I have 3 led lights that operate using 4.5 volts D.C.
Instead of frequent replacement of 1.5 volt AA batteries, I would like to use a
12V 5AH Sealed Lead Acid Battery.What would I need to step the voltage down ?
Very much appreciate your suggestions.
Andy
A linear regulator such as a LM317 or many of the buck converters or LED drivers offered by Linear Technologies come to mind. I'm sure that there are other options.
You could use a 12VDC to 120VAC inverter to power a 120VAC to 4.5VDC wall-wart. ;-)
the right size of resistor. how much current does the light use when fitted with fresh cells?
If you can't be bothered measuring divide 500 by the number of LEDs and use that many ohms, it should be close enough
-- Neither the pheasant plucker, nor the pheasant plucker's son. --- news://freenews.netfront.net/ - complaints: news@netfront.net ---
"Andy K"
** Have you tried wiring the 3 lights in series ?? 1.5V AA cells drop down to 1.1 V when almost flat - so the lights will still operate at 3.3VA 12V SLA is nearly flat at 11 volts.
Looks perfect.
... Phil
Ebay item 200964328225
Buck Converter Step Down Adjustable Power Supply Module Output 1.25-35V LM2596
$1.70
It is a buck regulator and will be a lot easier on the batteries than a resistor or linear voltage regulator. The down side is the delivery time.
Dan
Daym good observation!
It works so perfect that it looks like a homework problem.
12V SLAs run ~13.5V when fully charged and 4.5V * 3 would be ....13.5V.
For LEDs you need a >> constant current
How much current do the LEDs use? If it's low, a linear regulator or maybe just a series resistor would be OK.
People make little, inexpensive potted things that look and act like 7805 regulators but are actually buck switchers inside. That would be more efficient than a linear regulator. If your load current is low, a switcher isn't appropriate.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation
You could check the data sheet on the LM3404 chip from National Semi. I used the LM3404HV in a system with 10 LEDs in series at 300 mA, and it worked quite well. I'm not sure the
3404 (without the HV) will work down to 12 V, but I think it might. it needs a Schottky diode and a small inductor and low-Ohm resistor for current sensing. You also need a small capacitor or two, but that's it, a very simple circuit.Jon
o use a
e just
icient
a board with an LM2596
cost less than the chip alone from digikey
-Lasse
If you believe it's really the same chip.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
If your load current is low, a switcher isn't
I am not disagreeing with your statement, but would like to know the reason ing. Is it because with low current , higher efficiency does not save much power? So it takes a long time to recover the additional cost. That I c an easily see. But would like to know if there is any other reason.
Dan
oning. Is it because with low current , higher efficiency does not save mu ch power? So it takes a long time to recover the additional cost. That I can easily see. But would like to know if there is any other reason.
at low currents a switcher may not be very efficient because it takes curre nt to run it.
if the switcher needs say 10mA it wouldn't be very efficient if you only ne ed 10mA output current
-Lasse
Many switchers are quite efficient down into the double-digit microamps.
** It's so damn obvious, no-one else can see it ......
FYI:
Why is it that all the high falutin folk here reply to questions as if they were hypotheticals set on a test paper.
When the real situation is NOTHING of the sort.
.... Phil
The TI data sheet for the LM2596 shows it is about 80% efficient with 12 volts in and 4.5 volts out. That is a lot better than a voltage dropping resistor or a linear voltage regulator.
Dan
Switchers with 95% efficiencies at moderate currents is pretty easy to find. Even at 50%, you're winning.
On a sunny day (Sun, 09 Mar 2014 11:42:58 -0400) it happened snipped-for-privacy@attt.bizz wrote in :
But this chip (using it) has > 16mA quiescent current, so at 10mA load is not really effcicient. Its a 3A regulator, then it gets to 90% at high Vi.
I didn't look at that part but there are several synchronous bucks out there that have 10uA Iq. Some are better than others, when you actually want to power something. ;-)
It's also a good idea to size the buck to the load. One size doesn't fit all (that's why TI makes hundreds of 'em ;).
The Simple Switcher parts are good in instruments, because they don't spray all sorts of EMI around.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.