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From: Jim Pennino
Subject: Re: (Avionics) How can this circuit produce an "inductive surge"?
Date: Sat, 28 Sep 2002 16:18:26 +0000 (UTC)
Organization: Special Solutions, LLC
References: <firstname.lastname@example.org> <email@example.com>
NNTP-Posting-Date: Sat, 28 Sep 2002 16:18:26 +0000 (UTC)
User-Agent: tin/1.4.5-20010409 ("One More Nightmare") (UNIX) (SunOS/5.9 (sun4u))
In rec.aviation.owning Dan Thomas wrote:
> Jim Pennino wrote in message news:...
>> In rec.aviation.owning Dan Thomas wrote:
>> >> The inductive caused transients from things like turning off the starter
>> >> relay can easily be thousands of volts for a short period of time; i.e.
>> >> microseconds.
>> > I once measured the spike off the master solenoid at 600 volts. Some
>> > here are saying that the avionics are designed to withstand this; that
>> > is likely true. However, if the internal protection isn't working, the
>> > pilot is taking his chances.
>> >> 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
>> 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.
A silicon diode, which is what all modern diodes are, conducts ONLY if the
anode is about .7 Volts more positive than the cathode.
The diode is ALWAYS connected across the coil of the solenoid. The diode
conducts ONLY if the voltage across the coil reverses which is what happens
when the voltage applied to the coil is removed by opening the switch.
Go to the following URL for a simple explaination of how diodes work
then tell me how anything other than the above two sentances could
possibly be true.
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