Inductance Current 1

[Chezma]

The relation between voltage and current for an inductor is given by:
v = L . di/dt
I think this relation is valid only in steady state periodic cyclic operation and provided that the circuit contains a resistive element.
My question now :Why a resistive element should be contained in the circuit to make this relation valid ?

 

[Chezma]

Well,until now i have not get any clear answer,if the current contains ac and dc component,then to block the dc current component then we need a capacitor not resistor !

 

[Skogsgurra]

You have 'get' very clear answers. But you have not cared to even try understanding the math and physics in this quite simple system. I will not try to help you any further.

What can you do, Pete?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[VEBill]

OP should browse Wikipedia. There's enough detail there to fully understand how an ideal inductor works. No need to derive it all again here.

 

[desertfox]

Hi Chezma

If you want to study the inductor theory without resistance look at this site:-

http://www.allaboutcircuits.com/vol_2/chpt_3/2.html

desertfox

 

[electricpete]

Gunnar - I have the answer... Punt! (like the Chicago Bears did almost every time they had the ball today).

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[electricpete]

chezma - fwiw I agree with the others. These seem very basic questions. From what you described 23 Jan 11 11:36, your circuit simulator is giving the correct response. I would suggest to find a circuits book to study. Use your simulator... if the answer doesn't look like you expect then try to understand it. If you really think it's wrong, then you can post a full problem statement and simulator solution here and explain why you think it's wrong... most of the time the simulators are right.

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(2B)+(2B)' ?

 

[waross]

Try iterating sample circuit calculations with the resistance cut in half at each iteration.
Tip. Long before you get near zero inductor resistance the source effective resistance will become significant in the real world.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Chezma, In attempt to be helpful without being wordy, and to make sure we are all talking about the same thing, I have attached a solution of a series L/R circuit powered by a square wave voltage source Vs (see first plot for Vs(t)). The inductance value is L=1. Solution plots are provided for three different values of R: R = 0, R=0.2, and R=10 (as shown in the plot titles. )
R=0 has a dc component that does not decay
R=0.2 has a very noticeable decaying dc component
R=10 has no noticeable dc component (it decays before the 1st peak)

Compare it against your own computer simulator and mental simulator.

Note there is a "phase angle" inherent in the definition of Vs(t) that affects the dc offset results. I think if Vs had started with a half-width pulse instead of a full-width pulse, there would have been no dc offset.

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(2B)+(2B)' ?

 

[Chezma]

Thanks pete,that is a perfect answer to my question.
May i ask you where i can get a copy of the simulator software you have ?

 

[electricpete]

That software is a 10+ year old student version of Maple. It is currently around $99 for the student edition and $1,895 for the professional version. If you can get your hands on the student version of Maple, I would say it is a great investment... not so much for the numerical simulation but for the symbolic solution capabilities which encourage you to tackle a problem symbolically rather than numerically.

I think Mathcad and Mathematica are similar although I've never used them.

If simulation is what you're after, there are a lot of possibilities like Pspice, Matlab ODE routines like ODE45, free Matlab clones (Octave, Sci-lab) etc. I'm partial to using a general ODE solver rather than circuit solver myself. Maybe others will chime in on the simulator software question or direct you to a previous thread on that.

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(2B)+(2B)' ?

 

[Skogsgurra]

Nice, Pete!

I assume that the student version of Maple (and probably also the 'pro' version) has a minimum default value for the first order term. Without it, you get lots of questions like the one that caused this thread. In pure circuit simulators, there's a nanoohm-microohm built into the reactive components so you do not get the divide-by-zero message.

Or does Maple have any other mechanism that catches that?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

Chemza's post 23 Jan 11 11:36 described expected output for an R-L circuit. His earlier posts were not clear but I assume when he said the simulator wouldn't work he meant that it just wasn't giving the results he expected.

Maple have a number of options for dsolve... if you want a specific numeric solution method there are extra arguments to dsolve, but if you don't use any it uses a symbolic method and the solution is symbolic:
i.e. dsolve({subs({L=1,R=10},DiffEqn),i(0)=0},i(t)) outputs the following exact algebraic solution, just as if you had solved it by hand:

i(t) = 1/10 Heaviside(t) - 1/5 Heaviside(t - 1)

     + 1/5 Heaviside(t - 2) - 1/5 Heaviside(t - 3)

     + 1/5 Heaviside(t - 4) - 1/5 exp(-10 t + 140) Heaviside(t - 14)

     - 1/5 Heaviside(t - 5) + 1/5 Heaviside(t - 6)

     - 1/5 Heaviside(t - 7) + 1/5 Heaviside(t - 8)

     - 1/5 exp(-10 t + 100) Heaviside(t - 10) - 1/5 Heaviside(t - 9)

     + 1/5 Heaviside(t - 10) - 1/5 Heaviside(t - 11)

     + 1/5 exp(-10 t + 10) Heaviside(t - 1)

     - 1/5 exp(-10 t + 200) Heaviside(t - 20)

     + 1/5 exp(-10 t + 190) Heaviside(t - 19)

     + 1/5 exp(-10 t + 70) Heaviside(t - 7)

     + 1/5 exp(-10 t + 130) Heaviside(t - 13)

     + 1/5 exp(-10 t + 50) Heaviside(t - 5)

     + 1/5 exp(-10 t + 30) Heaviside(t - 3)

     - 1/5 exp(-10 t + 40) Heaviside(t - 4)

     + 1/5 exp(-10 t + 110) Heaviside(t - 11)

     - 1/5 exp(-10 t + 60) Heaviside(t - 6) + 1/5 Heaviside(t - 12)

     - 1/5 Heaviside(t - 13) + 1/5 Heaviside(t - 14)

     - 1/5 Heaviside(t - 15) - 1/5 exp(-10 t + 120) Heaviside(t - 12)

     + 1/5 Heaviside(t - 16) - 1/5 Heaviside(t - 17)

     - 1/5 exp(-10 t + 160) Heaviside(t - 16) + 1/5 Heaviside(t - 18)

     - 1/5 Heaviside(t - 19) + 1/5 Heaviside(t - 20)

     + 1/5 exp(-10 t + 150) Heaviside(t - 15)

     + 1/5 exp(-10 t + 170) Heaviside(t - 17)

     + 1/5 exp(-10 t + 90) Heaviside(t - 9)

     - 1/5 exp(-10 t + 20) Heaviside(t - 2)

     - 1/10 exp(-10 t) Heaviside(t)

     - 1/5 exp(-10 t + 80) Heaviside(t - 8)

     - 1/5 exp(-10 t + 180) Heaviside(t - 18)

The problem is not particularly challenging for a numeric solution. ODE solutions of d/dt(x(t),t) = f(x(t),t) fall into the categories of explicit and implicit.
The forward Euler method is an example of explicit:
X(t+h) = X(t) + f(X(t))
It will not fail as long as there is a solution f(X(t)) which does not blow up. An iterative explicit variable step size routine would decrease step size to the minimum around the step-changes in V(t), but wouldn't have any problems generating a solution.
Matlab's ode45 is a more common example of explicit method.

The backward Euler method is an example of implicit method:
X(t+h) = X(t) + f(X(t+h))
it is not a straightforward calc since X(t+h) is only implitly defined.
I think most circuit solvers use the trapezoidal method which is also an implicit method:
X(t+h) = X(t) + 0.5*(f(X(t) + f(X(t+h))
It is also an implicit technique since there is no explicit formula for X(t+h).
For an inductor
I(t+h) = I(t) + 0.5*(V(t) + V(t+h))/L
The problem is V(t+h) is not necessarily known without I(t)
If we solve for 2 variables I(t+h) and V(t+h) in terms of the variables I(t), V(t), we have a relationship between I(t+h) and V(t+h) in terms of known quantities (V(t), I(t), L). This resembles an impedance relationship in terms of h and can be collected together in an impedance matrix and inverted. I assume the inversion step is where the problem occurs in circuit solvers which use an implicit integration method. Again should be no problem for explicit methods.

Sorry if this is more than you wanted. If you have any insight on the error you were talking about I'd be interested to hear... I was guessing as to the cause (matrix inversion)



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(2B)+(2B)' ?

 

[Skogsgurra]

Yes. A little bit more than I can digest. But I think it is about the matrix inversion with a zero involved that causes the problem.

Looks like Maple handles it well.

Thanks.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

From the LTSpice_IV introduction:

"Another benefit of these new simulation devices is that convergence problems are easier to avoid since they, like the board level component the model, have finite impedance at all frequencies."

I think that says it quite well. Making all components having finite impedance at all frequencies - also DC. That means that there are no zero ohm inductors, among other things.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Chezma]

Is the super conductor already available in the market in a wire form ?
What that material is?
What about the cost ?

 

[Skogsgurra]

It has been available for a long time, but not commercially and in usable wire form. At least not to my rather limited knowledge. I was involved in a study back in the sixties where the use of superconductors in generators was tried at ASEA. The problem then was to keep temperatures down in the 0 - 4 K region as well as stopping resistivity from leaving zero when the conductor was exposed to magnetic flux.

Same problems today. Only that temperature can now be a lot higher. But still quite cold. Sensitivity to magnetic flux still there. Not good in a coil supposed to produce a magnetic field.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[VEBill]

Superconducting motors (including coils) are supposed to be installed in some 'near-future' ships:

2009 news: "[successful] full-power testing of the world’s first 36.5 megawatt (49,000 horsepower) high temperature superconductor (HTS) ship propulsion motor..."

See also: amsc.com

 

[waross]

One of my good friends spent his career working with NMR machines. The magnetic field was so strong and so far past the saturation point of iron that air core magnets were used.
These units would withstand the intense magnetic field but an upset in the field would often drive part of the coil into the ohmic conduction range. This would instantly avalanche and the whole coil would go into ohmic conduction. The liquid helium would boil and often a part of the safety valve would become impaled in the ceiling. The scientific name for this event is a "Quench". NMRs for student use may have several holes in the ceiling from quenches.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[VEBill]

The more-modern HTS require only liquid nitrogen.

There's an installation at Holbrook substation on Long Island, New York, running "138kV at 2400A ~574MVA".

 

[electricpete]

I was thinking about the difference between R-L circuit branch and L-only branch (driven by a sinusoidal voltage for example) to see what might be inherent in the circuit configuration that can create numerical difficulties. We know what the exact solutions look like: One important aspect is that any dc offset present on the L-only branch will remain forever whereas any offset present in the R-L branch will decay away.

Now what if we think about the computation error in one time-step as a dc offset for the next step... In the case of the R-L system, the effect of an error at time t will be no longer evident after awhile... for example after 10 time constants.. But the effect of errors in the L-only circuit remain forever for the entire simulation. So if I run the simulation for a very long time, the R-L circuit has only errors accumulated from the last 10 time constants for example, whereas the L circuit has errors accumulated from each step of the entire simulation.

The errors at each step come from local truncation error (approximating Taylor series with some kind of truncation used by the algorithm) as well as rounding errors. If the errors all tend to add in the same direction, then clearly the L-only case will accumualte more errors. But even if the errors vary randomly poisitive and negative, the standard deviation of the error will be much larger for the L-only case than for the R-L case (if we run the experiment many times, the error averages to zero but the spread of errors is much larger for the L-only case).

I did a small experiment to check if this would cause a problem (attached). It shows an L-only circuit driven by a 50hz sinusoid for 25 seconds (1250 cycles) using a 0.002 sec timestep. The closing angle and magnitude are set up such that the magnitude of the ideal current waveform varies between 0 and 2 (dc offset of 1). A Runge-Kutta 4th order method is used for simulation (see Runge-Kutta tab for equations from wiki... pretty simple). You can view in tab "main" the plot of the results between 24.5-25 seconds and the current magnitude is exactly between 0 and 2 as expected from ideal solution. In fact if you look at the time 25.000, we expect the value is exactly zero and we read the numerical value from the spreadsheet to be 7E-13. I think for most purposes it is not an error to concern ourselves with, even though we accumulated errors over a very long simulation.

Other than the above concern which is shown to be relatively minor, I can't see any other problems with this L-only configuration.

My best guess is that whatever reason Spice forbids this L-only configuration has more to do with the particular algorithm used by Spice than with problems inherent in the circuit configuration itself. But I don't know anything about Spice.


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