It seems to me that the calculations in the original post are not quite right.
> 1F = 1V * 1 Coulomb.
The Farad is a Coulomb per Volt or Q/V not Q * V
This starting point inserts errors in the calculations that follow.
For example:
> Thus 1F = 1V * 1 Amp Second.
The unit V * Amp * sec is equal to watts * sec = energy in joules. A Farad is
not a joule of energy.
Also note that the voltage drops in a capacitor as you draw current from it, so
you need some circuitry to keep the voltage up to keep your QRP rig happy. As
a result there is probably some practical level where the capacitor still has
energy left, but it can not be effectively used any more. (The rig would still
work for a while as voltage dropped, but in the example given the voltage is
considered constant at 12 Volts.)
A better approach to this calculation might be to start with the total energy
stored in the 5000 F cap that was charged to 12 V. The energy stored in a cap
is:
Energy = (1/2) * C * (voltage squared) where C is in Farads, V is in volts,
and Energy is in Joules
In the example that was used, the QRP rig takes a constant 12 volts dc at a 0.5
amp. This is 6 watts or 6 Joules per second. If this constant power could be
drawn from the capacitor until exhausted, then the time it would last would
directly follow by dividing the initial energy in the cap by the rate of energy
removal. However, the added circuitry mentioned above between the cap and the
rig would have losses, minimum useable cap voltage, etc.
73, Richard, KG4VOX
_______________________________________________
Elecraft mailing list
Post to: [email protected]
You must be a subscriber to post to the list.
Subscriber Info (Addr. Change, sub, unsub etc.):
http://mailman.qth.net/mailman/listinfo/elecraft
Help: http://mailman.qth.net/subscribers.htm
Elecraft web page: http://www.elecraft.com
