9v Battery to 5v

Hi Jim,

I don't know what your application is, but many devices that require 5V have
a 'window' of 4.5 to 5.5V. So you could either use 4 AA NiMH batteries
(4X1.2V=4.8V), or you could use a 9volt battery and a chip called a 7805
voltage regulator, which would regulate the 9volts into 5volts provided it
is connected correctly. The 7805 draws some idle current, so you would need
to measure that and figure out how long the 9v battery will last.

Joe

 

I have a circuit that needs 5v and i need to use batteries. How can i use a
9v battery and step is down to 5v?

thanks,

Jim

 

Using a 7805 will waste about 45% of the battery life.

If battery life is a big consideration, and noise isn't an issue, then use a
buck regulator to step down the voltage. You can buy switching regulators
that will take 9V down to 5V with pretty good efficiency. If you need
something small, and don't need a lot of current, and can build circuits,
try this one:

http://www.romanblack.com/smps.htm

It looks pretty easy to build, and is claimed to be 80 to 90% efficient.

Regards,
Bob Monsen

 

I have a circuit that needs 5v and i need to use batteries. How can i use a

If you're using a 9V battery, you obviously aren't interested in a high current
regulator. If you look at the LM78L05 (capable of supplying 100 mA), you need
a minimum input voltage of 7VDC, which isn't too bad. The 9VDC battery is
pretty much cashed by the time it drops down to 7.0VDC. You might have a
problem, though, with I(q), the quiescent current you need to run the regulator
apart from any load the regulator may be supplying. The LM78L05 has a typical
I(q) of 3 mA, with a maximum of 5 mA.

Obviously, that's going to limit your battery life, especuially if your load
uses less current than the regulator. If you have a low current application,
and you want to extend the battery life, you might want to look at some of the
low current LDO (Low Drop Out) regulators. One example is the LM2936-5.0,
which is good for 50 mA load max. It has a dropout voltage of 5.5V, but also
has an I(q) of 15uA for a 100uA load, 0.5mA for a 10 mA load, and 2.5 mA for a
50 mA load. If you figure a couple of bucks per Duracell, your LM2936 will pay
for itself with the first battery if you have a low current load.

Good luck
Chris

 

Insufficient information. Are you building one of these? Ten? 100K? Does
it need to run for a few hours? Days? Months? Do you need 10 mA? 100 mA?
1 A?

If it's a small, onsie-twosie project, MPJA has a closeout on a 5V
switcher that would run nicely off of 4x AAA or AA. It's perhaps not
"optimal" and it has a -5V output that you might not need but at under
$1 it's a pretty easy drop in solution that avoids the losses associated
with a linear regulator.

http://www.mpja.com/productview.asp?product=2308+PS

 

another vote for picking up 4 NiMh AAs if you have the room. 2000 ma/hr
batteries are very affordable now. Then regulate them if need be.

People can offer better info if you give more details of the project.