# Inductor in-rush current

The inductor equation V = L(di/dt) shows that inductors will resist instantaneous changes in current. Thus, they can be used as in-rush current limiters in power supplies, for instance; the in-rush current here is as a result of the filter capacitors in use. Yet, I've read that in-rush currents occurs with motors and transformers, which are primarily inductive, at start-up. So, which is it? Are the inductive parts really the cause of the in-rush or is it as a result of the capacitive elements of the devices?

Also, AllAboutCircuits has a page on inductor behaviour and a neon bulb, connected across an inductor, was used to illustrate the back-EMF generated by inductors in opposition to changing current. According to the article, the bulb does not turn on when the switch is first closed because the battery's 6 V is too low but when the switch is opened later, a high enough voltage appears across the inductor and the lamp lights up briefly. Since L is constant, this implies that the di/dt is somehow different when the switch was opened from when it was first closed. How is this possible when nothing was changed in the circuit; same switch, same power supply?

I've also seen Adafruit solenoid product descriptions talking about in-rush current needed to "charge up the electromagnet".

What am I missing here?

Unfortunately you are talking about many different types of inrush, all caused differently, so with different cures. Some 'caused' by inductors, some cured.

a) Motor inrush

Motors generate a back EMF when turning, and this cancels out most of the input voltage, leaving only a small net voltage to drive a current through the small motor resistance.

At standstill, there is no back EMF, so the normal supply voltage may drive typically 10x the rated current into the motor. The motor inductance is insignificant compared to the mechanical time constant of the motor. It's enough to level out PWM switching in the many kHz range, but not to cope with the seconds of acceleration.

Small motors just live with the inrush. Bigger motors need to use some form of controlled gentle starting.

b) Transformer inrush

Flux can be measured in volts.seconds. A transformer core has a maximum flux. It is designed to swing from -max to +max and back again. The transformer has zero flux before switch on. If you switch it on at the wrong part of the mains cycle, then instead of swinging between -max and +max, it will try to swing between 0 and +2max, which is obviously not possible. The large assymetrical current drawn due to saturation causes a net DC voltage in the winding resistance, which gradually shifts the flux to zero average over the next few cycles.

While some people say this is 'caused' by the transformer inductance, it is actually because the inductance falls when the core is saturated. This is generally mitigated by using a time delay fuse, that will stand the extra current for a second or so.

c) AC solenoid inrush

When a solenoid is un-energised, there is a large air-gap in the magnetic path, which means the inductance of the solenoid is low. When AC power is applied, typically the resistance of the coil will dominate, and a large current will flow. When the solenoid closes, the air-gap disappears, and the inductance increases by an order of magnitude or two.

In a well-designed AC solenoid, the inductive reactance will now dominate the solenoid impedance, causing the supply current to fall significantly. This fall in current happens automatically as a result of the changing magnetic circuit geometry.

d) DC solenoid, no inrush

As the supply is DC, the steady state current will be limited by the resistance of the coil, not by the inductance, whether large or small. The inductance will serve to slow the increase of current, the opposite of an inrush.

When energised, the smaller air-gap means less current is needed to supply the holding magnetic field. A special driver is sometimes used to supply a large pull-in current, which is then reduced to a lower hold-in current. This is done actively by the driver, not as a result of the solenoid's changing geometry.

e) Rectifier/capacitor switch-on inrush

In the first cycle, the supply has to charge the capacitors from zero. This can be handled by using a time delay fuse, and surge-rated diodes. The ubiquitous 1N540x series for instance are rated at 3A continuous, 200A half-cycle surge. Another way is to use NTC thermistors in series, or relay-shorted starting resistors. It's not practical to use a large enough inductance to limit the rate of current rise.

f) Rectifier/capacitor recharge inrush

Now this one can be mitigated by extra inductance. The capacitors are charged only when the input voltage exceeds the capacitor voltage, which might only be 10% of the time. This leads to a very peaky diode current waveform. A bit of series inductance, sometimes a discrete inductor, sometimes the transformer is wound to have finite leakage inductance rather than the minimum possible, will extend the current pulse. As the pulse starts, it limits the rate of rise. When the transformer voltage falls and the pulse would normally end, the back emf in the inductance adds to transformer voltage, keeping the pulse going while the pulse current drops to zero.

g) Filament lamp inrush

The resistance of a metal filament changes by more than an order of magnitude form cold to hot, so at switch-on, the current can be 10x the running current. This is handled with over design, or time delay fuses.

• @Andyaka I'm not sure everyone will get it, volts, flux and current are magic to most people, but I keep trying. Even tougher is what happens to the primary current when the secondary is half-wave rectified! But that's a different post. Jul 22, 2016 at 10:56
• You're right. It's dark magic, most of which I wont pretend I understand now. But from what I think I understand, it seems that the only times when in-rush happens with inductors is when there just isnt enough inductance either because of saturation in (2) or because there's no load in (1)? I'd appreciate it if you could explain what you mean by 'mechanical time constant' though. Also is it okay to say that the neon lamp didnt come on at first because its switch-on resistance was very low and it basically shorted the inductor thus preventing any real back-emf from being generated? Jul 22, 2016 at 13:53
• @TisteAndii Mechanical time constant => the time the motor takes to spin up, I'll add something to the answer about solenoids. Jul 22, 2016 at 15:31
• Thanks for clarifying. In other words, the large current that may be drawn by a solenoid isnt as a result of any "charge up" process but rather because after the field is set up, the resistance of the coil becomes dominant and is usually small. Also the motor's inductance is insignificant because the magnetic field is set up long before the motor achieves full speed, so a large current flows in the mean time due to the tiny motor resistance, which is what dominates the impedance now. And my assumption about the neon bulb is correct? Jul 22, 2016 at 16:17
• what changes in the neon example is that the switch is first closed, then open, that's a big change. Closing the switch lets a current flow in the inductor. When the switch opens, the current still 'wants' to flow, but has nowhere to flow, so the voltage rises to 'whatever it takes' to keep the current flowing. With a neon there, it lights the neon. Without a neon there, it would create a spark. Jul 22, 2016 at 17:56

The case is slightly different for motors vs. transformers.

In motors, the winding inductance is fairly low -- not enough to limit the current safely by itself. Only once the rotor is spinning and generating back-EMF is the current limited. Likewise, if an AC motor stalls, the current draw will be much higher than the normal operating current, even long after startup.

In transformers, the primary winding inductance only applies when there is no current in the secondary. If you allow induced current to flow in the secondary winding, that tends to cancel the flux in the core. Since inductance is caused by flux, the secondary winding effectively "short circuits" the inductance. In the normal configuration of a linear power supply, it feeds a bridge rectifier and a capacitor bank. The inrush is actually caused (primarily) by the capacitor bank. Once the capacitors are charged up, the secondary current drops to near zero, and the current is limited by the primary inductance.

• Secondary current DOES NOT cancel flux in the core except under very rare applications. Secondary current IS NOT induced; voltage is induced and current will flow dependent on load. -1 for bad terminology. Jul 22, 2016 at 10:18

An inductor has a limited current rise rate (proportional to input voltage). A motor, however, has a different characteristic, because when it is rotating, the inductive current rise rate is proportional to input voltage MINUS the so-called back EMF of the motor. That back EMF is the reason an unloaded motor draws small current, while a stalled motor draws large current. The back EMF is a generator effect, is zero until the motor achieves rotation.

Any motor starting at zero RPMs has a very high current requirement, pending its acceleration up to final speed. That acceleration is a different time constant than the inductive one, depending on mass of the rotor and torque. The raw inductance of an AC motor is rarely so high that it will draw acceptably low AC current when stalled, which is why motors (but not inductors in general) are often fitted with thermal cutout devices. Because something can get stuck and prevent the back EMF.

You are asking a lot of related questions, so I will try to summarise them somewhat. What your missing is the magnetic field built up by passing current through an inductor. It takes a finite amount of time for the magnetic field to build up and for the current to make its way through the length of the wire.

Combined with capacitors such as in switch-mode power supplies or any noisy device they combine to suppress much of the transient noise. LC filters are useful when a resistor cannot handle the current and causes power losses due to heat. Taken to extremes they can act as cross-over filters for your speakers.

Next is the neon bulb trick. Remember that an inductor returns back the current supplied to it when it is turned off. But what if that current has no return path such as a snubber diode? Then the voltage rises quickly until it finds a path, in this case the neon bulb. The current at 6 volts now returns and rises to 90 volts or more. If the bulb was not present the return path would be to arc across the switch contacts, slowly burning them out.

Don't be fooled by the "charge up the electromagnet" words. You are charging up the magnetic field of an inductor that is being used as an electromagnet-nothing more than that. An exception would be doing so with coils of wire and a very high magnetic field (1 to 2 Teslas) with the material to be permanently magnetized inside the coil. The current is on for only 10 to 20 uS but enough to do the job. Now you have made a permanent magnet whose field strength is a bit less than that used to magnetize it.

The permeability and composition of the material greatly affects how much of the magnetic field is absorbed permanently.

Regarding inductive load inrush current, you have to take into account the energy that is required to generate the magnetic component of the electromagnetic field.

At time zero, when you throw the switch, current flows into the circuit and the flow of charge through coils of wire sets up a magnetic field. The forming of the magnetic field takes energy from the circuit, and this energy has to be supplied by the power source.

In terms of the circuitry, the forming of the magnetic field will be seen as a power loss in the wires, and a voltage drop will occur as a result (an opposing electric field). This "power loss" can be accounted for by the power supply, which will provide additional energy to the circuit during this transient phase.

This additional energy can be viewed as a transient inrush current.

Then, at some time later, once the circuit has reached a steady state, you kill the switch. Now you have inductive loads with energy stored in their magnetic fields. The magnetic field acts like an inertia, trying to keep the current flowing. But the circuit is open, so there is nowhere for any current to flow. An opposing electric field is formed, and in the absence of an dissipative circuitry it will get larger and larger until eventually a spark occurs.

In practice the inductive loads usually have a dissipative circuit in parallel with them which serves to safely dissipate the energy that comes back to the circuit from the collapsing magnetic field.