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Basics question of Inductor to powering high voltage lamp

  1. Jan 27, 2013 #1
    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?
     
  2. jcsd
  3. Jan 27, 2013 #2

    sophiecentaur

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    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
     
  4. Jan 28, 2013 #3
    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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  5. Jan 29, 2013 #4

    sophiecentaur

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    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.
     
  6. Jan 29, 2013 #5

    NascentOxygen

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    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.
     
  7. Jan 29, 2013 #6

    sophiecentaur

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    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.)
     
  8. Jan 29, 2013 #7

    NascentOxygen

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    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. :wink:
     
  9. Jan 29, 2013 #8

    sophiecentaur

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    What was it that limited how fast you could 'wind it back'. ":wink:"
     
  10. Jan 29, 2013 #9

    NascentOxygen

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    Apparently the lab that day had issued me with a slow switch. :approve: 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.
     
  11. Jan 29, 2013 #10

    sophiecentaur

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    No one likes a smartypants. :smile:
     
  12. Jan 29, 2013 #11

    NascentOxygen

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    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. :wink:
     
    Last edited: Jan 29, 2013
  13. Jan 29, 2013 #12
    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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  14. Jan 29, 2013 #13

    davenn

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    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
     
  15. Jan 29, 2013 #14

    NascentOxygen

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    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.
    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! :smile:

    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.
     
  16. Jan 30, 2013 #15

    davenn

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


    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 ?

    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
     
  17. Jan 30, 2013 #16

    NascentOxygen

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    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.
     
  18. Jan 30, 2013 #17

    davenn

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    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
     
    Last edited: Jan 30, 2013
  19. Jan 30, 2013 #18

    NascentOxygen

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    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.
    Yes, inductor current never changes instantly, for to do so would require infinite voltage.
    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.
    Yes.
    Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.
    Yes.
     
  20. Jan 30, 2013 #19

    NascentOxygen

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    That was a follow on to what I'd written. I hadn't noticed your post.
     
  21. Jan 30, 2013 #20
    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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