Basics question of Inductor to powering high voltage lamp

[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?

That's the right idea. But because of the presence of the resistor, the inductor must develop a bit more than 65V. Some of its voltage is dropped across the resistor, to leave 65V for the neon.

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?

Yes, unless it is a superconductor. The particular value of resistance is determined by many factors, including the characteristics of the neon. The resistor is not crucial, but the neon may be damaged if the current is not limited to a value it is designed for. If you were worried, you could leave out the neon and let the spark jump across a narrow air gap between two wires.

I'm hoping you are about ready to sketch the waveform (current versus time) of the current in the inductor. The task is already half done for you: in the original video they have sketched the current as it builds up while driven by the battery. Now all you have to do is sketch the current as it winds back down to zero. Don't worry about getting the time scale exact, because you can't. We don't know the value of L or R, so it becomes a generalized sketch.

 

[sophiecentaur]

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?

If you aren't prepared to get down and dirty with some assumed values then you can do nothing but wave arms about. I have been trying to do this but no one seems interested. The 1Ω is fine to be getting on with but, unless you make some assumptions about the energy stored, you might as well say it will take a week for the neon to flash. It's even worse if you choose to ignore the Capacitance involved.
Engineers need to use numbers as well as Calculus.

P.S. What exactly is a "Voltage Path"?

 

[Jay_]

Voltage path is closing the circuit or in other words, not opposing the flow of current to the extend of behaving like an open.

 

[sophiecentaur]

Voltage path is closing the circuit or in other words, not opposing the flow of current to the extend of behaving like an open.

Ummm. Voltage does not follow a path; it is the potential difference between two points, which is independent of the path taken by charges flowing between the points. It is difficult to understand any argument that is using such 'alternative' usage of terms.

You are right when you say that any coil will have finite resistance. But the finite Capacitance in a circuit will also affect its behaviour.

 

[The Electrician]

Yes, unless it is a superconductor. The particular value of resistance is determined by many factors, including the characteristics of the neon. The resistor is not crucial, but the neon may be damaged if the current is not limited to a value it is designed for. If you were worried, you could leave out the neon and let the spark jump across a narrow air gap between two wires.

How can the resistor limit current? It has been pointed out over and over in this thread that at the instant the switch moves from A to B, the current in the inductor will remain unchanged. If there's a resistance in series with the inductor, then the inductor will cause the voltage to rise just the amount needed so that the current is unchanged, as you point out yourself in the post I'm quoting from:

That's the right idea. But because of the presence of the resistor, the inductor must develop a bit more than 65V. Some of its voltage is dropped across the resistor, to leave 65V for the neon.

 

[The Electrician]

Here are some oscilloscope captures of the behavior of an inductor under circumstances similar to the one being discussed in this thread.

I used a 1.734 mH crossover inductor. It is an "air core" inductor, with a DC resistance of 1.82 ohms.

I connected it to a power supply adjusted to cause a current of 3 amps DC to flow in the inductor. I then disconnected the power supply and monitored the voltage across the inductor. There was no neon lamp involved, nor was anything else other than the 100x scope probe connected to the inductor. Here's the voltage across the inductor at the instant of disconnecting the power supply; the two cursors are used to measure the self resonance frequency of about 555 kHz. Even though there's nothing connected across the inductor and theoretically the voltage should rise to infinity, the distributed capacitance of the inductor limits the voltage rise:

I didn't have a neon lamp handy, so I connected an approximately 100 volt zener diode across the inductor, with the polarity arranged to clamp the inductor voltage when the power supply was disconnected. Here is the scope capture of that event:

This iimage shows the leading edge of the rising inductor voltage as it approaches the clamping voltage; we can see that the voltage rises at a rate such that it takes about 50 nS to reach about 100 volts.

 

[sophiecentaur]

Interesting. From your results (the frequency of the oscillation), I conclude that the self capacitance of the circuit must be around 200pF. This is not an inconsiderable capacitance.
Also, the Q of the circuit is round about 2 (Looking at the decay rate on the trace), so that makes the resistance around 1.5k at that frequency. Quite a bit higher than the DC value but not surprising as it's an audio frequency core.
I could always be wrong in my arithmetic but I checked it once or twice.

 

[NascentOxygen]

How can the resistor limit current?

By Ohms Law. The resistor could be omitted, or set to a very low value, but this risks the current rising to a destructive level — exceeding the rating of the inductor while in switch position A, and/or exceeding the rating of the neon when in position B. Were the current to not be limited by resistance, then it would be crucial that the switch not being kept in position A beyond a certain duration. This is an unrealistic expectation to place on manual switching.

The point you raise suggests a further improvement: in addition to modifying the switching and the neon's placement, the resistor could be repositioned to be attached to the battery so it disappears from the circuit when the switch moves off position A. There may be practical considerations why some resistive damping is desirable while the inductor powers the neon, but I think the instructor probably felt that the circuit resistance should be lumped in with the inductor because at least some coil resistance is inescapable.

 

[The Electrician]

Interesting. From your results (the frequency of the oscillation), I conclude that the self capacitance of the circuit must be around 200pF. This is not an inconsiderable capacitance.
Also, the Q of the circuit is round about 2 (Looking at the decay rate on the trace), so that makes the resistance around 1.5k at that frequency. Quite a bit higher than the DC value but not surprising as it's an audio frequency core.
I could always be wrong in my arithmetic but I checked it once or twice.

You've raised an issue that I wasn't going to mention because it's an additional complication to the factors being discussed.

This inductor I used was not the sort of thing one would use in a switcher. It's a multi-layer solenoid wound on a nylon bobbin with 20 gauge wire; it has 10 layers in fact.

This kind of construction leads to considerable proximity effect loss and the real part of the impedance increases dramatically at higher frequencies.

Here's a plot of the magnitude of the impedance (green) and the real part of the impedance (yellow):

The frequency we saw on the scope is the natural resonance frequency, not necessarily the frequency of maximum impedance, and it's lower than what the analyzer shows because of the additional capacitance of the scope probe, etc. Anyway, at the frequency of maximum impedance, the real part is almost 30k ohms.

So, of course, the resistance causing the damping of the free oscillations is not the DC resistance, but a much higher value AC resistance. I wasn't going to mention this.

 

[The Electrician]

By Ohms Law. The resistor could be omitted, or set to a very low value, but this risks the current rising to a destructive level — exceeding the rating of the inductor while in switch position A, and/or exceeding the rating of the neon when in position B. Were the current to not be limited by resistance, then it would be crucial that the switch not being kept in position A beyond a certain duration. This is an unrealistic expectation to place on manual switching.

The point you raise suggests a further improvement: in addition to modifying the switching and the neon's placement, the resistor could be repositioned to be attached to the battery so it disappears from the circuit when the switch moves off position A. There may be practical considerations why some resistive damping is desirable while the inductor powers the neon, but I think the instructor probably felt that the circuit resistance should be lumped in with the inductor because at least some coil resistance is inescapable.

Here's what you said on the same issue in post #39:

"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"

This sure looks to me like you were talking about limiting the current after the switch moves to B, and is delivering energy to the neon bulb. At that time the resistor can't limit the current.

 

[Jay_]

Sophiecentaur

Current path might have been a more appropriate term. But I don't see why we have been going to circles on a simple query:(

 

[Jay_]

I came to understand the concept. But people here are themselves debating about it. If someone can clear this to me and explain what happens in the circuit at each time instant, I would appreciate.

I don't know the inductor working to the point to know who is saying what right.

 

[NascentOxygen]

This sure looks to me like you were talking about limiting the current after the switch moves to B, and is delivering energy to the neon bulb. At that time the resistor can't limit the current.

As a precautionary measure, I would include series resistance while adjusting circuit operation, then reduce the resistance in steps, until it was confirmed it could safely be eliminated (while preserving the essentially first-order operation). Neons can have their ratings exceeded.

 

[NascentOxygen]

I didn't have a neon lamp handy, so I connected an approximately 100 volt zener diode across the inductor, with the polarity arranged to clamp the inductor voltage when the power supply was disconnected. Here is the scope capture of that event:

Can you explain why the zener voltage didn't reverse polarity?

Along with inductor voltage, it would be instructive if you inserted a small current-sensing resistor in series with the inductor, to demonstrate corresponding features of the current waveform. It can be tiny enough that it won't measurably affect the voltage waveshape.

 

[sophiecentaur]

I came to understand the concept. But people here are themselves debating about it. If someone can clear this to me and explain what happens in the circuit at each time instant, I would appreciate.

I don't know the inductor working to the point to know who is saying what right.

What is it that you still want to know?
We have agreed that the induced emf is in the opposite direction when the switch opens. We have discussed time constants and damping (albeit, some of us don't seem to want actual values discussed or the importance of parasitics).
If you just remember dI/dt is what counts and use the values of the reactances to fine the rate of change of current then you can find the emf and the timescales involved.

 

[Windadct]

Hey Jay - the fact of the issue is that the case you presented is at the limit of "ideal" elements where many in the electrical field live and the theoretical. In this case the difference between an ideal switch (dt = 0 ... dI/dt is undefined) and a real switch where we can see through measurement and observation that dt > 0 for this inductive circuit ( similar issues with short circuit of a capacitor).
The amount if of instantaneous power available - when the switch is opened, can only be analyzed in the REAL world - there are no perfect switches and there are always parasitics (keepin' it real) - and I believe as shown in the scope traces, effects of just taking a measurement, or observing the case have a dramatic effect.
It is quite common here, in PF, that the discussion lives well beyond answering the original post - which in a way is unfortunate because the original poster ( you ) rightly assume there is more to the story than you understand. This post is a great case for this - there are people here with many ( dare I say all) levels of understanding, all (hopefully) looking to learn, but challenge and debate nonetheless.
My advice is offer some "Sympathy for the Devil" - no circuit is perfect - just like PF'ers.

 

[The Electrician]

Can you explain why the zener voltage didn't reverse polarity?

I'm sure it did, but the zener is a forward biased diode in the reverse direction, and the voltage across wouldn't show up when the scope is set to 100 V/div. Furthermore, almost all the energy in the inductor was dissipated in the zener by the time reversal would occur.

Along with inductor voltage, it would be instructive if you inserted a small current-sensing resistor in series with the inductor, to demonstrate corresponding features of the current waveform. It can be tiny enough that it won't measurably affect the voltage waveshape.

It's easier to just use a current probe.

 

[sophiecentaur]

I'm sure it did, but the zener is a forward biased diode in the reverse direction, and the voltage across wouldn't show up when the scope is set to 100 V/div. Furthermore, almost all the energy in the inductor was dissipated in the zener by the time reversal would occur.



It's easier to just use a current probe.

Depends what generation you are! Do you get a current probe in the box with Oscilloscopes these days? It's a long time since I bought one.

 

[NascentOxygen]

I'm sure it did, but the zener is a forward biased diode in the reverse direction, and the voltage across wouldn't show up when the scope is set to 100 V/div.

I assumed that initial offset was [somehow] due to the power supply, but you say it isn't? Can you account for it? (The blue line marks zero, does it?)

Furthermore, almost all the energy in the inductor was dissipated in the zener by the time reversal would occur.

Reversal occurs the instant that the power supply disconnects.

It's easier to just use a current probe.

Only if you have one. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

 

[sophiecentaur]

Reversal occurs the instant that the power supply disconnects.

How long has this thread been running? Yet you still use terms like "instant". Nothing is "instant" in electronics. The time involved depends upon the REAL values of all the components involved. Have you still not grasped that?

 

[The Electrician]

Reversal occurs the instant that the power supply disconnects.

Have a look at the capture where there was no clamping device. The first voltage excursion, which occurs just after the disconnection from the power supply, is positive going. I thought you were referring to the polarity reversal that occurs at 2.5 divisions, during the ringing.

I'll look into your "blue line" question when I get back home in a day or so.

 

[NascentOxygen]

Have a look at the capture where there was no clamping device. The first voltage excursion, which occurs just after the disconnection from the power supply, is positive going. I thought you were referring to the polarity reversal that occurs at 2.5 divisions, during the ringing.

No, I wasn't referring to the unloaded ringing. I refer to the clamped waveform on which I added the blue notation: at the vertical rise the inductor voltage reverses and it stays reversed for the duration of that steady 120V* level. I expected the zener's voltage prior to that vertical rise would be of opposite polarity to what it is after that point, and I can't account for it not changing polarity at that point.

I'll look into your "blue line" question when I get back home in a day or so.

Fine, no hurry, I think this thread will still be going strong.

* is the oscilloscope indicating your "approx 100V" zener is close to 120V?

What does the 64V figure on the foot of the display indicate?

 

[The Electrician]

* is the oscilloscope indicating your "approx 100V" zener is close to 120V?

Yes.

What does the 64V figure on the foot of the display indicate?

It's the trigger level, corresponding to the orange arrow on the right side of the screen.

 

[The Electrician]

Concerning the offset you asked about in post #69. I connected the power supply to the inductor with a pair of alligator clip leads, and performed the disconnect by simply touching and then "un-touching" one of the alligator clips. When doing this, one gets a spark every time, and usually you don't get a clean ring waveform. I had to do it a number of times to get the waveform I posted. Even though I was able to get a single disconnect without any bounce, there was still an arc. The offset voltage you asked about is the voltage of the arc. The true disconnect doesn't happen until the arc stops, then the inductor current is truly interrupted.

This capture shows what happens with a current probe monitoring the current. The orange trace is the voltage across the inductor, and the blue trace is the current measured with a current probe clamped around the external inductor lead. Starting at the left edge of the screen until about 3.8 divisions, the current is ramping toward zero, the alligator clip having been disconnected and arcing is in progress. At 3.8 divisions, the ramp stops and the current suddenly jumps to zero and the ringing at the self resonance frequency commences. The oscillation results from energy exchange between the distributed capacitance of the inductor and the inductance itself. This oscillating current and voltage occurs completely internal to the inductor and no sign of it is seen in the blue trace which is the current in the external inductor lead.

This capture shows the current in a loop of the wire comprising the inductor. I pulled a single turn of the wire from the middle of the inductor and clamped the current probe around it. Now we can see oscillations in the current because we are sampling the current internal to the winding of the inductor.

To get a better result, I got a couple of 500 volt power FETs (IRF740s) and connected them source to source, with the two drains available to interrupt the inductor current. This arrangement is necessary to avoid the clamping effect of the body diode which would occur if only a single FET were used.

This capture shows the current ramping up as the FETs are turned on and the power supply charges the inductor. The FETs are turned off at the middle of the screen.

See the next post.

 

[The Electrician]

In this capture we see at a shorter time scale the disconnecting of the power supply by turning off the FETs. The capacitance of the FETs is of the same order as the distributed capacitance of the inductor, and now we see quite a bit of oscillating current in the external lead of the inductor. The voltage oscillations (orange) are somewhat non-sinusoidal due to the nonlinear capacitance of the FETs.

Notice that there is no offset in the voltage as we approach turn-off from the left side of the screen. There is no arcing before the complete disconnect from the power supply.

 

[NascentOxygen]

You've abandoned the zener diode in all of these?

 

[The Electrician]

I wouldn't have used the word "abandoned". I just didn't have it in place because I was demonstrating things that couldn't be seen with it in place.