Inductor polarity

Question

Let's say an inductor has DC current going through long enough for it to completely build it's electro magnetic field.

If the source voltage then goes lower, the inductor will 'use' the energy stored in its EMF to keep the voltage at its original value as much as it can (the EMF will produce current having the same polarity).

Why then, when the current stops, does it produce a kickback with inverse polarity?

I guess I'm confused because I can't quite apply the right hand rule to both of these situations.

Answer 1

Your mistake is that you're thinking the inductor is keeping the voltage constant, while it is in fact keeping the current flowing through it constant. (Or, more precisely, a voltage across an inductor causes the current in it to change.)

If you have an ideal inductor with 1 Henry and connect it across an ideal voltage source with 1 Volt, the current through the inductor will rise at 1 Ampere per second. This means that the current will become arbitrarily high if you just wait long enough. After 100 seconds, 100 Ampere will flow through the inductor, and after 1000 seconds, the current is already at 1000 Ampere. It'll keep increasing forever. (At least with ideal components.)

So, if you apply a positive voltage to an inductor, the current through it will increase. Similarly, if you apply a negative voltage to the inductor, the current will decrease. (It'll keep decreasing past zero Ampere too, and become negative, if you keep the negative voltage connected long enough.)

If you want to stop a positive current flowing through an inductor, you have to decrease that current (by definition) until it reaches zero. Decreasing the current through an inductor, however, means applying a negative voltage to it. This works both ways: If something suddenly forces the current through an inductor to change, the inductor will generate the voltage necessary to make that happen.

This is the reason why the unit for inductance, Henry, is defined the way it is: Volt seconds per Ampere.

Inductance relates volt-seconds to current. In other words, to achieve a certain change in current in an inductor, you have to apply a voltage for some time.

So, if you decrease the current through an inductor very quickly, there will be a very large negative voltage across it. The volt-second product (voltage multiplied by the amount of time it was applied for) will be the same for the "charging" cycle (positive voltage across to the inductor, current going up) as for the "discharging" cycle (negative voltage across the inductor, current going down). As a result, the faster you discharge the inductor, the higher the voltage will be.

This is why inductors can be used to step up voltages (i.e. in a step-up converter). It's also why disconnecting an inductive load suddenly will generate a spark. (The rapid change in current causes a very high voltage across the inductor, enough to send the air into electrical breakdown.)

Answer 2

If you have a purely resistive load in series with your inductor, and provide a path for the current to return when the switch is turned off, you should see the behaviour you expect. The energy stored in the inductor's magnetic field will bleed off through the resistor until the field decays.

^{simulate this circuit – Schematic created using CircuitLab}

However, it's likely you're simply opening the switch to turn off the voltage, leaving no low-resistance loop for the current to return. When you open the switch, that switch now has two distinct terminals with different charge, separated in space by a dielectric. What component consists of two terminals separated in space by a dielectric?

In other words, when you open the switch, the circuit (left) becomes the circuit (right).

^{simulate this circuit}

Now, you have a series RLC circuit. The magnetic field in the inductor continues to move charges, and those charges build up on the inherent "capacitor" in your circuit. As energy bleeds from the inductor, it builds up in the capacitor, until all the energy in the circuit is now stored in the electric field of the capacitor -- which then starts driving current through the inductor the other way.

The capacitor terminals equalize as their charges move through the inductor, but this builds a magnetic field, which continues to pull charge from the capacitor and charges it again, this time with opposite polarity. As the capacitor voltage becomes more negative, the energy is transferred from the inductor, until the inductor is "empty" and the capacitor is "full".

But then the capacitor starts to discharge again...

Through each cycle, the resistance in the circuit converts some of the energy to heat, and the oscillations eventually die out.

The actual characteristics of the oscillation -- e.g. the period, the amplitude, the settling time -- depend on the specific inductances, capacitances and resistances in your circuit, including parasitic inductance, resistance, and capacitance inherent to all components and your breadboard.

In real circuits, designers will often add a flyback diode, connected antiparallel to an inductive load. This provides a low-impedance return path for an inductor, such that when it is switched off, the inductor's magnetic field will not induce damaging ringing or voltage spikes. The diode is sized to dissipate the expected current as heat without overheating or burning out.

Flyback diode circuit:

^{simulate this circuit}

Answer 3

Why then, when the current stops, does it produce a kickback with inverse polarity?

Charging. As the voltage source charges the coil, current flows out of the positive end of the source and then enters the coil creating a positive voltage at its input. The coil thus acts as a load ("chargeable current source").

Discharging. When we open the switch, the coil begins acting as a current source that must produce the same current in the same direction. For this purpose, the coil reverses its polarity and begins to increase its voltage until it achieves its goal. But because the switch is open, it never succeeds, and its voltage theoretically reaches infinity.

RL differentiator. Usually, there is a resistor in the circuit that limits the current. Here, for example, is my illustration of such an RL differentiation circuit, which is a snapshot of a Flash movie frame (you can watch it by downloading the exe file):

Answer 4

If the voltage changes, the current simply ramps to another value.

If the voltage changes by a huge amount, I guess you could call that a "kick", but who is really doing the kick? The voltage source or the inductor? The inductor's current just ramps around to whatever value it then settles on.

The phenomenon of "inductive flyback" doesn't arise from the application of voltage sources. Indeed we use inductors in switching power supplies, where fixed voltages are alternately applied, and thus the current simply ramps up and down; voltage and current are never unbounded but always well defined and tightly controlled.

Rather, it arises when the voltage source ceases to be so, i.e. the impedance goes suddenly quite high, and current begins ramping to zero (usually). For example, if we charge an inductor from a battery, then suddenly disconnect it: the impedance goes from a low value (voltage source) to high (~open circuit), and it is only this condition which permits the inductor to flyback at whatever natural response it has (usually a ringing wave, but the peak voltage of the initial swing is the important part here). Indeed, the voltage can be high enough that a path for current flow is found, after all -- the air gap in the switch might break down and arc over!

When we want to create such high voltages within a circuit (without using mechanical contacts), we again use a switch to go between low and high impedances; we can only get a (transistor) switch with up to some voltage rating, so we do need to be careful of that, but flyback ratios of >100x are feasible. We can also use a transformer to multiply the peak, hence for example, typical automotive ignition coils driven at 12V, with primary peaks up to 400V or so, and secondary peaks of say 40kV or more (a 1:100 or higher turns ratio).

As for polarity, that's simply the direction the current is ramping:

$$ V = L \frac{dI}{dt}$$

\$dI\$ and \$dt\$ are very small units of current and time (the slope of current with respect to time); when we're using square waves (as often the case in switching supplies), we can change dees to deltas and use the total changes during each phase of the cycle (since the slopes don't change perceptibly during that cycle). L is the inductance, and V the voltage. If we charge the inductor with a positive voltage, current is rising and eventually settles out (real inductors or sources also have some resistance, not captured by this equation). If we discharge the inductor, that's equivalent to saying we've reversed the terminal voltage.

Also, mind that this is only the voltage difference across the inductor itself, not an absolute voltage in a circuit. Voltages are always measured by difference, and it's merely a convention that we choose a ground to measure against, in circuit; the components themselves don't know this, and only know what's applied to their terminals.

Answer 5

Here is a simplified qualitative explanation. The voltage across an inductor gives rise to a magnetic field, which increases as the current increases. Then when the magnetic field collapses (because your switch opens), this induces a voltage across the inductor. This voltage is of the opposite polarity to the original voltage because the field is decreasing, whereas the original voltage caused the magnetic field to increase.

Answer 6

Here is a simulation that may help to understand what is going on - and that polarity reversal is not necessarily what happens:

^{simulate this circuit – Schematic created using CircuitLab}

I am modelling a copper coil as a 1Ω resistor in series with a 1H inductor.

At time = 0 the current through the coil is exactly 1A. At time = 100ms the switch opens inserting a 200mΩ resistor in series with the power supply.

You can see that the current just after the switch opens is still 1A (but starting to decrease over time), eventually it will get to 1V/1.2Ω = 0.833A.

In order to maintain the 1A the voltage across the coil R1 + L1 must decrease instantaneously by 200mV to make up for the 200mV dropped across R2. That voltage appears across L1. So the voltage across the coil does not reverse in polarity but rather drops a bit (the voltage across the inductance part L1 goes from 0 to -200mV).

Now imagine if we changed R2 to 1Ω- the voltage across the coil would drop to 0 when the switch was opened. It would then gradually increase to 0.5V. Anything more than 1Ω and the voltage would go as far negative as it needed to go (ideally) to maintain the 1A current if only for an instant. If R2 went to \$\infty\$ the voltage would have to go to negative \$\infty\$ to maintain the current.

In reality the switch would arc and burn up some energy, a semiconductor switch might break down, there might be some ringing from capacitance of the coil that's not modelled here etc. But it's perfectly possible in reality to get hundreds of volts from a few volts across the coil.