Please help me understand inductors

Marc Sugrue

Joined Jan 19, 2018
222
referencing this: https://www.allaboutcircuits.com/textbook/alternating-current/chpt-3/ac-inductor-circuits/

View attachment 189328

At the beginning of the current through the inductor, the opposition to the current (positive voltage) is at it's peak - I get that. But if there is opposition, then not as much current would flow, invalidating the assumption about the amount of changing current in the beginning? I mean how can changing current induce opposition that would prevent the current from changing in the first place?

Another thing that is confusing is that when reading about inductors, they always talk about that 90° phase shift between current and voltage. But that 90° shift pertains only to the voltage/current across the inductor, induced by the inductor, not the whole circuit at hand, right? After all, I can't just take a circuit's wire, wind it around my screwdriver a couple times and suddenly the current curve of the whole circuit lags the voltage curve by 90°?

Intuitively, and with the help of circuitlab.com, I come to the conclusion that inserting a very small inductor into the circuit would not be noticeable at first. Gradually increasing the inductor in size will slowly create a lagging current curve in the circuit, while also gradually attenuating the signal, i.e. less current overall / smaller current amplitude. By the time that the inductor is so large to induce a current lag of 90° in the whole circuit, it would be so large that practically no current would flow?

Every resource I consult keeps repeating the same paradigms over and over like "the inductor wants to keep current flowing" but that does not really help me much. I wonder if I just don't get it. How would I go about plotting a circuit's voltage and current curve, knowing that the voltage that can be dropped by the inductor at di/dt times henrys equals such and such volts?
Think of it like this. An inductor is a magnetic device and energy is stored in the inductor in the form of a magnetic field.

At some point (saturation) the inductor become just a piece of wire as the core can no longer store energy as its full so you have a limit to how much energy can be stored.

When you apply volts accross and inductor as long as the inductor isn't saturated the current in the inductor for a given voltage across it follows the the equation di = (V * dt) / L and the energy (joules) stored in the inductor is equal to J = 1/2*L* Isquared. Again there is only so much energy a given inductor can store before it saturates.

When the the inductor saturates the Voltage accross it become almost zero so based on this you can see that 0V accross the inductor results in no stored charge and its no longer acting as an inductor.

If you instantantanously stop the flow of current going through and inductor the collapsing magntetic field generates a EMF (voltage) which aims to continue the flow of current through the inductor. In an AC circuit where current is continually changing the inductor exhibits a characteristic called impedance or inductive reactance which is a apparent resitance caused by the changing magnetic field, this impedance is Xl = 2 * pi * F * L and is measured as Ohms. The phase shift caused by this inductive reactance causes the inductor phase current to lag due to the charging and discharging of the magnetic field impeding the current flow.

When combined with capacitors (which introduce positive phase shift) the phase elements can be pitched to cancel each other out. Where phase is important the L & C can be chosen to give very little phase shift. In areas where phase it isn't important (such as the noise portion filtered by a filter) the phase can be ignored as you don't care in this region. A common application example for this would be traditional power factor controller where a filter is designed to optimise the phase of the system current to keep it matched the mains power line voltage frequency using a L and C component.

Not sure if this helps
 
Last edited:

WBahn

Joined Mar 31, 2012
29,979
When you apply volts accross and inductor as long as the inductor isn't saturated the current in the inductor for a given voltage across it follows the the equation I = (V * dt) / L
This isn't quite right. It should be di = (V * dt) / L. The right hand side does not give the current, but the change in the current over the (differential) time increment given by dt.
 

sparky 1

Joined Nov 3, 2018
756
It can cause an unsettling feeling because the behavior of electromagnetic waves is not always intuitive.
If you let the math give you the answer going step by step then a class in physics (electricity and magnetism) there is more
laws and experiments that help in visual concept. It is more difficult to go about this process by reinventing the wheel.
I believe the discipline explained above is a step toward that goal also the 2D sine wave is just a helpful representation.
One door opens only to show there are ten more doors leading to various quantum theories. unsettling ? yep
 

Thread Starter

VooDust

Joined Oct 21, 2019
4
After 9 months, having read through "Grob's Basic Electronics" and parts of "The Art of Electronics" and followed the MIT's OpenCourseWare YouTube Semester about Calculus, it FINALLY clicked for me!

How can changing current induce opposition that would prevent the current from changing in the first place?
I realize I have been putting the carriage before the horse the whole time, so to speak. My way of thinking was: "I'll apply 5V to this inductor, to get the current to change. And the inductor will drop some of this voltage, I dunno, maybe 2V according to its inductance L." Of course this would invalidate Ohms Law, so looking at this equation again:

V = L * dI/dt

I always read this as: "The inductor will produce a voltage drop depending on the rate of current change through it.". Not that this were wrong per se, but in this case it's better to say: "A voltage applied across the inductor will let the current change at the specified rate." Faster for smaller values of L, slower for bigger values of L, as we can observe in real circuits. So of course, if I apply 5V to the inductor, the drop across it IS IN FACT AT 5V and, depending on the value of L, the current does change at a certain rate. And it starts at 0! (at least in my experiment).

Then I realized, and this really helped me with electronics: An equation describes two phenomena that are in balance, it does NOT declare causation!

And it's the current that is starting to increase through the inductor, that also runs through a resistor, either in series, or internal resistance of the power source. As this resistor begins to drop voltage, this voltage subtracts from the 5V I applied to the inductor, leading to ever decreasing values of dI/dt, and so on, thus creating the "exponential roll-off" of the current curve (or voltage curve, however you look at it).

Regarding the series resistor, a small resistor will drop less voltage for a given current through the inductor, thus the roll off curve will take longer, i.e. the inductor is allowed to ramp up current for longer. In contrast, a large resistor will very quickly drop the same amount of voltage that was applied (e.g. 5V) for a much smaller current, thus the roll off will be much faster. This is the exact opposite to the RC constant that we know and understand for capacitors, where larger resistors mean longer RC-time constants. This led me to understand why the time constant for inductors is L over R.

I read through all your comments again, and now they all start to make sense. Thanks a lot.
 
Top