$ from the new one, deliberately.
- $\tau$ with the source still present. Zero the sources to see the resistance the storage element really faces.
- Treating $5\tau$ as a law of nature. It is a 1% engineering convention; the exponential never quite arrives.
After exactly 5τ, a charging capacitor has reached its final voltage.
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It has closed $1 - e^{-5} \approx 99.3\%$ of the gap. Engineers call that settled, but the exponential never reaches the final value exactly.
Recap
- $v(t) = V_{F} + (V_{0} - V_{F})e^{-t/\tau}$ for every first-order circuit.
- $\tau = RC$ or $\dfrac{L}{R}$, with sources zeroed; 63% of the gap closes in one $\tau$, 99% in $5\tau$.
- $V_{0}$ from continuity in the old circuit, $V_{F}$ from the new one; check both ends.