Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Inductor question?

Status
Not open for further replies.

Electronman

New Member
Hi,

An inductor connected to a DC power source via a switch. Now consider you open the switch, What will happen to the current? I do know that we will have a spark due to current but I am not sure about the direction of the current while opening the SW?
Before now, I thought that when we make the SW opened, then the current direction will be reversed (I.e the inductor polarity will be reversed) but when I read about something called kick current or so I noticed that after opening the Sw the current tends to continues its direction (not reversed), right?
If so please tell me why when we connect a reversed biased diode parallel to the the inductors (specially relays) then the diode will be in direct biased when we open the SW????

Is there any reversed current inside the inductor when functioning or while opening?

Thanks.
 
Last edited:

Electronman

New Member
Do it has anything to do with Back EMF?

Damn inductors make me confused ever. lenz's law, farady's law, EMF, back EMF, Kick current, Free wheeling current, Voltage increase while opening the SW???!

Can somebody helps plz??!
 

dougy83

Well-Known Member
The current stays in the same direction when the switch is opened. The deal with inductors is that they oppose the change in current; so when you connect a voltage across one, the inductor resists the change in conducted current, but it gradually grows. When you disconnect the voltage, the inductor tries to keep the same current flowing, which results in the voltage across the inductor being of opposite polarity.
 

Electronman

New Member
Thanks,

Why keeping the current in the same direction changes the VOLTAGE polarity???

Why the voltage produced by this way is much more greater than the source volatge!?
 

MikeMl

Well-Known Member
Most Helpful Member
Maybe this will help.
 

Attachments

  • Snubber5.gif
    Snubber5.gif
    51.6 KB · Views: 309
Last edited:

crutschow

Well-Known Member
Most Helpful Member
Thanks,

Why keeping the current in the same direction changes the VOLTAGE polarity???

Why the voltage produced by this way is much more greater than the source volatge!?
It's not really mysterious. The inductor simply generates whatever polarity and voltage it takes to keep the current flowing in the same direction. Thus the connection that has the current flowing out will be positive with respect to the connection with the current flowing in. It somewhat resembles a battery at that point, except the voltage will go to whatever is required to keep the current flowing, at least until the inductive energy is dissipated.
 

smanches

New Member
To elaborate a bit on what crutschow said, when the switch is opened, the current can't flow, but there is still energy in the inductor. Since current can't flow, the energy is turning into voltage and will keep increasing the voltage until the current can flow again. This is why if you do not have a freewheel diode, you can get a spark at the switch or inductor. The voltage is high enough at that point where it can bridge an insulator (air gap or whatever) and cause the spark. It's just trying to make the current flow.

In short, the voltage from the back EMF of a coil will keep increasing until current can flow.
 
Last edited:

Warpspeed

Member
The current through the inductor initially remains exactly the same, even when the switch is opened.

Where does the current flow ?

The voltage rises until any stray capacitance is fully charged with all the stored inductive energy.
The circuit will then ring until all the energy is dissipated through various circuit losses.
Or the insulation breaks down at the weakest point, whichever occurs first.

The voltage across a coil reverses, because the inductor changes from being an energy absorber, to becoming an energy producer when the switch is opened.

It is something like a battery. Current reverses through a battery when you switch from charging it, to discharging it, but the voltage does not reverse.

With an inductor the voltage reverses when you switch from charging to discharging the stored inductive energy.

"Something" has to reverse when the direction of energy flow into a storage device reverses.

With a battery (or a capacitor) it is the current, with an inductor it is the voltage that reverses.
 
Last edited:

Electronman

New Member
Thanks,

Can you show it with a schematic? Yet I am not able to see why the current following at the same direction to the source will causes the polarity of the voltage to be reversed?!!
 

qdn

New Member
Refer to the attached schematic, the moment the switch is opened, V1[L1] will have higher potential than V2[L1]. The current will still be flowing from 2 to 1.
 

Attachments

  • sch.jpg
    sch.jpg
    14.1 KB · Views: 254
Last edited:

bountyhunter

Well-Known Member
The current stays in the same direction when the switch is opened. The deal with inductors is that they oppose the change in current; so when you connect a voltage across one, the inductor resists the change in conducted current, but it gradually grows. When you disconnect the voltage, the inductor tries to keep the same current flowing, which results in the voltage across the inductor being of opposite polarity.
It also causes an arc of electrons across the switch contacts the instant it opens. It may be a very small arc, but it has to do that because the current can not stop instantly.
 

phoenox

New Member
The equation for an inductor is V = L di/dt .

V - Voltage
L - Inductance
di/dt - The rate at which the current is increasing.

This means that if you apply a constant voltage to an ideal inductor the current will increase at a constant rate = V/L amps per second. Of course a real inductor has resistance that will limit the current eventually.

If you have a constant current going through an ideal inductor, the current's rate of change di/dt = 0, so the Voltage across the inductor is zero.

Then if you flip a switch so the current all stops immediately di/dt = -infintiy. for the moment in which the current stops. So .. V = L * - infinity = - infinity. For any real circuit the current will actually take longer than zero time to stop flowing. So if you have a 1 Henry inductor with 1 Amp flowing through it, when you switch it off and the current stops in 1/100th of a second, di/dt = 1 Amp / .01 seconds = -100 Amps / second (negative because the current is decreasing) .. So back EMF V = 1 H * 100 A/s = 100 V.

On the other hand, if there is a diode there, when the switch is closed the inductor current does not have to stop flowing in virtually zero time, instead it can flow in a loop through the diode and back through the inductor. Since the diode will restrict the Voltage to -.7 volts, the inductor current will slowly stop flowing at a
rate of di/dt = -.7 V / 1 H = -.7 A / s
 

Warpspeed

Member
This might still be difficult for some people to visualise.

So here is a mechanical analogy.

If you spin up a flywheel, and then suddenly release the drive energy, it just keeps spinning at the same speed. Nothing terribly dramatic happens.

If you compress a spring with some force, and then suddenly release that force, the spring flies back in the OPPOSITE direction with massive instantaneous speed. There is a very sudden and very violent release of stored energy.

The back EMF produced when current through an inductor is broken can be very violent, and it flies back in the opposite direction (voltage polarity), just like a released spring.
 
Last edited:

Electronman

New Member
Sorry but I think I found a problem with the direction of convectional current in a coming from a source and from a resistor (current seeker). Can somebody explain it to me please? does the positive side in a seeker (resistor or any passive element) is in opposite with a feeder (source)?

Thanks a bunch
 

Electronman

New Member
Please tell me if the inductor becomes a voltage source just after opening the SW?
Can I call it a voltage source just after SW is opened?
 

crutschow

Well-Known Member
Most Helpful Member
It's more accurate to call it a current source, with the voltage equal to the current multiplied by the impedance of any attached circuitry including the parasitics.
 

crutschow

Well-Known Member
Most Helpful Member
An ideal current source generates a current independent of the impedance it is driving.

A real current source, of course, has practical limits as to the maximum voltage it can provide when generating this current. An example of a such a current source is a power supply operating in the current limit mode. It will provide the set limit current up to the maximum output voltage of the power supply.
 

MrAl

Well-Known Member
Most Helpful Member
Hi Electronman,


If you look at the inductor's magnetic field you can see what happens.
When the inductor is fed a current, it's field builds up in one direction.
When the switch is opened, the field starts to collapse. The collapsing
field produces a voltage of opposite polarity to what the voltage *was*
before the switch opened.
The inductor becomes a power source, but it is often looked at as a current
source in order to simplify the calculations, and a perfect inductor works
very much like that anyway.

Here's what happens when we view the inductor as a current source as the
switch is opened...

A current source puts out constant current, such as 1 amp for example, and
the switch starting to open looks like a resistance say 0.01 ohms. 1 amp
through 0.01 ohms is not very much voltage, so the voltage is only 0.01 volts
for now, but the switch opens more and more as time goes by (this happens
gradually but it appears to happen instantaneously). As the switch opens more,
it's resistance might go up to 0.1 ohms, and now that same 1 amp means there
is 0.1 volts across the switch. A tiny fraction of a second later, the switch opens
more, and now maybe it has 1 ohm resistance, and so the voltage goes up to 1 volt.
Already the voltage has increased from 0.01 volts to 1 volt, but the switch opens
even more now, maybe the resistance goes up to 10 ohms, so now we end up with
10 volts across the switch. It doesnt stop there though. The resistance goes up
to 100 ohms, then 1000 ohms, then 10k, then 100k, then 1 megohm, then 10 megs,
then 100megs, then possibly even higher. As this happens, the voltage goes up
accordingly, to 100 volts, then 1000 volts, then even higher if the inductor's wire
insulation does not break down, but then something else starts to happen: the
voltage is up so high that the inductor can expel it's total energy in a very short
time, so by the time the switch is open fully all of the energy might be dissipated
and the voltage falls back down to zero (or oscillates if there is any self capacitance
in the inductor, which there usually is at least some).

It's also interesting to note the polarity of the voltage across the inductor while
all this is taking place as you already noticed...
As the inductor is charged from a source like a battery, the polarity would be positive
at the battery's positive terminal which say is wired to the switch. That means
the inductor terminal connected to the switch is positive as it is being charged.
As the switch opens however, the current flows through the switch and produces
a negative voltage at the switch instead of positive (it initially stays positive
because the resistance of the switch does not rise instantaneously, but it happens
so fast we often talk about it going negative as soon as the switch is opened).
In this way we end up with first a positive voltage at the switch and then a negative
voltage at the switch (the terminal that connects to the inductor).

So the real cause of the change in voltage polarity is because the field changes polarity,
but we can ignore that extra information if we think of the inductor as a current
source who's current flows in the same direction even when the switch is opened.

If you look at a circuit like this but replace the switch with a resistor of some value
like 100 ohms and keep the direction of current flow in the inductor the same as it
was just before opening the 'switch', you will see that the voltage across the
100 ohm resistor changes polarity, and so does the voltage across the inductor.

This is a principle that is used in many boost converters, to get a higher voltage from
a lower voltage using an inductor as the storage/converter element.
For driving relays however, this action works against us because the voltage may
become so high that it blows out the driving element (transistor), or starts to break
down the winding insulation and eventually causes a short in the coil. For this reason
we often use a diode to 'catch' that spike and shunt it so that the voltage can not
build up. A diode clamps the voltage to about 0.7 to 1.0 or so volts, so that's a
lot lower than 1000 volts which would blow out a regular 40v transistor.
Also, sometimes more than one diode is used in series to get that 'spike' to
subside more quickly. The level of the voltage across the coil has a profound effect
on the time it takes to dissipate the energy in that coil, and a higher voltage means
it dissipates faster, so sometimes a zener is used which will clamp the voltage to
some safe level like 10 volts which will not harm a regular 40v transistor.
 
Last edited:
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top