Driving LEDs from Battery Pack

If you don't need to dim them, go with JohnDG's advice.

If you do need to dim them, find a pulse-width-modulator circuit that runs off your dc input voltage.

id consider well placed resistors

Resistors, thought about that, but then won't I lose out when the battery has been going awhile, i.e. reduced lighting time on the leds?

The 'trick' here is to do a little math.

A) Find out the voltage drop of one LED when it is glowing the way you like, we'll call that V1.

B) Find out the current flow through the same LED when it is glowing the way you like, we'll call that A1.

C) Multiply this my the number of LEDs in the series, this is the total voltage drop of the chain, we'll call that V2.

D) Look at the voltage of your battery source, we'll call that Vs. (Voltage Source).

E) If Vs is less than V2, then you need very little current control, and a 1 Ohm resistor in series should suffice.

F) If Vs is greater than V2, subtract V2 from Vs, we'll call that V3.

G) Divide V3 by A1, the resulting number is the size of the resistor you need in Ohms.

H) Multiply V3 by A1, the resulting number is the total wattage the resistor will be dissipating as heat, make sure the resistor you select is as low a wattage as possible without going below this number.

Also, when choosing the resistor, select a 'carbon film' resister. These are the cheap, common resistors you find almost anywhere. When a carbon film resistor overheats, it 'burns out' like a fuse, thus protecting the LEDs from overvoltage.

I actually did this myself years ago when rewiring an old Christmas manger, replacing the old incandescent Christmas tree lights with LED lights, and had to control the current from the AC line to protect the bulbs.

As the OP stated, the LEDs are in parallel.

Sorry, didn't scroll down to that before I posted.

Parallel LEDs, that's tricky, if one starts to go, it'll hog the current until it dies, then the rest will start getting 'overfed' from the current that should have gone through the blown one, leading to the next failure.

Unless you current-limit each LED individually, it's a real mess.

Seems to work ok in the field though.

What I've seen as 'in the field' examples are series strips of LEDs set in paralel, with each strip designed to drop barely more voltage than the supply is providing, and with a current limiting resistor an one end of each strip. When one strip does, it doesn't affect the other strips because their resistors limit the current and prevent 'overfeeding.'

If you're talking about things like seven-segment LEDs with one resistor on the common, remember that those are typically fed by TTL or CMOS control circuits, which are limited in the amount of current they can supply per pin, and the control circuit's outputs are isolated from each other, a lack of current sink on e one pin doesn't equate to more current on the other pins. The control circuit's limitations act as the current limiting for the display.

Good point. Some of the installations are getting old, but leds are checked each spring and replaced as needs be. I shall set the current output to max out at 16x20 = 320mA.

LEDs are diodes. They don't conduct much until you reach the forward bias voltage (which varies, depending on color), at which point they conduct a lot. You need to control the current with a series resistance.

Yep. Here's one that looks to be just what the doctor ordered. (pdf)

Thanks for your help folks!

Going to set up a current limiter first off, as per picture.

Would like to know what the considerations are when chosing these transistors; hi/mid/lo Hfe, current rating etc. and approximate values of sense and Tr1 base-collector resistors. I'll be driving out 300mA at 3V6 max. Also, how am I calculating required volt drop across reference diode?

I'd just bung in an LM317.

Saves a lot of messing about.

Hi jdrbrown,

The forward voltage for a typical 5 mm white LED (in reality a blue LED coated with a yellow phosphor) is around 3.6 V @ 20 mA, more or less. A string of three of these white LEDs in series brings the total to 10.8 V @ 20 mA.

Now let's say, for instance, that your freshly-charged battery pack produces 12.0 volts. Subtracting the forward voltage of your string from 12.0 volts gives you 1.2 volts. 1.2 volts of 'headroom' to work with before the battery voltage drops below the string voltage and things start to grow dim. If you use some kind series regulator things get even worse.

Thing about regulators is that they need a little headroom themselves. This headroom is called the regulator's 'dropout voltage.' Once the difference between the regulator's input and output voltages falls below this voltage, the regulator 'drops out' of regulation - it can no longer do its job - hence the name.

Dropout voltages vary, some as low as 0.1 V to as much as 1.2 V for old-style, three-terminal regulators, and sometimes more. Whatever the dropout voltage, it takes away from the available headroom; here 12.0 - 10.8 = 1.2 volts. If the dropout is 0.3 V, for example - pretty typical - you have only 1.2 - 0.3 = 0.9 volts to work with before regulators stop regulating and your LEDs fade away.

The bottom line to all this is things stop working even while there is still plenty of juice left in your battery pack. Basically your circuit is skimming the cream off the top, leaving the milk behind with nobody to drink it. All that battery capacity (and weight?) for a mere 0.9 volts of headroom? See anything wrong with this picture?

What you really want is a circuit that supplies your LEDs with the right amount of current until that battery pack is nearly bone bluddy dry. Fortunately a lot of very clever people have tackled this problem before, and with considerable success.

One of the cleverest solutions (IMO) is called a Joule Thief and, best of all, it is simplicity itself. Basically it's a lot like the flyback circuit in old-style TVs, only with much lower output voltage.

The original Joule Thief could run a 3.6 V, 20 mA white LED from a single, 1.5V AA battery until the battery had less than 0.3 volts left on the gauge. Further improvements using JFETs dropped that limit to 0.1 V before the LED started to grow dim. Since its invention (80s, I believe) so many variations on this theme have appeared that it would be impossible to list them all, but this guy has come pretty close. You might want to take a look at his site.

My leds are in parallel. Shoulda mentioned that.

I've looked at current regulator diodes, both down in the 4mA range - can I use one to control a transistor's collector current to facilitate regulation at higher currents, e.g. 300mA through a FET?

Very difficult to control them with one device/circuit if they're in parallel. You're back to the case where failure of one or more LEDs (to open circuit) would mean driving all the current into the remaining ones (possibly cooking them). Also, unless they're well-matched, you'll probably get uneven fading as the battery volts drop.

If you can justify the cost, you'd be better to have one regulator per LED.

For parallel, I would consider voltage regulation. Give them all 3 V or whatever. They all get no more than that no matter how many are on the circuit.

A voltage-reference would be more suitable as FETs are 'voltage'-type devices - the current through the FET is controlled by the voltage on the gate, transconductance-style. Even if you use the sort sort of regulator you're considering, you're still going be lugging around a lot of unused battery capacity. I might also mention that, no matter how clever they are, linear regulators are basically glorified variable resistors. They dissipate power as function of the difference between their input and output voltage - even when they're acting as current regulators and, as everyone knows, resistors dissipate energy that could be used in better ways, such as lighting LEDs.

I strongly recommend you go take a look at that site. There are lots of ideas about ways to do things that you may not have considered. One thing is that most Joule-Thief circuits are happier with a lower input voltage and there are plenty of circuits on that site which accept 1.5 V and can drive multiple LEDs. Others can source up to a 1000 mA for those Luxeon-type high power LEDs. I don't know what sort of battery pack you've got, but you could certainly connect batteries in parallel as a sort of super-AA battery to power many of these circuits for a long time. Go take a look.

Nice.

Joule Thief sounds good but potentially risky with NiMH rechargeables?

Why would the circuit care about the battery chemistry? All it sees is a black box that supplies its power.

Think OP's worried about damage from deep discharge.

Ah. That would definitely be a concern with a Joule-Thief-type circuit. I haven't used NiMH batteries in situations where they would be deep-discharged, and so I've never run into this problem.

[rant]

In my own defense the OP could have voiced this particular concern directly instead of alluding to it by means of battery chemistry. We also found out rather late in the game that his LEDs are in parallel, not series. That makes a pretty big difference in how the problem is tackled. The OP mentioned NiMH but didn't actually say that his battery pack uses NiMH batteries, nor did he state its voltage. For my part it's irksome to get the facts piecemeal and have to divine the rest. Hey, if I'd wanted to pull teeth, I'd have become a dentist!

@OP: Put all the cards on the table, please? All of them?

[/rant]

My initial post was aimed to be concise. Can I help it if people make assumptions? Perhaps if I had said 'led ladder'?

So, 4x1.2V NiMH battery pack (charged from PV-array) driving 16x 20mA/3.2V leds in parallel via MOSFET on/off control (using PV-array as detector via comparator to indicate night-time).

Considerations are as follows;

I) Preventing over-voltage/over-current in first half-hour or so of turn-on.

II) Preventing battery wastage and potential led damage at lower voltages.

III) Getting more out of the batteries via the above, i.e. longer lighting time / quicker recharge.

I'm also looking at similar considerations at the charging end, i.e. not over-charging battery pack.

Thanks all for your help, time to start emulating and playing!