inductor current question

• posted

Hello,

ok, say you've got a DC voltage supply in series with a switch and an inductor, the inductor ties back into the power supply completing a circuit. ok, the switch is closed for a little bit then opened.... from what I understand this causes the inductor to produce a hell of a voltage across itself cause it doesn't like the drastic change in current. now the switch is open, the circuit it broken there's no loop for the current... my question is... is there current flowing? even though it's a broken line because of the switch? If the inductor is pushin current out of itself, is there current flowing through the inductor or just out one end of the inductor? I've heard you can really mess up a switch this way, with that high voltage across the inductor hanging on one end of the switch... but if the other end of the switch is floating, how does the switch get damage? I mean, there can't be any current flowing through the switch, can there? If one end is floating?

much thanks!

• posted

Here's the sequence:

Everything is at rest, and the switch is open.

Then you close the switch. Current starts ramping up in the inductor, because V = L * di/dt.

Now you open the switch. WHAM! The current in the inductor goes down really fast, so V goes up really high. At this point, one of two things are going to happen, based on things that aren't on your schematic: either that current from the inductor will go to charging up it's own native capacitance (doubtful, or at least doubtful that it'll be the only thing happening), or the voltage across the switch contacts -- which probably haven't opened all the way yet -- will be sufficient to make it arc.

It's the arc that damages the switch (and possibly other things in your circuit). It's the same effect that makes the old mechanical Kettering ignition system work, except there the spark is controlled and directed to the spark plugs.

It's that same urge to just make as much voltage as necessary to flow the current that damages transistors that turn coils on and off, unless you put in some sort of snubber or other device to absorb the energy stored in the coil.

And to make a boost switching converter.

```--
Tim Wescott
Wescott Design Services```
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Thanks Tim for the answer, so if I had a fuse with a transistor and inductor in series going to a power supply and the fuse blew... could the same logic you described above apply? The fuse blew and the transistor ended up dying here in this situation, but I couldn't figure out why the transistor died... if the fuse blew it isolated the transistor and inductor on one side of it and the power supply on the other... but maybe from what you said... when the fuse popped, the inductor voltage shot up high enough to cause an arc across the fuse and shot too much current through the transistor thus screwing it up?

thanks again for the insight

• posted

Most likely the transistor was off, and when the fuse sparked over the transistor got nailed with a high voltage transient.

```--
Tim Wescott
Wescott Design Services```
• posted

It's the rate of change of current in an inductor that determines the voltage. Some people use e=3DL * di/dt which is the inductance times the rate of change in amps per second. But I was always confused with that and just use LI =3DET which is the inductance (L) times the current (I) is equal to the voltage (E) times the time (T).

So, if you rearrange that into E=3DLI / T the voltage increases to infinity as the time goes to zero. It takes a long time to charge an inductor at low voltages. For example, it takes 1 second for the current to rise from 0 to 1 amp in a 1 Henry inductor with 1 volt applied. But the current can fall to zero very fast if the circuit is broken (switch open) which represents a very high rate of change (amps per second), and so a very high voltage.

But you could limit the voltage across the switch with a resistor in parallel. In the 1 Henry inductor case, if you add a 100 ohm resistor across the switch, the voltage would be lower (when opened) at 100 volts (e=3Dir =3D1*100 =3D100) and would last longer. The decay time would be about L/R for 63% decay, or maybe 10 milliseconds to fall to 37 volts, and another 10 milliseconds to fall to 14 volts, etc.

Probably not practical, just an example to get the idea.

-Bill

• posted

Assuming real components because the question is unanswerable if they are all ideal.

Some current flows in an arc between the switch contacta, some charges the capacitance of the switch, some is translated into eddy currents in the inductor, etc...

yeah, it's the arc that does it, you can see this yourself if you use the primary of small transformer for the incuctor and a 1.5V dry cell for the power supply and just bared wires for the switch (hold them by the insulated part)

compare with an equivalent resistor instead of the inductor.

• posted

Yes.

Yes.

An open switch has very high resistance, but it's not infinite.

The same current is flowing throughout the circuit.

Indeed.

Yes. The inductor ensures that opening the switch doesn't have any immediate effect upon the current through it.

Think of the switch as a variable resistor with two positions: a fraction of an ohm (closed position) and millions of ohms (open position).

[In practice, the resistance will drop significantly due to the switch arcing over and ionising the air.]
• posted

Mechanical switches will usually arc. A transistor can turn off fast enough that an inductor can ring naturally. With practical inductors and mosfet switches, one tends to get flyback ratios (Vpeak/Vsupply) in maybe the 20:1 sort of range.

The combination of core losses, eddy current losses, and charging of the capacitance of the coil and the fet limit peak flyback voltage. Classic car coil ignitions ran in that range too.

The condenser in a coil ignition reduced resonant frequency enough that the contacts had a chance to get open before the voltage built up enough to arc much. Its value was a compromise between that, and lowering the primary flyback voltage itself.

A HV fet will avoid the snubber, but a HV fet will tend to have high Rds-on. It's always something.

John

• posted

Transistors make excellent fuse protectors. ;-)

Cheers! Rich

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ok... what if, I had an inductor in series with a battery and a switch, in a closed loop, and I was able to break that switch open at the speed of light, the inductor would want to keep the current going, but it's an open circuit... what would happen there? Can the current flow through an open circuit? There's stored energy in the inductor right... and that energy is used to push that current out, to discharge the inductor coil.... but there's no where for the inductor current to flow... is it a law pf physics that the switch will arc no matter how fast you can open it?

• posted

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1- In real circuits, there is always parasitic capacitance to slow things down. So if the switch is opened - then closed - in a nanosecond, the inductor voltage won't have time to go very far. The current continued to flow, but through that stray capacitance rather than the switch. 2- In the fuse question, the fuse won't blow when the switch is off. After all, it's current that blows a fuse, and the current doesn't increase when you open the switch. If a high voltage develops as a consequence of the switch being opened, that similarly doesn't appear across the (still conducting) fuse. Of course if things are arranged such that a high voltage pulse causes an insulator to fail, leading to a fault current, then the fuse could fail - but due to this new fault current. 3- If you push hard enough (fast enough,...) on the simplified ideal component models, they will break. You will have to return to the electromagnetic field equations from whence they came.

Hope that helps...

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my question is, if you have an inductor, powersupply and switch in series, and the switch is closed, and then at light speed you pop that switch open all the way... can you open it fast enough so that there's no arc?

and if you can open a switch that fast, I know that the inductor is going to try and keep the current going, but it can't... cause it's an open circuit.... so what happens to the energy? I guess the inductor will experience an infinite amount of voltage across itself and crap its pants.

is it even physically allowable/possible to open a switch in my scenario so fast that there wont be an arc? Maybe my question ends here by someone saying... nope physics will not allow that, the electrons will hop through the air the instant that switch comes open... since electrons are moving at the speed of light anyway, you can't open it faster than that (I don't know if current moves at the speed of light I'm just throwing out a possible answer)

• posted

Sure. Put a capacitor across it. That's what the old points ignition did -- a lot slower than light, at that.

R1-L1 represents the inductance and a real inductor's DC resistance (not necessary, since superconductors exist), C1 represents parasitic self-capacitance (a necessity of the inductor taking up space). R2-C2 represents loss to radiation, which can be intentionally large (i.e., stick an antenna on it), or arbitrarily small (superconducting resonators achieve Q > 10^9, which is a better resonator than an ovenized quartz crystal).

The peak voltage, peak dV/dt and dI/dt at switch opening, resonant frequency and decay constant can all be evaluated exactly from circuit values -- first/second year EE stuff.

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Yes, if you have a circuit that can take the maximum possible voltage that the inductor produces as the current tries to fall to zero, then the stray capacitances and resistances will cause it to ring. Open switch, voltage goes up way high, sucking electrons out of the schtuff on one side of the inductor, and piling them up on the other (these are the plates of the stray capacitor) until the current stops flowing, then all those electrons on the one side of the inductor slosh over to the other side, through the inductor, and its resistance, until the stray capacitance is charged, and so on, back and forth, until the resistances dissipate all the energy that was stored in the inductor.

Hope This Helps! Rich

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No, no, no! NOTHING moves at the speed of light but energy. Electrons plod along at a few centimeters a minute.

CHARGE moves at the speed of light.

Hope This Helps! Rich

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If even.

Ho boy! Here we go again. ;-)

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Yeah, I don't actually know the real numbers, but I read something many, many years ago and don't remember the exact numbers, but it was surprisingly slow. Of course, if it's AC they just vibrate back and forth in place.

No! No! Please, let's not start some holy war. "Something" moves through the circuit at C * velocity factor - energy, charge, it's really irrelevant what you call it, but whatever it is, makes stuff work, so talking about specific terms, I'd rather avoid arguing about that probably as much as you'd also rather to. :-)

But we all know that the electrons themselves don't move much, OK? :-)

And really, for all we can see with the naked eye and stuff, it might as well be Faeries. ;-P

Cheers! Rich

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Then again, it would be interesting to know the speed of the electron stream in a CRT or whatever. In a supercollider, individual particles do approach C, but that's a whole nother thing anyway.

Being a tech who was raised on electron flow. I get a kick out of hobnobbing with the Conventional current gang, I like to ask, "So, in a CRT, how does the positive current know which pixel to jump out of and go down the throat of the electron gun, through the deflection system, and get an exact bull's-eye on the thermionic cathode? ;-)

Cheers! Rich

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-)

In a metal you can estimate the electron thermal velocity (v). mv^2 =3D kT. (m is the electron mass in the metal and k is Boltmanns constant.) And in most cases this turns out to be a lot faster than the drift velocity from any applied electric field.

George H.

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