> But this is very easily and cheaply fixed. =A0Just put a regular $.10 dio=
de
> reverse accross the power supply after the Schottky. =A0That will limit t=
he
> reverse voltage due to the Schottky leakage to one diode drop, which is
> quite unlikely to cause a problem. =A0It will have no effect during normal
> operation.

The reverse diode can itself  be a Schottky, limiting voltage to below
a typical silicon junction voltage.
It in turn contributes its reverse leakage but here it is just a small
extra psu load.

The MOSFET protector wins over the Schottky when the FET voltage drop
is below what can be reasonably achieved with a Schottky *AND* when
the voltage drop is significant to  the application.

In my lights I have 3 x AA NimH with a cuttoff voltage of around 1.1
V/cell. At the currents in use I may be able to get Schottky voltage
down to say 0.3V.

At 3 x 1.1 I have 3.3V available. Add a Schottky drop and with same
battery cutoff I have 3.0V available. It happens that that's close to
the minimum allowed by a circuit component (2.9V nominal), but that's
not the main issue. Here 0.3/3.3 =3D 9% of the energy is wasted. When
charged battery delivers about 3.75V (more for very brief while) so
loss is 0.3/3.75 =3D 8%.
When you want every % you can get for lighting, losing 8% for about
zero gain is very poor.
eg if a light will run for 5 hours on a full charge you get
100/(100-8) -1 (to show the thinking) =3D 25 additional lighting minutes
by leaving out the diode. Turn to 1/4 power when needed and you get
1.5 hours more still-useful light :-).

At 0.3V/ 200 mA a Schottky has 1,5 ohm effective resistance
!!!!!!!!!!!!!!!!!!!  It's not too hard to get a FET to improve on that
(although full enhancement at 3V Vgs is a little rare. )
.


                  Russell

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