Gary Emerson wrote:
Tinwings wrote:

Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.


If the batteries are both "good", but not charged equally, then the last
part is true. If a battery happens to go "bad" and won't take a charge
then the situation is different. Then the "bad" battery will just suck
down the extra power from the good battery and that power will not be
recovered.
Sorry, didn't read your post fully. You were only talking about the
very short period when the batteries are still connected while switching
from A to B. The nice thing about the diode setup is you'll never have
to fiddle with a battery switch again.