# Using 9V battery to power pair of LEDs?

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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:

Help would be appreciated.

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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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.

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Regards,
Bob Monsen```
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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```
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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```
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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.

• posted

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

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

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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.

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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

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