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Topic: Inductor SIzing For 5252F Solar Charger (Read 376 times) previous topic - next topic

Qdeathstar

Hello,

I am trying to make some back lighted numbers for my address, and i am using a solar charge with an 5252F IC. After putting everything together, the LEDS dont seem quite bright enough, so i want to change the value of the inductor to allow more current through.


Currently the board has a 41 microhenry inductor (yellow brown black silver).

I was reading this site:

http://artists.scitoys.com/three_volts

And it seems like currently, the inductor is outputting a value slightly higher than 50ma. I would like try several different values of inductors so that i can use the least amount of current to produce an acceptable amount of light, but i am unsure of how to calculate the current based on the inductor size.

Any help would be appreciated!

5252F datatsheet for reference
https://www.mikrocontroller.net/attachment/158139/QX5252.pdf



There is a formula on the data sheet, but it doenst seem to create sensibe results:

Pled = (2VIN/L)*10^-6

(2*1.5)/41)*10^-6 = .00000007

Even if that is amps, its not right... :-/
A creaking creeping shadow
stiff against the freezing fog
glares at a tickless watch.

Time has failed him -- all things shall pass.

ReverseEMF

#1
Oct 16, 2018, 03:02 am Last Edit: Oct 16, 2018, 03:35 am by ReverseEMF
I'm no expert on this, but since no one else seems to be chiming in, I'll offer this.
Currently the board has a 41 microhenry inductor (yellow brown black silver).

And it seems like currently, the inductor is outputting a value slightly higher than 50ma. I would like try several different values of inductors so that i can use the least amount of current to produce an acceptable amount of light, but i am unsure of how to calculate the current based on the inductor size.

https://www.mikrocontroller.net/attachment/158139/QX5252.pdf



There is a formula on the data sheet, but it doenst seem to create sensibe results:

Pled = (2VIN/L)*10^-6

(2*1.5)/41)*10^-6 = .00000007

Even if that is amps, its not right... :-/
On the third page there's a table of inductance values vs current.  According to that table, a 41uH inductor should result in around 80mA.

Why not just use the table? I has a nice list of inductor values and currents.


Also, it would help if you told us more about your circuit.  Like how many LEDs are you trying to drive per circuit?

The datasheet shows a couple of LEDs in parallel, but that's no a great way to do it.  At any rate, if you do use more than one LED in parallel, the current will divide across those LEDs.  So, if the 5252F is supplying 50mA, each LED will get 25 mA [in a perfect world -- but the world is not so perfect, so the division of current will not be so predictable -- BUT, the currents in the LEDs, even if not equal, will always add up to the current being supplied by the 5252F.

But, regarding the formula, the inductance is probably in Henrys, so it 41x10-6, thus:

(2*1.5)/(41x10-6))*10-6 = .073  or  73mW -- which sounds about right

If one LED with a forward voltage of 3V, then 73mW/3V = 24mA
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Qdeathstar

#2
Oct 16, 2018, 03:15 am Last Edit: Oct 16, 2018, 03:17 am by Qdeathstar
i am driving 15 leds in parallel, and i want to drive them at between 10 and 15ma each which would mean about 100-150ma. The table doesn't go that high. The chip is capable of up to 300ma
A creaking creeping shadow
stiff against the freezing fog
glares at a tickless watch.

Time has failed him -- all things shall pass.

ReverseEMF

i am driving 15 leds in parallel, and i want to drive them at between 10 and 15ma each which would mean about 100-150ma. The table doesn't go that high. The chip is capable of up to 300ma
I could be wrong, but it seems like you're courting disaster with so many LEDs in parallel.  Is their brightness even, or do some appear brighter than others?  I would suggest a serial resistor on each LED, like around 75Ω. 

Also, I played around with that formula, trying to get the same values as in the table, and I can't get any corroboration!

For instance, if we take the 56µH case, it says the input voltage [I'm assuming that what they mean] is 1.3V:
PLED = ILED * ELED = 2VIN/L  [where 'L' is inductance in µH]

So:

PLED = 2 * 1.3 / 56 = 46.4mW
 
If: PLED = ILED * ELED Then:  ILED = PLED / ELED

and assuming an LED voltage of 2V:

ILED = 46.4mW / 2V = 23.2 mA 

The table says 50 mA!!

46.4mW / 50mA = 0.93V

So, are they assuming an IR LED?!?

I would just extrapolate from what you know.  Run the thing with one LED, and a 220µH inductor and measure the current.  Then go from there.
"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

Qdeathstar

The leds seem to be evenly matched. Running them on my bench power supply they light evenly. I'm going to be under driving them so i shouldn't have to worry about a particular LED  being over driven.
A creaking creeping shadow
stiff against the freezing fog
glares at a tickless watch.

Time has failed him -- all things shall pass.

ReverseEMF

The leds seem to be evenly matched. Running them on my bench power supply they light evenly. I'm going to be under driving them so i shouldn't have to worry about a particular LED  being over driven.
Good point  :)
"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

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