MH's ideal coil and voltage question

[poynt99]

As i said to MH--i agree with this,and the reason i did not understand as to why you would say otherwise?
Quote: All inductors, whether ideal or real will have a voltage across them when there is current through them, regardless if the current is changing or not.

Apparently I was wrong. As you said, we all make mistakes.

In regards to the current flow lagging the voltage,i was referring as to why there is a delay in time for the current to rise to a maximum value after the voltage is applied to the inductor.
Is it not the CEMF produced by the changing magnetic field that causes this delay in current rise time?.

OK, I understand now. Your original wording did not seem to point that way.

As a general rule of thumb about inductors, they don't and can't change their current instantly, especially with high L/R ratio inductors. An inductor is a reactive component, which means it has an impedance whose value is dependent on frequency. With the simple addition of a series or shunt resistor, we can make high and low pass filters with capacitors and inductors because of this fact.

Inductive Reactance X_L=2PIFL, so with a fixed frequency and inductance, the inductive reactance "seen" across the terminals of the ideal inductor is X_L. We have gone over this before, but reactance (whether capacitive or inductive) is similar to a resistance, but at a given frequency. From the equation we can see that the higher the frequency, the higher is X_L. This is true of ideal and real inductors, but real inductors have an "impedance" which is inductive reactance + resistance.

Since your question is in regards to real inductors, we'll use a real one as an example. If we have a 1H inductor and it has a 1 Ohm parasitic resistance, we know that the tau is L/R, which is 1s. So in 5s the current should reach a maximum once we apply our voltage source (let's assume an "ideal" DC step source of 1V). From this we can conclude that the final current will be 1A after 5s.

So why does the current ramp up rather than go up instantly? The reason is because the inductor can be viewed as a frequency-dependent resistor (inductive reactance X_L). Since the edge of a square wave or a simple step voltage can be quite fast, depending on the source (10ns is not uncommon), we know that such an edge will contain many frequency components, possibly up to 100's of MHz. The industry "rule of thumb" for the frequency content (bandwidth) based on edge rise time is: BW=0.35/RT, where RT is rise time. So for an edge having a 10ns rise time, the bandwidth of frequencies contained in that edge are 0.35/10ns, which is 35MHz. If we plug this 35MHz into our inductive reactance equation, what do we get? Well, XL = 2PI x 35MHz x 1H = 219M Ohms. That is a pretty high "resistance" agreed? How much current is going to flow right at the start? The high frequency components die off with time so to speak until the inductive reactance is 0 Ohms, after around 5s in this case. So between t=0 and t=5, the reactance presented by the inductor to our step input goes from about 219M Ohms to 0 Ohms, at which time the parasitic 1 Ohm resistance limits the current to 1A.

That's how I view it anyway.

ETA: As a thought experiment, imagine you are simply using your variable DC power supply as the source voltage for our 1H and 1Ohm inductor. If you were at the control and you increased it from 0V slowly enough, would the voltage and current track perfectly? In other words, would there still be a delay with the current lagging the voltage? Why not try that on the bench?

[tinman]

Apparently I was wrong. As you said, we all make mistakes.
OK, I understand now. Your original wording did not seem to point that way.

As a general rule of thumb about inductors, they don't and can't change their current instantly, especially with high L/R ratio inductors. An inductor is a reactive component, which means it has an impedance whose value is dependent on frequency. With the simple addition of a series or shunt resistor, we can make high and low pass filters with capacitors and inductors because of this fact.

Inductive Reactance X_L=2PIFL, so with a fixed frequency and inductance, the inductive reactance "seen" across the terminals of the ideal inductor is X_L. We have gone over this before, but reactance (whether capacitive or inductive) is similar to a resistance, but at a given frequency. From the equation we can see that the higher the frequency, the higher is X_L. This is true of ideal and real inductors, but real inductors have an "impedance" which is inductive reactance + resistance.

Since your question is in regards to real inductors, we'll use a real one as an example. If we have a 1H inductor and it has a 1 Ohm parasitic resistance, we know that the tau is L/R, which is 1s. So in 5s the current should reach a maximum once we apply our voltage source (let's assume an "ideal" DC step source of 1V). From this we can conclude that the final current will be 1A after 5s.

So why does the current ramp up rather than go up instantly? The reason is because the inductor can be viewed as a frequency-dependent resistor (inductive reactance X_L). Since the edge of a square wave or a simple step voltage can be quite fast, depending on the source (10ns is not uncommon), we know that such an edge will contain many frequency components, possibly up to 100's of MHz. The industry "rule of thumb" for the frequency content (bandwidth) based on edge rise time is: BW=0.35/RT, where RT is rise time. So for an edge having a 10ns rise time, the bandwidth of frequencies contained in that edge are 0.35/10ns, which is 35MHz. If we plug this 35MHz into our inductive reactance equation, what do we get? Well, XL = 2PI x 35MHz x 1H = 219M Ohms. That is a pretty high "resistance" agreed? How much current is going to flow right at the start? The high frequency components die off with time so to speak until the inductive reactance is 0 Ohms, after around 5s in this case. So between t=0 and t=5, the reactance presented by the inductor to our step input goes from about 219M Ohms to 0 Ohms, at which time the parasitic 1 Ohm resistance limits the current to 1A.

That's how I view it anyway.

ETA: As a thought experiment, imagine you are simply using your variable DC power supply as the source voltage for our 1H and 1Ohm inductor. If you were at the control and you increased it from 0V slowly enough, would the voltage and current track perfectly? In other words, would there still be a delay with the current lagging the voltage? Why not try that on the bench?

First up-thanks for taking the time for such an in depth  answer,but most of that i already know.
I am after a more in depth answer-what is the mechanism  behind the gradual buildup/ramp up of the current. I am thinking that it is the building of the magnetic field that is the root cause of the current building or ramping up over time. This magnetic field that changes over time,cuts the conducting windings,and we get our CEMF. This is what impedes on the current flow. As this magnetic fields change over time starts to slow in a linear fasion,we also see the current ramp up in the same linear fasion. The current reaches its  maximum  value when there is no change over time of the magnetic field.

Am i on the right track here?

Thanks


Brad

[hoptoad]

snip...
Am i on the right track here?
Thanks
Brad

In one word, yes.
Cheers

[TinselKoala]

I've been waiting for that to come up!

The current, and the resulting magnetic field, are two sides of the same coin, so to speak. A mutual causal relationship.

[Magluvin]

We've gone over this a couple of times already, but we'll try to keep at it until it sinks in

An ideal voltage source with its voltage set to 0V is equivalent to an ideal piece of wire. Why then would it NOT allow the pre-existing circuit current to flow?

Hmm.  And the pre-existing circuit current, circuit being an inductor with a wire(ideal source at 0v) shorting its leads, continues to carry current at a constant level indefinately due to?? Like what are the specific mechanisms that make this happen?

Mags