Need a regulator or step-down circuit that steps a battery down from 7.2v to 4.7v - more info

OK, it is a Fisher CameraCorder I want to power from a 7.2v battery by converting to 4.7v. The battery that comes with
the Fisher tapeless camcorder is 4.7v Lithium ion battery rated at 720mAh. The battery can last for one hour continuous
use. I guess that is 720ma continually for one hour. Does this help you with the current draw? If not I can ask Fisher
or Sanyo. Thanks for your help.

The diode senerio looks extremely inefficent. How much current does the diodes draw on their own? I only want the camera
drawing most of the current when it is turned on. Again thanks to all for their responces.

 

Probably need to use a 5V regulator plus an additional diode to do what you
want. Nominal output of the available 5V regulators is 4.9 - 5.1 VDC. The .4
volt differential may, or may not cause problems. You can add resistors, or
diodes to drop that level or even a resistor and zener diode to clamp it
specifically at the 4.7 +- component tolerances. Simplest you'd seem to be a
regulator, series resistor, and 4.7 V zener circuit. Any circuit will
dissipate heat as it functions, that's he nature of what you are attempting
to do. By running the calculations you can determine what value resistor,
specific regulator, etc you will require for the project.

 

IdeaMan wrote:

[snip]

Don't use it. Blame that design on lack of sleep.

If you can tolerate 5.1 volts instead of 4.7, use
the 3 diodes in series with the battery that
someone else suggested. That setup will only
draw current when the camcorder is operating.

If you prefer 4.4v, make it 4 diodes...

 

You need more milliamp hours, not more volts. You dont say what the
rating of the 7.2 volt battery is. For a given technology, the
energy density is dependant on package size. I suspect you not buying
a thing by going to 7.2 volts. You'd probably be better to get 1 or
more spare 4.7 volt batteries.
Dan

Colorado Springs, CO
My advice may be worth what you paid for it.