Colyn,
1) My understanding is if the “B”
starting battery were nearly dead and tried to draw more than 20 amps during it’s
re-charging cycle, the alternator’s voltage drops slightly and then the
battery draws less re-charge current. Sort of self-policing. I know
my Cessna battery never seems to draw more than about 5 amps. Perhaps
someone can confirm/enlighten/correct me.
If the A side dies, and I engage the cross-tie
to power stuff from the B (backup) side, the total current load will be
supplied by both the B alternator and the battery (which attempts to makeup for
any alternator shortfall). Thus, I could continue flight with a 20 amp total
load indefinitely, or with a higher load (say 30 amps) by also slowly depleting
the battery(ies) which are supplying the extra 10 amps. My general intention
is to keep the “continuous” emergency/endurance equipment load safely
under 20 amps, and then let the batteries kick-in (if/as needed) for intermittent
stuff like flaps. Pilot heat, prop heat and de-icing, is a debate in
itself which probably calls for dual full size alternators among other things.
i.e. I perhaps can’t continue to fly with an iced prop, but I also can’t
land IMC without “some” navigational avionics (perhaps a
battery powered 496 handheld as a third redundancy level). Overall, say
no to ice.
Note, if the A side alternator fails both
the A and B batteries can still be used (and left on-line at the same time to
maximize reserve power). According to Bob Nuckolls (Aeroelectric
connection) there is really no case where lead acid batteries drain each other,
as these batteries all supply power at a voltage less than they accept it
(minimum charge voltage). i.e. one battery cannot charge the other.
Also note, in my simplified sketch, I have
an essential/endurance bus which is supplied from both the “A” and “B”
side via a diode on each. Thus, the essential bus gets power even if the A-side
fails and I haven’t yet activated the cross tie. Thus, the EFIS doesn’t
quit and have to re-boot. This also provides some additional EFIS power redundancy
in the event of a simultaneous A side and cross-tie failure. Normally, my
“A” side voltage regulator will be set to slightly higher than the
B side (~0.1 volt). Thus, power will normally flow into the essential bus
via the A side. Alternatively, I could have a manual switch on the B-side
lead or a normally closed relay on the B side lead which is dis-engaged by the A
side (open then A is hot). I opted for the diodes as KISS in this case
and am seeking to minimize the number of switches. However, a manual
switch would create some additional starting spike isolation. Perhaps a
normally closed relay which is powered (dis-engaged, opened) by the starter
relay. Perhaps the B side essential feed should be tied to the B-side
alternator switch, since it would normally be off during starting (providing
isolation).
I’m also considering activating the
alternators based on an oil pressure switch (one for each side as redundancy).
That way the alternators com on-line automatically with oil pressure (just
after engine start). Without oil pressure the alternators won’t
work long anyway {g}. Each could also be turned off by pulling the
circuit breaker on the respective voltage regulator. I’m undecided
on that, but I digress a bit…
2) If I understand your second question
correctly… I have a shunt and ammeter between the battery and the
bus/alternator (on both the A and B side). These are not shown on my
prior sketch. On the B side it is past the starter so the “starting
current” does not go though the shunt (a special high current event).
Otherwise, both sides are essentially setup the same. Thus, the ammeter(s)
show the current flowing into (or out of) each battery. With the
alternator on, this current shown is normally very low (just any battery
recharging). With the alternator off, the ammeters will show the “drain”
being placed by the electrical equipment that’s on at the time. This
setup dose not directly show the total load on the alternator. However,
it is indirectly available. If you want to know (approximately) how much total
load is being placed on the alternator at any point in time, you can look at
the ammeter with the alternator on (the current going into the battery) and
then momentarily turn the alternator off and see how much current flows from
the battery to the electrical stuff. The current flowing out from the
battery to the other stuff, was previously being supplied by the
alternator. So, the total alternator load is (was) the sum of the two.
Note: the needle (flow) will shift
from positive to negative, which just indicates current direction relative to
the battery. The total current being supplied by the alternator is (was)
the sum of the absolute values (disregarding the direction into/out of the
battery).
Some folks (manufacturers) favor putting
the shunt in the alternator B lead (between the bus/battery and the alternator).
That setup shows total alternator load directly, but is useless in an
alternator failure scenario, when you might want to really know total current
draw to preserve the remaining battery power. It also provides no
mechanism for determining whether the battery is charging (health) or draining
(unhealthy). Others favor putting the shunt at the head of the bus(es) i.e.before
all the electrical stuff. That setup shows total equipment load directly,
but provides no mechanism for determining total alternator load, nor whether
the battery is charging (health) or draining (unhealthy). Thus, the std
battery ammeter setup seems the best approach (as I understand it).
Rick
p.s. Now back to re-checking the end
connections on the big ground wire (single point of failure).
From: Lancair Mailing List [mailto:lml@lancaironline.net]
On Behalf Of colyncase on
earthlink
I wonder if anybody knows what happens to the 20 amp
alternator if you put greater than a 20 amp load on it?
Also, how do you figure out what the load is if you are
draining a fully charged battery at the same time?