W
Winfield
Michael said:Well, two batteries can deliver current at one time if they're at the
exact same voltage, but yes uninterrupted current flow is necesarry.
And yes, it'd be problematic if one battery started charging another,
especially at 500A (the batteries I'm using are rated for 20A
continuous, 30A for up to 10 seconds, though I'm not sure about how
they handle really large transients). It's unfortunate that I have to
throw some power away in a Schottky, but as far as I can see my
options are very limited, unless I'm missing something. Doing source
to source FETs will require incredibly accurate timing along with a
large capacitor to make uninterrupted, again unless I'm missing
something. Thus it seems like the Schottky + single FET is the best
solution?
Here's one other thought that just struck me: what if I had two FETs
source to source as discussed. However, what if I put in place a
switching scheme so that the top (connected to battery) switch was
switched separately from the bottom FET, and that the bottom FETs are
driven in such a way that only one or less of them can be on at any
given time. Additionally, there would be a Schottky in parallel with
each bottom FET. The idea is that by placing Schottkys in parallel
with the FETs hot swapping would be enabled, but the second FET could
be switched on a moment after the other bottom FETs were switched off,
decreasing power loss.
Does this make any sense at all?
Thanks,
-Michael
Yes, it makes perfect sense. Schottky diodes to insure
a continuous voltage availability (Fred's suggestion of
an inductor would do that as well, but would require you
to consider the flyback voltages from the stored energy
in the inductor during switching -- might be reasonable),
and the paralleled MOSFET switched on later, when it's
safe, to reduce heating.
The setup would be a bit different than the one in Fred's
drawing following your post.
.. F1 d___s SD1
.. -- BATT1 >--/---|___|--+---|>|---+---+-------o buss
.. | FET-A | | s___d | |
.. | | '--|___|--' | LOAD
.. | S1A | FET-B |
.. | S1B | o
.. | | |
.. | F2 ___ SD2 | GND
.. +-- BATT2 >--/---|___|--+---|>|---+---+
.. | FET | | ___ | |
.. | | '--|___|--' |
.. | S2A | |
.. | S2B |
.. | |
.. | F3 SD3 |
.. +-- AC POWER >--/-----------|>|-------'
.. |
.. GND
The primary issues are figuring out when to switch the "B"
MOSFETs (e.g., turn on a delay-time after the "A" mosfets,
but **only** if V_BattN is > than V_battM, and turn off
immediately with "A" mosfets), and devising the appropriate
gate-drive voltage (the same as for the "A" mosfet). This
extra trouble would save you a little more than 10.5 - 6.3
= 4 watts, call it 5 watts, over wiring 4 sections of two
60CTQ045 diodes in parallel. I dunno if it's worth it.
This looks so much more simple:
.. F1 d___s SD1
.. -- BATT1 >--/----|___|-----|>|---+-------o buss
.. | FET | |
.. | S1A | LOAD
.. | |
.. | F2 ___ SD2 | o
.. +-- BATT2 >--/----|___|-----|>|---+ |
.. | | | GND
.. | S2A |
.. | |
.. | F3 SD3 |
.. +-- AC POWER >--/----------|>|----'
.. |
.. GND