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# dum question: v = Ldi/dt4

## dum question: v = Ldi/dt

(OP)
Fellas,

I am having trouble understanding a basic concept.  What I would like to know is how, via what mechanism, the voltage spikes in an inductive load.  Lets say that you have an inductor hooked up to 240 VAC.  The voltage will raise according to the rate of change of current times the inductors value.  This obvious.  But how does the voltage increase if the voltage source is a transformer.  The transformer does not care what is load side of it, right (assuming you're not drawing a boat load of current)?  If that is the case, how, again via what mechanism, does the voltage increase?
-AT

### RE: dum question: v = Ldi/dt

The best explanation for this question is in the definition of inductance: the property of a conductor to limit the change in current.
The instantaneous change in current present in an AC circuit (or a DC circuit turning on or off) causes a change in the magnetic field which induces a 'back emf(or voltage)' to oppose the current change.
This voltage is measured across the inductance and has no relation to the applied voltage of the circuit, but rather to this change in current.

### RE: dum question: v = Ldi/dt

To take dandel's explanation further, look at the trasnformer voltage as a 'force' used to overcome the 'back emf' generated by the inductance to force the 'change' in the current through it.

In steady state the Voltage drop across a inductor is minimal, compared to when there is a sudden 'change' in the current. This also explains why they are effective as reactors to limit inrush or a short circuit current, while having negligible impact during normal operation.  During a 'sudden change' more back emf builds up, source voltage remains constant and the through current is limited.

### RE: dum question: v = Ldi/dt

Suggestion: v(t)=Ldi/dt may be approximated by v(T2-T1)~Ldelta(i)/delta(t)=L x (I2-I1)/(T2-T1) in some cases.
It is obvious if:
1. I2-I1 is very high, and
2. T2-T1 is very small (short IGBT switching time)
then
v(t) will be very high. However, if 1. is not so high and 2. is high, then v(T2-T1) will not be that high. This is the secret behind the zero voltage switching (ZVS), which results in I2-I1 very small or negligible, and there are no significant spikes and harmonics.

### RE: dum question: v = Ldi/dt

(OP)
thanks for all your replies, fellas.  Dan could you please go into more detail.  I almost have it.  IS this situation analagous to momentum?
-AT

### RE: dum question: v = Ldi/dt

Suggestion to lakevillethor (Electrical) Oct 9, 2003
Fellas,
I am having trouble understanding a basic concept.  What I would like to know is how, via what mechanism, the voltage spikes in an inductive load.  Lets say that you have an inductor hooked up to 240 VAC.  The voltage will raise according to the rate of change of current times the inductors value.  This obvious.
///Is it?\\\
But how does the voltage increase if the voltage source is a transformer.
///The voltage on the transformer secondary may only increase, if there is an induction load, which you mentioned above that the understanding of the increase is obvious.\\\
The transformer does not care what is load side of it, right (assuming you're not drawing a boat load of current)?
///Not quite. The transformer will see different voltage if the interrupted load is resistive:
Etr=I x R
If I is switched I=0 Amps, Etr=constant Volts, since R is an infinity. Therefore, there is no spike and no transient on transformer secondary side. However, there will some transient on the primary side of the transformer (similar to the switch phenomena explained below)

And different voltage, if the interrupted load by a switch S is inductive L with some circuit resistance R.
When time t=0 sec, i.e. in an instance when the switch resistance became Rs,
Uswitch=Etr + Uc = Etr(R+Rs)/R = I(R+Rs)
where
I is current flowing is the series circuit consisting of transformer secondary winding voltage Etr, Resistance R, Switch Rs, and Coil L.
Uswitch is voltage across the switch contacts
Etr is transformer terminal voltage
R is resistance of the circuit (coil L is assumed to have Rcoil=0 Ohms)
Rs is resistance of the opening switch path, i.e. variable resistance in Ohms,
Uc is variable voltage across the coil inductance, i.e.
Uc=Etr x (Rs/R) x exp{-[(R+Rs)/L]xt}, in Volts
t is time in seconds
If Rs equal to infinity is inserted in the circuit instantly with t=0 second,
then
Ratio=Uswitch/Etr would be equal to infinity.\\\
If that is the case, how, again via what mechanism, does the voltage increase?
///As shown above for the transformer secondary side.  The transformer primary side transient analysis involves the transformer electrical equivalent circuit and its secondary load circuit, which is more complicated case because of nonlinear magnetization branch impedance Zm.\\\

### RE: dum question: v = Ldi/dt

2

The analogy which could be drawn

v = L*di/dt
F = m*dv/dt
Voltage plays rols of force
current plays role of velocity
inductance plays role of mass

Now, what is the analogy for rapidly decreasing current flowing in an inductive circuit... accompanied by a spike in voltage?

Rapidly decreasing velocity of a moving mass.... accompanied by a spike in force.

The only confusion on first-glance is whether the spike is a cause of effect of the rapid decrease.  The answer imho is both.  Spike in voltage like spike in force is always matched by equal and opposite voltage or force from the remainder of the system.

### RE: dum question: v = Ldi/dt

equal in both cases.  not opposite in the case of voltage.

### RE: dum question: v = Ldi/dt

Suggestion: Kirchhoff's Voltage Law and Current Law will hold true since they are general laws.

### RE: dum question: v = Ldi/dt

(OP)
Thank you all for posting - this discussion is really helping me understand it throughly.  I always knew teh mathematical formulas - sadly this was enough to get my by school without having to understand it thoroughly.  Let me set up an example and see if this is exactly analagous to the voltage spike phenomenon.  Lets say you grab a hold of your friend and you shake him, slowly pushing him, and then slowly pulling him.  After you have pushed him, he begins to move away from you (bear in mind that you are keeping your hands on him the whole time).  As your arms extend and he begins to exert a force, pulling you with him.  Exactly at that time, you start to pull him towards you.  It takes more force to pull him towards you at this point because the momentum is carrying him the other direction.  This larger force you exert is exactly the same as the voltage spike.  lastly, is V = L Di/Dt like saying you cannot change velocity instantaneously in nature?  You guys are really helping me put everything together here.
-Andrew Thoresen

### RE: dum question: v = Ldi/dt

Basic rule of physics: Nothing happens instantaneously.
As far as the 'momemtum' analogy, imagine something of a certain mass traveling at a certain constant velocity. The greater the mass(inductance), the harder it is to change the velocity(current), or, for that matter, the direction of the velocity.

### RE: dum question: v = Ldi/dt

(OP)
thanks everyone.  I think I got it now.
-Andrew Thoresen

### RE: dum question: v = Ldi/dt

In a very general sense, what's going on is you're slowly storing up energy, and then quickly releasing it.

Inductors essentially add "electrical momentum" to a circuit.  This is very analogous to the momentum of water flowing through a pipe.

Have you ever heard of or experienced water hammer?  If you take a long long pipe, maybe a long coil of pipe, and run some water through it at a high speed, and then quickly shut off a valve at the far end, the water will violently slam into the end of the pipe, making a loud banging sound, sometimes with destructive force.  All this despite the pump itself not really generating all that much force.

Similarly, you can slowly pedal a bike up to maybe 20 or 30 miles an hour, without a great deal of force or effort.  But slam that bike into a brick wall, and a WHOLE lot of force is generated.  Enough to damage both you and the bike.  The force you exherted from your feet, though, would likely not have been sufficient to bend the bike frame.  Where did the extra force come from?

The key concepts here are conservation of energy, and the rate of energy release.

### RE: dum question: v = Ldi/dt

(OP)
Let me ask this question then...If you had an inductor and you hooked it up to a circuit (say 240 volts AC), would you initially get a voltage spike (higher than 240) across the inductor when you flipped the switch ON?  Moreover, since the current cannot change at one instant of time, is it possible that the rate of change multiplied by the inductance of the load would give a voltage higher than 240 volts AC across the inductor (even though 240 volts is the voltage that your pole mounted transformer is providing).  Based on what I have learned, it doesn’t seem that this would occur.
-AT

### RE: dum question: v = Ldi/dt

Using "ideal" components, if you put 240vac across the inductor, then the voltage across the inductor is 240vac.

More realistically, you'll actually usually get a current INRUSH with AC, until the magnetic field is stabilized.  This will actually lead to a voltage DROP.

### RE: dum question: v = Ldi/dt

(OP)
That's what I am saying; they are at the same node.  My boss told me different and that the voltage would increase as a function of the L di/dt.  I find that unlikely.

So, in short, there would be no spike when the breaker was flipped on.  The spike in voltage comes from the rapid decceleration then, right?  Electrical Pete said "Rapidly decreasing velocity of a moving mass.... accompanied by a spike in force."  Does he mean, a spike in the force the other direction.  If you're driving your car, and you slam on the brakes, the car slows down rapidly due to the brakes applying a large force in the opposite direction you're traveling.  Is this the spike in force he's talking about?

-AT

### RE: dum question: v = Ldi/dt

More accurately, there is often a voltage "notch" when you energize an inductor from an AC circuit.

Re "Does he mean, a spike in the force the other direction.  If you're driving your car, and you slam on the brakes, the car slows down rapidly due to the brakes applying a large force in the opposite direction you're traveling."  -- careful, it gets a little tricky determining the direction of the force.  Your car is actually putting a force on the brakes in the same direction, and the brakes are putting an equal and opposite force on the car.  With the bike analogy, is the bike putting a force on the wall or the wall a force on the bike?  Really, there's two equal and opposite forces, but the bike is where the energy's coming from.

### RE: dum question: v = Ldi/dt

(OP)
I don't know what you mean by voltage notch...I am going to talk my way through a step by step pocess of what happens.  Please correct me if I am wrong.  Okay, so you wire up your switch and your inductor and at time 0 you throw the switch.

1:  240 VAC is applied across the inductor
2:  The current begins to increase slowly through the inductor
3: the polarity switches on the 60 Hz signal
(here's where I am unsure of what exactly happens)
4: The voltage starts to rise on the 60 Hz, incoming signal.
5: back emf is produced (as the momentum of the current is still in the other direction)
Shit, I am lost.  I understand physics I just have a tough time grasping this concept because it is tough to visualize.
-AT

### RE: dum question: v = Ldi/dt

On energization, there is an INRUSH, not a slow increase in current, but rather a sudden high level of current.  That high level of current racing through the feeder wires causes a drop in voltage at the inductor.  You get a transient UNDERVOLTAGE condition, that's the notch I was talking about.

Once you're in steady state, the momentum of the current means that it will actually lag behind the voltage waveform -- there's a phase shift.  Mathematically, looking at V=Ldi/dt, since V is a sine wave, then I is a cosine wave (integral of sine).  That's essentially a phase shift.

For a mechanical analogy, imagine you have a spring in your hand, with a weight at the other end, and you bouncing the weight up and down using the spring.  Everything's going boing-boing-boing, etc.  And it's all going at the same frequency.  But you'll likely find that your hand and the weight do not reach the uppermost position simultaneously.  Rather, there's a time delay (or a phase shift) between when your hand and the weight reach the uppermost point.

### RE: dum question: v = Ldi/dt

(OP)
Peebee, I understand that the current lags the voltage in an inductive load.  But you just said it..."On energization, there is an INRUSH, not a slow increase in current, but rather a sudden high level of current."  Are you certain about the inrush of current?  If the inrush of current happens very quickly, then there is a high rate of change in current or di/dt is high because you went from 0 Amps to very high amperage.  Well, if the voltage is equal to L di/dt, and you have that major inrush, won't the voltage ping high for a small amount of time?  That is, if your theory is correct.

### RE: dum question: v = Ldi/dt

There is inrush, or at least there can be (depends on the exact phase angle of the voltage when you flip the switch on).

See http://www.physics.brandeis.edu/phys29a_2003/AC.pdf , page 252 (262 in the PDF file), Section 9.8.5., for a nice description of inrush currents.  Wherever it says "transformer", just think "inductor."

The di/dt is happening BECAUSE of the voltage you applied.  YOU applied the voltage, and as you connected it to the inductor, you DEFINED the voltage.  The voltage is the cause, the di/dt id the EFFECT.  The inrush current won't cause a voltage "ping", it is the RESULT of the voltage you applied.

### RE: dum question: v = Ldi/dt

On energization, the current builds up slowly. Remember, current through an inductor cannot change instantaneously. Mathematically, the current is the time integral of the applied voltage. Another way of saying this is that the inductor accumulates volt-seconds and the current increases in proportion to the volt-seconds received. An inductor behaves very much like a capacitor, but instead of storing ampere-seconds (coulombs), it stores volt-seconds.

### RE: dum question: v = Ldi/dt

I guess part of the problem is how you define terms like "sudden" and "slowly" (and "is").

As a power engineer, I tend to think of inrush as a rather instantaneous event that happens, like, BANG, when you turn something on, and then quickly decays out.  Viewed in terms of steady-state operation, it's "sudden".

In terms of the waveform, though, yes, it rises smoothly and essentially exponentially from zero.  Viewing the first few cycles, it is "slow".

### RE: dum question: v = Ldi/dt

lakevillthor:

Yes, if you quickly make and break the circuit of an inductor, it will cause a high voltage on open circuited inductor, the very phenomena used to advnatage in striking the starting voltage to 'start' a fluorescent lamp by switching a inductor called magnetic ballast. (its another matter that now a days electronic ballsts have taken over but performing the same task).  The initial strike voltage is in the range of 1500 to 2000V, while the supply voltage is rated for 120 or 220V.

### RE: dum question: v = Ldi/dt

Clarification to rbulsara's post:

You can get a voltage spike on opening the circuit.  But you won't get one on energization.

Rapid sequential make-break-make-break can add an overvoltage each time, up to something like 5x system voltage.

### RE: dum question: v = Ldi/dt

Suggestion to lakevillethor (Electrical) Oct 10, 2003 marked  ///\\\
Thank you all for posting - this discussion is really helping me understand it thoroughly.  I always knew the mathematical formulas - sadly this was enough to get my by school without having to understand it thoroughly.
///Some formulae in modern engineering and science are so much detachec from reality that are dificult to explain them in something real, e.g. various transformations.\\\
Let me set up an example and see if this is exactly analagous to the voltage spike phenomenon.  Lets say you grab a hold of your friend and you shake him, slowly pushing him, and then slowly pulling him.  After you have pushed him, he begins to move away from you (bear in mind that you are keeping your hands on him the whole time).  As your arms extend and he begins to exert a force, pulling you with him.  Exactly at that time, you start to pull him towards you.  It takes more force to pull him towards you at this point because the momentum is carrying him the other direction.  This larger force you exert is exactly the same as the voltage spike.  lastly, is V = L Di/Dt like saying you cannot change velocity instantaneously in nature?  You guys are really helping me put everything together here.
///It is highly commendable to seek thorough understanding of the mathematical relationships. However, the advent of software and software calculations changed the way the engineering and design are performed. Often, the most important part is to correctly interpret the input data for the software, correctly input data into software, which then guarantee the software correct output, if the software was properly verified and validated. This is a current trend. It means that to produce the required work, the software used formulae are not necessary to know, may be very difficult to understand and not really needed. However, it is convenient to know how to spot check the software output. That is where the understanding of relationships, analogies, proportions, etc. become good to know.\\\

-Andrew Thoresen

### RE: dum question: v = Ldi/dt

When you switch in an inductor, you instantaneously changing the voltage across it, not the current through it.  The current will depend on the integral of the voltage.  If the voltage is V·cos(wt), then the current is V·sin(wt)/wL.  wL is the inductive reactance XL.  Note that sin(wt) lags cos(wt) by 90°.

### RE: dum question: v = Ldi/dt

(OP)
Gentlemen,

Thanks for all your responses.  After going over this time and time again, one thing is clear...the voltage spike will occur when the circuit is opened.

One more question...In steady state conditions will an inductor act as a short circuit?  I know that an inductor acts a short circuit in DC sitautions, but I am talking about steady state AC here.  I think it does.
-AT

### RE: dum question: v = Ldi/dt

No, the inductor will not work like a short circuit in steady state AC. A short circuit has zero impedance. The inductor will have an impedance which will be a phasor quantity, i.e. will have a magnitude and a phase displacement. The phase displacement will be 90 deg. The magnitude will be (frequency x L), where L is the inductance.

P.S. You can see that for DC the frequency is zero, so the impedance is also zero, therefore the inductor acts as a short circuit.

### RE: dum question: v = Ldi/dt

(OP)
I didn't really articulate what I meant very well.  What you said is what I meant.
-AT

### RE: dum question: v = Ldi/dt

Suggestion to lakevillethor (Electrical) Oct 13, 2003 marked ///\\\
Gentlemen,
One more question...In steady state conditions will an inductor act as a short circuit?
///Not for AC, only for DC or the DC part of composite of AC and DC, e.g.
i=(Vmax/|Z|)x[sin(sin(wt+alpha-theta)-exp(-Rt/L)xsin(alpha-theta)]
The component with exp is decaying because of the inductance causing the short circuit to the DC and resistance, in the circuit.\\\

I know that an inductor acts a short circuit in DC sitautions, but I am talking about steady state AC here.  I think it does.
///No, it does not since XL = 2 x pi x f x L. For DC at steady state, the XL = 0 Ohms.\\\

### RE: dum question: v = Ldi/dt

The higher the frequency, the more a capacitor looks like a short and the more an inductor looks like an open circuit.

The lower the frequency, the more an inductor looks like a short and the more a capacitor looks like an open circuit.

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