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Subject: Re: (Avionics) How can this circuit produce an "inductive surge"?
Date: Sat, 28 Sep 2002 08:14:34 GMT
Organization: Posted via Supernews, http://www.supernews.com
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The starter can generate voltages of hundreds of volts on the battery
buss when the starter solenoid opens. It does this quite simply. The
starter is two series inductors with some combined inductance we will
call "L". The energy stored in the inductance is J=1/2 x I x I x L.
The windings have some small capacitance from the windings to the
rotor and stator iron case we will call this capacitance "C". The
energy stored in the capacitance is J = ½ x V x V x C.
When the starter relay is opened it interrupts a large current of
several hundred amps. If the relay contacts could open fast enough
(they can not ) then all of the energy stored in the magnetic field
of the coil would transfer to the electric field in the capacitor.
Now we add some reasonable numbers. If we assume that the inductance
of the starter is 1 milihenry and the current is 100 amps the energy
in the magnetic field is 0.5 x 100 x 100 x 0.001 = 5 joules.
Assume that the capacitance of the windings to the case is 0.01 micro
farads. If we assume no losses the energy stored in the capacitor is
equal to the energy that was just stored in the inductor. Then the
peak voltage on the capacitor will be square root of [5/ (0.5 x 0.01
E-6 )] = 31,000 volts !! The voltage will never get this high
because something will break down and arc which will dissipate the
energy. Normally what breaks down is the air gap between the slow
moving relay contacts. Lets say that the contacts open really fast
so that they break down at 400 volts. This is only a few thousandths
of an inch. We have now impressed 400 volts on the battery buss wire
with an arc that has a very fast rise time. What limits the current
of this pulse? The current is limited by the inductance of the
battery wire, the resistance of the arc and the effective series
inductance of the capacitance which is a LOT less than the total motor
inductance. The peak current can be HIGHER than the starter current
but only for a short period of time which is generally only a few
microseconds. However microseconds is enough to destroy
semiconductors. The voltage on the starter will ring like the Avon
lady. You will see a damped sine wave with a lot of arcing . When
the arc begins will determine whether you get a positive or negative
voltage spike impressed on the battery buss wire. If the battery buss
wire is long then you will get more voltage impressed on your avionics
unless you have wired the avionics directly to the battery terminals
with a separate wire. Batteries have a lot of capacitance and are
very low impedance. In general airplanes have longer battery buss
wires than cars. Cars generally tap right off the battery post
connector with a separate wire to run the electronics while airplanes
tap off at the master relay with several feet of inductive wire going
to the battery from the master relay. The inductive wire from the
master relay to the battery carries the starter current also. If you
want to see interesting results model this in Spice with switches
closing at different times to simulate the arc.
Airplanes also have an additional failure mode that cars do not have.
Airplanes have battery master relays. What happens if you open the
master relay before the engine stops? What you get is called "LOAD
DUMP". If your electrical load is more than what the alternator is
generating when the master is opened then generally nothing will
happen if the regulator is well behaved. However the regulator does
not regulate very well with out the battery to stabilize the system.
If however the battery is somewhat discharged and is soaking up a
large part of the alternators output when the master is opened and the
avionics load is small then the voltage on the avionics will rise to
possibly more than a hundred volts. This is due to the large
inductance of the alternators field winding. It takes several tens of
milliseconds for the field current to decay after the regulator tells
the alternator to stop charging. During this time the alternator is
putting out a CURRENT that is a multiple of the field current. If the
load is less that what the alternator is generating then the voltage
will rise to over 90 volts until the field current decays. A Cessna
(modified Ford) 60 amp alternator will generate more than 90 volts at
low load with full field (2 amps) if the engine is over 2000 RPM.
Here is a true story of what happens when you load dump an aircraft
A 1976 Cessna 172 has a interesting electrical system failure mode.
Lets say that your Cessna is hard to start when it is cold and you
grind on the starter a long time. This discharges the battery some
but the engine starts. You do not know it but the 60 amp alternator
circuit breaker is old and tired and its contacts have oxidized
somewhat so that they trip at 45 amps not the rated 60 amps. It is
night and you taxi up to the end of the runway. You have warmed up
the engine at slightly more than idle speed, just enough that the
alternator is generating the lighting and avionics loads but not
charging the battery much due to the low engine speed. Now you give
it full throttle and begin the takeoff run. Now the alternator is
generating its full rated 60 amps of current because it is cold and
the winding resistance is less when it is cold. The battery is
absorbing all the excess amps beyond the avionics and lighting loads
because it is still somewhat discharged. Now your tired alternator
output circuit breaker has only a 45 amp rating due to rusty high
resistance contacts. After a little while at 60+ amps the breaker
opens. Just after lift off the pilot sees a very bright red flash!
What the heck was that???!!!!! Fly the airplane first. The airplane
is flying and climbing so fly it back to the runway and land. I did
notice that the alternator was not charging on down wind but
fortunately did not notice that the breaker was tripped until I got
back on the ground. I pushed the breaker in but by this time the
alternator field had already fried from the 90 volts. A post mortum
of the system revealed that the alternator field was burned and open.
The alternator stator and diodes were fine. The regulator field pass
transistor was shorted as was the field relay contacts in the
regulator. The resistors and other transistors in the regulator had
been well cooked. The alternator output circuit breaker was measured
to trip at 40 to 45 amps not the rated 60 amps. The red over voltage
lamp was black and open circuit.
Here is what I think happened:
The alternator circuit breaker tripped when the alternator was
generating its full output of 60+ amps. When the breaker tripped the
output voltage of the alternator went to more than 90 volts. The
regulator is wired by Cessna so that it gets its field power from the
output terminal of the alternator which just went to 90+ volts. The
regulator field pass switching transistor was not rated to switch 90
volts so it dies shorted. Now the over voltage sense tells the field
relay in the regulator to open. The contacts were rated for 15 volts
not 90+ volts so they open and weld shorted by a big metal whisker.
Now the 90 volts put out by the alternator is applied to the field of
the alternator and the over voltage lamp. That is the red flash. The
5 amp field circuit breaker as wired by Cessna is in series with the
over voltage relay and feeds only the 0.1 amp coil of the field relay
in the regulator. The 5A field circuit breaker is NOT in series with
the alternator field! There is nothing to limit the field current of
the alternator except the resistance of the field winding which is
about 6 ohms. Now the 2 amp field is drawing 15+ amps so who knows
what the output voltage of the alternator is with 7 times normal field
excitation. All of the regulator and airplane wiring is published by
Cessna in the airplane service manual and charging manuals so look it
up and learn how not to wire up your alternator!
On Thu, 26 Sep 2002 14:44:34 +0100, Peter
>The normal procedure in light airfraft startup/shutdown is to have all
>possible avionics disconnected when the engine is being started or
>Moreover, one is normally advised to have the alternator field circuit
>broken at these times (there is a switch for this purpose).
>There are many stories going around of somebody's entire avionics kit
>getting blown up if this procedure is not followed.
>The alternator etc parts are more or less standard automotive parts.
>One could get into an interminable discussion as to why these issues
>were addressed in motor vehicles decades ago but still persist in
>brand new aircraft, perhaps loaded with avionics costing US$100k+, but
>I am mainly interested in finding out *exactly* what happens.
>I have not found anybody who is able to explain clearly what happens.
>The general circuit, certainly in the planes I have knowledge of, is
>I cannot see any way to generate an overvoltage surge in the +24V
>(+28V when the engine is running normally) output as a result of
>switching the engine OFF. At this time, as the engine RPM drops, the
>voltage regulator will pass an increasing current, until it runs out
>of steam. This should be a clean process - unless the regulator design
>is defective and it somehow passes a massive field current at the last
>However there appears to be more scope for an overvoltage surge when
>switching the engine ON (if the field circuit was operating). This is
>because the voltage regulator will be sitting there and passing its
>maximum current, for some seconds or more, and when the engine starts
>up, and the alternator spins up, the voltage regulator may not reduce
>the current fast enough.
You can not spin up the engine faster than the field can decay.
>There may be more obscure means of generating a NEGATIVE-going surge
>on the +24V rail but I can't really see how.
>I have also noted that on the aircraft I currently fly, the alternator
>is set up so that even with max field current, the normal regulated
>voltage (+28V) is not reached until one is up to 1200RPM, which is at
>or above what the engine immediately spins up to when correctly
>started. While this is an irritation (the battery is not being charged
>during taxiing around) it may have been done to minimise the risk of
>damage if the engine is started with the field connected.
The field connected during start only adds about 2 amps to the starter
load which is over 200 amps. Other than that this is not a problem
except on turbines which may hang start and over temp ($$$$) due to
the alternator load. The end of a turbine start is 2/3 of full speed
so the alternator is a large load at this speed. This is not true for
a piston engine start.
You have a large load on the alternator at idle. It will not output
as much as the load pulled until the engine RPM reaches 1200. You
could put a bigger HEAVER alternator on the aircraft, if it would fit
and you could get a 337 for this, that would supply this load at idle.
. It would work fine as long as the regulator was ok. If the
regulator died shorted the extra current would cause some excitement.
>Can anyone offer any clearly reasoned comments on this scenario?
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