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# Inductor i-v equation in action

We look at the inductor i-v equations and notice how important it is to give inductor current a place to flow. Written by Willy McAllister.
The inductor is one of the ideal circuit elements. Let's put an inductor's current-voltage equations to work and learn more about how an inductor behaves.

### What we're building to

• We explore the derivative form and integral form of the inductor $i$-$v$ equation:
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}$ $i=\frac{1}{\text{L}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}T}v\phantom{\rule{0.167em}{0ex}}\text{d}t+{i}_{0}$
• We create simple circuits by connecting an inductor to a current source, a voltage source, and a switch.
• We learn why an inductor acts like a short circuit if its current is constant.
• We learn why the current in an inductor cannot change instantaneously.
• When an inductor is connected to a switch, there is a paradox when the switch is thrown open. Where does the inductor current go?
• We show how to protect sensitive components from high voltages generated by an inductor.

## Inductor $i$‍ -$v$‍  equations

$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}$ $i=\frac{1}{\text{L}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}T}v\phantom{\rule{0.167em}{0ex}}\text{d}t+{i}_{0}$
These are the derivative form and integral form of the inductor equations.
$\text{L}$ is the inductance, a physical property of the inductor.
$\text{L}$ is the scale factor for the relationship between $v$ and $di/dt$.
$\text{L}$ determines how much $v$ gets generated for a given amount of $di/dt$.
${i}_{0}$ is the initial current flowing in the inductor, at $t=0$.

## Inductor voltage is proportional to change of current

When we learned about resistors, Ohm's Law told us the voltage across a resistor is proportional to the current through the resistor: $v=i\phantom{\rule{0.167em}{0ex}}\text{R}$.
Now we have an inductor with its $i$-$v$ equation: $v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$.
This tells us the voltage across the inductor is proportional to the change of current through the inductor.
For real-world resistors, we learned to take care that voltage and current don't get too big for the physical resistor to handle. For real-world inductors, we have to be careful the voltage and change of current don't get too big for the physical inductor to handle. This can be tricky, since it is very easy to create a very big change of current if you open or close a switch. Later on in this article we show how to design for this situation.

## Inductor and current source

The first thing we'll look at is an inductor connected to an ideal current source.
The current source provides a constant current to the inductor, $i=\text{I}$, for example, $i=2\phantom{\rule{0.167em}{0ex}}\text{mA}$. What is the voltage across the inductor?
The inductor equation tells us:
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
This says the voltage across an inductor is proportional to the rate of change of the current through the inductor.
Since the current source provides a constant current, the rate of change, or slope, of the current is $0$.
$\frac{di}{dt}=\frac{d2}{dt}=0\phantom{\rule{2em}{0ex}}$ (everybody knows $2$ doesn't change with time)
Therefore, the voltage across the inductor is:
$v=\text{L}\cdot 0$
$v=0$
If a constant current flows in an inductor, then $di/dt=0$, so there is zero voltage across the inductor.
Zero voltage means an inductor with constant current looks like a short circuit, the same as a plain wire.
Even if the current really big, like $100\phantom{\rule{0.167em}{0ex}}\text{A}$, if it is constant, the voltage across the inductor is still $0$ volts.

## Inductor and voltage source

Now let's connect an inductor to an ideal constant voltage source and see what the inductor equation tells us.
Let's be specific and say $\text{V}=3\phantom{\rule{0.167em}{0ex}}\text{V}$ and $\text{L}=10\phantom{\rule{0.167em}{0ex}}\text{mH}$.
If we plug these values into the inductor equation we get:
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
$3=10\phantom{\rule{0.167em}{0ex}}\text{mH}\cdot \frac{di}{dt}$
or, solving for $di/dt$:
$\frac{di}{dt}=\frac{3}{10×{10}^{-3}}=300\phantom{\rule{0.167em}{0ex}}\text{amperes}/\text{sec}$
That means the current through the inductor will have a rising slope of $300\phantom{\rule{0.167em}{0ex}}\text{amperes}/\text{second}$.
That is pretty amazing, but that's what the equation says. Needless to say, this is not a practical circuit. We just constructed it in our heads so we could see what happens with a constant voltage. If we build this circuit the current would ramp up until our real-world voltage source couldn't keep up with the demand for more current. But over a short time span, this is how real inductors work.
A constant voltage across an inductor results in a current with a constant slope.

## Inductor and a switch

Now we give the integral form of the inductor equation a try as we analyze this circuit with a switch.
The circuit has a voltage source in series with our $10\phantom{\rule{0.167em}{0ex}}\text{mH}$ inductor, plus a push-button switch $\left(\text{pb}\right)$. The top terminal of the inductor is at a constant $3\phantom{\rule{0.167em}{0ex}}\text{V}$ above ground. The name of the voltage across the inductor is ${v}_{\text{L}}$. Let's name the voltage across the switch: ${v}_{\text{pb}}$. This is also the voltage of the bottom terminal of the inductor.
We press the pushbutton at $t=0$, which completes the circuit and allows current to flow. Let's figure out the current $i$ through the inductor, this time using the integral form of the inductor equation.

### Before the switch is pressed

We'll assume the initial current through the inductor is zero: $i\left(0\right)=0$ because prior to $t=0$ the switch was not pushed and the circuit was open.

### After the switch is pressed

We press the pushbutton switch at time $t=0$.
At the moment we press the switch, ${v}_{\text{pb}}$ goes to $0\phantom{\rule{0.167em}{0ex}}\text{V}$. $+3\phantom{\rule{0.167em}{0ex}}\text{V}$ from the source is now connected across the inductor, and current begins to flow. The current starts at $0$ and gradually grows while inductor integrates its voltage according to the inductor equation:
$i\left(t\right)=\frac{1}{\text{L}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}t}v\left(x\right)\phantom{\rule{0.167em}{0ex}}\text{d}x+i\left(0\right)$
The limit $t$ on the integral is the amount of time the switch has been held down. As long as the switch remains pressed, the inductor integrates (sums up) the voltage and the current continues to rise.
We can fill in the variables we know, $\text{L}$ and $v$:
$i\left(t\right)=\frac{1}{10\phantom{\rule{0.167em}{0ex}}\text{mH}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}t}3\phantom{\rule{0.167em}{0ex}}\text{d}x+0$
This integral evaluates to just $x$. We evaluate $x$ between the limits $0$ and $t$.
$i\left(t\right)=\frac{3\text{V}}{10\phantom{\rule{0.167em}{0ex}}\text{mH}}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}x\phantom{\rule{0.167em}{0ex}}{|}_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}t}$
$i\left(t\right)=\frac{3\text{V}}{10\phantom{\rule{0.167em}{0ex}}\text{mH}}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}\left[\phantom{\rule{0.167em}{0ex}}t-0\phantom{\rule{0.167em}{0ex}}\right]$
$i\left(t\right)=\frac{3\text{V}}{10\phantom{\rule{0.167em}{0ex}}\text{mH}}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}t$
This is the equation of a line, valid while the switch is pressed. The slope of the line is:
$\frac{3\phantom{\rule{0.167em}{0ex}}\text{V}}{0.010\phantom{\rule{0.167em}{0ex}}\text{H}}=300\phantom{\rule{0.167em}{0ex}}\text{amps/second}$
As long as the switch is closed, the current in the inductor increases $300$ amperes every second. All that energy gets stored in the inductor's magnetic field.
As an explicit example we will use this in a moment: $0.002$ seconds $\left(2\phantom{\rule{0.167em}{0ex}}\text{ms}\right)$ after the switch is pushed, the current will rise to $300\cdot 0.002=0.6\phantom{\rule{0.167em}{0ex}}\text{amps}$ or $600\phantom{\rule{0.167em}{0ex}}\text{mA}$.
The current gradually rises (or falls) as an inductor integrates its voltage over time.
This is the same result we got using the derivative form of the inductor equation.
We should probably let go of the switch at some point.

### Release the switch

Let's say we release the button at $t=2\phantom{\rule{0.167em}{0ex}}\text{ms}$ and the switch opens. Let's find out what happens by reviewing the derivative form of the inductor equation:
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
When we release the push-button, we expect the current to instantly change from $600\phantom{\rule{0.167em}{0ex}}\text{mA}$ to $0\phantom{\rule{0.167em}{0ex}}\text{mA}$. But wait a second. This means $i$ changes from a finite value to $0$ amps in $0$ time.
The derivative of current, $di/dt$, is $\left(0-600\right)/0$, or infinity!
The inductor equation predicts $v$ will be infinite! Can that happen? No, it cannot. The current in an inductor cannot change instantaneously because it implies an infinite voltage will exist, which isn't going to happen. This reluctance to change is because of the energy stored in the inductor's magnetic field.
The current in an inductor does not (will not) change instantaneously.
We have a puzzle on our hands. We opened the switch while current flowed in the inductor. The open switch means there is no place for current to flow. What happens to the inductor current that insists on flowing?

### What happens in an ideal circuit?

This is a mess. We have all the current and voltage we want, but this situation breaks the ideal models because we created an impossible fight: the current has to be zero and finite at the same time. We've reached the limits of our ideal models. This makes my head hurt.

### What happens in a real-life circuit?

When the switch opens at $t=2\phantom{\rule{0.167em}{0ex}}\text{ms}$, we expect the current to change from $600\phantom{\rule{0.167em}{0ex}}\text{mA}$ to $0\phantom{\rule{0.167em}{0ex}}\text{mA}$ in $0$ time. Let's not be greedy. Let's say it's okay for the switch to take $1\phantom{\rule{0.167em}{0ex}}\mu \text{sec}$ to go from closed to open. The voltage we would see across the inductor is:
${v}_{\text{L}}=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
${v}_{\text{L}}=10\phantom{\rule{0.167em}{0ex}}\text{mH}\cdot \frac{\left(0-600\phantom{\rule{0.167em}{0ex}}\text{mA}\right)}{1\phantom{\rule{0.167em}{0ex}}\mu \text{sec}}=10×{10}^{-3}\cdot \frac{-600×{10}^{-3}}{{10}^{-6}}$
${v}_{\text{L}}=-6,000\phantom{\rule{0.167em}{0ex}}\text{volts}\phantom{\rule{0.167em}{0ex}}!!$
The inductor voltage gets huge! The $+$ terminal of the inductor is $+3$ volts above ground. The $-$ sign on ${v}_{\text{L}}$ means the negative terminal of the inductor is $6000$ volts above the positive terminal.
That makes ${v}_{\text{pb}}=3+6000=+6003\phantom{\rule{0.167em}{0ex}}\text{volts}$.
Does this really happen?
When the voltage gets this high what actually happens is a spark arcs across the air gap between the push-button contacts. The energy in the magnetic field is released in a brief bright burst of light. In fact, if you want to generate a spark, this is one of the best ways to do it.
In a real-life version of our example circuit, the spark starts at a around $3000$ volts. If you have a physically tough switch, it can withstand the spark. But if the switch is fragile (like you use a transistor as a switch), there's a good chance the high voltage will destroy it.
Our paradox: How can there be a finite inductor current at the same time as an open circuit? In real life, the paradox is resolved by the inductor winning and the open circuit losing. The open switch was compelled to be not open for the duration of the spark.

You may stop reading here if you want. You have a good understanding of how the two forms of the inductor equation work. This next optional section describes how to design around the voltage spike from an inductor.
To get the most from this description it helps if you are familiar with how a diode works. A diode conducts current in one direction but not in the opposite direction.

How do we avoid having our circuit destroyed by the voltage spike from an inductor? When designing a circuit with a switched inductor, we think ahead and make sure current always has a place to flow.

## Give current a place to flow

The problem we face with inductors is they don't like sudden open circuits. Here is one way to deal with the design challenge: provide an alternate path for current.
If we add a diode in parallel with the inductor, it solves the voltage spike problem in a neat way. The diode provides a path for inductor current when the switch opens, preventing sparks and damage.
The first thing to notice is the direction the diode is facing: its forward current arrow is pointing up. Current will only flow up through the diode.
Before the switch is closed, there is no current flowing anywhere, so there is $0$ volts across the inductor and diode. ${v}_{\text{pb}}$ has a value of $3\phantom{\rule{0.167em}{0ex}}\text{V}$.

### Close the switch

When the switch is closed, the current flows down through the inductor and switch, just like it did without the diode:
What is the voltage across the diode when the switch is closed?
${v}_{\text{L}}=\phantom{\rule{0.167em}{0ex}}$
$\phantom{\rule{0.167em}{0ex}}\text{V}$

Is the diode forward biased or reverse biased?

What is the approximate diode current?
Diode current $\approx \phantom{\rule{0.167em}{0ex}}$
$\phantom{\rule{0.167em}{0ex}}\text{A}$

### Release the switch

Now we release the push-button switch and it opens. The diode does something clever for us. Before, when there was no diode, the opened switch caused ${v}_{\text{pb}}$ to spike up to a huge positive voltage.
With the diode in place, when the switch opens, there is a big $di/dt$, which makes ${v}_{\text{pb}}$ rapidly head towards a positive value, just like it did with no diode.
When ${v}_{\text{pb}}$ rises above $3$ volts plus about $0.7$ volts, what happens to the diode?

The diode provides a path for the inductor to let its current continue to flow, without the need to spark across the switch contacts. The diode's $i$-$v$ characteristic prevents the voltage from getting much greater than this. ${v}_{\text{pb}}$ goes up to $3.7$ volts or perhaps a little bit higher. The diode clamps the voltage at this value, preventing the dreaded arc. Everybody is happy.
In real life, the inductor current, $i$, circulates around through the diode for as long as needed until the resistance of the inductor wiring dissipates the energy as heat. The diode prevents a major inductive voltage spike and protects the surrounding components.

## Summary

The current in an inductor does not change instantaneously.
When its current is constant, an inductor looks like a short circuit.
Be careful making circuits with an inductor. A sudden changed in current, like a switch thrown open, breaking a current path, that means the derivative of current, $di/dt$, can become very large. The inductor equation tells us there can be a large voltage generated across the inductor.
One way to deal with potentially destructive inductor voltage is to design a path for the current, so you don't get a large $di/dt$. We showed how to add a diode to provide a current path and clamp the inductor voltage to an acceptable value when a switch was thrown open.

## Want to join the conversation?

• Why closing the switch (open circuit to close circuit) is not a sudden change in current • Closing the switch causes a sudden change in voltage. The inductor current just after the switch changes is always the same as just before. This is true for both directions of switch movement (open-to-close, and close-to-open).

Starting with the switch open... The current right before the switch closes is 0, and it is 0 the moment after the switch closes, too. Ideal switches are perfectly happy to carry 0 current in either the open or closed state.

When we open the switch, we use the same reasoning, BUT, in this case there is an epic clash of concepts: The inductor model goes to battle with the ideal switch model.
The inductor says: the current is the same before and after, no matter if current is 0 or non-0. The ideal switch model says there can be any current when closed, but only 0 current when open. Who wins when these ideal superhero models clash? In Ideal Land, the answer is a momentary infinite voltage across the switch. In Real Life, the inductor wins (the current stays the same), and the switch gets a spark between its contacts as a reminder of who's the boss in real life.
• How long does it take for an inductor current to dissipate around the diode-inductor loop once the switch is opened (is it an inverse logarithmic rate)? Is the rate of dissipation correlated to the magnitude of the current? How much does the dissipation rate vary based on the quality of the diode? • Hello Dr.

It depends on the inductance, current, voltage drop across the diode, and the resistance of the circuit. Resistance includes the intrinsic resistance of the inductor as well as the remainder of the circuit.

This is an important question because inductors are often found in electromechanical devices such as relays and solenoids. These devices will not turn off until the current has be reduced to some minimum. Know that there are tricks to speed up this process...

Regards,

APD
• If inductor is connected to constant voltage source, will current keep on increasing forever ?
I thought that inductor resists change in current, but it itself is increasing current by 300 A per second. I don't understand, how these concepts work together ? • An inductor resists sudden changes of current (instantaneous di/dt implies the existence of infinite voltage, which doesn't happen). It tolerates gradual changes.

Think about inductor current as an analogy to inertia or mass. If you set a bicycle wheel spinning fast and you try to grab it with you hand and bring it to a stop instantaneously, it will give your hand a giant jolt and it will still rotate a little more. That experiment is analogous to trying to stop and inductor current in zero time. If you put your hand on the tire and let it rub against you the tire will slow down and eventually stop.
• Does same phenomena also happens with capacitors ? • We can resolve the ideal case, right? (Release the switch: What happens in an ideal circuit?) In an ideal world, an open circuit is just an infinite resistor. It becomes an RL circuit. The voltage graph isn't exactly an exponential curve because R=∞, so V approaches Ie^Rt/L = 0. There is no contradiction because of infinity. • Hello Alexander,

Capacitors keep voltage constant. Inductors keep current constant.

Suppose the inductor has been in circuit a long time. The flowing current has caused energy to be stored in the inductors magnetic field. Now lets open the circuit. Release the switch!

The circuit will attempt to make R = ∞. The current will attempt to go to zero. But wait, the voltage across an inductor = Ldi/dt. This is a problem. Know that voltage wind every time. The voltage will instantaneously rise until it finds a way to make the current equal to what it was before the switch was opened. Remember inductors keep current constant.

This is usually seen as an arc. The plasma between the switch contacts will maintain the current. If the "switch" is a semiconductor device such as a transistor it will arc internally leading to its destruction.

Keep reading and you will be introduced to the "flyback diode." This device provides a non destructive bath that allows the current ton continue.

Regards,

APD
• I have a question to the little quiz in the text. "What is the voltage across the diode when the switch is closed?" The right answer: 3V. The second question: "What is the approximate diode current?" The right answer: almost 0. I do not understand how can a diode have 3V with almost no current flowing through it? • What happens if we opened push button by keeping gap fair enough. Voltage is high, but gap too is more, so spark wouldn't take place. In that case will voltage stay in wire, stored as magnetic energy in inductor, as it has got no way to release that energy...
(1 vote) • The energy in an inductor is stored in its magnetic field, which released its energy back into the circuit in the form of current (not voltage). If the gap is bigger than a millimeter, the voltage will be forced higher until it's high enough to form an arc (spark), thereby allowing the current to keep flowing.

Remember this sparking happens when the switch goes from closed to open. So the gap starts at 0mm (full contact across the switch) and changes to open something like 2-3mm. In between you visit all the gaps between 0 and 3mm, so there is always a narrow gap for part of the time.   