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I hope this is not too trivial a question.

I know that a freewheeling diode is connected across an inductor to eliminate the sudden voltage spike that occurs across an inductive load due to interruption of supply current or sudden reduction.
(This voltage spike will damage the switches present in the circuits).

$$V = L\cdot{\mathrm dI\over \mathrm dt}$$

What I don't understand is: why I never hear about freewheeling diode in the case of capacitor?

Since in the capacitor \$I = C\cdot \mathrm dV/\mathrm dt\$, a large voltage change at its ends produces a huge current ⇒ it also needs a freewheeling diode to not burn out everything in series with the capacitor.

Thank you in advance for your valuable clarifications

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7 Answers 7

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why I never hear about Freewheeling diode in the case of capacitor

In the case of charging a capacitor, the device is called an inrush limiter circuit. When wanting to avoid discharging too much current from a capacitor, a simple resistor (or the internal ESR of capacitor) may do the job. Other more complex circuits may be used in more extreme situations.

Diodes usually won't help but, the internal resistance of a forward biased diode may reduce the inrush current in some cases.

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No the logic is wrong, a capacitor does not need a free-wheeling diode, because when you stop applying current to a capacitor, it holds the voltage it was charged to.

If you apply a voltage step to a capacitor, there will be a high peak current that may destroy switches with overcurrent. But a diode does not prevent it from happening.

A resistor to limit the current could be used.

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To shunt the high voltage spike from an interrupted inductor, you need a parallel device that becomes more conductive with more voltage applied across it.

  • A diode.

To inhibit the high current spike from a switched (being connected) capacitor, you need a series device that becomes less conductive with more current through it.

  • A current limiter (in one of its many realizations).

The analogy with a diode are still obvious, which can be also seen by a symbol of one of the possible current limiters, the constant current diode:

enter image description here

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  • \$\begingroup\$ Just wondering, did you mean to write "a series device that becomes less conductive with more current applied to it" (which, being the dual, would fit in with the rest of the sentence)? \$\endgroup\$ Commented Jun 25 at 18:04
  • \$\begingroup\$ There's no "high current spike from a switched capacitor". A charged cap that is abruptly switched off will provide its voltage, and will cause a back-flow of the current into rest of the circuit.. If you want to stop current flowing back out of 'interrupted' coil, you have to add a diode in-direction of the current that shows up (reversed wrt to normal charging flow). And, if you want to stop propagation of voltage from a cap, you have to block it through something with hi-Z, like, a diode, but opposed to the flow FROM the capacitor (so, a diode, in direction same as the charging flow). \$\endgroup\$ Commented Jun 25 at 21:14
  • \$\begingroup\$ @TannerSwett good idea. It comes to the same effect, but due to the "duality" you mentioned, it makes more sense. \$\endgroup\$
    – tobalt
    Commented Jun 26 at 17:27
  • \$\begingroup\$ @quetzalcoatl If you connect (not interrupt) a capacitor to a low impedance source of different voltage, then - yes - a large current spike will result. \$\endgroup\$
    – tobalt
    Commented Jun 26 at 17:29
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It may (or may not) help to think of the diode as another switch.

When you switch the voltage across an ideal inductor from one voltage to another the current increases or decreases linearly with a slope proportional to the applied voltage, without limit. If you have 0V across the inductor the current continues indefinitely. In reality the inductor is not very ideal and there is some resistance so the current dies out.

In order to switch from one voltage to another you need two switches. With an inductor we may be able to use a diode as one of the switches since the polarity reverses across the inductor when the switch opens. It's not the only way though, we can also use another transistor and we sometimes call that "synchronous rectification" in a switching power supply.

When you switch the current into an ideal capacitor from one current to another the voltage increases or decreases linearly with a slope proportional to the applied current, without limit. If you have 0A into the capacitor, the voltage continues indefinitely. In reality the capacitor will tend to leak off through itself or some non-ideal element.

To switch a current source from some current to 0 you also might need two switches, one to short the current source or the voltage would increase without limit, and one to disconnect the capacitor. You might be able to use a diode for the latter, shorting the current source with an active switch.

This is a rare requirement though since we usually have constant voltage sources rather than current sources available, and we have to make the latter. It is one way to make a desirable signal- a very linear and extremely fast voltage ramp since we're just steering the current from the current source rather than trying to turn it on and off, and the latter takes time.

schematic

simulate this circuit – Schematic created using CircuitLab

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Because excessive voltage is what kills components(e.g. oxide layer in MOSFET, arcing over a component, ...). Whereas with current the capacitor discharges based on the current the load consumes, not what the source provides.

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    \$\begingroup\$ That's a bit poor analogy, because if you have e.g. a battery, power switch, and the capacitor, when you flip the switch on, the capacitor tries to charge with all the current that is available. In theory infinitely high current, in practice just very high and limited by ESR of the battery, switch and capacitor. \$\endgroup\$
    – Justme
    Commented Jun 24 at 16:01
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The large current surge can certainly be a problem for the capacitor, but to avoid it a parallel diode, like the freewheeling diode, would not help (and a diode in series would block all current!)

So what you need to address this problem is an "inrush current limiter" which could be just a resistor. In electronic circuits like lamp drivers for LED lamps or fluorescent lamps, this is often a "fusistor", which means that it can also act as fuse if the current surge lasts too long, indicating there is something broken in the circuit.

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I'd say the 'symmetry' logic is not fully applied here. The diode at inductor is added to take care of the situation when the current through the inductor (tries to) decay, the inductor (tries to) sustain the current, but this extra current flips the voltage to negatives wrt original flow, magnetic field discharges, etc. For a capacitor, an identical applicable situation would be: capacitor was held at certain voltage, but then the voltage (tried to) decay, so capacitor (tried to) sustain the voltage, this extra voltage flips the current to negatives wrt original, electric field discharges...

It's just that for a capacitor, what we get is a some reverse-current flowing out from cap at its max-charge voltage, limited by circuits resistance, and that current is usually not a problem. At the same time, for an inductor we get some reverse-voltage produced from inductor's max-flux current, boosted by circuits resistance, and this reverse-voltage often can be high enough to be catastrophic for unprepared circuits.

I can however imagine some old transistors or early ICs, that didn't have any protection diodes (remember early ICs VERY easily destroyed with static discharges?), that could be fragile to any reverse bias, and that connecting their outputs directly to a charged capacitor, and then removing external power could cause reverse current flow, reverse-biasing, and some damage.. I don't know if there were any real-world cases..

Anyways, in such case, "just" add a diode, but in series, so most most of the reverse-current does not form and most of (sustained) voltage decays at it.. Just like for a coil, we'd add a diode in parallel, so reverse voltage does not form and most of the (sustained) current decays at it..

Please note that I do not argue with in-rush current for a suddenly-charged capacitor. I just think that the term "free-wheeling" phrase for a capacitor should be considered for the same situation as for a coil: When the component is energized/charged (device gained 'inertia'), and when then the 'charging' abruptly stops (and then 'inertia' manifests and tries to dissipate back into the circuit). Both inductor and capacitor fit that description and situation, but the exact effects are flipped w.r.t current/voltage.

Talking about huge-inrush-current (near-zero voltage at terminals, despite high R of the cap) for a suddenly-connected uncharged cap is symmetrical to nearzero-inrush-current (highvoltage at terminals, despite low R of the coil) for an suddenly-connected unmagnetized coil ;)

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