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In the book The Art of Electronics the author writes the following when trying to explain the inductive kick:

enter image description here

And then he mentions when the switch is opened abruptly, the inductor tries to keep current flowing and damages the switch if no flyback diode is used. And that differential equation formula shows that large peak in voltage.

So far I understand his point. But the same logic applies when the switch is closed because dI/dt again can be large.

It seems like something is missing here to differ between what happens to dI/dt in case of switch being closed comparing to the switch being opened. How can we have a more clear insight about this? Why theres is no large voltage in case of the switch is being closed? How can the difference between opening and closing be mathematically demonstrated?

user1245
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    The act of closing the switch forces a 'sharp' edge of current into the inductor. This edge is kicked back into the closed switch. The power source must absorb the kick. That is why sometimes you see a Zener diode or TVS or MOV being used as both a snubber and a kick absorber. –  Jun 28 '18 at 01:29
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    i dont understand – user1245 Jun 28 '18 at 01:51
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    @Sparky256. Completely wrong, on closing the switch in a circuit feeding an inductor, the current will be zero and increase at the rate LR. You can't force current into an inductor. When you open the switch in a circuit with current flowing through an inductor the current will (for a very small time) be exactly what it was with the switch closed. You therefore get a voltage sufficient to breakover any resistance, hence the very high voltages. – Jack Creasey Jun 28 '18 at 02:04
  • @JackCreasey wrote *"You can't force current into an inductor"* - That's a bit too strong. Consider the counter example of an ideal current source connected across an ideal inductor. – Alfred Centauri Jun 28 '18 at 02:59
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    @JackCreasey, the last two sentences seem to contradict each other. If, for a very small time, the current will be *exactly* what is was with the switch closed then why is there a voltage sufficient to breakover etc. ? By definition, the voltage across the inductor depends on the rate of change of current and if the current is *exactly* what it was, it isn't changing. – Alfred Centauri Jun 28 '18 at 03:05
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    @atomant When the switch is closed, the current is zero just prior, but a non-zero voltage may be applied by that act to the inductor which may initiate a current increasing from zero. No abrupt current change is possible. Almost by definition it isn't abrupt. Also, there is no magnetic field in the inductor. When the switch is opened, though, there may be an attempt at an abrupt change in current. So it is only here, in the middle of an active current being abrupted by the switch, that a high voltage induced by a collapsing magnetic field is possible. Because... *there is a magnetic field*. – jonk Jun 28 '18 at 04:11
  • @JackCreasey. If you had read my comment more carefully, at no point did I say anything about current *through* the inductor, which can only increase at rate LR. The sharp rising edge is reflected back to the switch, just like a miss-matched antenna. –  Jun 28 '18 at 04:45
  • @Sparky256 ...you said " forces a 'sharp' edge of current into the inductor." ...that cannot happen. – Jack Creasey Jun 28 '18 at 05:03
  • @jonk. Thank you for clarifying the issue. A better wordsmith than me.`No abrupt current change is possible.` That 'abrupt' part is reflected back to the switch until RL takes over and current flows. By abrupt I mean a very short term transient NOT going through the inductor. –  Jun 28 '18 at 05:03
  • @JackCreasey. My bad. I should have wrote "tries to force". –  Jun 28 '18 at 05:05
  • @AlfredCentauri ...when you open a switch there will be a spark (no matter how small), and that plasma will conduct ...if it cannot support the current flowing in the inductor, the field will start to collapse and the terminal voltage of the inductor rise. ….however you explained it nicely in your answer. – Jack Creasey Jun 28 '18 at 05:05
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    @Sparky256 It's no more than having a mental vision and finding words. Heart of the matter isn't anything more complicated than that there is *NO* magnetic field in one case, and supposedly is *SOME* magnetic field stored in the space-time vacuum in the other case. It's a NONE vs SOME thing. Couldn't be easier. Not really word-smithy. Just obvious when you think about it. Equations provide quantities but they don't help understand. When you remove what is supporting the storage of energy in a vacuum, that energy must act. No surprise. Understanding the switch is less relevant to the OP's Q. – jonk Jun 28 '18 at 05:39
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    Datapoint: " ... the inductor tries to keep current flowing ..." -> FALSE. The inductor **KEEPS** current flowing. If there is nowhere to flow to the voltage WILL instantaneously rise to infinite. Long before Vinfinite is reached there will always be SOMEWHERE to flow to. In a very well insulated system the current may rtransfer into the stray capacitance such that 0.5.L.i^2 = 0.5.C.V^2. | It is easy to get hundreds of volts from a low voltage circuit when an inductor circuit is 'opened'. – Russell McMahon Jun 28 '18 at 08:46
  • https://electronics.stackexchange.com/questions/288380/energy-stored-and-lagging-of-current-in-a-inductive-circuit/288384#288384 – G36 Jun 28 '18 at 13:57

3 Answers3

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An open switch is an open circuit. It enforces a rule: I = 0.

A closed switch is a short circuit between its terminals. It enforces a rule: V = 0 (between the terminals).

You can see that these situations are fundamentally different. One has a rule about current and the other has a rule about voltage.

When you close the switch it doesn't do anything to force an instant change in current in the wires that connect it, while when you open the switch it does force an instant change in current.

(Of course nothing in life is ideal, neither the switch nor the inductor, so when you really open a switch in series with an inductor if you want to know the actual behavior, you must consider a more complex model than ideal devices. Include the inter-winding capacitance of the inductor, and the arcing behavior of the switch in your model if you want to find out the actual voltage developed by the switch opening)

The Photon
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There is no difference. For the exact same reasons (can't have infinite dI/dt), you can't instantly change from no current to some current when you close the switch.

Jashaszun
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  • but why in the case of closing no large voltage is induced? – user1245 Jun 28 '18 at 01:52
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    @atomant By closing the switch you apply a voltage step across the inductor. Presumably (or let's assume) the inductor current is zero on closing the switch, and V = L*di/dt. So the current changes as the integral of the voltage, which is a step meaning the current ramps linearly from zero until the inductor saturates or the switch is opened. – John D Jun 28 '18 at 02:23
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    I actually disagree with Centauri when he writes to Photon saying that Photon got right to the heart of the matter. And I think that while Jashaszun's answer is rather too short and lacks context, it does highlight the central point that the current is *already* zero prior to the switch closing. So closing it doesn't change, nor imply a change, in the current. It's not as though there is some huge current "just waiting to get through" the moment the switch closes, after all. So I just wish Jashaszun had taken more time to provide a better context. The OP should read this though, and think. – jonk Jun 28 '18 at 04:04
  • @jonk, (just being honest here) this answer doesn't make sense to me. As I read it, the *reason* given for the answer "There is no difference" is "can't have infinite dI/dt". But, in the ideal circuit element context (which is AFAIK the context of the question), an ideal closed switch with non-zero current through *must* instantaneously change the current through to zero when opened (otherwise, it is not ideal). Conversely, an ideal open switch with non-zero voltage across *must* instantaneously change the voltage across to zero but the dI/dt is determined by the circuit, not the switch. – Alfred Centauri Jun 28 '18 at 14:11
  • @AlfredCentauri If you want to take everything as ideal components the switch is an ideal spark gap, this will fire and the current will decay through the discharge. In practice as mentioned elsewhere capacitance (in the surroundings of the inductor) will absorb some of the energy in some cases or else some protective device will operate or the switch will arc. Ideal components will still arc over at infinite voltage. – KalleMP Jun 28 '18 at 17:49
  • Jashaszun, the difference is not in the current handling but as mentioned elsewhere there is the stored energy to contend with. In one scenario stored energy has to be discharged without violating the instantaneous current change rule and in the other scenario there is no stored energy yet, it gets charged up at a rate determined by the component values (ideal or real) and imposed voltage. – KalleMP Jun 28 '18 at 17:52
  • @KalleMP wrote *"If you want to take everything as ideal components the switch is an ideal spark gap"* - that's just plain nonsense. – Alfred Centauri Jun 28 '18 at 18:02
  • @AlfredCentauri ? How do you model two closely spaced conductors? This obviously refers to the opened switch. The closed switch is just a conductor. Please elaborate on the nonsense as it is not plain to me. – KalleMP Jun 28 '18 at 18:19
  • @KalleMP, from ideal circuit theory, an ideal switch is an *ideal* open circuit (\$i = 0\$ for *any* voltage across) when open and an ideal short circuit (\$v = 0\$ for *any* current through) when closed. As I've done in my answer, and as is done throughout the literature, singularity functions, e.g., unit step \$u(t)\$ are used to mathematically treat ideal switches. See, for example, [this screen shot](https://i.stack.imgur.com/wmaT8.png) from "Electrical Circuits: An Introduction" (courtesy of Google books). – Alfred Centauri Jun 28 '18 at 18:35
  • @KalleMP, I'll quote from the book: *"It will be appreciated that the representation shown in fig. 6.20 refers to an ideal switch; that is, one which has infinite resistance before closure, zero resistance after closure, and for which the transition between these states is of infinitesimally short duration. **An absence of inductive and capacitive effects is also implied**."* – Alfred Centauri Jun 28 '18 at 18:42
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How can the difference between opening and closing be mathematically demonstrated?

For simplicity, assume that we have the following ideal series connected circuit elements: a constant voltage source with voltage \$V_S\$, a switch, and an inductor with inductance \$L\$.

If we assume the switch is open for time \$t \lt 0\$ and that the switch is closed at time \$t = 0\$, the voltage across the inductor is just (by KVL)

$$v_L(t) = V_S\,u(t)$$

where \$u(t)\$ is the unit step function. The current is then

$$i_L(t) = \frac{1}{L}\int_0^td\tau\,V_S = \frac{V_S}{L}t\,u(t)$$

which is a ramp function.

Now, instead, stipulate that the switch has been closed for some time before time \$t = 0\$ and that the initial inductor current is \$i_L(0-) = I_0\$ when the switch opens at time \$t = 0\$. It follows that the inductor current is given by

$$i_L(t) = \left(I_0 + \frac{V_S}{L}t\right)\left(1 - u(t)\right)$$

and so

$$v_L(t) = L\frac{di_L}{dt} = V_S\left(1 - u(t)\right) - LI_0\delta(t)$$

That is, the voltage across the inductor is \$V_S\$ for \$t\lt 0\$, zero for \$t\gt 0\$, and an impulse (infinitely large for infinitesimal time) at the time the switch opens.

Alfred Centauri
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