One LED, 6 Volts, what's more efficient?

[Revolvr]

Hi all,

I need to drive one white LED (3.1v) at 20mA, using two 3V lithium button batteries for 6V (CR1025, 30mAh, 2.9 to 2.8V most of the useful life). I have very little room for components.

My design choices would be:

1) Just hook up the LED. It will be overdriven a while - but is the internal resistance of the batteries is enough to limit the current?

2) Put in series a 100 or 150 Ohm resistor.

3) Use an LM317L (TO-92 package) in a constant current configuration with a smaller resistor to regulate the current to 20mA as long as the batteries can handle it.

4) Use something like a 78L05 to drop the voltage to 5, with or without a current limiting resistor.

I don't have room for something fancier like a buck converter.

If I wanted the longest useful life - say to the point the LED is only getting 15mA, which of these options would be the best?

And no this isn't a homework assignment, although it'd be a good one.

cheers,

-- Dan

 

[Russlk]

Sure sounds like a homework problem to me. A very basic one also.

 

[Sceadwian]

Ever heard of a Micro Photon Light? Overly expensive brand named attached to two 2016 coin cells in series with a white/blue LED. I'm not sure how the 2016 compares to the 1025 but looking at the package size I found on the net I'm willing to bet the 2016 is able to source more current. You should be fine running the LED right off the cells.

 

[audioguru]

1) Just hook up the LED. It will be overdriven a while - but is the internal resistance of the batteries is enough to limit the current?

Energizer doesn't spec the internal resistance in their datasheet for their CR1025 cell.

2) Put in series a 100 or 150 Ohm resistor.

Try it and watch the LED brightness fade.

3) Use an LM317L (TO-92 package) in a constant current configuration with a smaller resistor to regulate the current to 20mA as long as the batteries can handle it.

An LM317 uses 1.25V for current regulation plus its input must be at least about 2V higher than its output. So the minimum battery voltage is 3.1V+1.25V+2V= 6.35V. Not enough voltage even when the battery cells are new.

4) Use something like a 78L05 to drop the voltage to 5, with or without a current limiting resistor.

A 7805 needs an input voltage of at least 2V higher than its output. You don.t have anywhere near 7V.

 

[dch222]

I would go connect the batteries in parallel and connect the LED directly across them. You are unlikely to over-drive the LED if it only takes 20mA at 3.1V. This will give you the most efficient solution as there will be no resistive losses. I don't know much about white LEDs, but should think you would still get reasonable brightness at 2.8V.

If you don't feel happy about leaving out the series resistor, use something low like 10 ohms.

 

[ericgibbs]

hi revolvr,
You can purchase a 150mA low voltage drop out regulator
equivalent to say a 78L05. These require just 0.25v to 0.5 across the regulator to maintain +5Vout. Cost less $1.0

Type: TS2950CT 5.0
380mV drop out , 150mA
Price 58 pence UK.
Maplin Code: N69CA

Regards
EricG

 

[audioguru]

How long do you expect the tiny battery cells to last with such a high overload? Their capacity is rated at 30mA/h when the load current is a whopping 64uA. At 20mA then the battery might be dead in a few seconds.

 

[kchriste]

At 20mA then the battery might be dead in a few seconds.

The OP would have better luck building something like the "joule thief" and powering it from a 1.5V "N" cell or something:
**broken link removed**

 

[Sceadwian]

More like minutes, but still.

 

[Oznog]

Yeah you can hook it up straight. Microlights do it. The internal impedance of the 2016 is enough for the LEDs they use. The 1025 is smaller still and will have higher impedance. But honestly the mah aren't there to power the LED, not even considering the impedance issues!

You can't use the batteries in parallel, 3V won't turn on a white or blue LED well at all.

 

[audioguru]

I saw the expensive and tiny CR1025 battery cell in a store. It is so thin that there is nearly no chemicals to make any power. Nearly half of its tiny amount of power will be wasted by heating its internal resistance.

 

[Revolvr]

So here’s the real problem and why the constraints are as they are. Over the Holidays everyone in the family got these “stocking stuffers” – miniature Coleman Lantern LED key fob lights.

Now the original designers decided to use 2 cheap 1.5V batteries, either GP189’s or 389A’s. So with 3V they couldn’t use a bright white LED, but had to use a very dim amber/yellow LED that was completely worthless.

So I took one of the lanterns, found 2 3V button batteries that would fit close enough – the 1025’s, a bright white LED and fit a 100 Ohm resistor thanks to a Dremel. Works great – bright and useful light.

And yes the 1025’s each cost as much as the entire lantern – but what’s a few bucks among family.

Probably won’t last long, but may be similar to other typical overpriced key fob LED lights that use 2 2016’s.

What led to the OP was my interest in finding a more efficient longer lasting way to drive the LED given these constraints. I figure I could drill out enough space to add something in a TO-92 package. Would regulating the current give me a longer useful life?

I think the answer is no. The difference is constant current regulation vs. constant resistance. With a fixed resistor the LED is initially driven over 20mA, brighter and dies a slow death. Constant current regulation with a 317T is essentially a variable resistor changing the voltage – on top of a fixed 1.25V drop. I still waste energy, but as heat instead of extra light. Further when the 317 drops out of regulation the LED dims very quickly – faster than with just a resistor at the same voltage. Although on a bench that drop occurs at a supply of around 5V suggesting the 317 regulating current only needs the 1.25V extra to regulate, not 2 more on top of that.

I was really looking for a mathematical expression that would allow me to compare the regulated and non-regulated case. Cheaper than an actual test which would cost me 4 batteries.

Although I couldn’t find internal resistance specs for the 1025 I did find a 30-40 Ohm spec on a very similar battery. So I decided to drop the 100 Ohm resistor too.

The only thing that might really extend the life a bit would probably be a boost converter something like the Joule Thief mentioned. This might also let me keep the cheaper 1.5V batteries. Tiny as it is, not sure I could make it fit but I may give it a shot.

Cheers and thanks for the thoughts

-- Dan

 

[audioguru]

The LM317 has a typical dropout voltage of 1.5V at 20mA. Then it is dropped out and fails to regulate.
With a current of only 2mA when the LED is still glowing the dropout voltage of an LM317 is probably 1.375V.
The current regulating resistor needs 1.25V in addition to the dropout voltage.

 

[Hero999]

Simple, use a 5V low dropout regulator and a resistor.

You really need to use a bigger battery though, a Joule Thief and an AAA battery would be perfect, a rechargeable 1.2V 1000mA battery should power your LED for >12 hours.

 

[audioguru]

One AAA Ni-MH cell weighs 9.3 times more than two CR1025 Lithium cells. It is very much bigger.

 

[Nigel Goodwin]

One AAA Ni-MH cell weighs 9.3 times more than two CR1025 Lithium cells. It is very much bigger.

You don't think '9 times as heavy' might have been near enough?

 

[Revolvr]

One AAA Ni-MH cell weighs 9.3 times more than two CR1025 Lithium cells. It is very much bigger.

Even more important than weight, an AAA is 8.6 times taller than two 1025's.

Inside the lantern there is room for perhaps 6mm depth, 12mm diameter.

A picture would help.

 

[Hero999]

What about AAAA batteries then?

 

[audioguru]

It looks like a very small lantern. How long was the LED bright? Could you see it dimming?