No, I'm saying a linear constant current source is no more efficient that a resistor because a 1F capacitor charged to 1V will store 0.5J, yet it will take 1J to charge from 1V source powering a linear constant current regulator.
Sorry, I probably didn't explain the formula I previously posted well enough.
Ein = Energy provided by the power supply = 1J
E(c) = Energy stored in the capacitor. = 0.5J
The size of the capacitor or current makes no difference, if you look at the graph of it charging, the area on the top side of the graph is still equal to the area on the bottom.
In practise, a linear regulator will be less efficient than a resistor because it will have a certain drop out voltage.
Hi again,
Oh ok great, that is what i found too.
I also looked at a switcher that can maintain the 'charging' voltage a small
amount above the cap voltage though a small sense resistor and found that
the power lost in the resistor is greatly reduced. Without trying too hard
for example a 0.1v voltage differential and a 0.1 ohm resistor over 1 second
(assuming a quick charge time) uses only 0.1 J. To contrast, the cap stored
0.5 J in that time.
That's basically how i charge my Li-ion batteries too for the last few years
or so. The other nice thing about a switcher is that it can often take a wide
range of input voltages without too much difference in operation. My charger
for example can work with wall warts of 9vdc to about 24vdc, and even will
still charge (slower) with a regular non regulated 6vdc wall wart. Thus, i
can grab any wall wart i have laying around (with same connector) and
use it to charge with. Of course that means it will also accept 12vdc from
the automobile too, which makes it even more versatile.
To get this to happen a regular switcher only needs to be (carefully) modified
to regulate current instead of voltage, and then voltage when the current
drops.