REFLECTOR: How to protect the battery bus

[Terry Miles]

Dave,

Thanks you are absolutely right.  There is no resistance to flow until you
have flow.  I didn't make that clear.  Better to have said that you have a
known resistance potential in this big long battery to starter circuit in
the airplanes we are trying to wire up.  I was trying to toss out what I had
learned from Skytec and from my Penn State hob-nobbing regards starter
circuit issues.  

The 6 volt circumstance in my note was just a hypothetical number of what
might be the actual line voltage felt under a heavy mechanical load placed
on an electric motor.  It was intended to exemplify the concept of voltage
sag under an amp demand sufficient to pull the line v down, and that certain
other devices (as was mentioned in another post) can feel the effects.   

I thought Brian's points were good too.  I failed to make clear that 30 foot
run includes the (+) and the (-) route those little electrons must make, so
30 feet is a reasonable scratch number for all of us to use, and that any
terminals, connectors, etc the electrons have to transit just add to the ohm
count.  I struggle about how many facts to include when I have thought to
contribute based on how a guy phrases his circumstances and his question.  

My suggested experiment about the jump start falls in that same category.
Brian's was a better suggestion.  I am just a Saturday afternoon home
builder, not an avionics tech, or engineer by trade or training.  I am glad
we can all flavor the soup here.  Thanks to both of you.  We all benefit, or
at least I sure do.  Don't ever stop making improvements to anything I toss
out there on how to get these puzzles solved and into coherent working
systems.  

Terry



-----Original Message-----
From: reflector-bounces at tvbf.org [mailto:reflector-bounces at tvbf.org] On
Behalf Of Dave Philipsen
Sent: Saturday, February 17, 2007 7:08 PM
To: Velocity Aircraft Owners and Builders list
Subject: Re: REFLECTOR: How to protect the battery bus

Terry,

The general idea of your discussion is correct and maybe I'm being a 
little picky here. Just clarifying a minor point...the voltage that the 
starter sees at its input lug prior to engagement will be essentially 
the same as the battery voltage. With no load on the circuit there is 
virtually no voltage drop even if resistance is introduced (except for 
the miniscule drop caused by the load your voltmeter puts on the 
circuit). Try measuring the voltage of a 12.4 volt battery through a 
10,000-ohm resistor sometime and you'll see that without a load it's 
still 12.4 volts.

The calculation you used to find the voltage drop of the wire uses a 300 
amp load in the calculation and therefore doesn't apply to the voltage 
'prior to engagement'. The voltage drop is directly related to and 
occurs when the load is applied. Also, if you had a superconducting wire 
with virtually no resistance to electrical current, then your voltage 
measured at the starter, when engaged, would be the same as measured at 
the battery posts.

If you calculate the wire to have 1.4 volts loss at a certain amperage 
and you only measure a voltage of 6 volts at the starter while it is 
engaged, where is the rest of the voltage drop? If all of your numbers 
used for calculation are correct, then the extra five volts lost are due 
to internal resistance of the battery. Oddly enough, if you use 4awg 
wire as opposed to 2awg, even if the starter suffers due to the 
increased voltage drop of the wire, your other electronic equipment may 
not suffer as large of a drop during starting because it's presumably 
being fed by a wire on the *other* side of the larger voltage drop (away 
from the starter yet nearer to the battery).

Terry Miles wrote:
> Here's some related physics and math if you want it. Your battery has 
> a cranking amp limit. If the starter motor is asking for more than 
> what the battery has in cranking amps, then what happens is the 
> voltage starts to sag. Regards voltage demands and line loss: 2awg 
> looses .156 ohms per 1000 feet. If you had 30 feet for example. That 
> is .03 as a fraction of 1000. .156 times .03 is .00468 ohms lost. 
> Given a demand of 300 amps you multiply 300amps times .00468 ohms and 
> get 1.4 volts lost into the length of the wire. This is the voltage 
> the starter sees at its input lug prior to engagement. Say you have a 
> 12.4 battery, it can only deliver 11 volts to the starter. When the 
> starter is engaged the V will sag to say 6 volts aprox. That's why any 
> electronic equipment running at this point will drop off line since 
> most of that stuff needs about 8 volts or they shut off due to their 
> own internal protection circuits, but will pop back on when minimum 
> input voltage is sensed.
>
> If you are using 4awg, the volt loss to due to line length is greater. 
> If you are using 00awg the volt loss will the lesser. This does not 
> include any extra ohms lost to mechanical connectors, or other devices 
> in the line-especially diodes....
>
>

-- 
Dave Philipsen
Velocity STD-FG
N83DP


_______________________________________________
To change your email address, visit
http://www.tvbf.org/mailman/listinfo/reflector

Visit the gallery!  www.tvbf.org/gallery
user:pw = tvbf:jamaicangoose
Check new archives: www.tvbf.org/pipermail
Check old archives: http://www.tvbf.org/archives/velocity/maillist.html