Second-order RLC circuits
導入済み| English | 意味 |
|---|---|
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | natural frequency |
| damping/ˈdæmpɪŋ/ | 減衰 |
| overdamped | 過減衰 |
| critically damped | 臨界減衰 |
| underdamped | 不足減衰 |
| Q factor | Q値 |
Why does it ring?
- Close a switch on a charged capacitor in series with an inductor. The current cannot jump, and the capacitor cannot empty at once, so energy swings back and forth between them.
- Whether it rings, creeps or snaps back depends on two numbers. This lesson finds them and reads the response from them.
The second-order equation
- Series RLC by KVL gives $L\dfrac{d^{2}i}{dt^{2}} + R\dfrac{di}{dt} + \dfrac{i}{C} = 0$. Two energy stores mean a second derivative.
- Two numbers describe it: the natural frequency 固有频率 $\omega_{0} = \dfrac{1}{\sqrt{LC}}$ and the damping 阻尼 $\alpha = \dfrac{R}{2L}$. The roots are $s = -\alpha \pm \sqrt{\alpha^{2} - \omega_{0}^{2}}$.
A series RLC has L = 10 mH and C = 1 µF. What is ω₀, in rad/s?
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$\omega_{0} = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{10^{-2} \times 10^{-6}}} = \dfrac{1}{10^{-4}} = 10\,000$ rad/s.
Three regimes from one comparison
- Overdamped 过阻尼 ($\alpha > \omega_{0}$): two real roots, two decays, no ringing. Critically damped 临界阻尼 ($\alpha = \omega_{0}$): the fastest return with no ringing.
- Underdamped 欠阻尼 ($\alpha < \omega_{0}$): a decaying oscillation at $\omega_{d} = \sqrt{\omega_{0}^{2} - \alpha^{2}}$, the ringing you see on a scope.
Series RLC with α = ω₀ is
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The roots $-\alpha \pm \sqrt{\alpha^{2} - \omega_{0}^{2}}$ merge into one repeated real root, the border between ringing and creeping.
Match each condition to the response.
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With no resistance nothing removes energy, so the circuit rings for ever at $\omega_{0}$.
Worked example: pick the regime
- $L = 0.2\,\text{H}$, $C = 0.05\,\text{F}$: $\omega_{0} = \dfrac{1}{\sqrt{0.01}} = 10\,\dfrac{\text{rad}}{\text{s}}$. With $R = 4\,\Omega$, $\alpha = \dfrac{4}{0.4} = 10$: critically damped.
- With $R = 2\,\Omega$, $\alpha = 5 < 10$: underdamped, ringing at $\omega_{d} = \sqrt{100 - 25} = 8.66\,\dfrac{\text{rad}}{\text{s}}$.
R = 4 Ω, L = 0.2 H, C = 0.05 F. The regime is —
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$\omega_{0} = \dfrac{1}{\sqrt{0.01}} = 10$ and $\alpha = \dfrac{R}{2L} = 10$: equal, so critical. Halve R to 2 Ω and it becomes underdamped with $\omega_{d} = 8.66$.
The Q factor
- The Q factor 品质因数 $Q = \dfrac{\omega_{0}}{2\alpha}$ measures how long the circuit rings. Critical damping is $Q = 0.5$; below it, no ringing.
- In the example, $R = 2\,\Omega$ gives $Q = 1$. A high-Q circuit, such as a radio tuner, rings for many cycles.
L = 0.2 H, C = 0.05 F and R = 1 Ω. What is the Q factor?
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$\alpha = \dfrac{1}{0.4} = 2.5$ and $\omega_{0} = 10$, so $Q = \dfrac{10}{5} = 2$: underdamped, ringing for a few cycles.
Worked example: the overdamped exponents
- $R = 8\,\Omega$: $\alpha = 20 > 10$. The roots are $-20 \pm \sqrt{300}$: $-2.68$ and $-37.32$ per second.
- Released from 1 V with no current, $v_{C} = Ae^{-2.68t} + Be^{-37.32t}$ with $A + B = 1$ and zero initial slope. So $A = \dfrac{37.32}{34.64} = 1.077$, $B = -0.077$.
The fast root dies within a tenth of a second; the slow root, nearer zero, owns the tail. That is the dominant-pole rule: the root closest to the origin sets how long the response lasts.
Series RLC with R = 8 Ω, L = 0.2 H, C = 0.05 F: the two natural exponents are
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$\alpha = 20 > \omega_{0} = 10$: $s = -20 \pm \sqrt{300}$. The slow root −2.68 owns the tail (the dominant pole).
Three common slips
- $\omega_{0}$ for $\omega_{d}$. The ringing you see is at $\omega_{d}$, slightly below $\omega_{0}$.
- One initial condition. Two energy stores need two: the capacitor voltage and the inductor current.
- Comparing the wrong things. Compare $\alpha$ with $\omega_{0}$, not $R$ with $\omega_{0}$.
An underdamped circuit rings at exactly ω₀.
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It rings at $\omega_{d} = \sqrt{\omega_{0}^{2} - \alpha^{2}}$, a little below $\omega_{0}$. With $\alpha = 5$ and $\omega_{0} = 10$ that is 8.66 rad/s.
Recap
- $\omega_{0} = \dfrac{1}{\sqrt{LC}}$, $\alpha = \dfrac{R}{2L}$; one comparison picks the regime.
- Underdamped rings at $\omega_{d} = \sqrt{\omega_{0}^{2} - \alpha^{2}}$; $Q = \dfrac{\omega_{0}}{2\alpha}$, with critical damping at 0.5.
- The slow root owns the tail; two stores need two initial conditions.