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?

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    \$\begingroup\$ 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. \$\endgroup\$ – Sparky256 Jun 28 '18 at 1:29
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    \$\begingroup\$ i dont understand \$\endgroup\$ – atmnt Jun 28 '18 at 1:51
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    \$\begingroup\$ @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. \$\endgroup\$ – Jack Creasey Jun 28 '18 at 2:04
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    \$\begingroup\$ @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. \$\endgroup\$ – Alfred Centauri Jun 28 '18 at 3:05
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    \$\begingroup\$ 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'. \$\endgroup\$ – Russell McMahon Jun 28 '18 at 8:46

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)

  • \$\begingroup\$ +1, this answer gets right to the heart of the matter. \$\endgroup\$ – Alfred Centauri Jun 28 '18 at 3:08

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.

  • \$\begingroup\$ but why in the case of closing no large voltage is induced? \$\endgroup\$ – atmnt Jun 28 '18 at 1:52
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    \$\begingroup\$ @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. \$\endgroup\$ – John D Jun 28 '18 at 2:23
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    \$\begingroup\$ 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. \$\endgroup\$ – jonk Jun 28 '18 at 4:04
  • \$\begingroup\$ @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. \$\endgroup\$ – Alfred Centauri Jun 28 '18 at 14:11
  • \$\begingroup\$ @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. \$\endgroup\$ – KalleMP Jun 28 '18 at 17:49

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.


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