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Using 9V battery to power pair of LEDs?

I have a pair of clip-on high brightness white LED "computer lights" each normally powered by 5V from a USB port on the computer. Each is wired in series with a 15ohm 1/8W resistor. I'd like to gang them together independent of the computer and use a 9V battery for power. Where I'm getting confused is calculating what resistance to use to drop the 9V to a suitable value to handle both of them. I don't want to change the resistors tied to the LEDs; they're encapsulated and encased in a metal ferrule.

If you haven't seen these gadgets, there's a picture at:

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Help would be appreciated.

Marty

Reply to
Marty
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Hello Marty,

I would use a regulator, preferably a little buck switcher. 9V batteries are rather wimpy and their voltage begins to sag early on. Personally, I'd avoid 9V batteries altogether and rig up a boost converter around a couple of AA cells. AAs are also cheaper.

Regards, Joerg

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Reply to
Joerg

I'm guessing you can connect them in series. The LEDs probably have 3.4V forward voltage, and so with a 15 ohm resistor, they will draw about 100mA of current each. So, at 9V, two together will draw (9-6.8)/(30) = 73mA, which is probably good enough.

--
Regards,
  Bob Monsen

It is the theory that decides what we can observe.
    Albert Einstein (1879 - 1955)
Reply to
Bob Monsen 

--
Assuming that the [white] LEDs have a drop of about 4V and, as you
say, a 15 ohm resistor in series with each them, with a 5V supply
across them: 

      +5V
       |
   +---+---+
   |       |
 [15R]   [15R]
   |       |
 [LED]   [LED]
   |K      |K
   +---+---+
       |
      GND

the current in each LED would be:

          Vs - Vled     5V - 4V
     I = ----------- = --------- ~ 67mA
             RS           15R

That seems a liitle high, since most LEDs are rated to carry 20mA.

Are you sure your numbers are right?
Reply to
John Fields

John Fields wrote in news: snipped-for-privacy@4ax.com:

Actually,white light LEDS have an average Vdrop of between 3.4V-3.6V. I have a LED flashlight that has a .5W LED,so it would draw about 150ma,the Luxeon 1W LEDs draw ~300ma(at ~3.5V). I also have "bright-white" LEDs with max I of 25ma,a batch of 10 measured Vdrops of 3.5V to 3.9V,most in the 3.4-3.6V range,only one measured

3.9Vdrop. 
--
Jim Yanik
jyanik
at
kua.net
Reply to
Jim Yanik

Thanks all for the comments. Meantime I found that I could hook up both LEDs to my single available USB port using a Y connector and it didn't complain about an overload. Will leave it that way until I put together the independent battery operated version.

Reply to
Marty

FWIW, my current home project is to make an LED flashlight from a 9V battery and two white LED's. I'm getting about 21 hours of operation at 20 mA. Don't know if this would change your mind about using a battery (are your LED's to be on continuously?), but I just thought I'd mention it to you.

Mark

p.s. I'm not using resistors, but a two-transistor current regulator as shown at

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Scroll down to the section titled "LED tester", about 80-90% of the way down the page.

Reply to
redbelly

Thanks for the reference. That tester circuit will serve my purpose nicely since the light will be used only intermittently and moved to different positions so I can better see some LCD readouts that unfortunately aren't backlighted.

Reply to
Marty

You're welcome.

Since you might use such a circuit, I'll also mention that I use 2N3906 transistors, and get a slightly lower Vbe (0.65V) than what that website says for a 2N4403.

You'll have an extra voltage drop of 0.3V per 15-ohm resistor to contend with, so that will reduce the useful range of the battery. My tests show that current drops to 75% of 20 mA at a battery voltage of

7.2 V; for your circuit with 30-ohms of series resistance, the 75%-point becomes 7.8V.

Mark

Reply to
redbelly

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