How to get CC through LEDs as battery discharges?

[kyle]

Hello,

I've built a simple circuit using 15 amber, 3000mcd, 2.0 forward voltage LEDs wired in parallel to be powered by 2x AA batteries. I figured that I would use NiHM AA and the Vs would be 2.4v. 15~20mA gives me the best brightness. I used 1 2.2ohm 1/4w resistor to limit the current going to paralleled LEDs. Few issues that I found out is that as voltage drops on the batteries the less current flows and LEDs get dim. Also I can't use 2x Alkaline AA, e.g. 2 x 1.62v = 3.24v, since current there is too much for the 2.2ohm resistor, I = 1.24v/2.2ohm = ~0.5A, that's around 0.7w of power dissipation. Since I was using a SPDT switch, I soldered extra 2.2ohm across switch's 1&3 pins giving me 4.4ohm, so I ended up with 2.2ohm/off/4.4ohm modes. That way I use 4.4ohm mode when alkaline AAs are inserted and flip to 2.2ohm mode once their voltage drops.

I have plenty of 4401 NPN and 4403PNP transistors lying around, and only 1 LM324 OPAMP, I would not prefer to use that since I still have to learn about OPAMPs. How could I build a circuit that would provide constant current to 15 LEDs as voltage drops so I could get the same brightness and use up all battery charge? I would like them to use up to 250mA or ~16.7mA each and make it work for battery voltage range of 2.0~3.3v.

I know that cut-off voltage for most batteries is 0.8v. Would a LED with 2.0 forward voltage get lit by 2x0.8v voltage source?

I was thinking about using NPN BJT to drive those 15 LEDs but as Vs drops so does the current going to the base of transistor  .

Thanks guys, I hope you can help me to build a better desk lamp!
Kyle.

[peter.mitchell]

ideally you'd want a small boost converter with a low voltage minimum input, and then a small switch mode driver too.
Limited components makes things more difficult though.

[amyk]

ideally you'd want a small boost converter with a low voltage minimum input, and then a small switch mode driver too.
Limited components makes things more difficult though.

A boost converter doesn't require much in the way of parts; it's the regulation that takes a lot more components.

[Emil]

You could use an LM317 as a constant current regulator.

If you google "constant current LM317" you will find many examples of this circuit.

Basically it's just a resistor connected between the Output and Adjust pins. The load is connected to the Adjust-pin. The current will be 1.25/R.

It will not work with voltages <1.25V, but that should not be a problem for this application.

I think 1V would be a more reasonable cut off voltage for alkaline cells.

[peter.mitchell]

ideally you'd want a small boost converter with a low voltage minimum input, and then a small switch mode driver too.
Limited components makes things more difficult though.

A boost converter doesn't require much in the way of parts; it's the regulation that takes a lot more components.

Indeed...

Does OP have a toroid or something suitable to wind an inductor on?

Edit: could take the parts cost all the way down and try it with cat5 and a big nut (eg, bolt +nut)

[kyle]

Thanks for the replies! I do have a 24awg and 30awg magnet wire, cat5 wires, and plenty of toroids cores/shells that I can salvage. I might even have the right Henry value choke. Ah and boost/buck converters, I got to learn about these one day after I finish a book that I'm currently reading: "Electronics Self Teaching Guide by Kybett, Boysen". Joule-thief, that's really interesting. I also do have many LM317s, however buck converter sounds cooler since it uses inductor, and it would be a good DC/inductor lesson for me.

I wonder if there was a clever way to maybe use some kind of voltage divider on a base of NPN BJT to accomplish similar task?

So far I'm leaning toward Joule-thief circuit to drive 15 LEDs that I have. Any suggestions or circuit that you would recommend?

[PA0PBZ]

I wonder if there was a clever way to maybe use some kind of voltage divider on a base of NPN BJT to accomplish similar task?

There is, but the problem is that you loose the voltage that drops over Ri:

[mariush]

You could probably use something like this:

http://uk.farnell.com/on-semiconductor/nsi45020at1g/ic-led-driver-45v-0-02a-sod123/dp/1794973
http://uk.farnell.com/on-semiconductor/nsi50010yt1g/ic-led-driver-50v-0-01a-sod123/dp/1794979

[Memphis]

, if you want something really easy for this problem, try this: How to build a Joule Thief. 

[SoftwareSamurai]

Although it's not CC, it does keep an LED well lit.  NCP1402

[vlf3]

You don't need voltage boosters, just low ohm resistors for each LED or series string of LEDs...

At 3.0 volt nominal AA battery; with a voltage drop for the LED of 2.0 each thus, 1 volt / .06 ohm = 16.6 mA... your problem is to find or make-up this .06 ohm value, being non-standard.

LED's in parallel is a very bad technical practice... each LED act's as a zener diode, and will set it's working voltage in your case, at 2.0 volts +/_ .  Any additional LED's placed in parallel, will individually draw more or less current away from the first, to your last LED; due the manufacturing tolerance variations.

 
Standard practice is to have each LED with it's own ballast resistor value, as the above calculation... in some respects a higher working voltage would be better, (using a standard ballast resistor value).

Series LED's would be that much better, e.g. 5 x require a total of 10 volts, feed with a ballast resistor value of 12.0 ohms for that string of 5 = 16.6 mA; you then parallel the 3 strings to give the 15 LEDs; powered from 12 volt battery having a 1 volt nominal capacity above the 10 volt LED regulation; otherwise a PSU is an option, so the total current would be about 49.8 mA.

In fact you could sub-divide a further set of series LEDs to 3, by 5 paralleled, at the expense of more current being drawn though.

 
A voltage booster needs to be given an input to output current capable of delivering the total power required, in addition to minim loss and high efficiency, from the device generating the extra boost voltage; this is particularly so in your case, with 15 LEDs required.

[kxenos]

Since you're powering the thing with a battery you don't want a linear solution because you want to avoid the associated power losses. A pretty easy switch-mode solution would be either a ready made step-up module module from e-bay (around $3) or a self-made one using an IC like MC33063 for which there are spreadsheets to help calculate the components values. You can have a 50mA 12V output that will go to 3 rows of 5 led each. Total V drop would have to be 10V then. So you must place a resistor for each row as mentioned. For example, you can place a 0.1Ohm 0.5W resistor in each row. To calculate the power consumed in the resistor you will have Pr = (Vout-Vled)*I --> Pr = (12-10)*0.1 --> Pr = 0.2W. The consumption will be (50mA*12V)/0.8 = 0.75W. (0.8 comes from 80% typical efficiency). So with a 3V battery you will need 0.75/3 = 250mAh or 250mA current for every hour of operation. Of course that is an aproximation because as the batteries will discharge, the current will increase but you get the idea.

[SoftwareSamurai]

@ vlf3
The reason I pointed out the NCP1402 is because it can maintain an output voltage and current high enough to keep an LED lit even when the input voltage drops down as low as 0.3V. Also, the NCP1402 is internally limited to 350mA output. I've breadboarded an NCP1402's output with an LED straight to ground (nothing else): It runs quite well and stays on for a very long time with only a single AA battery.

@Kyle
Check out the NCP1402. I think if you add a white Cree XLamp LED you'll really be amazed at how well it performs on batteries, especially for just providing light as in a desk lamp.

[vlf3]

@ SoftwareSamurai... that's fine, I have no problem with that; however my concern is based on the proper use and technically correct application for a ballast resistor for each LED, or a series of LEDs.

Mass paralleling of LEDs will produce design limitations, and a high current; a correct design for this desk light as I have proposed, will provide the correct biasing for these LEDs, at a much lower current requirement e.g. the series 5 LEDs will only draw the 16.6 mA x 3 = 49.8 mA total, as opposed to the arrangement being tried by kyle.

Their is so much lazy use of LEDs, with the common misunderstanding how to apply series, or parallel, without a ballast resistor and working voltage head-room; this will only produce a short lived life, for the LEDs.

[Rick Law]

You don't need voltage boosters, just low ohm resistors for each LED or series string of LEDs...

At 3.0 volt nominal AA battery; with a voltage drop for the LED of 2.0 each thus, 1 volt / .06 ohm = 16.6 mA... your problem is to find or make-up this .06 ohm value, being non-standard.

LED's in parallel is a very bad technical practice... each LED act's as a zener diode, and will set it's working voltage in your case, at 2.0 volts +/_
....Standard practice is to have each LED with it's own ballast resistor value.....

Kill two birds with one stone.  0.1ohm is standard and it is not expensive.  I have seen metal-film version for just $2 for a 100.

So:
- Ten 0.1ohm in parallel, now you have a 0.01ohm.  This is for all the LED's to connect to.
- Each LED is series with another two 0.1ohm arranged in parallel.  The two 0.1ohm in parallel gives you 0.05ohm, so each LED has it own 0.05ohm current limiter.
- All (0.05ohm + LED) are parallel to each other, and connected directly to the 0.01ohm (the 10 in parallel).

So, now you are not running each LEDs in parallel naked (without it's own current limiter).  Each has its own current limiter;  No need for booster; and with multiple resisters in parallel, the 1% metal film resistors will probably do a lot better than 1%.

That should work.

I actually would be lazy enough to just use the 0.05 ohm (from the pair of 0.1ohm) and forego the 0.01ohm (from the 10 0.01ohm in parallel) altogether.  0.01ohm is practically contact resistance and wire resistance.  Skipping them should not affect the current much.

Rick