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Joined: Jun 2005
Posts: 8
Z
zed Offline OP
Junior Member
Hello all,
Could someone please explain why there is such a large,fast voltage from a collapsing magnetic field.
Why is the build up of the field slower and less "violent" than the collapse.
Kindest regards.
Zeddy

Joined: Jun 2005
Posts: 8
Z
zed Offline OP
Junior Member
And how does one calculate the voltage that could occur.
Zed

Joined: May 2005
Posts: 177
J
Member
Zed, I'll assume you're talking about the transient that occurs when an energized inductor (such as a relay coil) has just been disconnected from its power source. Let me start with a little theory.

If you understand capacitance, then inductance is simple -- the two are highly symmetrical. In the same way that a capacitor stores energy as an electrostatic field, an inductor stores energy as a magnetic field.

The rate at which the voltage across a capacitor changes is proportional to the current passing through it:

dV/dT = I/C

e.g. a 1 amp current flowing through a 1-farad capacitor will build a voltage across the capacitor linearly at the rate of 1 volt per second. A capacitor is therefore said to resist any change in voltage across it, but the current through it can change rapidly.

In the same way, the rate at which the current in an inductor changes is proportional to the voltage impressed across it:

dI/dT = V/L

e.g. 1 volt impressed across a 1-henry inductor will build a current through the inductor linearly at the rate of 1 ampere per second. An inductor is therefore said to resist any change in current through it, but the voltage across it can change rapidly.

(Note: Electronics guys normally use "E" for voltage, but physics guys use "V" to differentiate it from Energy "E".)

The energy stored in a capacitor is proportional to the square of the voltage across it:

E = (C x V^2) / 2

The energy stored in an inductor is proportional to the square of the current through it:

E = (L x I^2) / 2

Isn't that a nice symmetry?

Now, to get to your questions:

The slow build-up of the field is due to the controlled voltage applied across the inductance. When you energize a relay coil by connecting a DC voltage across it, the current starts building up at a predictable rate (initially dI/dT=V/L). Eventually, the current reaches a steady state determined by the applied voltage and the coil resistance, and the coil is fully "charged" at that current value. The time constant at which this slow buildup occurs is the ratio of the coil inductance and the circuit resistance (t=L/R).

When this charged inductor suddenly finds itself disconnected from the voltage source, it tries hard to do what it's intended to do: Keep the current flowing through itself at a constant rate. Its terminals immediately rise to whatever voltage is necessary to achieve that current.

As an example, suppose a relay coil draws 1 amp while energized. If we disconnect the power source and immediately connect a 100 ohm resistor across the coil, the coil will produce -100 volts across the resistor, and that voltage will decrease exponentially over time as the coil discharges into the total circuit resistance (the coil resistance plus the external resistor).

Notice two things about that description:

1) The voltage across the resistor is negative compared to the energizing voltage because the inductor is discharging instead of charging.

2) I didn't mention what the energizing voltage was because it doesn't make any difference! It could be 6 volts or 240 volts. The only thing that affects the discharge voltage is the current and the resistance.

It's just like a capacitor that's been charged to 100 volts. If you then remove the charging current and connect a 100 ohm resistor across it, it discharges at -1 amp into the resistor, decreasing exponentially to zero. It doesn't matter what the charging current was.

So, what peak voltage do you get when you disconnect the coil and there's no suppressor of any kind? It's infinite! Theoretically.

In practice, assuming there's no arc or insulation breakdown, the inductance resonates with the stray capacitance in the coil, producing a high-voltage oscillation that decays as the energy gets dissipated in the coil resistance and through other losses. Since the voltage is high, the discharge rate is very fast. Also, since the stray capacitance is unspecified, it's impossible to predict the peak voltage.

This is comparable to taking that charged capacitor, placing a dead short across its terminals and measuring the peak current -- Again, it's theoretically infinite, but limited in practice by the stray inductance and circuit resistance.

Was that more than you wanted to know? [Linked Image]

(edit: time constant is ratio L/R, not product LxR)

[This message has been edited by John Crighton (edited 07-25-2005).]

Joined: Feb 2005
Posts: 693
L
Member
John, that was an outstanding explanation.

Bravo!


Larry Fine
Fine Electric Co.
fineelectricco.com
Joined: May 2005
Posts: 177
J
Member
Just giving a little back to the forum when I can. Thanks, Larry.

Joined: Oct 2000
Posts: 2,721
Broom Pusher and
Member
I'll second the Outstanding Explanation, John!!! [Linked Image] [Linked Image] [Linked Image]

With no doubt, Zed's questions are answered - and very understandable too!

Scott35


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Joined: May 2001
Posts: 717
G
Member
John,
Absolutely BRAVO!!!!!!

A simple explanation (meaning, of course, one which even I can understand) is worth it's weight in GOLD!! Especially to an instructor, hope you don't mind, but I stole it...... [Linked Image]

Thanks a BUNCH,

George

Joined: Oct 2004
Posts: 135
B
Member
I'm impressed. I've always known this happens with collapsing fields but have never received such a clear explanation. Thank you.

Joined: May 2005
Posts: 177
J
Member
Golly, maybe I should turn it into a screenplay...

Thanks for the kind words, everyone.

=== Pop quiz! ===

Assume a "perfect" capacitor (infinite leakage resistance) and a "perfect" inductor (zero winding resistance).

To keep energy stored in the capacitor indefinitely, you leave the terminals open.

How do you keep energy stored in the inductor indefinitely?

(Counter-intuitive, isn't it?)

Joined: Jun 2005
Posts: 8
Z
zed Offline OP
Junior Member
Thank you for your fine explanation John, and thanks also to Scott for an earlier question, it is very much appreciated.
Kindest regards
Zed

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