"Perfect" LED current limiting challenge

[ian]

Dale......... Here's where you're wrong.....

The current limiting circuit is only needed if the PIC goes wrong and doesn't control the current. But it is not an efficient way of driving the LED, and is only a backup in case of failure. Once again, the efficiency of ALL suggestions provided thus far is less than 50%. I can achieve 80% without the help I've received here so far.

No, I don't need a 2nd PIC if the PIC is not overly taxed.

Nor is my circuit or application a secret, it's just that I've stated clearly what I want and that's what I want. What I don't want is people who aren't capable of producing such a circuit telling me to change what I want so it suits their capabilities.

In terms of EFFICIENCY, I have defined the parameters of the EFFICIENCY I'm loking for. I want maximum efficiency from the circuit. Cost and component count are secondary, although I don't think this needs to be that expensive.

WHAT IS SO FREAKING HARD TO UNDERSTAND???????
Is it sooooooo impossible or foolish to come up with AN ORIGIONAL CONCEPT AND DESIGN?????
I'm getting the feeling you guys are all just cut and pasters. Sure, there's a vast knowledgebase here, and that counts for something. But are there ANY ORIGIONAL THINKERS HERE???????

IT IS A SIMPLE CONCEPT.
******MAXIMUM EFFICIENCY FOR DRIVING A LED FROM A BATTERY.*******

There are MILLIONS of applications for this. You guys are making me cry.

<small>[ December 20, 2005, 07:07 AM: Message edited by: ian ]</small>

[jimandy]

ian says

I'm getting the feeling you guys are all just cut and pasters.

Ah, a personally addressed criticism is understandable, but a blanket indictment of the entire forum community sounds like something bordering on psychosis.

[ian]

Another non answer to my challenge.........
35 and counting..........
"Your're not psychotic if they're really out to get you."

<small>[ December 20, 2005, 07:16 AM: Message edited by: ian ]</small>

[jimandy]

Hmmm... shades of the recent "making hydrogen out of water & DC" topic debate, But being the dunce I am about the technicalities of this subject, I offer the following link that appears on quick reading (to me at least) to have some relevance.
http://www.edn.com/article/CA446990.html
What da ya think?

[Chris Smith]

The solution is there, even your braille hand book agrees with me.

Funny how you dont do the math to find out, yet you continue on like some hidden miracle still awaits your magic question?

The only problem here is you cant read while the answer sits before you solved, answered, and finished. Your journey is over, you just dont know it yet?

A battery voltage of 3.6 volts, a resistor value of 10 ohms, and a forward led voltage of 3.2 volts will have a resistor loss of only 6% over all current draw.

Beat that for stability, current loss, and reliability, and don’t keep changing your emphasis on what it is you really are looking for.

Two parts, minimal current wasted, reliability, and cost, ....and all have been met here and no wheels needed to be reinvented. Perhaps you were trying to make a candle out of ten parts or more instead of buying a wax candle for 15 cents?

<small>[ December 20, 2005, 08:25 AM: Message edited by: Chris Smith ]</small>

[Joseph]

Ian:

posted December 20, 2005 06:51 AM
--------------------------------------------------------------------------------
Dale......... Here's where you're wrong.....

The current limiting circuit is only needed if the PIC goes wrong and doesn't control the current. But it is not an efficient way of driving the LED, and is only a backup in case of failure. Once again, the efficiency of ALL suggestions provided thus far is less than 50%. I can achieve 80% without the help I've received here so far.

Incorrect. The idea I mentioned can obtain about 90% efficiency with a 20ma LED. More LEDs allow higher efficiency since if the oscillator and driver are comprised of a circuit built around the 74C14, the 500ua draw of that circuit becomes a smaller percentage. Efficiencies of over 95% are not hard. Even the straight inductor Steven mentioned without the capacitor can get close to this efficiency, but switching losses tend to be higher.

[Joseph]

Chris, your idea does look very practical. My LED flashlight actually does it that way.

[ian]

Thanks Joseph, I'll do some experimenting with that.

Chris, your concept has 2 flaws. First of all I can't change the battery to suit the application and even if I could the battery voltage will drop below the forward drop of the LED before it's useful life is used up. Secondly, various LEDs have various voltage drops, your suggestion is only good if the exact LED has an exact forward drop and matches the battery.

jimandy, the URL you supplied is fine but I'm trying to drive white LEDS with 4 "D" batteries. When the battery drops below 6V there isn't enough voltage differential for that circuit to work wasting most of the batteries' useful life.

My primary design is 4 "D" cells into white leds, but coloured LEDs are an option. But this is such a universal concept and design I'd like to see an efficient circuit for varying battery voltages and LED's. This isn't so crazy since the voltage of a battery varies widely over its useful life AND the forward drop of a LED also varies widely depending on lot and colour.

<small>[ December 20, 2005, 10:23 AM: Message edited by: ian ]</small>

[Will]

80% efficiency ?? 20 mA at 3.1 volts divided by 12.5 mA at 9.0 volts looks to me like 55% efficiency

[ian]

I guess reading isn't your strong suit Will.
77.5% efficiency. 20 mA at 3.1 volts divided by 12.5 mA at 6.4 volts (peak voltage of 4 "D" cells).

[Chris Smith]

Ian

Batteries voltages come in 1.2v [floating nicad voltage] , 1.44v [peak charged nicad], 1.5 standard, 2.0 lead acid etc,.... and others things like the use of a Super cap which allows you to chose any voltage what so ever, including in between voltages while delivering the full allowed current draw.

My super cap is one farad in capacity and its two "quarters" thick in size, the size of one quarter in diameter.

More charge in there than any nicad for its size or bigger, and it has a natural discharge limit that doesn’t require any dropping resistors when you set the voltage right.

Six of these in parallel, would be six farads and the size of one “C” cell standard battery or less, and voltages up to 5.5 volts, or less of your choosing.

Caps don’t lose their current discharge as the voltage drops, unless you stress them to their maximum limit.

No matter what value voltage or current you have for a led you will find a battery or a super cap close to the operating voltage so that a minimal loss is had through the use of a resistor.

I have one flash light that uses two button batteries, with NO dropping resistor because the battery can only deliver 40 mills max at a time.

You wont get any closer to efficiency than choosing the right voltage for a led then trimming it from there out. All the standard forward voltages for all the standard LEDS can be met with the right battery or cap combination.

[Mike6158]

I just wanted to post something so that I could click the email notification button

[rshayes]

Your 6 volt battery will actaally vary from somewhere over 6 volts down to about 4 volts. Using a simple current limiting resistor with a 3.5 volt LED would give very poor current control. As the battery discharged, the current would vary over about a five to one range.

If the LED current is 20 milliamps into a 3.5 volt load, the output power is 70 milliwatts. For 80 percent efficiency, the losses should be less than 17.5 milliwatts.

At 6 volts, the drain from the battery would be 14.6 milliamps, increasing to 21.9 milliamps at 4 volts. If a flyback type of converter is used, the duty cycle would be around 50 percent at 4 volts. Peak current would be about 80 milliamps. Base drive would be about 4 milliamps for 50 percent of the time. Base drive power required would be 8 milliwatts. Saturation losses in the transistor would be about 4 milliwatts. Using a current transformer for proportional base drive would reduce the base drive requirement to about 1 milliwatt.

With a shottky barrier rectifier on the secondary, the rectifier loss would be about 6 milliwatts.

The power handling circuits would have losses between 11 and 18 milliwatts.

If the current in the inductor is shut off at the same value on each cycle, the energy transferred per cycle is constant. If this cycle is repeated at a fixed rate, the power delivered is constant. Since the voltage across the LED is nearly constant, constant power implies constant current.

The control can be done with a dual comparator such as the LM393. One section can be used as an oscillator with the other used to sense the inductor current. A CMOS flip-flop can be toggled by the oscillator and reset by the currrent sense comparitor. The LM393 will need about 4 to 6 milliwatts.

Efficiency in the 70 (without proportional drive) to 80 (with proportional drive) percent range should be possible.

[Will]

" I Guess reading isn't your strong suit " - You ignorant clown - I read your portion about the 9 volt battery - More to the point - how come you said 80% efficiency first time then, after I gave you a tutorial on how to calculate it (I remember from last year or earlier this year, that you didn't know the difference between Watts and Watt-seconds) you got it to 77.5% ? - No need for a reply - the answer is fairly obvious

[Chris Smith]

In your own words.......“I need a battery operated very efficient(bordering on perfect) circuit for diving a LED but it has to be safe”...........

Still cant beat the correct voltage to start with for a low current loss, size, and reliability.

And If your using 6 volts now, don’t tell me you cant change to a lower, smaller, and more efficient lower battery count.

Six volts with a 150 ohm resistor and a 3.2 Vled @ 20 ma has 60 MW dissipated in the led and 50 MW wasted in the resistor. [54% efficient]

Drop the battery voltage to 3.6 volts and use a 20 ohm resistor with a 3.2 Vled @ 20 ma, and the led uses 58 MW and the resistor only wastes 7 MW [ 89% efficient]

3.2 [3.3] volts is 97%+ efficient [1 x 2 volt cell, + 1 x 1.2 volts cell]
3.6 volts is 89% efficient [3 x 1.2 volt cells]
4.0 volts is 79% efficient [2 x 2 volt cells]
4.8 volts is 66% efficient [4 x 1.2 volt cells]

6.0 volts is 54% efficient

A 3.6 volt battery pack like the one in my phone [600ma] will run the LED for an average of 25 hours or more and for the size you cant top that. If your already using 4 D cell batteries you can use that space for about 3 or more of these battery packs with plenty of room for more, making a long life, reliability, efficiency and the “perfect circuit” all being met.

Any thing else is inferior to this simplistic approach.

And regardless of your led choice, you can find the appropriate battery pack to suit its Vled and minimal loss. Any Vled from 1.5 to 4 volts [and more] can be matched to the right battery combination.