> With a single capacitor this would be true , if you would 
> charge a
> 1 uf cap from the 20 volt and subsequently charge a  3 uf 
> cap from
> this charge you end up with 5ish volts at a fairly high 
> efficiency only
> the loss in the two switches and the series resistance of 
> the caps
> would matter

No, alas.
What you say about which resistances matter is true, but the 
consequences are not what you expect.

To amend Watergate's 'Deep Throat's advice - 'follow the 
current'.

ie work out what the current MUST do to get to its 
destination and what must happen to it along the way. 
Realise that a zero resistance switch is an infinite current 
switch.

If you posit zero loss connections then when you parallel 
two capacitors you get infinite current. The losses tend to 
be high :-).

If you posit a finite resistance, even very small,  then you 
can do the arithmetic to derive actual currents and losses 
and the result is "not good" [tm].

As a demonstration of the outcome.
Charge a capacitor C to Voltage 2V.
Discharge into an equal cap C so combined voltage is now V.
Original energy is 0.5CV^2 = 0.5 x C x (2V)^2 = 2V^2C.
Final energy is 2 x 0.5xCxV^2 = V^2C.
*HALF* your energy has vanished!!!!
Q: Where to?
A: Into the necessary resistance that allowed you to carry 
out the energy transfer.

Caps CAN be charged resistively at good efficiency by 
limiting the voltage excursions to a small percentage of the 
voltages involved. People make very efficient convertyers 
and multipliers doing tghis. But, discharge or charge the 
cap resistively by a substantial percentage of it's 
greatests voltage and efficiency will drop badly.

My suggestion re caps which are charged in parallel and 
discharged in series used physical space division to perform 
energy translation.
ie Charge C+C in series to 2V. Now parallel them so you get 
2C at V. An energy analysis will show that for ideal 
components no energy is lost when changing from 2V to V.


        Russell












>
> Peter van Hoof
>
> ----- Original Message ----
> From: Apptech <apptech@paradise.net.nz>
> To: Microcontroller discussion list - Public. 
> <piclist@mit.edu>
> Sent: Tuesday, December 4, 2007 5:44:20 PM
> Subject: Re: [EE] Challenge II:: LED driver
>
> Comment on one aspect:
>
>> It also does not need
>> to be an efficient buck converter, PWM is fine.
>
> This alas is a fallacy.
> If you use PWM to provide a given output current at a low
> voltage from a higher voltage then the output power will
> indeed be I x Vlow, as desired, BUT the net power 
> dissipated
> will be I x Vhigh and the difference will be dissipated 
> just
> as with a linear regulator.
> If this sounds unintuitive (and it usually is) work 
> through
> the arithmetic while ensuring that ALL real world effects
> are accounted for. eg if you apply 20 volt to a 5 volt
> capacitor to charge it to 10 volts account for either some
> series resistance and finite charging time OR zero
> resistance and infinite currents.
>
> Note that infinite current^2 x R tends to spoil your day.
>
> ONLY with an energy transformer will you get a net 
> reduction
> in power dissipation.
>
> Unfortunately.
>
>
>        Russell
>
> -- 
> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist
> -- 
> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist
> 

-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist