My past experience with this says that my way works better.

Here's why:  think of the transistor as being an=20
emitter-follower.  The emitter will faithfully reproduce the voltage=20
that is on the base, except that it is one Vbe lower.

Now look at what happens when there is significant ripple on the=20
collector.  First, look at it without the transistor.  The capacitor=20
will float to the *average* voltage on the input.  That is: about=20
half-way between the peaks and valleys of the ripple.

Now add the transistor back into the equation.  Assuming a light load=20
on the collector, the base voltage will drop slightly because of base=20
current - but not very much.

When the input voltage drops below the average voltage on the=20
capacitor, the transistor can't supply current.  You see ripple on the emit=
ter.

Adding a large-value resistor in parallel with the capacitor drops=20
the base voltage slightly and now the transistor always supplies=20
current.  No ripple.

I have always set the base to be about 1 Vbe below the expected=20
valley voltage on the input.  I get nice, clean DC at the transistor emitte=
r.

dwayne


At 04:13 PM 3/6/2015, Bob Blick wrote:
>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.
> >
> > What Bob drew should work, except that it is missing a resistor in
> > parallel with the capacitor on the base of the transistor.  The base
> > voltage needs to be at least 1 Vbe lower than the lowest point in the
> > ripple on the incoming power supply.
> >
> > dwayne
> >
> >
> > At 10:17 AM 3/4/2015, Bob Blick wrote:
> > >Hi Joe,
> > >You could use a one transistor pass element to multiply a capacitor. L=
et
> > >me see if I can find an example on the web... nope. here's napkin art.
> > >
> > >Bob


--=20
Dwayne Reid   <dwayner@planet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
www.trinity-electronics.com
Custom Electronics Design and Manufacturing

--=20
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