RL circuit practice

[Ratchit]

Do you disagree with my calculations? using Z= √{ R^2+X(f)^2} and solving for L using Rs(DCR) ?

No, I do not. I did not disagree when you displayed that derivation in post #32, and I do not do so now. After all, it was the same type of calculation I espoused in post #29.

You cannot equate voltage and then 100Ω =2πf*L as implied.

I never did say or imply that 100 ohms = coil reactance. I said in post #29 that 100 ohms = coil impedance. Right or wrong, the coil DC resistance was established in post #6 as 13 ohms. There is strong evidence that this value is not correct, but we cannot be certain until another measurement is made.

Ratch

 

[Tony Stewart]

Ratchit,
You post #29 does not mention how to include DCR in the calculations, In fact no mention is made of it. Maybe just an assumption on your part that he ought to know.
But for pedantic nature, I defer to my results.

 

[Ratchit]

Ratchit,
You post #29 does not mention how to include DCR in the calculations, In fact no mention is made of it.

Correct, I rightly assumed that everyone involved in AC circuits knows that impedance consists of resistance and reactance. That is "staple goods" that should not need to be explained for this problem.

Ratch

 

[Tony Stewart]

Answering with the source of the measurement error
and proper calculations can be acknowledged.

Making false assumptions is the crux of Murphy's Law.

 

[Ratchit]

Answering with the source of the measurement error
and proper calculations can be acknowledged.

Making false assumptions is the crux of Murphy's Law.

I said before, and still do, that the OP's method was solid. A mistake in measurement is not the same as a false assumption.

Ratch

 

[Tony Stewart]

Ratch , let me be perfectly clear, your instruction was vague and did not include how to exclude DCR nor did it give any insight as to real-world values that I referenced. In fact you suggested a lower R for a load, which would make his errors greater.

Also the fact he could not understand why his Simulation did not match real measurements which should not be a problem for this simple case. So be clear again. this was a false assumption on your part to expect he knew how to answer his own question.

with respect,

Tony
EE since 1975

 

[Ratchit]

Ratch , let me be perfectly clear, your instruction was vague and did not include how to exclude DCR .

I think you mean include, not "exclude". I also did not tell the OP how to use algebra and arithmetic. Since in talked about resonance and Q in his earlier posts, it can be rightly assumed that he knew what impedance is and how to use it in calculations.

Ratch

 

[Tony Stewart]

No I meant exclude for 2 simple reasons;

1) Subtraction by solving for L
√{X(f)²+DCR²}=100Ω at equal voltage point for OP's original question.
or L² = (100² - DCR²) /2πf

2) ignoring DCR by make R load bigger like 10k instead of 100 instead of smaller ( recall he could not generate higher f.)

 

[Ratchit]

No I meant exclude for 2 simple reasons;

1) Subtraction by solving for L
√{X(f)²+DCR²}=100Ω at equal voltage point for OP's original question.
or L² = (100² - DCR²) /2πf

2) ignoring DCR by make R load bigger like 10k instead of 100 instead of smaller ( recall he could not generate higher f.)

Which would require a higher frequency, possible AC loss, and maybe a small skin effect, both of which cannot be determined directly. Why not stick with a known measureable quantity like DC coil resistance and go from there? You are right, I did not suggest that method with differential voltage measurements.

Ratch

 

[Tony Stewart]

What AC loss?

There are no parasitic losses well below SRF.

The only losses with AC are the same with DC ... the DCR which was ignored.

 

[Ratchit]

What AC loss?

There are none well below SRF.

The only losses with AC are the same with DC ... the DCR which was ignored.

Don't be too sure about that. Lossy coils depend on their permeability, not their self-resonance. The DC resistance was not ignored, at least not be me.

Ratch

 

[Tony Stewart]

the standard practice for rating Inductance is where L drops 10% at max rated current. So permeability and current saturation effects are included. But I agree they can be more lossy with rising F at high current.

In this question, the current limits were not at risk until you suggest using a smaller load resistor.

~fini~

https://www.digikey.com/product-detail/en/B78108S1105J/495-5550-2-ND/652579
Epcos example with 1mH, DSR=20
Every part has expected Series Resonant Frequency (SRF) which rises with lower L values.
Standard linear parts.

 

[Ratchit]

the standard practice for rating Inductance is where L drops 10% at max rated current. So permeability and current saturation effects are included. But I agree they can be more lossy with rising F at high current.

In this question, the current limits were not at risk until you suggest using a smaller load resistor.

~fini~

Small resistor should not be a problem. Just lower the voltage for less current.

Ratch

 

[Tony Stewart]

Suggesting a smaller R does not solve the confusion. It only adds error to his results. OK ?

 

[Ratchit]

the standard practice for rating Inductance is where L drops 10% at max rated current. So permeability and current saturation effects are included. But I agree they can be more lossy with rising F at high current.

In this question, the current limits were not at risk until you suggest using a smaller load resistor.

~fini~

View attachment 96399
https://www.digikey.com/product-detail/en/B78108S1105J/495-5550-2-ND/652579
Epcos example with 1mH, DSR=20
Every part has expected Series Resonant Frequency (SRF) which rises with lower L values.
Standard linear parts.

Yes, but what is this supposed to prove?

Ratch

 

[Ratchit]

Suggesting a smaller R does not solve the confusion. It only adds error to his results. OK ?

There is no confusion in method. Only in measurement. How does it increase error?

Ratch

 

[Tony Stewart]

read post #1. Even he knew

tried with 1 kOhm and 2 kOhm resistors too.
It seems that the gap between theoretical/real frequency narrows as R gets bigger.

Of course , false math, but then vague responses.

 

[Ratchit]

read post #1. Even he knew

tried with 1 kOhm and 2 kOhm resistors too.
It seems that the gap between theoretical/real frequency narrows as R gets bigger.

He was not using the differential voltage (DV) method at first. The DV method I suggested eliminates the problem with a finite source voltage impedance.

Ratch

 

[Tony Stewart]

the correct calculation to factor out DCR was not considered , explained until my posts and thus low values of load R negatively affect error, significantly.

If you wish to be hopelessly pedantic, then be explicit about the sources of error
; scrupulous, precise, exact, perfectionist, punctilious, meticulous, fussy, fastidious, finicky;


Your advice to equate DM voltage on each part was good as one criteria to lead to a solution.

But it lacked the reasons for details on source impedance, and more importantly DCR was essential for this method to work.

Whereas if one discovers all the sources of R and included in the geometric equation, it can be done at any practical frequency or voltage ratio. Using equal DM voltage allows you to ignore the source impedance. Anyway you look at it, the math I pointed out is essential and was overlooked.

 

[MrAl]

Hello there,

In short, a simulation will never match real world measurements unless a more accurate inductor model is used in the simulation, and even then it's hard to pin down perfectly.
An air core inductor would be better for use for study here, at least to start.
Here is some more information about this...

This is a simple problem when the inductor is linear, for example with an additional series resistor R1 and calling the inductors internal series resistance Rs, the voltages across R1 and the physical inductor are equal when the following equality is satisfied:
w^2*L^2+Rs^2=R1^2

and solving for w^2 we get:
w^2=(R1^2-Rs^2)/L^2

and obviously with both w and L always positive that means we have:
w=sqrt(R1^2-Rs^2)/L

so:
2*pi*f=sqrt(R1^2-Rs^2)/L
f=sqrt(R1^2-Rs^2)/(2*pi*L)

That's the frequency where both the voltages become equal, and the inductor voltage is measured physically right across the inductor (so the inductor series resistance adds a little to the inductor voltage), and as long as the voltage from the source is measured right at the top of the first resistance R1 it does not matter what R1 is as long as we know the value.

There is a catch of course, and that is that most inductors that are more than a few uH are going to have a core that is not air, but something that was chosen to have a high enough permeability to get more inductance from the same number of turns and that not only makes it smaller it also increases the efficiency. This brings in a HUGE problem in the analytical nature of the problem unfortunately, because now the inductor is non linear and the degree of non linearity is very hard to specify most of the time because of the variables involved and the core model we choose to use.

For a linear core model the above is valid, but for any other core type it is not valid because for one the excitation current is no longer a sine wave even with a pure sine wave input. It would look more like a wide needle like wave, where we get higher peaks with lower currents to ether side. That's because as the current changes, the permeability changes, and as the permeability changes the inductance itself changes. The above would change then at the very least to:
f=sqrt(R1^2-Rs^2)/(2*pi*L(i))

where now we see that L is a function of current, and we assume that we know the function that yields some sort of averaged L value over the entire time period for that average current level.

It's even more complicated than that because L is also a function of time:
f=sqrt(R1^2-Rs^2)/(2*pi*L(i,t))

and so now we have a much bigger problem in trying to determine the averaged value of L for example.

In reality though, using even the simpler anisotropic BH curve we see the inductance start out low for low currents, then increase to some max for intermediate currents, then decrease again for high currents. Even if we choose to stay on the lower end of the curve we will see the inductance start from some low value and increase to some higher value, and during that portion of the curve it can change a lot.

If we use the real BH curve assuming we could know that exactly for the given temperature, we would see the inductance increase as the current increased and then start to decrease again for increasing current, and then when the current starts to decrease the inductance would increase again but more slowly.

Obviously if we dont know the inductance we cant know the waveshape, although the current usually looks like a very non sinusoidal wave.

This last problem brings in the problem of what the voltage might look like, because if the current is very non sinusoidal then the voltage could be too. If the voltage is non sinusoidal then we cant use a meter to measure the voltage we have to use a scope.

In a simulation the current and voltage will look sinusoidal because the inductor is most likely a linear inductor, but if you would like to see what it looks like for a non linear inductor then you need to choose a model which models some of the characteristics of the core material as well.
If you want to stick with a linear model in the simulation then it would be better to use an air core coil so you get at least some results that might be close, as long as the other effects dont start to take a toll, like for example the AC resistance of the wire itself goes up with frequency so the equivalent series resistance Rs could go up significantly in the real world inductor as frequency goes up. This AC resistance would not be the same as the inductive reactance but would act more like an increase in Rs as frequency went up. For the approximate averaged analysis you can see this complicates things too:
f=sqrt(R1^2-Rs(f)^2)/(2*pi*L(i,t))

where as you can see Rs is now a function of frequency f.

To add still yet another difficulty, if there is an y DC offset present in the drive waveform then we have to contend with that too, meaning the equation gets more complex again:
f=sqrt(R1^2-Rs(f)^2)/(2*pi*L(iac,Idc,t))

where the inductor is now a function of both the instantaneous AC current and the DC current as well as time.