I suggest a LF50CV, Very Low Dropout Voltage Regulator. You can get it fro=
m Mouser. Part no. 511-LF50CV It draws a very small amount of current=
and will still provide 5.0volts out until the battery voltage gets down to=
5.2 volts. Make sure you put an ON/OFF switch between it and the battery s=
o that it draws no current when you=E2=80=99re not using it.
If someone wants a schematic I can draw one up right quick.
Gene
From: =
.com] On Behalf Of Gregory O'Drobinak
Sent: Friday, September 17, 2010 3:50 PM
To:
Subject: Re: [Nature Recordists] Re: Sound Devices updates the USBPre
Flawn:
Be very careful with stuff you find on YouTube! I don't believe that soluti=
on
will work at all in an application where you need 500 mA @ 5v to power a ra=
ther
expensive I/O unit. The 9v battery will not be able to provide enough curre=
nt to
power the USBpre for very long; it will die quickly. Then the Zener trick i=
s
only useful for peripherals that draw a small amount current since it opera=
tes
in shunt regulator mode (don't even try it if you value your USBpre!).
You really need a proper circuit that uses an IC built for this purpose, in=
either a series regulator circuit (wastes power & generates lots of heat) o=
r a
switching regulator circuit (gets the most out of your battery, can be very=
efficient). The reason why commercial power supplies seem expensive is beca=
use
of the engineering needed to produce efficient power conversion that has
fail-safe features to protect the powered device.
I would look at Maxim or National Semiconductor or Texas instruments. They =
make
good power conversion ICs and have lots of application notes to help you ou=
t.
Take care,
Greg
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