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Induced voltage.

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alphacat

New Member
Hey,
I simulated in Pspice a RL circuit, where R was very close to zero - pico ohms.
I wanted to see that VL was lower than Vsource, since the induced voltage is in the oppostie direction and therefore should decrease VL.

Why was VL=Vsource, and not smaller?
 

dougy83

Well-Known Member
Post the schematic.

I would guess you series connected a dc source, an R & an L. The voltage across L will be Vsource at t=0, as there is no current flowing. As the current builds up in the inductor, VL will decrease, VR will increase (as VL + VR = Vsource). You would need to use a transient simulation to see this.

The inductor will only induce a voltage when you try to reduce the current you're supplying to it (the inductor will try to keep the current the same which is why the voltage is opposite to the applied voltage).
 

alphacat

New Member
Thank you very much for your help.

here is the simulation:


How does the size of R affect the induced voltage?

I also saw that when R->0, then the current never changes sign, why is that?

Its all got to do with how R's size affect IL, could you explain it please?
 

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MrAl

Well-Known Member
Most Helpful Member
Thank you very much for your help.

How does the size of R affect the induced voltage?

I also saw that when R->0, then the current never changes sign, why is that?

Its all got to do with how R's size affect IL, could you explain it please?


Hello there,

You did not see the current go through zero because you did not run
the simulation long enough. Unfortunately, with a resistance so small
like that you may need to run it very long in order to see the current
actually go through zero. If you instead increase the size of R a bit
you will see it go through zero as you probably expect.
The reason for this behavior is because when we do a transient
analysis with a sine source and an inductor and resistor there is
really another source in there that we dont see, and that is a
unit step function. Thus, the sine source is two sources in one,
not just a sine source like this:
v=u(t)*sin(t)
In order to get around this the sine source would have had to be running
in the circuit for all time since t=-infinity so that when we get to t=0
(where most analysis starts) all the effects of the u(t) source have
had time to dissipate. To simulate this, simply run for a much longer
time and look at the current wave after much time has past, then compare
to what it looks like around t=0.
Another way around this sometimes works is to set the initial current
in the inductor to some non zero value before starting the analysis
so that the u(t) part is canceled out at t=0.

We can ordinarily do an AC circuit calculation that involves only a sine
source but the answer we have to remember is the answer we get
after the circuit reaches steady state and does not represent the
entire response of the circuit...
I=amplitude(V/Z)
where
Z=j*w*L+R

The current i think you are looking for is I above.

Oh yeah, another way to get around the u(t) problem is to use a cosine source
instead of a sine source. You'll get the same response except it will be phase
shifted by 90 degrees.
To do this, you may have to set your sine source to have a phase shift of +90
degrees, or pi/2 radians because your software may not have a cosine source.
 
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alphacat

New Member
MrAl
Thank you very much!

1.Could you explain why physically the current will eventually cross the zero line?
I've calculated the full response of current to a sin(wt)u(t) input Vsource, and if I assumed that there was no initial current stored in the inductor, then the response was periodical, therefore I dont see why should it change after a long time if its periodical.

2. Moreover, could you explain please why doesnt the induced voltage cause VL to be less then Vsource, when R is very very small?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,

1. You have to think about the exponential part of the response as well as the
periodical part. The exponential responds to u(t) in a way similar to how this
circuit would respond to a dc source...think about that for a little while. Also,
a periodical response can ride on a dc level so it may not go through zero.
This circuit however should eventually go through zero.
Before i start posting equations i would like to see how you arrived at your
response for this circuit as i think that would be more beneficial. Try to
show as much work as possible.
Also, i suggest that you actually try this experiment with increasing R slightly
and looking at the response for longer time periods.

2. The frequency domain equation for the voltage across the inductor is:
VL=V*j*w*L/(j*w*L+R)
and if we take the limit as R approaches zero we get:
VL=V
so it looks to me that the voltage with R=0 will be equal to the source
voltage. With R slightly greater than zero the voltage will be very slightly
less than the source voltage. It may be hard to notice a difference in a
spice program with very small R so what you can do is start with a higher
value R where you can see a difference, then make R half of what it was
and see how that affected the voltage. You can then assume that if you
keep halving R you will keep seeing the same basic change until R gets so
small that you can no longer see a difference.
 
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alphacat

New Member
Hey MrAl.
I understand from your answers that you very like math :)

1.
I did see in Pspice that when I increased R, then the current did eventually crossed the zero.
I'm interested in knowing why physically the current at the beginning is not crossing the zero, and what makes if after a certain amount of time to cross the zero.

2.
The induced voltage doesnt decline the inductor's voltage?
The term self-inductance tells that if a current changes in the inductor, than the inductor's voltage changes to, and I assume its due to the induced voltage.
So how come the induced voltage doesnt decline the inductor's voltage?
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,

1. I wanted to see how you solved this circuit yourself so i can better answer
your question mathematically. It would help if you posted your solution and how
you arrived at it.
One way to see this phenomenon with a sine source is to look at the current
wave with a circuit that has been running for a very long time. You will note
that the wave looks like a minus cosine wave. Now a minus cosine wave is
negative as the voltage (the sine wave) goes through zero, but this cant be
possible at t=0 when the sine wave just starts up [due to the u(t) part] so
we end up with approximately the SAME wave shape, only here it starts out
at *zero* instead of *minus*. This immediately boosts the current above zero
to a level that it would not attain if it had started out at the minus level
as a steady state response would have.
This means you should look at the steady state response at a time when
the voltage wave goes through zero and see what the current is doing,
then try to imagine that same current wave only when it has to start out
at zero amps because we know that at t=0 the voltage AND current are
BOTH zero (unlike the steady state response which never does that).
What ends up happening is the current ramps up much farther than in
the steady state response, and hence the current offset during the
first quarter cycle of the voltage wave.

What makes it eventually cross zero is that this extra dc offset dissipates
through the time constant of the circuit in the exponential part of the response
that i talked about in a previous post. This can take very long if the resistance
in the circuit is very small.

2. The inductors voltage is lower than the source voltage with some R in the
circuit. If you look at the equation:
VL=V*A/(A+R)
for some constants A and R (R being the resistance), you will see that VL
is lower than V when R is greater than zero.
Think about it this way: How can the voltage across the inductor be lower
than the source voltage if the source voltage is connected directly across
the inductor??
 
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alphacat

New Member
Hey.
Thank you very much for the very detailed answer.

1.
Here is my solution:
iL'(t) + τ*iL(t) = (1/L)sin(wt)u(t)

Vsource = sin(wt)u(t)
i(0-) = initial condition.
τ=L/R.
γ=R² +(Lω)²

iL(t) = i(0-)/e^(-t/τ) + [(R/γ)sin(ωt) -(Lω/γ)cos(ωt) + Lω/γ]u(t).

I assume that i(0-) equals zero.
As you can see, if i(0-)=0, then there is no exponential part in the solution.

If you take R->0 then iL(t)= 1/(Lω)[1-cos(ωt)]u(t), and the current will always stay possitive.

"What makes it eventually cross zero is that this extra dc offset dissipates
through the time constant of the circuit in the exponential part of the response
"

But if i(0-)=0, then there's no exponential part in the response.
What causes the DC offset to be created?
its voltage offset or current offset?



2.
I'm not interested in mathematical equations, I understand the mathematics but not the physics, and for an electrician, its no good.
How do you explain that the induced voltage doesnt decrease the inductor's voltage?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,

1.
First, your solution is not correct for iL(t) but i can not tell what you did wrong
because you didnt show enough work.
Lets do this another way which is a little simpler:
Start with the frequency domain equation for i:
i(s)=v(s)/(s*L+R)
now since v(s) is Vpk*sin(wt) we'll convolve the Laplace for sin(wt) with Vpk/(s*L+R):
i(s)=Vpk*w/((s^2+w^2)*(s*L+R))
Next, we take the inverse Laplace Transform and we get:
i(t)=Vpk*[w*L*e^(-t*R/L)/K+R*sin(wt)/K-w*L*cos(wt)/K]
where
K=R^2+w^2*L^2

From this expression for i(t) we can plainly see that the exponential part is
not zero but approaches zero at some time t long after t=0. We also see that
when t=0, the exponential part exactly cancels the cosine part and the sine
part is zero, so at t=0 the current i is also zero as we knew.

If you show some more work (as much as possible) we might be able to narrow
down what went wrong in your calculations. BTW, the expression above for i(t)
is verifiable using a circuit analysis program im sure.

More intuitively, as i tried to explain before, is that the current shoots up to
an abnormal level when the sine source is suddenly switched on at t=0. This
happens because the current is not at its normal level at t=0 (the normal level
would be much negative when the sine source goes through zero). Because
the current is already higher than it should be at zero degrees (it should be
negative but it's actually zero), the sine source first quarter cycle makes it go
even higher, and that is where the temporary offset comes from. At that
point it is wayyy too high, and it takes several cycles to get back to its
normal dc level (it's normal dc level being zero).
Interestingly, many larger inductor type devices produce quite a bit of audio
noise in the form of a sort of buzzing when they are first switched on. The
reason is because of this temporary offset, and the noise stops or reduces
significantly once enough time has passed (the exponential). Also interesting
is that if you switch the device on at the right time, the wave looks more
like a cosine wave and you dont get any buzzing at all (or very little) because
the offset does not occur with a cosine wave. It takes several tries to get
this to happen though, because the line has to be switched at the very peak
of the voltage wave.


2.
I told you before that if you connect a sine source directly across an inductor
that the inductor voltage must be equal to the sine source voltage unless of
course there is some internal resistance in the source, which i assume we are
assuming does not have any internal impedance.
Did you mean something else by that question such as the instantaneous
voltage perhaps?


ADDED LATER:
Following what work you did show, i was able to narrow it down to the calculation
you did. Apparently you did not solve the differential equation properly. I think
if you review your solving technique you will come up with a slightly different
form for the solution but it will still have an exponential part and it will yield
the correct answer for i for any time t. I suspect the error came in when you went
to solve for i in the last step.
I will post this alternate solution if you want to take a look at it.
 
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alphacat

New Member
I cant thank you enough,
Its awsome to understand this.
I understand now that this abnormal shoot-up happens because of the continuance of the energy, in particular the magnetic energy, of the inductor, is it correct?

Now that you mention this, I do have an inductive heater that I turn on occasionally and it does produces a large noise on power-up.
What in this shoot-up makes this noise to happen?
Is it the anticyclicality of the current at the beginning? is it the temporarial over-current?

2.
I'm not quite sure about whether the induced voltage affects the inst. voltage or the rms voltage.
I just know that I learned that in an inductor, an induced voltage is created in the opposite direction of the creating voltage.
In our example - Vsource-R-L circuit - the creating voltage is as you said:
Vcreating = Vsource * Lω / (Lω + R).
I expected that the real voltage of the inductor will be
(Vcreating * √2 - Vinduced), that is if the induced voltage affects the instantaneous voltage.
I just expected the induced voltage to have an influence on the inductor's voltage.
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,

The shoot up happens because the current is not where it normally would
be with a steady state response and instead is much higher than that
already (zero is higher than any minus number). The positive sine just
takes it even higher. Im not sure if you can say it has to do with the
energy though, because that starts off at zero, but perhaps you can
clarify your statement a little if you like.

What causes the noise is again the higher current. If the normal peak
is say 10 amps and the turn on takes us up to 15 amps peak (for a time)
that's 50 percent higher. Higher amplitude current means higher amplitude
vibration (usually in the laminations) which means louder noise.
If you feel like trying this, turn it on, then off for a few seconds, then back
on, then off for a few seconds, etc., and see if you can get lucky a few times
and turn it on so that it is quiet during that one turn on.

Im still not really sure what you are asking about the inductors voltage decreasing,
because in a parallel circuit all the voltages are equal. Maybe you are saying
that the induced voltage is opposite so it should be less than the source?

The reason i present these equations is because from these we can get a lot
of information about how something works just by doing a few calculations
or many calculations with the same equation to simulate a time response.

BTW, i would like to see you go over your differential equation solving and
see that you get the correct solution so you can see for yourself how it is done.
 
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alphacat

New Member
Thank alot Mr. Al.

I learned my mistake in the differential equation.
The solution for: iL'(t) + (L/R)*iL(t) = (1/L)Vsource(t), where Vsource(t) = sin(ωt)u(t),
Is iL(t) = iLzsr(t) = Vp(t) + Vh(t) = Acos(ωt) + Bsin(ωt) + Ce^(-Rt/L).

My mistake was that I considered Vh(t) = C instead of Vh(t) = Ce^(-Rt/L).

After placing Vp(t) + Vh(t) inside the differential equation, with the condition that iL(0-)=iL(0+)=0, you reach the solution you've reached.

What it the physical reason that For very low R, it takes enormous time for the inductor to discharge from its given DC current offset?

Is it because that small resistor causes larger change in current, which causes the induced voltage to resist harder for current changing?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,

Oh i am happy to hear you got this to work out. Isn't it great to get the solution
to these problems?

I believe that the physical reason for the time it takes is because with a perfect
source and perfect inductor (no R) there is no place for that extra energy to
go, but with at least some R the energy has a place to dissipate. With larger
R the energy dissipates faster because of I^2*R. Maybe this is what you were
trying to say earlier on?
It helps to look at the exponential part to see the mechanism for this, which is:
e^(-t*R/L)
which is responsible for the dc offset. As this term decays to zero, we are left with
only sine and cosine terms, which is more what we expect with a sine source.

Interestingly, try this same experiment with a cosine source instead of a sine source.
 

alphacat

New Member
Yes, its a great feeling to understand this, it really feels awesome.

I really loved your explanation about why when R->0, the energy doesnt dissipates.

As you said, I now fully understand that when the voltage source is cos(t)u(t), together with the initial condition that i(0-) = 0, which derives i(0+) = 0, the current starts off right where it would normally be in its steady state, therefore there's no DC offset applied on the current.
I had a great teacher :)
 
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