# Driving LEDs with a battery pack

• posted

Hello,

I'm trying to drive some LEDs with standard batteries (eg. from a battery holder with 3 AAAs in it). Should be easy ... right?

There's various colors of LED which need from 2.2 to 3.3 volts at around 25mA and I want to drive maybe half a dozen from the battery pack.

I can calculate the right resistor for any give voltage, no problem, but how do I deal with the wide range of voltage over the lifetime of the battery. With a brand new battery the voltage is around 4.6V but as it discharges it goes down to about 3.3V (with about 10% battery left). If I pick resistors which work at 3.3V then there's far too much current when the batteries are new (I measured 60mA on some of them and they get warm to the touch so I'm guessing that's bad)

So:

a) How delicate are LEDs? Is 60mA going to burn them out?

b) If it is, what's the simplest/smallest circuit which will give me (eg.) 3.3V @ 150mA from a set of AAA batteries? Size is important as I want to pack it into a small space.

I did some Googling and tried a 3.3V Zener diode to drop the voltage but it only dropped the voltage by about 0.2V. I'm guessing the reason for that is something to do with the the load current being quite high which makes the Zener resistor very small (two or three ohms).

I also looked at voltage regulators but is seems a 3.3V regulator needs a higher starting voltage than the batteries can provide.

• posted

1. What are the specs on your LEDs? If they're generic/Ebay unmarked units, limit the current to a max of
20ma.

1. Use a resistor value that limits the current at maximum battery voltage. Brightness will drop off some as the batteries age, but probably not much.

2. Google for joule thief This is a circuit that boosts battery voltage with a minium number of components. The most common place to find a pre-built unit is the common solar walkway light. Many of these use a single NiCd or NiMH battery and use a joule thief type circuit to boost that voltage to the 3+ volts needed for a white LED.

You can buy a kit for \$10:

John

• posted

I've built these and they work, great. Not hard to make, either. But....

This link comes up quickly:

I read through it. And the description under "How it works" is just a little bit too simplified for me. Perhaps the shortest possible explanation that retains important details is: the BJT turns on via R1 allowing current through the collector winding to rise along a ramp. During the on-time, rates of flux change in the base winding actually aids the battery voltage a little in keeping the BJT on still harder. At some point, though, the fast rise in collector current demands more base current than can be supplied and the BJT tries to limit the collector current. In doing so, the collector winding goes through zero volts across it and reverses its voltage to continue driving the current through the LED. But the rate of flux change in the base winding also then reverses, opposes the battery supply, and acts to turn the BJT completely off. While the BJT is off, the collector winding's current (and thus, the LED current) declines on a ramp of its own. As it does, the voltage across it required to drive the LED also drops. As that drops, so does the rate of flux change in the base winding and thus it's voltage opposing the battery. Eventually, the induced voltage isn't enough and the BJT is able to begin turning back on. As that happens, the voltage across the collector winding reverses again and the base winding starts aiding the battery, once again, snapping the BJT into fuller on conditions. And the cycle repeats.

There are some additional details. As the collector current rises, so does Vce. As Vce rises, the voltage across the collector winding declines gradually, which results is lower rate of flux change and that softens the supporting base current at the very time when more base current is required due to the rising collector current. So the BJT doesn't turn off suddenly, but "rolls quickly into" the turn off.

Hobbyist view, anyway.

Jon

• posted

It sounds like you just need a simple boost converter to boost your battery voltage up to a voltage that will drive the LED stack with the highest forward voltage (3v3 in your case). Choose the current limiting resistors accordingly for each LED stack to give about 20mA flowing through each LED (25mA seems a bit high for a standard LED, but check the datasheet).

Just Google 'boost converter'. LT and Maxim do them in DIP packages so are easy to solder. Other parts mentioned in later posts refer to winding transformers, which can get a bit tricky. A standard boost converter uses an off the shelf inductor.

You can get clever and use a different boost converter for each LED (LED brightness is a function of the current going through them, not the voltage across them), but if you dont mind throwing away some power, powering them all off one common rail is much simpler.

```--
Bill Naylor
www.electronworks.co.uk```
• posted

Ebay specials...

Quite a lot in my tests. If I select resistors for "brand new battery" then for most of the life of the battery I see about half brightness. This would be ok for an on/off indicator but the stuff I'm making is decorative for a procession so I want them to be as bright as possible. I'm not worried if the life of the LED is reduced from 100,000 hours.

I found a ferrite bead in an old PC and lashed on up. It works but isn't very bright (maybe only a quarter of 20mA brightness).

• posted

I'm not necessarily trying to run off a single battery. Do you think I should be?

ge

Isn't it basically the same thing...? Maybe not if I should be googling for "current limiting circuit" instead of "voltage limiting circuit".

OTOH, I might give the Zener diodes another try (because it's easy/small/cheap) with an eye to putting one diode per LED (or maybe one for every two LEDs if that works).

I'm just waiting for the shops to open because my ham-fisted experiments made the magic pixie smoke come out of the one I had...)

• posted

you may be able to use a miniature light globe as a current regulator. Look for something like a 15 - 25 ma globe, with a voltage rating of say

3 - 7.5 volts. Wire it in series with the LED. As the battery voltage reduces the globe dims, as the lamp filament cools it's resistance lowers and hence somewhat compensates for the sagging voltage. (such odd light globes are often supplied for illuminated switches and sometimes for Wein bridge oscillators

Alternatively, connect a JFET (plus a low value resistor)as a constant current source. Most JFET's will not work perfectly with so few volts to play with, but it will be better than nothing.

For something sophisticated you could use almost any of the charge-pump IC's to double the voltage and give you enough volts for a proper constant current source or as a simple voltage doubler for when the battery / led combo is just too low.

• posted

Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or I=V/R or R=I/I. (and P = V*I)

"V" is analogous to pressure and too much pressure blows holes in things. "I" is probably the same as you concepts of a river current. "R" is a measure of how much a substance resists current flow - think of water pressure at one end of a clean pipe - water goes though quickly because there is no resistance in the pipe and lots of current flows, next stuff the pipe full of tightly packed sand - this has high resistance - little current flows. If you increase the water pressure then more current flows and with enough pressure the whole thing blows to bits and lets out the pixi water.

Something like your case:

you want 20 milliamp (0.02 of an amp) supplied from a 4.5 volt battery to flow through an LED. The simple application of R=V/I gives the wrong result

4.5v /.02A = 225 ohms (nothing wrong with the formula - just wrong application)

because there is no account for the necessary voltage drop accross the LED. Assume in this case it is 2.2 volts if you used a 220 ohm resistor you would find the current is too low at something around 10 ma. This

2.2 volt drop is somewhat constant regardless of current thought the LED (in water terms it is a weir across the river). If you had to use the exact voltage drop of the LED you would lash up the above values and measure the voltage across the LED.

In this case:

R = (4.5v - 2.2v) / 0.02A = 120 ohm.

To check for power requirements of the resistor (so the pixi smoke stays inside the resistor)

P = (4.5v - 2.2) * .02A = .046 watts - any small sized resistor will be fine.

As you have noted the battery voltage drops over time. If you want just a simple resistor solution you cope for it this way. Find the maximum current the LED can RELIABLY withstand this is probably 20 milliamp and calculate R using the new battery voltage.

Hope some of that was useful

Maybe not if I should

• posted

Was the battery voltage you used __below__ the on-voltage for the LED? (In the joule thief thing, that's the way it is supposed to be.)

Jon

• posted

Yes, I used a single battery...

I also tried two batteries to see if it would get brighter (it didn't)

• posted

Yes, I heard about that one...

I was under the impression that if you fixed one of the values in part of a circuit (volts or amps) the other value would sort of figure out what it was supposed to be (assuming the power supply can provide it)..

So if I figure out a way to fix the current at 20mA the voltage will sort itself out and I don't need to worry about it.

The specs on the LEDs aren't very exact so the way I did it was to put the LED in series with a variable resistor (a ten turn wire-wound pot - quite fine tuning), set up my multimeter to measure current, dial "20mA" using the pot ... then take it apart and measure it.

Not very scientific but I didn't see any other way to get exactly 20mA using the loose specs of the LED.

The batteries give between 3.6 and 3.9 volts for most of their lifetime so I was designing around an average value of 3.75V.

See this chart, which seems quite accurate:

The problem is that brand new batteries give out 4.7V for a while and this makes the LEDs get warm (not hot, just warm to the touch). I discharged two sets of batteries through it and got peak currents up to 45mA but nothing bad happened.

My idea was to use a Zener diode to take away the extra current at the beginning but I didn't figure out all the resistors before the pixie smoke escaped. I'll get another one tonight and try again...

Yes, of course. I'm a complete newbie at this...all I know is what I've read on the web over the last few days.

• posted

No (no, no, no, no). If you fix two the other takes care of itself.

Yep, sounds good.

Not much good if the LED blows while the battery is still at 3.9 volts

Zener diodes don't do that. You could make it work but the cost would be horrible power consumption and bad efficiency.

• posted

Like this?

A few days ago I was thinking like this:

"It's just a battery and an LED, how hard can it be...?"

• posted

A lot like that. If you put a resistor between the gate and source you have some control over what the current limit is.

It's not hard. If you just want to hook up an LED then just a resistor works fine.If you need constant brightness, maximum life, maximum reliability or maximum efficiency then it gets more complicated.

• posted

The circuit diagram just below that heading does not relate to the heading at all.

• posted

This is true if all the components follow Ohm's Law. Unfortunately, LEDs (and most other semiconductors) don't follow Ohm's Law.

If you apply a low voltage (below 1.9 volts or so) to a green LED, very little current will flow. As you increase the voltage slowly, at some point (2.1 volts or so for a green LED) the current will start to increase rapidly, and the LED will light. If you attempt to increase the voltage over ~2.5 volts, the current will increase to a point where the magic smoke comes out of the LED.

If you carefully read the specs for your LED, I expect that the 20 mA rating is the maximum recommended current. The LED will work fine at much lower currents, but won't be quite as bright as at 20 mA. I normally aim for about 10 mA current in common LEDs - they are bright enough for most applications at that current, and the current is low enough that I don't have to bother checking the specs on the particular part I'm using to see if it is OK (and it makes the math easier!) I also assume that red, yellow and green LEDs all drop about

2 volts, although I know that red and yellow are lower voltage than green (but blue and white want 3.3 to 3.6 volts, I think.)

As I said above, a LED isn't really very fussy about the actual current. You're being much too scientific about it.

If the maximum current rating of the LED is 20 mA, you should design for 20 mA or less at the maximum supply voltage.

```--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca  ```
• posted

No, this is not true. Ohms law has three variables, two of which must be defined to calculate the third.

(remainder snipped)

• posted

3.9 is only 4% more than 3.75...

...only until the battery voltage drops down a bit (which should happen in the first 10% of the battery life).

• posted

The brightness should also be affected by the base resistor value, too. Did you try changing it to a lower value -- say about half? I think that should lower the frequency, leave the duty cycle about the same, and increase the peak current (not quite by double) -- with the net effect of raising the brightness a little -- assuming the transistor and LED can handle the peak. If that works as expected, you may consider trying even lower values.

Jon

• posted

Yes (yes, yes, yes, yes). He's talking LEDs. If he controls the current at 20 mA, the LED will be fine. The LED holds the voltage to Vf.

Ed

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