Inductor Voltage produced, frequency affect

[zesla]

Hello guys,

I ahve a question for you please.

I have got a coil. it has an indacutance of 10uH. The input source voltage across this coil is 10V. what is the voltage across the coil jus after openning the key or turning the source off at frequencies of say 1000Hz and 10KHz?
the inductor has no ESR.

Thanks so much.

P.S1: Actually I would like to see the frequency affect on the Voltage produced after oppening the key or turning the source off....
Can we say that the frequency of the source has any affect on the amplitude of the voltage produced?

P.S2: I know that changing the frequency would change the impedance of the inductor. but what about the volatge change as I said?

 

[rumpfy]

Zes,
These questions make one think too much!
You have drawn a coil with no resistance and no self capacitance and this is not what is found in practice. Due to the conservation of energy, the coil will attempt to produce a voltage of infinity across itself.
Keeping this in mind, the application of an alternating voltage across the coil will produce an alternating current in the coil. In actual fact the voltage across the coil is proportional to the rate of change of current. So the application of an alternating voltage across the coil will set up a current which will produce an induced EMF which will be equal in magnitude and opposite in sign to the applied voltage. Of course, this means that the current fed into the coil will be zero. That this is so is because you have assumed a perfect inductor. If the applied voltage is say e=e^ x sin wt, then the current which produces this voltage will be e w cos wt.
If you open the switch, the circuit will attempt to maintain the current flow and so the coil voltage will rise to infinity.
In practice, the coil has self capacitance and resistive losses, so when the coil is at steady state, the current applied to the coil will be some value. When the switch opens, the coil voltage will rise,(depending on where in the cycle the switch opens) and the energy in the coil inductance will be transferred to the self capacitance of the coil. The self oscillation that takes place produces a ringing at a particular frequency given by f=1/(2 ∏ (LC)1/2) [f equals 1 over two pi root lc]. Due to the losses in the coil, the amplitude of the ringing will decay exponentially.(the so called logarithmic decrement)
The coil current when the coil is driven is dependent on the source voltage frequency, but when the key is opened, the ringing frequency is controlled by the coil characteristics.
This effect is used extensively in circuit design. For example, the horizontal deflection system of a television display has a period of 64 microseconds (CCIT 625 line system) The line frequency is 15,625 Hz. The system design allows a flyback time of 14 microseconds so there is 50 uS of scan followed by 14 uS of flyback. The self inductance of the deflection components (deflection unit and Horizontal output transformer) is adjusted by design so that it rings at around 50 Hz. This ringing frequency returns the electron beam to the left side of the CRT in the required short time ready for the next scan line.
It would improve your understanding if you are familiar with differential calculus.
Hope this helps.

 

[ericgibbs]

hi zesla,
This is what LTSpice shows.
I am assuming that you are saying the 10V source is a DC source,???

The switch operates on/on in 1usec.

Lenz's Law EMF= - L * [di/dt]

EDIT.
Rereading your post. parts of it are ambiguous

Is its the frequency of the source voltage that can be 1KHz or 10KHz.??

Also on which angle of the current sine wave are you switching.??

 

[zesla]

Zes,
These questions make one think too much!
You have drawn a coil with no resistance and no self capacitance and this is not what is found in practice. Due to the conservation of energy, the coil will attempt to produce a voltage of infinity across itself.

actually me and my freind want to see if the rate of the frequency has any affect on the voltage produced across an inductor? For instance does a signel inductor connected to a sine source will produce different back voltages by changing the frequency of the sine source? or the amplitude of the volatge produced would be the same for different frequncies?
Then what If the signle coil happens to be a transformer having a primarry and a secondary winding?

Keeping this in mind, the application of an alternating voltage across the coil will produce an alternating current in the coil. In actual fact the voltage across the coil is proportional to the rate of change of current.

Right, as I see thats when the frequency comes up, so acording to what you are saying the more frequency mean the more RATE of changes at the frequency and hence the more voltage wiould br produced across the inductor (but more frequency means the more impednace for the inductor as well and of cource means the less current will pass through the inductor). So the other question which comes up is:
Is the voltage produced across an inductor related just to the RATE of current change or it has something to do with the AMOUNT of current passing through the inductor too?

So the application of an alternating voltage across the coil will set up a current which will produce an induced EMF which will be equal in magnitude and opposite in sign to the applied voltage.

Why the EMF (Back EMF you meanm right?) here is EQUAL to the applied voltage but not more???Is that means that any inductor connected to a sine source (like the mains) in series or in parallel will produce an emf eaual to the volatge applied to it?
I thought that the alternating volatge for an inductor acts something almost like opening a key connected to an inductor and would cause the EMF the raises up?

Of course, this means that the current fed into the coil will be zero. That this is so is because you have assumed a perfect inductor. If the applied voltage is say e=e^ x sin wt, then the current which produces this voltage will be e w cos wt.
If you open the switch, the circuit will attempt to maintain the current flow and so the coil voltage will rise to infinity.

I agreed, specially I am agreed with the red one, But what is the difference of openig a key and the alternating volatge applied to such a circuit having no key? for the alternating volatge The volatge reaches to zero for some portions times (which is almost and somehowequal to opening the key of the circuit when connected to a source) and for some portions of time it reaches to MAX (which is almost and somehow equal to closing the key), right? SO the EMf should rises up more than the applied volatge when the coil is connected to an alternating voltage withough opening the key. Why it doe not?

In practice, the coil has self capacitance and resistive losses, so when the coil is at steady state, the current applied to the coil will be some value. When the switch opens, the coil voltage will rise,(depending on where in the cycle the switch opens) and the energy in the coil inductance will be transferred to the self capacitance of the coil. The self oscillation that takes place produces a ringing at a particular frequency given by f=1/(2 ∏ (LC)1/2) [f equals 1 over two pi root lc]. Due to the losses in the coil, the amplitude of the ringing will decay exponentially.(the so called logarithmic decrement)
The coil current when the coil is driven is dependent on the source voltage frequency, but when the key is opened, the ringing frequency is controlled by the coil characteristics.

What is the ringing frequency please?! Is it the frequency caused by the key?

Thanks so much

 

[MrAl]

Hello guys,

I ahve a question for you please.

I have got a coil. it has an indacutance of 10uH. The input source voltage across this coil is 10V. what is the voltage across the coil jus after openning the key or turning the source off at frequencies of say 1000Hz and 10KHz?
the inductor has no ESR.

Thanks so much.

P.S1: Actually I would like to see the frequency affect on the Voltage produced after oppening the key or turning the source off....
Can we say that the frequency of the source has any affect on the amplitude of the voltage produced?

P.S2: I know that changing the frequency would change the impedance of the inductor. but what about the volatge change as I said?

Hi,


This question is not really complete because it suggests a practical circuit (not theoretical) yet it does not specify all the needed practical elements which would be indicative of a theoretical question. This makes answering this question correctly not possible.

First, in theory when we say that something happens at t=0 we use simply zero (the 'number' zero which is 0 on the keyboard). When we say that something happens just before zero, we use 0- which is just a zero followed by a minus sign. And when we say that something happens just after zero we use 0+ which is just a zero followed by a plus sign.

What you are asking here is what happens at t=0+, which is an infinitesimally small time after t=0.

For an pure inductor as indicated in your schematic that has been energized by even the smallest energy from a voltage source that has been allowed even a small time period to charge, when the switch is opened the inductor voltage would jump up to infinity. This is because the switch opens and there is no load for the inductor energy, so the voltage shoots up high. It would go up to infinity which of course can not happen in the real world. But what actually happens in the real world depends *highly* on other things which are not shown in your schematic. Thus, this question can not be answered until these other elements and quantities are specified.

For a few examples...
An inductor with one kind of core would response entirely differently than one with another kind of core. A lower sat value would mean lesser voltage for example.
An inductor with one value of parallel capacitance would response entirely different than one with another value. A lower cap value would mean higher peak voltage (if you are after the peak that is).

Also, if you assume *ANY* capacitance then the correct answer is that the voltage is exactly the same as when the switch was first opened. So if the voltage was 10v just prior to opening, the voltage at t=0+ would also be 10v, but it would change soon after that. This is because of the rule that the voltage across a capacitor can not change in zero time without the application of an infinite current.

But the question posed as is can only be taken as a theoretical sort of trick question. As soon as the switch is opened the inductor voltage jumps up to infinity since there are no limits imposed by the element definitions.

 

[zesla]

hi zesla,
This is what LTSpice shows.
I am assuming that you are saying the 10V source is a DC source,???

The switch operates on/on in 1usec.

Lenz's Law EMF= - L * [di/dt]

EDIT.
Rereading your post. parts of it are ambiguous

Is its the frequency of the source voltage that can be 1KHz or 10KHz.??

Also on which angle of the current sine wave are you switching.??

Thanks, Maybe my above post would clarifythe topic?

What are 1K and 10K in your pic please?

How can I calculte the volatge you have reached at your pic?
Can I reach to such a high current through the inductor by a normal low current battery?

 

[MrAl]

Thanks, Maybe my above post would clarifythe topic?

What are 1K and 10K in your pic please?

How can I calculte the volatge you have reached at your pic?
Can I reach to such a high current through the inductor by a normal low current battery?

Hi,


You had better read post #5 before you do any simulations. You can not do simulations of that circuit without knowing more of the variables which could easily change the results by 1000 times.

 

[zesla]

Hi,


You had better read post #5 before you do any simulations. You can not do simulations of that circuit without knowing more of the variables which could easily change the results by 1000 times.

Yea, I know what you are meaning, but I have asked my main problems at post #4.

Thanks

 

[MrAl]

Hi again,


The ringing is caused by the opening of the switch, at least for a practical coil. It is caused because the inductor and it's equivalent parallel capacitance makes up an oscillator. Thus, the circuit oscillates at that frequency instead of the applied frequency. Thus happens unless the switch opens at a point where the excitation current happens to be zero, and then there is no response for that particular opening of the switch.

 

[rumpfy]

zes,
let me ask one question;
if a sinusoidal voltage wave is given by e = E(max) sin wt, can you say what is the first derivative (de/dt) of this function?.
This is not a trick question but an attempt to find out how far an explanation for you can go.

 

[zesla]

Hi again,


The ringing is caused by the opening of the switch, at least for a practical coil. It is caused because the inductor and it's equivalent parallel capacitance makes up an oscillator. Thus, the circuit oscillates at that frequency instead of the applied frequency. Thus happens unless the switch opens at a point where the excitation current happens to be zero, and then there is no response for that particular opening of the switch.

But what is ringing freq really? Is it the impulse response seen across a coil or the connected key when opening key ?

thanks

 

[zesla]

zes,
let me ask one question;
if a sinusoidal voltage wave is given by e = E(max) sin wt, can you say what is the first derivative (de/dt) of this function?.
This is not a trick question but an attempt to find out how far an explanation for you can go.

Yea i am familiar with derivation:
It would be:
e'= E (max) W cos Wt

So please kindly use both math and explanation to solve this problem (post #4) I am dealing with for years

 

[MrAl]

But what is ringing freq really? Is it the impulse response seen across a coil or the connected key when opening key ?

thanks

Hi,


If i understand your question correctly, the ringing comes from the step response of the inductor in parallel with the capacitance (and some damping resistance present in almost all circuits). It's the step response of the three passive elements in response to the inductors current flow at the time the switch is opened. It will be a damped sinusoidal and will last until it damps out completely or the switch is closed again. There may be energy in the capacitance too which affects the waveform as well, but it will most likely be much smaller and less significant than the inductor energy. There may also be a time when the inductor current is zero when the switch opens and possibly a small amount of energy in the capacitance. In this case the response would be smaller because of the lower initial energy.

The way to completely evaluate this is to analyze the circuit with an initial current in the inductor and an initial voltage across the capacitance, or skip the cap voltage and just estimate based on the initial current in the inductor.

 

[zesla]

MrAl.
Thanks but before responce to your last post, plz let me know why the alternating volatge is not acting almost like a key so that incrase the EMF of an inductor? In post #4 I told that have a problem with this, furthermore I want to see the affect of the rate of the source freq on the amplitude of the EMF produced by an inductor.

Thanks.

 

[MrAl]

Hi,

Then what i suggest that you do next is turn your attention to an small network with an inductor, a parallel capacitor, and small series resistor. This will form a tuned circuit. As the excitation signal comes closer to the natural resonance of the LCR circuit, you'll see a larger response from the network. This is without a switch in the circuit too.

Once you see how that works then you might want to add the switch back into the circuit. And dont forget that you can not do this without the parallel capacitance because without that you just have a pure inductance and that would just draw more current at lower frequencies and you would not be able to get a clear picture of what happens when the switch is opened because again that's not possible without considering the other circuit elements.

 

[rumpfy]

Zes,
Kirchoff says that the sum of emf's around a circuit loop will be zero.
For your circuit, the terminal voltage of your coil will be equal to the supply voltage when the key is closed.
From Lenz law e =-l di/dt.
If the supply voltage is a constant amplitude, the the coil voltage will be constant and the rate of rise of current will be constant.
Note that Lenz law has a minus sign. This sign says that the voltage produce in the coil is of the opposite polarity to that of the voltage which caused it. Doing a Kirchoff shows that the sum of the voltage around the loop is zero.
So, while the switch is closed, there is a constant rate of change of current through the coil. That is, the current increases linearly with time, and continues to increase, and in your case, because there is no resistance, the current increases to infinity.
If at some point, the switch is opened, then the rate of change of current becomes infinite (dI/0 = infinity). The voltage across the coil becomes infinitely large.
A change to your circuit to short circuit the coil when the switch opens, will result in the terminal voltage of the coil becoming zero. In the is case, the rate of change of current will be zero; that is, the current will be maintained. So, with a perfect coil, once a current is flowing through a coil, by short circuiting its terminals, the current will be maintained forever. This principle is used in magnetic memory devices which operate under superconducting conditions.
In the above, we used a step voltage to drive the coil. If we used a source of alternating voltage, the with the switch closed, the coil terminal voltage would again be equal to the source voltage. So, for e = E sin wt, the coil voltage would be E sin wt = - L * di/dt . So dI/dt = (E sin wt)/L. Integrating gives; I = - (E/(L*w)) * cos wt . Since cos wt = sin(90-wt), the we can say that; I=(E/(L*w)) * sin(wt-90).
Thus, with a sine wave source, the current lags the voltage by 90 degree.
You can substitute your constants for L and wt and calculate current.
Part of your question is about transformers. In a perfect transformer, the transformer simply transforms the load side impedance into a different impedance at the primary side. The transformer is only an impedance changing device. to deal with a transformer, change the load impedance by the ratio (N1/N2)squared.
In practical circuits, the transformer has leakage inductance between the primary and secondary and this leakage inductance is used for timing effects and other application requirements. For example, the Kettering spark ignition coil has a loosely coupled secondary winding. When the coil current is interrupted, the primary voltage rises to say 250 volt. the turns ratio of the coil then multiplies this voltage to say 20,00 volts and generates the spark.if the primary and secondary were closely coupled with no leakage inductance, the primary voltage would rise to only 24 volt or so and the turns ratio of the coil would need to be greater to get a spark. Note on the newer cars that the coil structure looks different. These modern coils are designed to have closer coupling than the old Kettering types.
Hope this helps.