I'm modifying a 14.4v flashlight to use an array of LEDs. The voltage of the white LEDs is 4.0v, and total draw is 0.145mA (measured).
This means dropping 10.4 volts, which means using a 72 ohm resistor (180 &
120 in parallel). This resistance evaporates 6-plus watts which is valuable battery power turned into useless heat.Would using a small voltage regulator in stead of resistors waste less power? How does one determine the comparative power consumption figures for regulators?
Thanks,
-- DaveC me@privacy.net This is an invalid return address Please reply in the news group
On Sat, 18 Mar 2006 10:57:43 -0800 in sci.electronics.basics, DaveC wrote,
Right. That would be very bad.
Not if it is a straight linear regulator. It would waste exactly the same amount of power to drop the same voltage as a resistor.
The obvious simple thing to do would be to group your LEDs into groups of three in series, operating at 12V with a much smaller drop across the resistor or regulator. Total current draw would be 1/3 of operating them all in parallel.
A switching regulator works by controlling the power instead of wasting the excess, and is probably what you really want, whether you use series groups or single LEDs. But, that's a little more complicated than a linear regulator.
Nope.
You could however either rearrange how the batteries are connected, or place the LEDs in series to get the total voltage drop closer to 12..14V so that a series resistor would dissipate less.
There are also several white LED drivers available, but mostly the problem is to get the voltage /up/... Still, if you google for "white LED driver" you might be able to find one that works up to 15V. A quick look gave the LM3590 inductor based step down, but it is good only up to ~12V input... A LM2575M-ADJ switching regulator might do, too, although it eats a lot of power just to run itself.
- Jan
Dave, somebody else already suggested this, but I will reiterate, your best bet is to put three led's in series, then calculate the resistor value that will give you 145 mA.
I find your voltage and current both to be odd values, the current being so low you would never need to change the batteries!
That said, if you are using 1.5 volt cells why not leave a few out and make an adapter to fill the extra battery space which would hold your added resistor to complete the circuit.
Temecula CA.USA
Chris is right. I suspect Dave meant 145 mA using maybe 5 or 6 LEDs in parallel since most ultra-bright LEDs have a max current rating around 30mA. Two strings of 4 LEDs will draw around 60Ma and will require a 40 ohm current limiting resistor that will dissipate 144 mW of power or slightly more than a single LED.
Of course, operating LEDs at the maximum current rating is not recommended since this will seriously shorten the life of the LED.
When you say an array of LED's, I would imagine each parallel string of multiple LED's should have its own 10 ohm resistor in series with it to balance the voltage between strings, then add your one main current limiting resistor.
Oh try this calculator,
I don't speak engineer-eze, I'm just a garage hacker.
Good Luck,
Temecula CA.USA
That's probably 0.145 A or 145 mA, isn't it? Since each LED uses about
20 mA that would correspond to 7 LEDs in parallel.Right, but the solution is simple: Put 3 LED in series, than they would draw 20 mA at 3 * 4 V = 12 V and you need to dissipate 2 V at 20 mA, thats 100 Ohm at 40 mW. Standart 1/8 W resistors would be perfectly fine.
Since you have 7 such LED you use 2 such chains in parallel, and end up with one left over. Of course, you can buy another 2 LEDs and use 3 chains. Total current would be 40 or 60 mA, respectively, at 14 V that's
560 or 840 mW. The 2 (3) resistors would be frying about 14% of the total energy, rather than 71% as in your current design.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.