Capacitors, inductors and energy
Introduzido| English | Significado |
|---|---|
| capacitor/kəˈpæsɪtə/ | capacitor |
| capacitance/kəˈpæsɪtəns/ | capacitância |
| inductor/ɪnˈdʌktə/ | indutor |
| inductance/ɪnˈdʌktəns/ | indutância |
| flyback diode/ˈflaɪbæk ˈdaɪəʊd/ | diodo de roda livre |
How a camera flash works
- A phone's camera flash collects energy for a second and releases it in about a millisecond. No arrangement of resistors can do that: resistors only burn energy.
- Two new elements store it. That one change is what makes circuits change with time.
The capacitor: current follows the slope of voltage
- A capacitor 电容器 stores charge on two plates separated by an insulator: $q = Cv$, with capacitance 电容 $C$ in farads. Differentiating gives $i = C\dfrac{dv}{dt}$.
- It carries current only while its voltage changes. At steady DC, $\dfrac{dv}{dt} = 0$, so the capacitor is an open circuit.
A 220 µF capacitor ramping from 0 to 10 V in 50 ms has $\dfrac{dv}{dt} = 200\,\dfrac{\text{V}}{\text{s}}$, so it carries $i = 220\,\mu\text{F} \times 200 = 44\,\text{mA}$ during the ramp, then zero.
The voltage on a 10 µF capacitor rises steadily at 100 V per second. What current flows, in mA?
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$i = C\dfrac{dv}{dt} = 10 \times 10^{-6} \times 100 = 0.001\,\text{A} = 1\,\text{mA}$, constant because the slope is constant.
The inductor: voltage follows the slope of current
- An inductor 电感器 stores energy in the magnetic field of a coil: $v = L\dfrac{di}{dt}$, with inductance 电感 $L$ in henrys.
- It develops voltage only while its current changes. At steady DC, $\dfrac{di}{dt} = 0$, so the inductor is a short circuit.
At DC steady state, what are a capacitor and an inductor equivalent to?
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Steady DC means dv/dt = 0 and di/dt = 0, so i_C = 0 (open) and v_L = 0 (short).
Combinations are mirror images
- Parallel capacitors share one voltage, so their charges and capacitances add. Two 47 µF capacitors in series give $\dfrac{47 \times 47}{94} = 23.5\,\mu\text{F}$.
- Inductors do the opposite: in series they add, in parallel they combine by the product-over-sum rule. The storage laws swap the roles of $v$ and $i$, so the resistor rules swap too.
A 2 µF and a 3 µF capacitor are connected in series. What is the combined capacitance, in µF?
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$\dfrac{2 \times 3}{2 + 3} = 1.2\,\mu\text{F}$. In parallel they would give 5 µF.
Stored energy, and the rate that matters
- Energy depends only on the present state: $w_{C} = \dfrac{1}{2}Cv^{2}$ and $w_{L} = \dfrac{1}{2}Li^{2}$.
- The amounts are small; the rate makes them dramatic. That is why large capacitors are discharged before anyone touches them.
A 100 µF capacitor at 12 V holds $\dfrac{1}{2} \times 100 \times 10^{-6} \times 144 = 7.2\,\text{mJ}$. Released in 1 ms that is 7.2 W; in 1 µs it would be 7.2 kW. A 20 mH inductor at 3 A holds $\dfrac{1}{2} \times 0.02 \times 9 = 90\,\text{mJ}$.
A 100 µF capacitor is charged to 12 V. How much energy does it store?
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w = ½Cv² = ½ × 100×10⁻⁶ × 144 = 7.2 mJ. Forgetting the ½ doubles it; using v instead of v² gives 1.2 mJ.
Stored quantities cannot jump
- Capacitor voltage cannot jump (that would need infinite current); inductor current cannot jump (that would need infinite voltage). This continuity is the key to every switching problem.
- Interrupting an inductor is therefore dangerous: stopping 1.5 A in 1 ms through 0.1 H gives $v = 0.1 \times 1500 = 150\,\text{V}$, the arc across the opening contact.
A flyback diode 续流二极管 across the coil gives the current a circulating path. The current then dies away through the coil's own resistance, and the spike shrinks to about one diode drop.
Why can inductor current not jump to zero when a switch opens?
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v = L di/dt — driving i to zero in ~0 time needs an enormous di/dt, hence an enormous voltage that arcs to keep the current flowing.
Match each element to how it behaves.
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Continuity of the stored quantity decides what happens at every switching instant.
Three common slips
- Parallel capacitances added as reciprocals. That is the series rule; parallel capacitances simply add.
- "A capacitor blocks DC" at all times. It blocks steady DC; while the voltage changes it carries current gladly.
- Units in the energy formula. Use farads and volts: $\text{F} \times \text{V}^{2} = \text{J}$. Convert µF and mH first.
A capacitor connected to a DC supply carries no current at any time.
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While it charges, its voltage changes and it carries current $i = C\dfrac{dv}{dt}$. Only once the voltage is steady does the current fall to zero.
Recap
- $i = C\dfrac{dv}{dt}$ and $v = L\dfrac{di}{dt}$: capacitors remember voltage, inductors remember current.
- Steady DC: capacitor open, inductor short. Energies $\dfrac{1}{2}Cv^{2}$ and $\dfrac{1}{2}Li^{2}$.
- Capacitor voltage and inductor current cannot jump; never open an inductor's path suddenly.