# LED driver circuit help needed

Please help me figure out which inductor to use from this datasheet. Sorry - it's in Chinese as this is an Asian sourced IC and not sold in the US. (Use google translate if necessary - but, the meaning is clear enough from the digrams and tables. )

The formula for the inductor, given on page 3 is as shown below:

I'd like to use a single energy efficient LED like this (I'm open to other white LED suggestions)

I'm using 2 alkaline AAA batteries.

Question 1
Based on the formula given above, should Vin be

• 2V fully discharged point of 2 x AAA Alkaline cells?

or

• 3V (= 2 x new Alkaline AAAs cells) to avoid over current?

Question 2: to calculate P for LED, I use 5mA at 2.9V = 145 mW - which is normal conditions for LED (page 2 from LED datasheet) - good assumption?

I came up with 580nH - is this correct?

• No, I'm not going to use Google to translate. It is your job to properly present information relevant to your question. You have forgotten that you are coming here asking for a favor from a bunch of volunteers that are not obligated to help you. Too much hassle, moving on to a better written question. – Olin Lathrop Jul 24 '14 at 18:17
• Look at page 3, it's in English - all you need to know is there. – Jim Jul 24 '14 at 18:37
• It's a sad and sorry day when people let Olin & followers throw their weight around like this in order to "teach newcomers a lesson". This is bullying and arrant nonsense. If Olin or others got this as an exam question and could not understand it they would get a failing grade - be it an EE exam or an English exam. What sort of community are we running here? – Russell McMahon Jul 25 '14 at 14:24
• OK folks - hands up? How many people find this question unclear and do not understand what is being asked? If so, is your EE knowledge poor or is your English language knowledge poor? The question was edited from the original and THEN closed in the form you see now. Sure, it could be tidier, but the question should be clear to any person of modest EE or English language skills. If your English language or EE skills are so poor that you can't handle this you should get help rather than penalising others. – Russell McMahon Jul 25 '14 at 14:28
• @Funkyguy Op is uncertain whether he is "in the ballpark" as he is applying a formula from a Chinese data sheet without any feel for whether he has got the answer right. I and maybe you have an idea of how the formula was derived so can assess he has a correct result. | I strongly suspect that Olin's complaint is not with the validity of the question but that some other language reading is suggested and/or that he has not brought the formula out into the question. One of those concerns is valid. How to address such things is moot and this is Olin's consistent manner, as he has explained. – Russell McMahon Jul 28 '14 at 21:08

Yes. Your application of the formula is correct. I can provide part numbers for better LEDs than that. "Soon" ... (just stumbled out of bed to complete a report)

At about 15 mW a 'good' modern LED will give >= 150 l/W or about 2.25 lumen. That's enough to illuminate 1+ sheet of A3 paper to good readable level with colour well discerned. If illuminating a small enclosed space it should be more than you need.

The formula (from above) is

$$P = {V_{in}^2 \over L \cdot 10^-6}$$

or

$$L = {V_{in}^2 \over P \cdot 10^{-6}}$$

$L$ in henries, or remove $10^{-6}$ and use $L$ in μH.

With this sort of IC the brightness varies approximately with the square of the input voltage so you get a power ratio of 9:4 for voltage changing from 3V to 2V.

• A 2:1 difference is hardly noticeable to the eye if you look at one light at a time.
• Look at two side by side and you can distinguish down to the 20% - 50% difference range.
• "Wall wash" with two LEDs side by side and you can typically see differences down to 10%.

LED: Look at Cree ML-E to start. There are much better now. More on that soonish.