# Basics question of Inductor to powering high voltage lamp

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
Gold Member
2020 Award
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

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?

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sophiecentaur
Gold Member
2020 Award
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
Staff Emeritus
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
Gold Member
2020 Award
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
Staff Emeritus
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
Gold Member
2020 Award
What was it that limited how fast you could 'wind it back'. ""

NascentOxygen
Staff Emeritus
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
Gold Member
2020 Award
No one likes a smartypants.

NascentOxygen
Staff Emeritus
No one likes a smartypants.
Oh, come on now. Enough of that talk! I'm sure lots of people like you. :tongue:

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

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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

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davenn
Gold Member
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 havent misread that ;)

Dave

NascentOxygen
Staff Emeritus
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
Gold Member
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 isnt 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 thats going to be high enough to strike/light the neon ?

just wanna clarify this for myself too :)

Dave

NascentOxygen
Staff Emeritus
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
Gold Member
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 thats aimed at ??

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

Dave

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NascentOxygen
Staff Emeritus
is not this delayed current flow still part of that which flowed from the battery ?
Um, delayed 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 isnt 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 thats going to be high enough to strike/light the neon ?
Yes.

NascentOxygen
Staff Emeritus
not sure who or what particular comment thats aimed at ??

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?

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sophiecentaur
Gold Member
2020 Award
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
Gold Member
Um, delayed 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.

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 Im not, we have already established that

Dave

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
Gold Member
2020 Award
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
Staff Emeritus