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:

enter image description here

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?


  • 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?

  • 5
    \$\begingroup\$ 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. \$\endgroup\$ – Olin Lathrop Jul 24 '14 at 18:17
  • 1
    \$\begingroup\$ Look at page 3, it's in English - all you need to know is there. \$\endgroup\$ – Jim Jul 24 '14 at 18:37
  • 5
    \$\begingroup\$ 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? \$\endgroup\$ – Russell McMahon Jul 25 '14 at 14:24
  • 2
    \$\begingroup\$ 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. \$\endgroup\$ – Russell McMahon Jul 25 '14 at 14:28
  • 1
    \$\begingroup\$ @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. \$\endgroup\$ – 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

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


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

\$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.

  • \$\begingroup\$ Thanks, Russell! how about this one? digikey.com/product-detail/en/XHGAWT-00-0000-00000HXE3/… \$\endgroup\$ – Jim Jul 24 '14 at 21:01
  • \$\begingroup\$ Also Russell, about inductor, since 15mW might be a bit too much with limited power, maybe 10mW or so is better, for example digikey.com/product-detail/en/B82432T1824K/495-1984-1-ND/723973 for 820uH (it's not nH, but uH!). Will that work? \$\endgroup\$ – Jim Jul 25 '14 at 0:07
  • 1
    \$\begingroup\$ Look at Samsung LM561B - datasheet & Digikey pricing . Best bin gives 165 lumen/ Watt at 150 mA and will give better again at under 5 mA. Digikey pricing in cents is 48 31 18 for quantity 10 100 1000. There are cheaper and probably even better when I get to them. You do not need this much power. \$\endgroup\$ – Russell McMahon Jul 25 '14 at 22:44
  • 1
    \$\begingroup\$ The part Digikey list above appears to be the top flux bin (aka best) available. Also excellent (to my eye) colour - R1-R6 chromaticity bins on page 6 of datasheet. This is on a somewhat daylight side of warm-white but not too blue. Notionally 5000K - and tightly clustered for colour (not that it's liable to matter in most cases). Probably about 190 lumen/Watt at 5 mA ! Maybe 200 l/W on a really cold night :-). That's a nice LED. \$\endgroup\$ – Russell McMahon Jul 25 '14 at 23:13
  • 1
    \$\begingroup\$ @Jim The 820 uH inductor has a horrendous claimed resistance in the data sheet. 27 Ohms. If true then at say 10 mA peak current voltage drop = 270 mV which is a significant loss. Peak current in inductor is >= Vout/Vin x Iout /DC where DC is duty cycle ton:(ton+toff) ratio. If this is say 2:1 then Ipk at 5 mA LED current >= 3V/2V x 5 mA x 2 = say 15 mA so maybe 400 mV peak voltage loss in inductor. Would probably still run OK. Just annoyingly lossy. A better part in 1k up in China would be under 10c and maybe under 5c. \$\endgroup\$ – Russell McMahon Jul 25 '14 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.