In article ,
Roy Lewallen wrote:

Tom Ring wrote:
. . .
1 week is 168 hours which gives us 403 ampere hours.
. . .

which at 13.6 volts is about 5480 watt-hours. This is the energy storage
requirement.

The OP asked about using a capacitor. The energy stored in a capacitor
is C * V^2 / 2. For capacitance in farads and voltage in volts, the
result is joules, or watt-seconds -- you need 5480 * 3600 ~ 20,000,000
joules in round numbers. So suppose you wanted to store this same amount
of energy in a capacitor, and you had a switching regulator which would
handle 50 volts maximum input voltage. Solving for C, assuming it's
charged to 50 volts,

C = 20,000,000 * 2 / (50^2) = 16,000 farads.

This assumes you can get all the energy out of the capacitor, which
requires your regulator to work down to zero volts. But you'll get 3/4
of the energy out of it if your regulator cuts off at 25 volts, 7/8 if
it cuts off at 12.5 volts, etc.

How does that sound?

Roy Lewallen, W7EL

well i guess it sounds like i wont have any 16,000 farad cap's stacked
up in my myplace anytime soon, but i understand the math behind it
now thanks

perhaps i'll just get a small cap to handle a brief second keydown dip
smooth things out a bit

thanks