From: Dan_Thomas_nospam@yahoo.com (Dan Thomas)
Subject: Re: (Avionics) How can this circuit produce an "inductive surge"?
Date: 28 Sep 2002 11:41:14 -0700
References: <firstname.lastname@example.org> <email@example.com>
NNTP-Posting-Date: 28 Sep 2002 18:41:14 GMT
> > >> To make matters worse, when you turn off an inductive load such as the
> > >> starter relay/solenoid, the polarity of the induced voltage is the opposite
> > >> polarity of the applied voltage.
> > > No, it's the same polarity. An inductor resists an increase in
> > > current, and resists any decrease by giving the flow a shove as the
> > > field collapses. The result is a spike of the same polarity. The diode
> > > across the Cessna solenoids doesn't allow any current flow during
> > > normal operation, but shorts the spike. Drawing it on paper makes it
> > > clear.
> > Yes, drawing it on paper makes it quite clear the voltage reverses and
> > what happens.
> > The cathode of the diode is connected to the (normally) positive side of
> > the inductor, the anode to the negative side. Under normal operation, the
> > diode is back biased and no current flows through it.
> > When the switch is opened, the voltage across the inductor reverses, the
> > diode is now forward biased and limits the voltage to one diode drop, or
> > about .7 Volts. The initial magnitude of the current through the diode will
> > equal the steady state current that was flowing before the switch was opened.
> > If the voltage didn't reverse, the diode would never do anything. The diode
> > across the inductor is an ordinary diode, not a zener diode or anything
> > else.
> > Without the diode and with an ordinary switch, the voltage instantaneously
> > tries to go to infinity (which doesn't happen in the real world, but
> > it gets large fast) and at some point the switch contacts arc and you have
> > a very large negative voltage across your bus.
> > If you still don't believe it, get any basic electronics book and look up
> > inductive loads and diode protection. I suggest "The Art of Electronics"
> > by Horowitz and Hill.
> The operational flow through the solenoid coil is from ground to
> the positive. The diode is connected as you say, so there is no flow
> through it. At shutdown, the spike flows in the same direction, but
> instead of going through the battery it flows through the diode to the
> coil's ground terminal and back into the coil.
> An inductor wouldn't be an inductor if it did the opposite.
Further to my own post here, I see now where the disagreement is
coming from. While the current flow at shutdown is in the same
direction through the coil during normal operation, it is seen by the
bus, and therefore the rest of the electrical system, as a negative
spike at the coil's hot terminal. Forgive my misunderstanding. This is
one of the limitations of a text-only NG. I wish we could draw