Is this LED hookup correct?

I am a beginner and I would appreciate if someone confirmed that this is correct:

There are 2 ways of hooking up a larger number of superbright LEDs from a battery:

1) Hook up as many LEDs in series to get the current (20mA) you need. Add a resistor if necessary (if the current is slightly too high and adding an additional LED in series would make the current too little)

2) Use a current regulator like an ML317. Design it so that even when the batteries are close to discharged the output voltage is still high enough to power the LEDs at full brightness. If the batteries are fully charged the ML317 makes sure that the current stays within limits.

Correct so far? Since I am constructing a light panel for a video light I would like the brightness to be 100% even if the batteries are draining. This makes me lean for option 2. However doesn't #2 use much more power than #1 since the ML317 dissipates the extra power when the batteries are full? I am confused. I would like equal brightness regardless of battery charge and yet the batteries must be able to last for a while. What should I do?

Thanks very much for your suggestions. Karotto

Reply to
karotto
Loading thread data ...

For fixed-power cases, a string of LEDs with an LM317 works fine. I use that for 99 UV LEDs in an exposure box.

For battery powered, you'd probably be better off with a boost regulator, perhaps one of the specialty "led driver" ones. They boost a range of input voltages up to a higher fixed voltage such that the string of LEDs gets a constant current, without needing a load resistor. Very efficient, but a bit more complex.

Reply to
DJ Delorie

Constant current is the only option that will give you what you want. An IC designed to drive leds from batteries is your best option - but you don't say how much the voltage will fluctuate.

I favor two transistor current regulators with an NPN transistor in the ground return. The circuit drops about 1.2 volts. A suitable (safe) resistor will often drop more.

Bob Pease has a current regulator that only drops sixty millivolts or so. Takes one transistor and one constant current IC.

Reply to
default

Um. The OP specifically mentioned the use of batteries. ("I would like the brightness to be 100% even if the batteries are draining.")

Just so you know.

Jon

Reply to
Jon Kirwan

Photography of some kind?

What battery system are you using?

How equal? What's the criteria here?

Jon

Reply to
Jon Kirwan

I will be using some kind of recahrgable battery (like NiMeHi or Li ion, or Nicad). Well, I just don't want the lights to be very bright at first and an hour later when the batteries are at 50% to be a lot dimmer. Thank you.

Reply to
karotto

I saw. He mentioned batteries and a string of LEDs in the same post, so I covered both cases.

Reply to
DJ Delorie

A 'system' includes at least the fully charged voltage rating and its ampere-hour value, I'd imagine, for each of the chemistries in mind. Do you have these figures handy?

In other words, you don't want a "lot" of variation (unspecified, so open to interpretation) from your eye's point of view but won't care at all if some other device sees a huge difference?

What bothers me about this, is that I have varying standards depending on my own use so it is hard to know what you mean. For example, I may accept a fairly wide range of optical output when using a flashlight in the deepest part of night for walking down my driveway, but may find it irritating if the flashlight is low when using it to light up some small parts on a circuit board while using a magnifying glass. It's really hard to know what you want without having either some specification or else a specific use to which we can relate our own experiences.

Jon

Reply to
Jon Kirwan

In the absence of an active regulator, the resistor is always necessary. The number of LEDs is chosen such that the total voltage drop is below the supply voltage (75%-80% of the supply voltage is typical). The resistor is then chosen to give the desired current: I = (Vsupply - Vdrop) / R.

The main problem here is that Vsupply can vary significantly (particularly when running from batteries), and Vdrop may be subject to process variation, so the difference needs to be large relative to the variation if you want to maintain a reasonably constant current.

For non-trivial power levels, an active constant-current regulator is preferable, as it avoids the need to leave significant "headroom" between the supply voltage and the LEDs' voltage drop.

Reply to
Nobody

It's tricky, maybe even dangerous, to drive LEDs in constant-voltage mode.

The most efficient way is to use a constant-current switching regulator. It will also use the battery down to the last drop of charge.

Lots of people are making switching-regulator ICs specifically for driving LEDs.

John

Reply to
John Larkin

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.