How inductor voltage spikes do not break the conservation of energy?

Thread Starter

babaliaris

Joined Nov 19, 2019
158
I've seen people talking about inductors and how to use diodes in order to protect them when the current suddenly becomes 0.

My professor explained this phenomenon using the following statement and maths:
"Imagine you have a big current that suddenly becomes zero. This basically means that the rate of change of the current \( \frac{di(t)}{dt} \to \infty \).
So if you have an inductor \( \lim_{\frac{di(t)}{dt} \to \infty}v(t) = L \frac{di(t)}{dt} = \infty \) . In other words, you get a large voltage drop across the inductor in a small dt time (thus creating a spike)."

I understand that, but how can an inductor that in one instant of a time it was under a 10V source supply suddenly get a voltage drop across the inductor in the range of Kilo volts? Where does this huge energy come from? The supply is only 10 volts, how can it exceed it?

It seems to me that it breaks the conservation of energy.
 

Papabravo

Joined Feb 24, 2006
19,578
It comes from the energy in the magnetic field. The energy in the field is strong enough to create the large voltage, but only for a short period of time. The potential energy in the inductor is:

\( E_p\;=\;\frac{1}{2}LI^2 \)
 

nsaspook

Joined Aug 27, 2009
10,683
The stored electrical energy is transformed from the mainly magnetic field component to a mainly electric field component. It's simple power ('rate of energy transfer per unit time') = voltage * current relationship. No extra energy is created as there is only one EM field entity (electromagnetic field tensor) with various frames of reference.
 

ZCochran98

Joined Jul 24, 2018
221
Voltage is not energy, but is related to it. Let's do a little math here: if we know instantaneous power is defined as:
\[P(t) = V(t)I(t)\]
We have the inductor equation:
\[V(t) = L\frac{dI}{dt}\]
Substituting that in and doing the math, we get:
\[P(t) = LI(t)\frac{dI(t)}{dt} = \frac{d}{dt}\left(\frac{1}{2}LI(t)^2\right)\]
Then, knowing the definition of power as \(P = dE/dt\) (where \(E\) is energy), we can say the instantaneous energy in an inductor is
\[E(t) = \frac{1}{2}LI(t)^2\]
This energy is stored in the magnetic field of the inductor and is independent of voltage. If you have no current (i.e.: no kinetic energy for electrons), then you will have no energy stored in the magnetic field. The voltage spike carries/stores no energy, so energy is conserved (keeping in mind the action of applying energy - providing current - and removing it is a non-conservative process, so the energy is conserved on both sides of the operation: there's energy in the field when there's current, there's no energy in the field when there's no current).

And where does the massive voltage come from when you remove the current suddenly, if the supply only provides 10 V? That's the reactive voltage, or, in this case, the voltage from the field itself collapsing. It's not from the supply, but from the magnetic field collapsing and, in the process of collapsing, creating a back-EMF via Faraday's law:
\[|V| = \left|\frac{d\phi}{dt}\right| = \left|A\frac{dB}{dt}\right|\]
I'm using magnitudes because I really don't care about the sign of the voltage at the moment. Here, \(A\) is the area of a slice of the field (roughly-speaking) and \(B\) is the magnitude of the magnetic field created by the current in the inductor, with a few assumptions to simplify the math. So, the voltage spike does not come from the power supply, but from when the magnetic field you previously had in the inductor suddenly drops down to zero.

Edit: these back-EMFs are why a lot of motor drivers have large inductors in them, in order to prevent these large voltages from generating currents that would damage in the rest of the circuit.
 

Thread Starter

babaliaris

Joined Nov 19, 2019
158
To tell you the truth I feel a little embarrassed now since I totally forgot about the energy that Inductors store as a magnetic field :p

Already knew the math you guys showed me. I feel stupid now.
 

crutschow

Joined Mar 14, 2008
31,123
Below it an ideal simulation to show the conservation of energy in your inductive spike scenario:

When the 1A inductor source current is suddenly stopped at 100μs (blue trace), the voltage across the resistor goes to 100kV due to the instantaneous 1A going through it (green trace), and the peak power in the resistor goes to 100kW (yellow trace).
But the integrated total energy dissipated in the resistor, as shown in the small Waveform window, is 500mJ which, as expected, is the same as the energy stored in the 1H inductor for 1A of current (½ LI²).

1670028099014.png
 

Thread Starter

babaliaris

Joined Nov 19, 2019
158
But the integrated total energy dissipated in the resistor, as shown in the small Waveform window, is 500mJ which, as expected, is the same as the energy stored in the 1H inductor for 1A of current (½ LI²).
So basically the stored energy of the inductor escapes into the environment through heat in dt time?
 

crutschow

Joined Mar 14, 2008
31,123
So basically the stored energy of the inductor escapes into the environment through heat in dt time?
Yes.
For example, If the voltage is high enough to form and arc, such as from mechanical contacts opening, then it will be dissipated there.
If it's from an unprotected transistor turning off, then in may be dissipated in the shorting of the transistor junction. :eek:
 

Thread Starter

babaliaris

Joined Nov 19, 2019
158
So this is why we were told as kids to not unplug electronic devices like TVs directly out of the socket but instead turn them off the right way. :p
 

WBahn

Joined Mar 31, 2012
27,880
So basically the stored energy of the inductor escapes into the environment through heat in dt time?
Not necessarily, it is merely transformed from energy stored in the magnetic field to energy in some other form. Heat is just one option. The higher voltage can be used to, for instance, boost the voltage in a switch-mode power supply.
 
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