# I don't understand inductors

#### Lesaje

Joined Feb 6, 2024
9
I'm reading Art of Electronics without background and in 1.5.1 there is an example

In Figure 1.52A the left-hand side of inductor L is alternately switched between a dc input voltage Vin and ground, at some rapid rate, spending equal times connected to each (a “50% duty cycle”). But the defining equation V =L dI/dt requires that the average voltage across an inductor must be zero, otherwise the magnitude of its average current is rising without limit. (This is sometimes called the volt-second balance rule.) From this it follows that the average output voltage is half the input voltage (make sure you understand why)
I don't understand why. To be fair I don't understand anything at all. I treat current like a river and voltage like a difference in height between different parts of the river flow. As I understand inductor saves energy when you make changes in current flow, and releases it in form of voltage when the flow is steady (no matter how large it is). Am I dumb?

#### BobTPH

Joined Jun 5, 2013
8,921
Forget your analogy. Electrical current is not water flowying downhill.

Do you understand this equation?

DI / Dt = V / L

Explain in your own words what it means.

#### Lesaje

Joined Feb 6, 2024
9
Explain in your own words what it means.
Well, when there is a large change in current in a small amount of time, there will be some large voltage (across inductor, I guess?) even if your coil have small inductance, and if your coil have large inductance, then voltage will be enormous.

#### crutschow

Joined Mar 14, 2008
34,386
An inductor is rather like the mechanical inertia of a mass, which stores energy equal to 1/2 mv² where m is its mass and v is its velocity.

An inductor stores energy equal to 1/2 LI², where L is the inductance and I is the current through the inductor.
This current will stay constant if there is a connection with no resistance connecting the two ends of the inductor winding (as in a superconducting magnet) giving no change in its stored energy, similar to a mass moving through empty space.
A voltage across the inductor will increase or decrease that current, depending upon the voltage polarity, as per the equation in post #2, causing the stored energy to increase or decrease per the above equation.

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#### Jony130

Joined Feb 17, 2009
5,487
The inductor in the water analogy is just like a free-running turbine (water wheel) with the flywheel attached to the shaft. The heavier the flywheel, the greater the inductance of the coil.
So the water wheel needs time to accelerate to full speed determined by the river current. During this phase, we store the energy in the flywheel (in the inductance). But when we stop to water flow.
The water wheel cannot stop immediately. And now the water wheel "generates" the pressure ( we release previously accumulated energy).

As you should know the voltage across the coil is proportional to the rate of change in the current.
V = L * dI/dt

When the inductor current increases (charging phase)the rate of change is positive di/dt > 0, thus, Voltage across the inductor must be positive because L times a positive number yields a positive voltage.
But when the current inductor current starts to decreasing (discharging phase), the rate of change is negative di/dt < 0, so the voltage across the inductor must be negative because L times a negative number yields a negative voltage.

Take a look here:

#### MrAl

Joined Jun 17, 2014
11,447
I'm reading Art of Electronics without background and in 1.5.1 there is an example

I don't understand why. To be fair I don't understand anything at all. I treat current like a river and voltage like a difference in height between different parts of the river flow. As I understand inductor saves energy when you make changes in current flow, and releases it in form of voltage when the flow is steady (no matter how large it is). Am I dumb?
Hello,

The inductor works for current like a capacitor works for voltage. In fact, they are considered duals.

It may take you a little while to think in terms of current rather than voltage.

When you connect a voltage across an inductor, the current ramps up to some level. If you lower the voltage, the current stays the same for a short time.
You should show the circuit too, but it sounds like a buck circuit. In that case, the inductor current ramps up and down a little. You can kind of think of it as a sort of current regulator, although the average current could change. In a buck circuit though the average does not change unless something else changes first, like the load resistance or the input voltage.

To understand the inductor better you may have to resort to the incremental definition if you have not yet had any calculus.
This definition is:
v=L*di/dt

where 'v' is the voltage across the inductor, 'L' is the inductance, 'di' is a small increment in current through the inductor over a small time period 'dt'.

This may be harder to imagine, but let's solve for the current increment di and see what happens...

Solving that for di, we get:
di=v*dt/L

What this means is that the current 'di' will increase slightly the over small time 'dt' when the voltage changes.
Let's first assume L=1. That gives us:
di=v*dt

Now, when the buck circuit is turned on, the voltage 'v' across the inductor becomes equal to the supply voltage Vcc because the output is zero and has not risen yet. Now let's say the voltage 'v' is 10 volts and wait for a time duration of t=0.1 seconds. The current is calculated from above:
di=v*dt
di=10*0.1
therefore
di=1 amp.

Now let's assume there is a capacitor and resistor on the output, the cap filters and the resistor is the load.
Since di went up to 1 amp, that means that the capacitor charges and the resistor draws some current.
After another time t=0.1 seconds has passed, the cap may have charged up to 0.5 volts for example, and so now the resistor draws some current.
It does not stop there though because the inductor still has 9.5 volts across it, with the output being 0.5 volts and the input still 10 volts.
Now what happens after another time period t=0.1 seconds...
di=v*dt
di=9.5*0.1
di=0.95 amps

So we see the change in current went down a little. The cap is going to charge to a higher voltage now too, and the resistor draws more current.
Now for simplicity, let's say the output voltage rose to 1 volt. Now we have 9v across the inductor so after another time t=0.1 seconds we have:
di=v*dt
di=9*0.1
di=0.9 amps

So the change in the inductor current went down again.
With each time interval t=0.1 seconds, the capacitor charges more and more, and the inductor current goes down. After a longer time we reach a point where the voltage on the output is higher than the input (the pulse goes low). Since the voltage across the inductor now goes negative, that means the current will go down.
Eventually the current goes up a little when the pulse goes high, and goes down a little when the pulse goes low. If the current went up by 0.2 amps then it would go down by 0.2 amps. The average current would be whatever the resistor on the output draws. If it was 10 amps for example, the current would ramp up to 10.1 amps and then ramp down to 9.9 amps, then back up to 10.1, etc., etc.

I hope this gives you some idea what is going on here. If you like we can go through an actual example with a known filter cap value and resistor value.
We can look at the way you can calculate this in more detail. There are standard methods for doing this in this incremental form that includes the cap and load resistor, and it's not that difficult to do.

#### BobTPH

Joined Jun 5, 2013
8,921
Well, when there is a large change in current in a small amount of time, there will be some large voltage (across inductor, I guess?) even if your coil have small inductance, and if your coil have large inductance, then voltage will be enormous.
That is part if it.

But what happens if you put 10V across a 1 Henry inductor? What does the current do?

#### Lesaje

Joined Feb 6, 2024
9
But what happens if you put 10V across a 1 Henry inductor? What does the current do?
I guess it will go up to 10A? But I don't know how fast it will happen.

#### BobTPH

Joined Jun 5, 2013
8,921
I guess it will go up to 10A? But I don't know how fast it will happen.
Then you don’t understand the equation. Why would it ever stop going up, assuming ideal components?

Do you know what dI/dt means? Does it say anything about how fast the current goes up?

#### Lesaje

Joined Feb 6, 2024
9
Then you don’t understand the equation. Why would it ever stop going up, assuming ideal components?
I don't understand how current can be infinite.

#### ericgibbs

Joined Jan 29, 2010
18,826
hj
Copper wire has resistance
E

#### crutschow

Joined Mar 14, 2008
34,386
I guess it will go up to 10A?
The only thing that limits the current is any resistance in the circuit, including the inductor's.
The value of inductance only determines how fast the current increases.
But I don't know how fast it will happen.
That's given by the equation in post #2.
If you don't understand that, then you need to do a little study of basic math.

#### morzh

Joined Jan 18, 2017
38
I'm reading Art of Electronics without background and in 1.5.1 there is an example

I don't understand why. To be fair I don't understand anything at all. I treat current like a river and voltage like a difference in height between different parts of the river flow. As I understand inductor saves energy when you make changes in current flow, and releases it in form of voltage when the flow is steady (no matter how large it is). Am I dumb?
I am an EE by trade, but a physicist by the education. In every course on electricity I looked at, they did use "electro-mechanical analogy".
Here it is:

L (inductance) is M (mass)
C (capacitance) is the Spring, namely its 1/K (where K is the stiffness of the said spring) or compliance C (which is 1/K)
I (current) is velocity (speed).
V(voltage) is Force (where it acts as the source of action) or spring displacement (charge of a capacitor).
Q (Charge) is Displacement X (of the spring).

The differential equation of the LCR circuit looks then exactly as the equation of the movement.
So are other euqations:

Stored energy of the inductance L with current I : L * (I^2) / 2
Compare the kinetic energy of the moving mass M with velocity V: M* (V^2)/2
(I use ^ for the power of, that is Square in this case).

Stored energy of a capacitor C at the voltage V: C * (V^2)/2
Stored energy of a compressed (streched) spring: K * (X^2)/2

Etc.

Therefore, the behaviour of the inductance (can't change the current instantaneously) is the same as the behaviour of the mass (inertia: can't change the speed instantaneously).

So on, so forth.

#### nsaspook

Joined Aug 27, 2009
13,231
I am an EE by trade, but a physicist by the education. In every course on electricity I looked at, they did use "electro-mechanical analogy".
Here it is:

L (inductance) is M (mass)
C (capacitance) is the Spring, namely its 1/K (where K is the stiffness of the said spring) or compliance C (which is 1/K)
I (current) is velocity (speed).
V(voltage) is Force (where it acts as the source of action) or spring displacement (charge of a capacitor).
Q (Charge) is Displacement X (of the spring).

The differential equation of the LCR circuit looks then exactly as the equation of the movement.
So are other euqations:

Stored energy of the inductance L with current I : L * (I^2) / 2
Compare the kinetic energy of the moving mass M with velocity V: M* (V^2)/2
(I use ^ for the power of, that is Square in this case).

Stored energy of a capacitor C at the voltage V: C * (V^2)/2
Stored energy of a compressed (streched) spring: K * (X^2)/2

Etc.

Therefore, the behaviour of the inductance (can't change the current instantaneously) is the same as the behaviour of the mass (inertia: can't change the speed instantaneously).

So on, so forth.
They used electro-mechanical force mass energy work equation equivalencies, not analogies. Physics models are mechanical but the closed-loop water analogy is not a mechanical equivalent to EM. Gravitation is a much better analogy and mechanical model.

https://www.scirp.org/journal/paperinformation?paperid=95087

#### ericgibbs

Joined Jan 29, 2010
18,826
Hi @Lesaje
Maybe this simulation response of a 1H Inductor when a 10V pulse is applied, will help you understand.
Note, the final current through the inductor depends upon the resistance of the inductor winding.

E

#### Lesaje

Joined Feb 6, 2024
9
With each time interval t=0.1 seconds, the capacitor charges more and more, and the inductor current goes down.
Thank you! I knew calculus since I have background in math, but I didn't know that in physics there is units of measurement for derivatives too, I thought it dimensionless quantity. It starting to come all together now.

#### MrAl

Joined Jun 17, 2014
11,447
Hello again,

Quite common for the translational system analogy is the Force Current mechanical/electrical analogy and the mass becomes a capacitor (force to current, mass to capacitor).
For the rotational system analogy, it's the Torque Current analogy (torque to current).
For fluid flow it's the volumetric rate of flow to current.
For the thermal system it is the heat flow rate to current.

There's just one thing to note here though. These analogies are usually used to go from a mechanical system to an electrical system because electrical systems are considered easier to analyze. I'm not sure that it makes any sense to go from an electrical system to say a rotational system or even a translational movement system.
Maybe to show the analogy, but in practice I can't think of a reason to go from a circuit with a Resistor and a Capacitor and Current source to a Friction, Mass, and Force system. Usually this goes the other way, from mechanical or fluid or thermal system to electrical. We can then even use a circuit simulator to solve these other problems too.

#### MrAl

Joined Jun 17, 2014
11,447
Thank you! I knew calculus since I have background in math, but I didn't know that in physics there is units of measurement for derivatives too, I thought it dimensionless quantity. It starting to come all together now.

Hello again,

Oh that's great. Then you may be able to appreciate some of the better math behind some of this stuff.

For example, a resistor R in series with a capacitor C and driven by a voltage source E.
The voltage vC across the cap is E-vC, and that means the current through R is iR=(E-vC)/R.
Since one definition of the capacitor is i=C*dv/dt and 'i' flows through both the resistor and capacitor, we can equate the two:
C*dv/dt=(E-v)/R
where we let v be the voltage across the capacitor same as vC.
Now divide by C and get:
dv/dt=(E-v)/(R*C)
and there we have our differential equation, which we can also call an ODE.
There are many ways to solve this, most notably using the Laplace Transform, but numerically there are also a lot of ways to solve this.
One of the simplest is the Euler Method:
y[n+1]=y[n]+h*dv/dt
which simply means the next calculation for 'y' is obtained from the previous value of 'y' (which was y[n]) and the time increment 'h' and the derivative we calculated above. The time increment 'h' has to be small though, so it could take a lot of iterations to get to the final time period which would be 'h' time the number of iterations. It's easy to imagine that if we made 'h' equal to 0.1 seconds and we wanted the solution at t=1 second, it would take 10 iterations to get the result, but if we made 'h' equal to 0.001 seconds, it would take 1000 iterations to get to the desired result. Of course with 'h' made equal to 1e-6, it would take a million iterations, something we would do with a computer program not by hand. For modern computers though, this would be done in a heartbeat or less.
Of course as the system becomes more complex, the number of ODE's increases, and then we have to turn to more complex solution methods. It could take a lot of time to complete even with a fast computer if the system is complicated enough, but often there is no way to solve it any other way. The Laplace Transform method comes in handy, but you have to be able to write the equations first which can get pretty involved for big systems.
Current simulators use numerical methods almost exclusively, and it also means they can solve a lot of types of systems.

It's a similar task for an inductor we just use the definition of that:
v=L*di/dt

#### MrAl

Joined Jun 17, 2014
11,447
I don't understand how current can be infinite.
Hello again,

Current being infinite is a mathematical tool only and is not considered to exist in real life, so you do not have to worry about this right now. It can become infinite in a limit because then usually something else becomes finite (like zero) and the whole thing becomes solvable. Without this tool sometimes it would be harder to solve or to understand.

A simple illustration is one way we might set the initial current through the inductor when we first start a circuit analysis with an inductor.
The definition of the inductor is:
v=L*di/dt

and solving for di we get:
di=v*dt/L

Now with a fixed applied voltage 'v' and fixed inductor 'L', we would need to set 'dt' to some particular value. Say 'v' was 1v and L was 1 Henry and we wanted 1 amp through the inductor at t=0. Well, we'd have to wait for 1 full second before the current would reach to 1 amp (di), but we need it to be 1 amp at t=0. Since we don't want to introduce negative time periods into the mix, we need a way to get the current up to 1 amp without waiting for 1 second or any other finite period of time.
If we introduce the concept of an impulse, we can have the current 'di' set at t=0 which we might call t=0+ (the plus sign meaning an infinitesimally short time after t=0). An impulse is a very short pulse that approaches an infinite amplitude but only lasts for an infinitesimally short time. This combination, combined with the desired final level of current, sets the initial current in the inductor to the level we need like that 1 amp.

This gets a little deeper into theory and there are other ways to do this too, so you don't really have to worry too much about this right now. I just wanted to give you some idea how an infinite current or voltage works, and that although it can be used in theory (there's no reality to worry about except in the end solution) it is considered to never exist in real life. This means you are likely to never see an infinite current source in real life but in theory you could.

Oh, another simple example is when we already have a current flow through an inductor which comes from say a voltage source, and then we suddenly disconnect the voltage source. The definition is again:
v=L*di/dt
and because we open circuited the inductor the current wants to go from some level like 1 amp to 0 amps in zero time. The change is thus 1 amp, and with a 1 Henry inductor we end up with:
v=1*1/0
because dt=0, or so it tries to be anyway.
This results in an infinite voltage 'v', which of course is in theory only.
In real life, there are physical constraints that kick in such as insulation resistance and distance between inductor leads. This insulation can break down and destroy the inductor over time, or the two inductor leads can arc over. That means the voltage can go up to a very high level like 1000 volts or more, but then something else gives out so that a path for the current is formed. That means the voltage stays under some finite level in real life.

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