# Discharging an inductor

I wanted to know if it's practically possible to use inductors the same way we use capacitors , in the sense of charging it using a charging circuit then discharge in another circuit.

I thought it could be possible but then i read on the web that when an inductor is disconnected the magnetic field starts collapsing inducing a very high voltage, wouldn't this voltage breaks down any transistors used in the switching from the charging to the discharging circuit? If i understand this correctly, this only happens when there is no way for the current to go so voltage builds up, but then again there is a small split of second between the switching between the 2 circuits, will that cause such a problem?

Edit:

If I used this circuit to charge and discharge the inductor, will there be any voltage spikes?

• Current flow at "turn off" CANNOT be interupted. Voltage will rise with opposite polarity until current is continuous. If there is a "circulating diode" the V will be Vfwd of diode and voltage is small.If a zener is used power in zener is I x Vz. If a resistor is uised V rises to V = I x R. If there is no formal load the voltage will rise and rise and rise and .... until Iturnoff is caused to flow. In typical inductors there is leakage capacitance and E=1/2LI^2 is converted to E=1/2CV^2. If C is small, V is large. You can get 100's of volts easily with an eg 12V coil. Aug 11, 2017 at 14:28
• Energy storage is possible BUT inductors tend to be physically large at high iunductance compared with capacitors. eg 1A in 1H gives E=1/2 x L x i^2 = 1/2 x 1 x 1^2 = 0.5J. A 1F supercap at 1V contains the same energy. Aug 11, 2017 at 14:42
• The sole Digikey inductor rate at >=1A and >= 1H costs $73. A 1F, 2.7V "supercapacitor" starts from$1.22 in unit quantity (and is vastly smaller than the inductor). Aug 11, 2017 at 14:44
• There is no switch in your circuit, the battery is upside down. R1 and L1 will pass 12 mA. Load and D1 will have 0 mA. Turn off the grid when taking screen-shots. I fixed your paragraphs and capitalisation. Aug 11, 2017 at 16:50
• @Transistor so sorry, i forgot to put the switch, the switch is supposed to be after R1,disconnecting the charging circuit Aug 11, 2017 at 19:43

I find it helpful to think of capacitors and inductors to be complimentary.

simulate this circuit – Schematic created using CircuitLab

Figure 1. Ideal and imperfect components.

• Capacitors store energy in an electric field. Inductors store energy in a magnetic field.
• A capacitor holds energy when open circuit. An inductor holds energy when short circuited.
• Capacitors lose energy through parallel leakage resistance. Inductors lose energy through series resistance.
• Capacitors "like" to keep the voltage across them constant. Inductors like to keep the current through them constant.
• When a capacitor is short circuited the resultant current is very high. When an inductor is open-circuited the resultant voltage is very high.

... when an inductor is disconnected the magnetic field starts collapsing inducing a very high voltage, wouldn't this voltage breaks down any transistors used in the switching from the charging to the discharging circuit?

Yes it would but there's a simple solution:

Figure 2. A simple buck converter. Source: All About Circuits.

In Figure 2 S is the transistor switch similar to that mentioned in your question. When it is switched the inductor tends to maintain current in the direction of I. Since the right side of L is "held" by C and current is to keep going then the left side of L goes negative to try to maintain current. Whe the voltage reaches -0.7 V D starts to conduct and maintains the current through L keeping it "happy" and avoiding a transient high voltage.

You will see this arrangement more commonly in snubber diodes on relay coils.

simulate this circuit

Figure 3. A typical relay control circuit. Without D1 the inductance of the relay coil would generate a large negative voltage on switch off. This would be likely to destroy Q1 . The diode limits the negative excursion on Q1 collector to -0.7 V.

• i made an update can you check it and tell me if it works please? Aug 11, 2017 at 16:41

Disconnecting inductor is the same like shorting a capacitor in the way they are similar.

A perfect inductor, if it would be shorted, would store energy. But this is only possible for superconductors. In reality losses on wire resistance waste the energy quite quickly.

But not instantly. In fact in switching power supplies this is exactly what inductor does: charges from the source and discharges to the load. Just fast enough so losses are small (and for several other reasons)

• can you explain more about how are inductor used in power switching? thank you Aug 11, 2017 at 14:20
• Just Google boost dc/dc or buck dc/dc.
– user76844
Aug 11, 2017 at 14:23

Superconducting inductive energy storage has been used commercially to some extent, so it can be considered to be "practically possible." It appears that they probably need to be of the megawatt hour or tens of megawatt hour scale to be practical. Even at that scale, commercial viability is questionable. For additional information look at the Superconducting magnetic energy storage Wikipedia article and the references cited there.

Lets consider storing Energy in both a capacitor and an inductor.

For a capacitor we charge it to some voltage by storing a charge between its plates and we we can easily show $E = \frac{1}{2} C \cdot V^2$. in principle if we can take a charged capacitor out of one circuit and put it in another we can make use of this Energy in another circuit or in the same circuit at a later time. One common use for this is to detect when power is failing and use this to store last settings and ensure a clean shut down on products with microprocessors in them. We can't store energy in a capacitor forever however as real capacitors have leakage and will eventually self discharge.

For an inductor we store energy in a magnetic field and we can easily show $E = \frac{1}{2} L \cdot I^2$ To store this energy having charged it we need to keep the current flowing so need to place a short across the inductor. Because real inductors have resistance this makes them less useful than capacitors for storing energy long term as they tend to be big and lose this energy relatively quickly. However inductors are used to store energy in several circuits. One obvious example is a buck type switch mode power supply

simulate this circuit – Schematic created using CircuitLab

In this circuit we apply a positive voltage at V1 greater than the output. This causes the current in the inductor to increase, ramping up. When V1 disappears or goes negative current continues to flow in D2 and ramps down. The inductor is continually storing and releasing energy to provide a DC output voltage.

Yes, inductors can be used to store energy. That's the basis for many switching power supplies, just to mention one example.

However, the problem with storing energy in a inductor is that the current has to be kept circulating. Our current technology makes that quite lossy for long term storage. The inevitable resistance of whatever conductor is used to make the inductor will dissipate the energy rather quickly.

There have been such things as inductors using superconductors. Those can store energy "long term", but such things are difficult and expensive to do with our current state of technology.

• Sorry if i wasnt more clear , but what i meant by storing is immediate charge and discharge Aug 11, 2017 at 20:28
• @user: Then there are many examples. Switching power supplies are common ones. Aug 11, 2017 at 20:34