At 17:17 17 February 2016, Steve Koerner wrote:
I think the right way to look at this is to take a gander at the spec
sheet=
of the referenced part. At the top of the sheet is a table showing the
re=
sistance that the breaker introduces for varying current ratings. It
shows=
for example that a 5A breaker introduces 0.03 ohms. Next, consider the
p=
eak load of the device that you're powering. If it's a radio, for
example,=
look at its transmit current. Let's say the peak current is 1 amp.
Then
=
the voltage drop across the breaker is: V =3D iR =3D (1)(.03) =3D 30 mV.

=
That's insignificant. The actual effect on battery life is that it will
mo=
re likely be extended by the introduction of the additional load; it's
just=
that the drop out point for the connected instrument will be 30mV lower.

=
For a lithium battery with a fairly sharp fall off, I think you'll
normally=
end up net positive to battery life by the introduction of the small
serie=
s load (as long as the connected instrument works down to the steep part
of=
the battery discharge curve). =20

And, BTW, fuses have resistance too -- it's the same game with a fuse.

I was using ETA 106-P30 trips that have internal resistance of 0.9ohm for
the 1a and 0.05 for the 5 amp. The problem was my Becker radio which only
likes to work at 12V+. The 5amp breaker dropped the voltage by over 1V on
transmit which for SLA batteries made it unworkable. A standard blade fuse
had much less impact.

I now use LifePo4 so this is now a mute point. However one point to note
with trips is the trip time. Mine were 10 seconds at 2x rated load.
Standard fuse trips much faster.

Jim