> I need to power a uP circuit with 3.3V @ 6mA from a 4700uF capacitor
> charged to 12V. The circuit has to run until the capacitor 
> discharges.
> Have tried the Roman Black's SMPS circuits, but could'n go over 50%
> efficiency at this 6mA current output. Even the Zener diode current 
> is
> excessive...
>
> Any ideas? Low cost and part number is a constrain.

Depending on how important those two things are relatively 'my' GSR 
circuit may be useful.
I presently have a version giving over 90% efficient driving a Luxeon 
1 Watt LED from 10 volts.
Getting this efficiency takes 'a little care' (tm)  :-)

Parts count is not as low as using any number of off the shelf 
controller ICs.

At 3.3V out a Schottky flyback diode alone at 0.2V will account for 
about 0.2/(3.3+0.2) =~6% efficiency loss.
A FET could be used as a synchronous flyback diode in the GSR if 
desired at the cost of added driver complexity.

Can you give us more details about how low the cost needs to be, what 
"low part number means" and approximate production volumes.

I may be interested in developing you a GSR (or some other solution) 
to a predefined spec as a paid job (payment only on meeting 
specification) if other alternatives didn't meet your need. This would 
only make economic sense if volumes were large. Advice on doing it 
yourself with a GSR is (up to a point :-) ) free.

The main advantage of the GSR is liable to be low component cost as it 
uses 3 or 4 low cost transistors and no controller IC. The 
disadvantage is higher parts count than an integrated IC controller. 
In an SMD implementation the board area may not be significantly 
larger than for an IC solution. .

Another highly successful alternative is the use of an eg 74C14 hex 
Schmidt inverter as a controller. I have a design which draws under uA 
providing voltage controlled no load output. This is liable to be 
physically larger than a "proper" IC smps based solution but with far 
lower component cost.

BOTE calculation indicates that the 4700 uF / 12V max storage 
capacitor is liable to provide about 6 seconds of storage with a 
linear regulator and approaching 15 with a very good smps. Tripling 
the storage capacity would give you the same result with a compact 
linear solution and MAY be a superior solution depending on other 
factors that you haven't discussed. (Cap volume, cap already exists, 
...)



        Russell


Linear time:

8 volts drop, 6 mA load.
T = CV/I = 0.0047*8/.006
~=6

SMPS

Energy available = 0.5 x C x (Vmax^2-Vmin^2)
Vmax = 12V
Vmin = 5V say
0.5 x 0.0047 x (12^2-5^2) ~= 280 mJ
Power out = 3.3V x 6mA = 19.8 mW
Time = 280/19.9 ~= 14.






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