Collapsing fields and coil polarity, freewheel diode, .

[mmalf_mi]

When a DC coil (relay) is denergized, shouldn't the collapsing field cause a current that would oppose the change in current? Wouldn't that induced current be in the same direction as when the coil was enegized? Not seeing how a freewheeling doide gets forward biased. If I apply Lenz Law, current will flow in a direction that will oppose the force that caused it. The collapsing field caused it (denergizing the coil), the current direction to oppose the collapsing field would be the same current direction that caused an expanding field (energizing the coil). Shouldn't there be a voltage spike across the coil with the same polarity as when the coil field was energized and steady state?

Please shed some light on this EET.

 

[NoTime]

Welcome to PF.

You generally seem to have the right idea.
However, note that when the current going to the coil is turned off the coil becomes a power source.
So with the external drive terminal 1 is at a higher voltage potential than terminal 2.
With the drive turned off terminal 2 is staying at the same (for illustration) voltage (because the coil continues the current) while terminal 1 is now at a zero potential.
The relative polarity across the coil reverses.

 

[mmalf_mi]

NoTime,
Thanks for your response. Is this the same as when the secondary voltage of a transformer is 180 degrees out of phase with the primary and the secondary has no load? We read a voltage because the induced secondary circuit is not closed, but electrons have still moved to one side of the wire's end (the - side means excess of electrons), and the other end now has fewer electrons (the + side). Lenz Law would still prevail, the collapsing field would try to oppose the decreasing current by boosting the current in the same direction as when the switch was closed on the coil. However, there is not a closed circuit for current to flow, it just moves electrons to one end and creates a potential difference.

I got this from another website. I did not think it was a good technical response:

"However, when the switch is opened, the coil's inductance responds to the decrease in current by inducing a voltage of reverse polarity, in an effort to maintain current at the same magnitude and in the same direction."

Great to have people to go to. Thanks again.
mmalf_mi

 

[NoTime]

For a transformer secondary with switched DC.
When the switch is closed you will get a voltage in one polarity.
When the current in the primary reaches the limit defined by E/R then the secondary voltage is zero.
Just what you get when the the switch is opened depends on what the primary circuit is (with the switch open), but the initial voltage pulse will have the opposite polarity.

 

[berkeman]

Here's the flyback topology that I think NoTime is referring to:

http://en.wikipedia.org/wiki/Flyback_converter

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