Basics question of Inductor to powering high voltage lamp

[Jay_]

I was trying to study the inductor role in Flyback converters and all, they seem to rely on the principle that the inductor is first energized while the switch is closed and then, it acts as a voltage source itself when the switch is opened.

All that is good, but I am confused with the POLARITY.

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE5803

The simulations here seem to show all the current directions from negative to positive. Are they talking about the flow of electrons?

Either ways, I don't see when the inductor gets 65volts across it to power the neon lamp.

This is what I understood: First the inductor has a voltage of 10 when the switch is closed with A, then it goes reducing to zero. After the switch is closed with B, the voltage is formed across the inductor to power the neon-lamp.

But all the currents shown are negative to positive?

Can someone clear me on this? I am aware of the equations V = L(di/dt). So when the constant current becomes zero, voltage is there across the inductor, right? But the polarity?

 

[sophiecentaur]

Assume a positive supply voltage, a series L, a switch and a load connected to ground.
When you (try to) break the circuit, the voltage across the inductor will change from a small forward voltage (due to the finite resistance of the winding) to a high reverse voltage. This high, inverse voltage will be such as to maintain the current that is flowing out of the inductor into the circuit it feeds. It will provide a (positive) voltage at the switch terminal that is much higher than the original value. If a high voltage neon has been connected between L and ground, the voltage will then be enough to strike the neon. The more current the load takes, the higher the voltage when the switch breaks the circuit - which you said when referring to Ldi/dt

 

[Jay_]

When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?

In the page of the time constant, it has 10 V just when the switch is closed. Then it goes reducing till zero as time passes.

After that the next page has these two situations (shown in the attachment). It seems to shift its polarity once the connection is made. Even after that it seems to be showing current going from negative to positive!

So which is the correct polarity of voltage at the inductor, and what about the current direction?

 

[sophiecentaur]

Your circuit is not the one I was discussing. 'My' switch and neon are the other side of the inductor, so my wording was not appropriate.
In your circuit, in order to maintain the same current through the resistor, when the switch is opened, the volts at the 'top' of the inductor will, as you show, go negative with a large amplitude.
Could your confusion about the direction of the current be to do with the fact that the inductor is an energy source (like a battery) whilst the voltage spike is being formed. For current to flow out of it, through the resistor, the potential that end must be maintained and this is achieved by virtue of the emf, generated in the inductor. When the battery is connected, the negative end of the battery is connected to ground. Whilst the inductor is functioning, the ground connection is replaced by the neon , which will only strike when there is 65V across it. Point B will be 65V negative for this to happen. (total volts on inductor will be 75V - note that on my circuit, you only need 55V across the inductor as there is already 10V from the supply)

You also mention the timing of all this. There will be finite capacitance in the circuit (mostly the self capacitance of the coil) and this will introduce a time constant which will cause a finite time for the voltage reversal. If the neon were replaced with a capacitor, you would have a series of oscillations as the capacitor discharges back through the inductor - and so on - until the energy is dissipated in the load resistor.

 

[NascentOxygen]

When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?

The high voltage develops the instant that the current begins to change significantly. Because when current changes rapidly, di/dt is large and there's the large voltage you are asking about.

Let's simplify things even more, using just a battery and the inductor. You connect the battery across the inductor, and current flows, small at first but increasing with time. You then decide to disconnect the battery. Instant removal of the battery would cause the current to instantly fall to zero. But that means di/dt would be infinite, so an infinite voltage would appear across the inductor terminals. An infinite (or, at least, very high) voltage here would jump across the gap of air between the inductor terminals and the battery you are valiantly trying to drag off the inductor, this voltage revealing itself as a blue spark. And if you do the experiment, that's exactly what you'll see. The faster you try to remove the battery from the inductor, the bigger the spark that results.

The spark represents current flow, meaning the current you tried to interrupt and cause to drop to zero is actually not instantly falling to zero — some current manages to continue flowing by jumping across the air gap. Therefore, di/dt is not infinite (and negative), but it's still a large value.

If you have a neon connected across the inductor terminals, current will preferentially flow through the ionised neon gas rather than jump through air.

 

[sophiecentaur]

Here's a question for 'the student' - What happens if there is no spark or 'neon'? (If the insulation is good enough and there clearly can't be infinite voltage.)

 

[NascentOxygen]

Well, this student says that if that's the case then you haven't rapidly interrupted the current, but rather gradually/steadily wound it back to zero.

 

[sophiecentaur]

What was it that limited how fast you could 'wind it back'. ""

 

[NascentOxygen]

What was it that limited how fast you could 'wind it back'. ""

Apparently the lab that day had issued me with a slow switch. The mechanical separation of its contacts was slow (and, indeed, smooth and gentle) enough that di/dt was limited during the separation process to a value that induced no voltage large enough to ionize any air across the separating points.

 

[sophiecentaur]

No one likes a smartypants.

 

[NascentOxygen]

No one likes a smartypants.

Oh, come on now. Enough of that talk! I'm sure lots of people like you.

P.S. I didn't say I believed the instructor's claim of the switch generating no spark.

 

[Jay_]

sophiecentaur

My confusion lies with the direction of current.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.

The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction

 

[davenn]

sophiecentaur

My confusion lies with the direction of current.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.

The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction

The confusion is that you are using conventional current flow positive to negative
where I suspect sophicentaur is using electron current flow which is negative to positive

so when its negative at the top of the inductor, current flow is anticlockwise from top of inductor down through neon through resistor and back to inductor

hopefully I haven't misread that ;)

Dave

 

[NascentOxygen]

My confusion lies with the direction of current.

With the switch set to position A, current flows through the inductor from top to bottom. As expected, the top is + because it's connected to the battery + terminal. We agree on this.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.

So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!

What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.

Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.

 

[davenn]

Have read a number of times and trying to make heads and tails of it ??

...
When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

is not this delayed current flow still part of that which flowed from the battery ?
if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?
therefore you are not going to have current flowing through the neon as the voltage isn't high enough to strike/light the neon ?

Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.

So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!

What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.

Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.

isnt the voltage/current spike going to be caused by the collapse of the magnetic field
and its going to flow in the opposite direction ?
its that spike that's going to be high enough to strike/light the neon ?

just want to clarify this for myself too :)

Dave

 

[NascentOxygen]

The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.

 

[davenn]

The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.

not sure who or what particular comment that's aimed at ??

I agree, and we do know the history :) so not sure what point you are making ?

Dave

 

[NascentOxygen]

is not this delayed current flow still part of that which flowed from the battery ?

Um, delayed[/color] current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.

if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?

Yes, inductor current never changes instantly, for to do so would require infinite voltage.

therefore you are not going to have current flowing through the neon as the voltage isn't high enough to strike/light the neon ?

I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.

isnt the voltage/current spike going to be caused by the collapse of the magnetic field

Yes.

and its going to flow in the opposite direction ?

Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.

its that spike that's going to be high enough to strike/light the neon ?

Yes.

 

[NascentOxygen]

not sure who or what particular comment that's aimed at ??

That was a follow on to what I'd written. I hadn't noticed your post.

 

[Jay_]

I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?

Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

 

[sophiecentaur]

I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?

Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

Firstly, what does the table in the diagram refer to? I don't think the heading "time constant" means what it says. Is it supposed to be 'time'? Could you explain what it all means? This looks more like a h/w question which supplies the information that you need to solve it. Are you expected to plot a graph of the volts across the L and extrapolate to the high volts you need (75V) after the graph has crossed the Y axis and intercepts -75. It's a strange circuit because it requires more volts from the L than necessary - if it were connected as I described a few days ago - you'd only have to get to 55V. But the question setter was after an answer from what he gave you, I suppose. Plot the graph and see what you get. When you are given data which makes no sense to you, a graph can often help.

To work out the "when" question (precisely), we need to know more than the information given. As I mentioned way back, the rate of change of current will be due to the Inductance value and also the parasitic Capacitance involved. You basically have a resonant circuit which has current flowing through it and, ignoring any other paths in the circuit, when you open the switch, the current in the Inductor will divert into charging the capacitance and the resonant circuit will ring at a frequency given by 1/2∏√(LC). There will be losses which will cause the oscillation to decay, of course and which will limit the peak voltage. You don't know those losses so you don't know when the volts will reach the 75V you need for your neon to strike. Neither do you know the value of the parasitic C but you could work out what you want if you knew (or could measure) the R and C values.
If you wanted to, you could put a capacitor across the L which would swamp the parasitic one and then you would have an idea of the (much lower resulting) resonant frequency and that would tell you how long before the peak was reached.
But all this may not be relevant to the answer you want. If you want a more useful answer to homework problems then I suggest you make it more clear what it's all about and the post in the right forum - where you won't find people batting your problem about in quite the same way. You can see a "no homework" message at the top of the list of topics in each forum. It tells you what to do.

 

[davenn]

Um, delayed[/color] current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.

poor choice of words ... was referring to the brief period of time the current still flows

Yes, inductor current never changes instantly, for to do so would require infinite voltage.

OK can accept that

I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.

No I'm asking you to clarify as I cannot follow your expalanation :)

there is NO closed loop, its an open cct at the neon, but you said the current is flowing up through the neon...

When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

So if that "residual" current is still flowing, and we know its not enough to strike/light the neon so as said I believe its an open cct.

I want you to tell me how you have current flowing in an open cct please :)

Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.

no I am not, we have already established that

Dave

 

[Windadct]

The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.

I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.

 

[sophiecentaur]

The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.

I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.

The rate of change of current is limited, ultimately, by the stray Capacitance.
But the data (table) given in the question is what will yield the answer to it. All the rest is arm waving because we do not know enough of the details about the actual circuit. The slope of the graph gives the rate of change of the volts - i.e. it will give you the rise time of the pulse.

 

[NascentOxygen]

I am lost in all this discussion of so many posts.

I'm sorry to hear that.

Also, at the various [strike]time constants[/color][/strike][/color] times[/color] the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

The inductor gets 65V across it at no time during the intervals you tabulated. The switch is obviously fixed in position A during all of that time.

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

The neon gets a high voltage the very instant the switch is moved out of position A. The voltage across the neon will increase to whatever it takes to strike the neon. The 65V here is obviously a characteristic of that particular manufacturer's product. If the neon happened to not be present, then I would expect to see a blue spark [/color][/size] leap through the shortest path of air between the electrical ends of the coil.

 

[NascentOxygen]

So if that "residual" current is still flowing, and we know its not enough to strike/light the neon so as said I believe its an open cct.

Current always flows in a closed loop. The instant that it ceases to flow from the battery, it begins to flow in the neon, so there is no moment when there is not a closed path for inductor current.

I don't believe you can claim "we know its not enough to strike/light the neon". Basically, it definitely is enough, because if it isn't then current would instantly stop, and if current in an inductor stops flowing di/dt would be infinity and an infinite voltage would develop. But before it hits infinity the neon strikes, current flows, and di/dt is not infinity, so the infinite voltage never develops and during that crucial time when the switch opens inductor current never ceases.

I want you to tell me how you have current flowing in an open cct please :)

There is no current flowing in an open circuit. There is no open circuit.

 

[sophiecentaur]

I'm sorry to hear that.

The inductor gets 65V across it at no time during the intervals you tabulated. The switch is obviously fixed in position A during all of that time.

The neon gets a high voltage the very instant the switch is moved out of position A. The voltage across the neon will increase to whatever it takes to strike the neon. The 65V here is obviously a characteristic of that particular manufacturer's product. If the neon happened to not be present, then I would expect to see a blue spark [/color][/size] leap through the shortest path of air between the electrical ends of the coil.

Read the question with the attached image. Do you not think that the exercise is to draw a graph of the given data and to extrapolate to the 65V? This isn't a practical problem at all - or even much of a theoretical one. If it were theoretical then there would be actual values of inductance etc. given.
All this chat we have been giving is overkill and not relevant to the real question.

 

[sophiecentaur]

Current always flows in a closed loop. The instant that it ceases to flow from the battery, it begins to flow in the neon, so there is no moment when there is not a closed path for inductor current.

I don't believe you can claim "we know its not enough to strike/light the neon". Basically, it definitely is enough, because if it isn't then current would instantly stop, and if current in an inductor stops flowing di/dt would be infinity and an infinite voltage would develop. But before it hits infinity the neon strikes, current flows, and di/dt is not infinity, so the infinite voltage never develops and during that crucial time when the switch opens inductor current never ceases.

There is no current flowing in an open circuit. There is no open circuit.

This is not true. With a perfect switch, there is still some Capacity in the circuit, into which charge will flow as the volts increase. Only when the Neon has struck, can current flow in it. There is always a Delay, due to the LC time constant. Please don't confuse the OP by making unfounded assertions and over-simplifying the situation. He asked for the time involved and, as I have already made clear, the data can be used to find that time. It's a college exercise.

 

[NascentOxygen]

Read the question with the attached image. Do you not think that the exercise is to draw a graph of the given data and to extrapolate to the 65V?

The instructional video? It was followed by 3 questions, and I got all 3 correct. I never saw anything about extrapolating to 65V. Sorry, have no idea what you are referring to.

This exercise must be like poetry, it brings different things to different people!

This isn't a practical problem at all - or even much of a theoretical one. If it were theoretical then there would be actual values of inductance etc. given.

The circuit is not practical. The switch shouldn't switch in the neon, the neon must be kept across the coil at all times or the switch contacts will glow brighter than the neon. But the student wouldn't know that, so picturing the switch as an ideal switch serves the purpose.

 

[NascentOxygen]

There is no capacitance shown in the schematic, so I relegated it to secondary effects. You can picture as much or as little capacitance as you choose. A perfect switch has no capacitance, and the same goes for the rest of the circuit. If capacitance was to be a consideration, I would expect it to be shown or dotted in or alluded to in the accompanying text.

The question I first addressed concerned the polarity of the flyback voltage. This is a very basic question.

 

[sophiecentaur]

There is no capacitance shown in the schematic, so I relegated it to secondary effects. You can picture as much or as little capacitance as you choose. A perfect switch has no capacitance, and the same goes for the rest of the circuit. If capacitance was to be a consideration, I would expect it to be shown or dotted in or alluded to in the accompanying text.

The question I first addressed concerned the polarity of the flyback voltage. This is a very basic question.

The polarity question is straightforward, of course. The timing is another matter. It can either be calculated by knowing a lot more information than is supplied OR by using the numerical data supplied.
I just went to the trouble of plotting the data and is clearly is a plot of the volts when the switch is Closed and shows the exponential decay in current build up . The ratio of those volt readings, as a fraction of the next, is about 2.7, throughout the period - so it really does have the time intervals in units of time constant.
Problem is that that time constant has nothing to do with the time constant involved when the switch is opened. One thing I was right about was the advice to draw a graph before trying to come to conclusions. - but I did hope the OP would have done that.
Even more , I have just found that (power point) link and read it in full. Quantitative data is only shown for the switch on situation and they are totally non-committal about the timing involved between when the switch is opened and the neon strikes. They are not actually saying 'instantaneous' and nor are they saying what causes delay or how to work it out.A perfect switch has no capacitance and will instantly (on a timescale compared with other effects in a practical / inductive circuit) go to infinite resistance. Any inductor will have self capacitance and it is this that introduces a finite delay and subsequent 'ringing'. This can always be seen on relay windings, driven by a transistor switch, for instance. After the switch goes open circuit, NO current can flow 'around the circuit' but charge flows into the self capacitance of the coil (or anywhere else it can find; the switch capacitance is unlikely to be more than a very few pF, however). The time constant involved is much much shorter than the time constant for energising the coil, though. Any simpler treatment of the topic has to involve the words 'instantaneous' and 'infinite' - which are not very helpful. Neither is that PP presentation very helpful because it peters out without giving any quantitative help about what actually makes the neon strike.

Annoyingly, they quote a conventional car ignition system as an example of this effect. We all know that, without a suitable 'condenser' in place, it just won't work properly. Same goes for a Rumkorf Induction coil.

 

[davenn]

I really don't understand NascentOxygen's responses

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

2)switch opens from A and closes to B
He's saying there is still current from the battery that's flowing through the coil for a brief time

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?
So WHEN does the neon strike ?

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...
b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?Dave

 

[Jay_]

Sophiecentaur

"Firstly, what does the table in the diagram refer to? I don't think the heading "time constant" means what it says. Is it supposed to be 'time'? Could you explain what it all means?"

The time constant of an inductor is T = L/R. So it basically shows the values of voltage across the inductor at T, 2T, 3T etc. At the 0T (beginning), it says the value of voltage across the inductor is the highest = 10 V (battery voltage). After that point it keep reducing.

Now when the switch is moved from A to B, there are two immediate scenarios.

1. Switch is between A and B, not connected to either.

2. Switch is closed with B.

What exactly happens to the voltage across the inductor in these moments and why?

"You don't know those losses so you don't know when the volts will reach the 75V you need for your neon to strike. Neither do you know the value of the parasitic C but you could work out what you want if you knew (or could measure) the R and C values."

So that means this circuit makes no sense without a capacitor? I can understand that by V = L(di/dt), the quick switching from A to B will cause the top of the inductor to have negative voltage. Could you refer to the diagram I attached (again), and tell me why current won't flow in the manner shown?

If a HIGH voltage is suddenly induced when switch closes with B, what makes us control this value? How do we know the switching won't cause a current of 100 V instead of 65V to damage the neon?

" If you want a more useful answer to homework problems then I suggest you make it more clear what it's all about and the post in the right forum"

I just want to know how an inductor works with DC source! Thats all sir! I know it causes an impedance to an AC voltage, but its applications with a DC source is not something I understand which is why I thought I'd ask here.

But all these posts are themselves confusing, if someone (just one person!) could explain the circuit operation step wise (this happens first, this happens second, this happens third etc) I would get a better understanding. Thanks.

 

[Jay_]

NascentOxygen

"The inductor gets 65V across it at no time during the intervals you tabulated. The switch is obviously fixed in position A during all of that time."

Okay, that is understood the voltage reduces gradually until its zero (like a short wire) across the inductor. But what controls the development of voltage across the inductor to be 65V? Also, what about the direction of the current. From the link it seems like the direction is opposite to what I have shown (with black arrow in Paint), but current always goes from positive to negative. I know about the electron flow and all, but by convention its positive to negative, why would we make it different here?

 

[Jay_]

There is something else in the link.

When the switch is at between A and B ('floating'), the inductor has positive polarity on "top". The moment the switch closes with B, it gets a negative polarity on "top".

Is this how it happens? Again, if someone could explain what exactly controls the voltage developed on the inductor (at a particular value like 65 V) and what exactly is the current direction, I could get an understanding and end this thread which has been pulled way longer that I would have thought it to go.

 

[NascentOxygen]

I really don't understand NascentOxygen's responses

Then let's call him back and demand that he explain himself! http://imageshack.us/a/img10/9783/whiponionheademoticon.gif


First, the question seems to come from Circuit Principles 101, and concerns a simple first order system. So it's not meant to invoke a treatise on inductive non-idealities. Further, the circuit appears to have been dreamt up by someone who himself doesn't have a good grasp of inductors. Well, that's my conclusion, because had he constructed the circuit he would have discovered that it has not a hope in hell of doing what he obviously expects it to.

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

Not much to argue about there.

2)switch opens from A and closes to B

Ha! There's the problem. And it's a big problem. Unless the switch goes from A to B without opening the circuit for even one-squillionth of a second, then the inductor will dump its energy inside the switch and the party is over and we may as well all go home. (Remember, -L.di/dt. We must not interrupt the inductor current.)

So I gave the author the benefit of the doubt by assuming that he really means for the neon to be connected across the inductor at all times, and for the switch to do nothing more than disconnect the battery. Fair enough? Because if we don't assume this then there is not much left for students to analyze.

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?

Why would an ideal neon have significant capacitance? The question doesn't say to consider it is a capacitor as well as a neon, so why complicate it? Let's use a tiny ideal neon globe, it has no capacitance.

So WHEN does the neon strike ?

We are using an ideal neon. It [apparently, since the question doesn't say otherwise] strikes at practically its holding voltage, so we have no need to complicate the issue by separately considering the striking and the holding voltages.

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...

It couldn't, for a couple of reasons. One I pointed out above. Another rather major one being that the neon is not even in the circuit until the switch is safely in position B! (But I'm confident that students of Circuit Principles 101 would not notice there's an issue with the switching, so I wouldn't anticipate they would baulk here.)

b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?

That. (But I'm uneasy with your emphasis on reversed current flow. I thought we agreed that the inductor current does not reverse? Ever — this being a first order system. No current in any element here reverses.)

Hope that has cleared some things up.

 

[NascentOxygen]

what controls the development of voltage across the inductor to be 65V?

That is wholly determined by the characteristic of the neon. We can infer from the data that the author of the question is using a neon which maintains 65V across its terminals regardless of the current though it. Were you to replace it with a 95V neon, then the inductor would increase its voltage to 95V, at which point the neon would prevent the inductor from increasing the voltage further. The inductor's voltage just keeps increasing until current punches through the neon.

Also, what about the direction of the current. From the link it seems like the direction is opposite to what I have shown (with black arrow in Paint), but current always goes from positive to negative.

In wires and resistors, yes. But that does not apply to inductors. The current (magnitude and direction) in an inductor is unrelated to the voltage at that moment. You know v=L.di/dt , so it's the derivative of the current that determines an inductor's terminal voltage.
Or, expressing current in terms of voltage, i(t) = (1/L)⋅_∫[/size]v(t).dt

 

[sophiecentaur]

I really don't understand NascentOxygen's responses

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

2)switch opens from A and closes to B
He's saying there is still current from the battery that's flowing through the coil for a brief time

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?
So WHEN does the neon strike ?

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...
b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?


Dave

There is no certain answer to this. The only data available refers to energising the coil and not discharging it.
But you can make an educated guess about it. Assume that the demo needs to be impressive, so let's assume you have 1J stored. The energy stored is given by
W = l²L/2
With 10A flowing, that requires an L of 20mH.
Including the capacitances of the switch, neon and self capacity of the coil (the majority), would give a total C of around, say 100pF.
Two reactances will give a resonant frequency of
f =1/2π√LC
which gives around 100MHz and this means that the neon would strike at or before the first peak of oscillation, which would mean a time of around 2.5ns. I can't help thinking that the self capacitance would be greater than this - which would make the time correspondingly longer.
Once the neon has struck, everything changes, of course.

The value of the original R would be 1Ω. (10A 10V) and the time constant for this and the 20mH inductor would be L/R, which = 20ms.
Interesting to compare the two times involved - even if they are only back of fag packet calculations. Feel free to find gross errors in my arithmetic.

 

[NascentOxygen]

There is something else in the link.

Which link would that be?

When the switch is at between A and B ('floating'), the inductor has positive polarity on "top". The moment the switch closes with B, it gets a negative polarity on "top".

My contention is that, had the author here intended that the top of the inductor spend any time floating, even for the most fleeting of moments, then his schematic would clearly show that. The way to show this would be a 3 position switch, one having an intermediate position interposed between A and B, let's designate it position N. In position N the top of the inductor would be connected nowhere. The switching would be from position A, to position N, then to position B. Such a switching sequence is not shown, so I can only conclude that in the operation he envisaged it doesn't happen that way.

Is this how it happens? Again, if someone could explain what exactly controls the voltage developed on the inductor (at a particular value like 65 V)

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

and end this thread which has been pulled way longer that I would have thought it to go.

Oh, I guess you must be new around here? Welcome to Physics Forums!

You'll always get your answer, in the end. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon10.gif

 

[sophiecentaur]

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

Oh, I guess you must be new around here? Welcome to Physics Forums!

You'll always get your answer, in the end. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon10.gif

I agree that the switch action needs to be instant - a justifiable approximation. This is why I said the switch doesn't need to be a changeover and the neon could just as easily be permanently wired to make the analysis easier.

It's a 1Ω resistor! (10V 10A) and will obviously act as an energy loss mechanism but it will hardly affect the rise time of the first cycle of oscillation. It's the neon that dissipates the energy significantly - being the higher resistance - as W = I²R and its on resistance will be a fair bit higher higher than 1Ω. So it will limit its own current and damp the resonance once it has struck
I just wish somebody would acknowledge that the Capacitance of the circuit is vital if you want a quantitative clue about its operation and that the coil has the most C. Presumably at least some readers have been involved with some practical effects of self resonance of inductors.

 

[Jay_]

NascentOxygen

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

Okay. If the resistor didn't exist, would it keep the neon from preventing the inductor to increase its voltage at the neon's rated value, like you mentioned? If so, what is the value of this resistance?

That is wholly determined by the characteristic of the neon. We can infer from the data that the author of the question is using a neon which maintains 65V across its terminals regardless of the current though it. Were you to replace it with a 95V neon, then the inductor would increase its voltage to 95V, at which point the neon would prevent the inductor from increasing the voltage further. The inductor's voltage just keeps increasing until current punches through the neon

So if my neon was like 1000 V and required a lot of current to punch through it. In that case would the inductor's voltage keep increasing till it punches through the 1000 V neon? Where exactly would the inductor get all this energy from?

----

The link I was mentioning is the one at the OP. In that clearly, at the N position (as you have described), the voltage at the top of the inductor is shown as positive.

----

Question on switching

Does the voltage developed across the inductor depend on how quickly we switch from point A to B? Or is it still as NascentOxygen says keep increasing till the neon prevents it from increasing?

 

[Jony130]

I strongly advise you to read this
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF (from page 22 Understanding the Inductor)


sample

Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’
So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps into try to keep the current down to its initial value(zero).
So we apply ‘Kirchhoff’s voltage law’ to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm’s law).
As time progresses, we can think intuitively in terms of the applied voltage “winning.” This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff’s voltage law).
That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.
Why does the applied voltage “win”? For a moment, let’s suppose it didn’t. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other — and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.
We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday’s/Lenz’s law). And this “allows” for the additional drop appearing across the resistor, as per Kirchhoff’s voltage law!

 

[sophiecentaur]

That quotation, above, is dodgy because Kirchoff 2 doesn't apply for non conservative fields. Also, phrases like
"The induced voltage now steps in to try to keep the current down " are anthropomorphic and don't help a proper understanding.

The only way to predict what will happen in any circuit is to write the differential equation out and solve it! There is no 'explanation' that ignores all relevant components and parasitic components - including the dominant resistances and capacitances. Any arm waving description involves a 'bootstrapping' argument which is not very useful.

 

[Jay_]

Jony130

It says the page cannot be displayed.

Sophiecentaur

But Kirchoff's laws are used in RLC circuits too - with the laplace taken and then, converted back to time domain form in the end right?

----------

Another thing I don't understand - isn't an inductor as good as a short circuit when dealing with a DC source. So why would there even be 10V across the coil in the beginning, shouldn't it just be zero? Or does it go to say that the DC source in reality is actually passes a zero current first and then its value increases and so a di/dt exists?

 

[Jony130]

Try this link
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF

And inductor in DC circuit act just like a piece of wire (short) in steady state condition.
Not in the transient condition. So after the time equal to t = 5*R/L . The coil acting more like a short circuit.

And before we proceed to AC analysis kept in mind that when you connect the coil to the DC voltage source, at first step the voltage across the coil reach its maximum value equal to Vsource, but the current is equal zero amps. And when current in the coil reach his maximum value the voltage across the coil reach zero volts.
So we can say that current in the coil lags the voltage.

This imagine will help you understand behavior of the coil using this equation:

V = L * dI/dt

when we replace DC source with AC source.

So as you can see "the larger" the rate of change of a AC current (AC signal Frequency).
The "larger" the voltage drop across the coil. So less voltage is seen by the load.

 

[sophiecentaur]

Jony130

It says the page cannot be displayed.

Sophiecentaur

But Kirchoff's laws are used in RLC circuits too - with the laplace taken and then, converted back to time domain form in the end right?

----------

Another thing I don't understand - isn't an inductor as good as a short circuit when dealing with a DC source. So why would there even be 10V across the coil in the beginning, shouldn't it just be zero? Or does it go to say that the DC source in reality is actually passes a zero current first and then its value increases and so a di/dt exists?

Firstly, look at this link, which will tell you why K2 is not always applicable. I think this applies in your inductor case.

The data given tells you there is zero volt drop across the L when there is 10V across the R.

 

[NascentOxygen]

Okay. If the resistor didn't exist, would it keep the neon from preventing the inductor to increase its voltage at the neon's rated value, like you mentioned? If so, what is the value of this resistance?

Some series resistance must be present, otherwise the inductor will dump all of its energy in a brilliant flash and likely exceed the neon's rating.

So if my neon was like 1000 V and required a lot of current to punch through it. In that case would the inductor's voltage keep increasing till it punches through the 1000 V neon? Where exactly would the inductor get all this energy from?

Immediately the battery is disconnected, the inductor current must find another route. If it were to find no new route, then current in the inductor would cease instantly, making di/dt = infinity, so an infinite voltage would appear across the inductor. But, as you'd expect, any neon will begin conducting well before the voltage reaches near infinity. (If the neon was not included, then you could expect to hear a high voltage discharge sizzle through the air around the inductor.)

The energy stored in the inductor is the energy that the battery delivered to it while the switch was in position A for current to build up in the coil. This stored energy has a value = ½LI², where I is the value of current at the moment the battery is disconnected.

The link I was mentioning is the one at the OP. In that clearly, at the N position (as you have described), the voltage at the top of the inductor is shown as positive.

All I see is position A and position B. It's a 2 position switch.

Does the voltage developed across the inductor depend on how quickly we switch from point A to B?

For the situation I'm discussing, the switch instantly jumps from A to B. It is either in position A, or it's in position B. There can be no intermediate state for this exercise.

Or is it still as NascentOxygen says keep increasing till the neon prevents it from increasing?

That.

 

[sophiecentaur]

@NascentOxygen
Unless you are prepared to use some actual values (other than zero) for the other components in a real circuit then the circuit is not real. This is what I have been attempting to do, in order to answer the OP question of "When?"
That 1Ω resistor is fine, although it would be a bit low if anyone wanted to do an actual experiment / demo. Would you care to choose a different value of energy stored from the 1J I chose in my example? Would you also care to suggest a different value of total parasitic C? There would be an answer for your choices too. You do realize (?) that, without some Capacitance, the circuit is totally unreal and will produce an infinite, instantaneous leap of volts - clearly nonsense.
You can't skirt round the issue of possible component values. There's no point in just talking vaguely about a neon tube. Neons, along with every other components, have VI characteristics and there are examples available in spec sheets. Once it has struck, a real neon will have a resistance value in the order of tens of Ohms (it's not a dead short). Before it has struck, it is a pretty good open circuit.
We have to stop arm waving and start to do something more constructive, with some values assigned to components, if we want proper answers.

 

[Jay_]

Firstly, look at this link, which will tell you why K2 is not always applicable. I think this applies in your inductor case.

The data given tells you there is zero volt drop across the L when there is 10V across the R.

Thanks. That was revealing. But consider this: If I connect both my voltmeter probes at point "d", I am measuring Vd - Vd, which means I am measuring the potential of one point with respect to the very same point.

Logically, a place can have just one potential, yet the difference being non-zero tells me that it has two values?! That's like an object existing in two places wouldn't it?

 

[Jay_]

Jony130

I got the pdf, I will give it a read soon. Thanks :)

NascentOxygen

So let me see if I got this right. If V = L(di/dt) and the switching is instant, so the inductor will try to reach infinite voltage. But the neon gives the voltage a path to flow when it reaches 65, and so the voltage build up on the inductor stops? Would that be right?

"Some series resistance must be present, otherwise the inductor will dump all of its energy in a brilliant flash and likely exceed the neon's rating."

Even a coil has some resistance right? So what I was asking is: Is there a particular resistance value (a minimum) for this condition of "dumping all its energy in a brilliant flash" to not happen?