Took me some time to understand your idea. 

http://www.tnt-audio.com/clinica/regulators_noise3_e.html

A picture of what it is suppose to be(based on your
description.)

John

--- Olin Lathrop <olin_piclist@embedinc.com> wrote:

> Jinx wrote:
> > For example, enter 10 (volts in), 4.7 (zener
> voltage), 10 (mA load)
> > and click Solve -> 265ohms 0.1W resistor, 0.1W
> zener
> 
> Yes, but the resistor will drop 2.65V all the time
> when the load is 10mA,
> even when the input voltage is the 5.0V it's
> supposed to be.  That means the
> PIC Vdd will be 2.35V with 5V in.  The 10F will
> operate down to 2V, so maybe
> that's OK, but the OP should at least be aware of
> this limitation.
> 
> If this is not acceptable, you could try a 5.6V
> zener with emitter follower
> as a crude regulator.  That requires one additional
> NPN transistor from the
> plain shunt regulator described above, but will only
> drop about 700mV when
> the input voltage is below about 5.7V.  Above that
> it will regulate
> reasonably well to 5V up to the C-E voltage limit of
> the transistor.  The
> extra $.03 for the MMBT4401 or whatever will be
> worth it in many
> applications.
> 
> By the way, 10mA is quite high for a 10F itself,
> although the whole circuit
> could easily draw that.  The OP needs to provide
> some real specs.
> 
> 
>
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