Hi Dwayne,

Beg to differ. Even if the base were tied to the collector, the
transistor would not saturate, because the base is one Vbe higher than
the emitter. That's where it is at zero load.

Depending how close they are to their maximum ratings, most transistors
will saturate with less than 0.2 volts C-E. By using just one resistor
it is easy to tailor the resistor value so the output voltage is as high
as possible while still keeping just above the dropout point, through a
large variation in current. As the load increases, the base current does
too, so the resistor drops more voltage and it all stays clean.

Best regards, Bob

On Fri, Mar 6, 2015, at 02:54 PM, Dwayne Reid wrote:
> HI there, Bob & Joe.
>=20
> What Bob drew should work, except that it is missing a resistor in=20
> parallel with the capacitor on the base of the transistor.  The base=20
> voltage needs to be at least 1 Vbe lower than the lowest point in the=20
> ripple on the incoming power supply.
>=20
> dwayne
>=20
>=20
> At 10:17 AM 3/4/2015, Bob Blick wrote:
> >Hi Joe,
> >You could use a one transistor pass element to multiply a capacitor. Let
> >me see if I can find an example on the web... nope. here's napkin art.
> >
> >Bob

--=20
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