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 ?

There is this article describing how a capacitor behaves in a similar fashion: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-capacitor-equation-in-action

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?

That happens when the diode is reverse biased. The 3 volts is applied in the reverse direction (3V on cathode side).

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...

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.

i don't quite understand how is it possible for the diode to get forward biased voltage? In this case, the upper side voltage of the inductor is way above the bottom, and i agree Vpb could reach high level due to air-resistance, yet that doesn't mean the direction of VL is upside down. so, how does this diode perform protection function? thank you.

Check how you are thinking about the inductor voltages. The *top* terminal of the inductor is stuck at +3 volts. The *bottom* terminal of the inductor is the one that shoots up to some very high voltage. That mean the diode becomes forward biased because the diode symbol arrow is pointing in the direction from the high voltage (bottom of inductor) towards the low voltage (top of the inductor).

When calculating the inductor current in the example above gave 300A/sec. I work in a factory where we have several DC motors with their field windings supplied from an unregulated 3 phase full-wave rectifier with a smoothing capacitor on the output. There is a bit of ripple on the supply, but it is effectively DC and is pretty stiff. The current through our field windings is constant, so where does the theoretical depart from the reality in this case. I should mention that there is a resistor in series with each field, with a wiper so field current can be adjusted. Thanks!

A DC motor accepts a DC power supply and converts it to AC using either a _commutator_ (for a DC motor with brushes) or a _motor controller_ (for a "brushless" DC motor). In both cases the current going through the windings of the motor are periodically turned on and off (or reverses direction) depending on the rotational speed of the motor.

Why is there no extreme rise in voltage when we connect the current source to the inductor? Considering the equation for the voltage across the inductor, a sudden change in current provided by the source will drive the voltage to a high number.

The first thought experiment in the article is a Current Source connected to an inductor, with the question "what is the voltage?". Very intentionally there is no switch in this circuit. That's because I would have to deal with your valid question at the very beginning of the article (and I didn't want to do that yet). The issue of suddenly switching an inductor current is covered in the following section where a voltage source and switch are connected to an inductor. There, it shows that a sudden change of current indeed drives the voltage to a high number.

Main content

Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response

Inductor i-v equation in action

Google Classroom

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

In this article:

We explore the derivative form and integral form of the inductor $i$ ‍- $v$ ‍ equation:

$v = L \frac{d i}{d t}$ ‍ $i = \frac{1}{L} \int_{0}^{T} v d t + i_{0}$ ‍

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 = L \frac{d i}{d t}

i = \frac{1}{L} \int_{0}^{T} v d t + i_{0}

These are the derivative form and integral form of the inductor equations.

L

is the inductance, a physical property of the inductor.

L

is the scale factor for the relationship between

v

and

d i / d t

L

determines how much

v

gets generated for a given amount of

d i / d t

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 R

Now we have an inductor with its

i

v

equation:

v = L \frac{d i}{d t}

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 = I

, for example,

i = 2 mA

. What is the voltage across the inductor?

The inductor equation tells us:

v = L \frac{d i}{d t}

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{d i}{d t} = \frac{d 2}{d t} = 0

(everybody knows

2

doesn't change with time)

Therefore, the voltage across the inductor is:

v = L \cdot 0

v = 0

If a constant current flows in an inductor, then $d i / d t = 0$ ‍, 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 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

V = 3 V

and

L = 10 mH

If we plug these values into the inductor equation we get:

v = L \frac{d i}{d t}

3 = 10 mH \cdot \frac{d i}{d t}

or, solving for

d i / d t

\frac{d i}{d t} = \frac{3}{10 \times 10^{- 3}} = 300 amperes / sec

That means the current through the inductor will have a rising slope of

300 amperes / 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 mH

inductor, plus a push-button switch

(pb)

. The top terminal of the inductor is at a constant

3 V

above ground. The name of the voltage across the inductor is

v_{L}

. Let's name the voltage across the switch:

v_{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 (0) = 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_{pb}

goes to

0 V

+ 3 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 (t) = \frac{1}{L} \int_{0}^{t} v (x) d x + i (0)

The expression inside the integral,

v (x) d x

looks a little odd.

x

is doing a favor for us by playing the role of instantaneous time, while

t

is the total time the button has been held down. The reason we need

x

(or any other variable name besides

t

) is because we want variable

t

to be the upper limit of the definite integral.

x

will disappear when we compute the definite integral, leaving us with current as a function of total time.

x

is called a dummy variable of integration.

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,

L

and

v

i (t) = \frac{1}{10 mH} \int_{0}^{t} 3 d x + 0

This integral evaluates to just

x

. We evaluate

x

between the limits

0

and

t

i (t) = \frac{3 V}{10 mH} x |_{0}^{t}

i (t) = \frac{3 V}{10 mH} [t - 0]

i (t) = \frac{3 V}{10 mH} t

This is the equation of a line, valid while the switch is pressed. The slope of the line is:

\frac{3 V}{0.010 H} = 300 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

(2 ms)

after the switch is pushed, the current will rise to

300 \cdot 0.002 = 0.6 amps

600 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 ms

and the switch opens. Let's find out what happens by reviewing the derivative form of the inductor equation:

v = L \frac{d i}{d t}

When we release the push-button, we expect the current to instantly change from

600 mA

0 mA

. But wait a second. This means

i

changes from a finite value to

0

amps in

0

time.

The derivative of current,

d i / d t

, is

(0 - 600) / 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.

Inductance is analogous to mass or inertia in a mechanical system. The energy in the magnetic field of an inductor doesn't allow the current to change instantaneously, just like the mass of the car tends to resist changes in velocity. A car cannot be stopped or started instantaneously. An inductor is basically the electrical version of Newton's First Law of Motion (the Law of Inertia): A body in motion tends to stay in motion.

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 ms

, we expect the current to change from

600 mA

0 mA

0

time. Let's not be greedy. Let's say it's okay for the switch to take

1 μ sec

to go from closed to open. The voltage we would see across the inductor is:

v_{L} = L \frac{d i}{d t}

v_{L} = 10 mH \cdot \frac{(0 - 600 mA)}{1 μ sec} = 10 \times 10^{- 3} \cdot \frac{- 600 \times 10^{- 3}}{10^{- 6}}

v_{L} = - 6,000 volts!!

The inductor voltage gets huge! The

+

terminal of the inductor is

+ 3

volts above ground. The

-

sign on

v_{L}

means the negative terminal of the inductor is

6000

volts above the positive terminal.

That makes

v_{pb} = 3 + 6000 = + 6003 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.

Although air is normally a very good insulator, a high voltage can break down air. When the electric field is about

3000

volts across an air gap of

1

millimeter, air becomes partially conductive. If the voltage goes higher, the air breaks down completely and becomes conductive. The electric field is strong enough to rip electrons away from oxygen and nitrogen molecules. That's when an electrical spark or arc bridges across the gap.

Souce: Wikipedia Electrical breakdown

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 don't need to be afraid of inductors, but you should respect them. Inductors are energy storage devices, just like capacitors are energy storage devices. If you release the energy all at once, there might be a boom!

Current sources and inductors may seem unusually hazardous, but no more so than other devices. This mostly because of our lack of familiarity with current sources, since they are not as common as voltage sources. Most of us know it's a bad idea to put a short across a voltage source. Don't stick a fork into a wall plug or you get a ton of current. It is also a bad idea to short out an energized capacitor, since, in the short term, a capacitor acts like a voltage source.

Current sources are the opposite of voltage sources. They do fine with a short circuit, but they don't like to be open circuited. An inductor with a flowing current acts a lot like a current source. If you open an inductor that's storing a lot of energy, you get a ton of voltage. It may seem strange, to think an open circuit can cause problems, but it does when inductors are involved. The inductor and switch circuit discussed here is the most common place where it becomes a significant issue.

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_{pb}

has a value of

3 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_{L} =

V

Is the diode forward biased or reverse biased?

What is the approximate diode current?

Diode current

\approx

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_{pb}

to spike up to a huge positive voltage.

With the diode in place, when the switch opens, there is a big

d i / d t

, which makes

v_{pb}

rapidly head towards a positive value, just like it did with no diode.

When $v_{pb}$ ‍ 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_{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,

d i / d t

, 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

d i / d t

. 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?

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mnjkmr398
Posted 8 years ago. Direct link to mnjkmr398's post “Why closing the switch (o...”
Why closing the switch (open circuit to close circuit) is not a sudden change in current
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(7 votes)
Answer
- Willy McAllister
  Posted 8 years ago. Direct link to Willy McAllister's post “Closing the switch causes...”
  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.
  Button navigates to signup page
  (20 votes)
doctor_luvtub
Posted 7 years ago. Direct link to doctor_luvtub's post “How long does it take for...”
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?
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(3 votes)
Answer
- APDahlen
  Posted 7 years ago. Direct link to APDahlen's post “Hello Dr. It depends on...”
  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
  Comment on APDahlen's post “Hello Dr. It depends on...”
  (4 votes)
manavpandya31
Posted 5 years ago. Direct link to manavpandya31's post “If inductor is connected ...”
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 ?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “An inductor resists _sudd...”
  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.
  Comment on Willy McAllister's post “An inductor resists _sudd...”
  (3 votes)
shreyansh.baid1995
Posted 8 years ago. Direct link to shreyansh.baid1995's post “Does same phenomena also ...”
Does same phenomena also happens with capacitors ?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Willy McAllister
  Posted 8 years ago. Direct link to Willy McAllister's post “There is this article des...”
  There is this article describing how a capacitor behaves in a similar fashion:
  https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-capacitor-equation-in-action
  Button navigates to signup page
  (2 votes)
Alexander Wu
Posted 7 years ago. Direct link to Alexander Wu's post “We can resolve the ideal ...”
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.
Button navigates to signup pageComment on Alexander Wu's post “We can resolve the ideal ...”
(2 votes)
Answer
- APDahlen
  Posted 7 years ago. Direct link to APDahlen's post “Hello Alexander, Capacit...”
  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
  Button navigates to signup page
  (3 votes)
Alexey
Posted 6 years ago. Direct link to Alexey's post “I have a question to the ...”
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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “That happens when the dio...”
  That happens when the diode is reverse biased. The 3 volts is applied in the reverse direction (3V on cathode side).
  Button navigates to signup page
  (2 votes)
manavpandya31
Posted 5 years ago. Direct link to manavpandya31's post “What happens if we opened...”
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...
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(1 vote)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “The energy in an inductor...”
  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.
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Chambers Wong
Posted 7 years ago. Direct link to Chambers Wong's post “i don't quite understand ...”
i don't quite understand how is it possible for the diode to get forward biased voltage? In this case, the upper side voltage of the inductor is way above the bottom, and i agree Vpb could reach high level due to air-resistance, yet that doesn't mean the direction of VL is upside down. so, how does this diode perform protection function? thank you.
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(1 vote)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Check how you are thinkin...”
  Check how you are thinking about the inductor voltages. The top terminal of the inductor is stuck at +3 volts. The bottom terminal of the inductor is the one that shoots up to some very high voltage. That mean the diode becomes forward biased because the diode symbol arrow is pointing in the direction from the high voltage (bottom of inductor) towards the low voltage (top of the inductor).
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BigJim
Posted 5 years ago. Direct link to BigJim's post “When calculating the indu...”
When calculating the inductor current in the example above gave 300A/sec. I work in a factory where we have several DC motors with their field windings supplied from an unregulated 3 phase full-wave rectifier with a smoothing capacitor on the output. There is a bit of ripple on the supply, but it is effectively DC and is pretty stiff. The current through our field windings is constant, so where does the theoretical depart from the reality in this case. I should mention that there is a resistor in series with each field, with a wiper so field current can be adjusted. Thanks!
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(1 vote)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “A DC motor accepts a DC p...”
  A DC motor accepts a DC power supply and converts it to AC using either a commutator (for a DC motor with brushes) or a motor controller (for a "brushless" DC motor). In both cases the current going through the windings of the motor are periodically turned on and off (or reverses direction) depending on the rotational speed of the motor.
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Anton
Posted 4 years ago. Direct link to Anton's post “Why is there no extreme r...”
Why is there no extreme rise in voltage when we connect the current source to the inductor? Considering the equation for the voltage across the inductor, a sudden change in current provided by the source will drive the voltage to a high number.
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(1 vote)
Answer
- Willy McAllister
  Posted 4 years ago. Direct link to Willy McAllister's post “The first thought experim...”
  The first thought experiment in the article is a Current Source connected to an inductor, with the question "what is the voltage?". Very intentionally there is no switch in this circuit. That's because I would have to deal with your valid question at the very beginning of the article (and I didn't want to do that yet). The issue of suddenly switching an inductor current is covered in the following section where a voltage source and switch are connected to an inductor. There, it shows that a sudden change of current indeed drives the voltage to a high number.
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Electrical engineering

Course: Electrical engineering > Unit 2

What we're building to

Inductor i‍ -v‍ equations

Inductor voltage is proportional to change of current

Inductor and current source

Inductor and voltage source

Inductor and a switch

Before the switch is pressed

After the switch is pressed

Release the switch

What happens in an ideal circuit?

What happens in a real-life circuit?

Give current a place to flow

Close the switch

Release the switch

Summary

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Inductor $i$ ‍- $v$ ‍ equations