Using 9V battery to power pair of LEDs?

Question

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

Joerg · Accepted Answer

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

Bob Monsen · Answer

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.

John Fields · Answer

---Assuming that the [white] LEDs have a drop of about 4V and, as yousay, a 15 ohm resistor in series with each them, with a 5V supplyacross them:       +5V       |   +---+---+   |       | [15R]   [15R]   |       | [LED]   [LED]   |K      |K   +---+---+       |      GNDthe current in each LED would be:          Vs - Vled     5V - 4V     I = ----------- = --------- ~ 67mA             RS           15RThat seems a liitle high, since most LEDs are rated to carry 20mA.Are you sure your numbers are right?-- John FieldsProfessional Circuit Designer

Jim Yanik · Answer

John Fields  wrote in news: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 Yanikjyanikat

Marty · Answer

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

redbelly · Answer

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.

Marty · Answer

Thanks for the reference.  That tester circuit will serve my purposenicely  since the light will be used only intermittently and moved todifferent positions so I can better see some LCD readouts thatunfortunately aren't backlighted.

redbelly · Answer

You're welcome.Since you might use such a circuit, I'll also mention that I use 2N3906transistors, and get a slightly lower Vbe (0.65V) than what thatwebsite says for a 2N4403.You'll have an extra voltage drop of 0.3V per 15-ohm resistor tocontend with, so that will reduce the useful range of the battery.  Mytests show that current drops to 75% of 20 mA at a battery voltage of7.2 V; for your circuit with 30-ohms of series resistance, the75%-point becomes 7.8V.Mark

Using 9V battery to power pair of LEDs?

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