resistance values for LED's

Hi Matt,

There won't be a problem at all except the voltage regulator has a standby current, and the 78XX regulators unfortunately are a bit heavy on juice. Their standby current is from around 5mA to 12mA, which is quite a lot considering that is almost half the LED's current, so you'd be drawing 50% extra current that goes to waste.

The low standby regulators or better known as supervisory regulators are more expensive and their current capabilities limited.

I'm waiting for a component I've ordered to do a test, if you're interested I'll post you a diagram, simple linear regulator with <100uA standby current and up to 1.5A or 5A supply depending on the regulator used (if it works like I suspect it should).

The only difference between a low standby linear and a switchmode is the in / output differential voltage of the linear result in some losses in the form of heat, which is less in a switchmode. In MOST applications a switchmode wrt complexity and cost just doesn't justify it. The losses is mW, one probably won't even notice it.

 

One note on those Cree leds or whatever they are called. They run friggin hot, so they must be properly heatsinked. Their current consumption is 300mA or more or they work really shitty, besides bloody expensive.

Tbh, I get with a small cluster of those 20mA LED's at 20 000mCa ea as much light as one of those Cree spot lights and at less current. Remember the 20mA jobbies have a much lesser voltage drop over them so you can run them is series ie 2 - 2 in series, so for 300mA I can run 30 of those lesser LED's and get better light distibution depending on how you arrange them.

If I want to make light instead of a spotlight, then CF's are still king.

 

Sorry, one note on the 78XX regulators -

They are fixed, and the valueas available are 5V, 6V, 8V, 9V, 10V, 12V, 15V and 24V ones. That's it, they don't go below 5V. To go below 5V you have to use an adjustable regulator (sucks eh) like the LM317 or LM338 (nice one).

The voltage devider that determine the adjustable's output is the problem, they then have as much standby current as the 78 series.

 

Boy I must be a bit thick. I still don't understand!!

CDK saya 1000 ohms. Billy doc gave a fabulous explanation on it all and says I need a 540/560 ohm Yet I have been using brown ,green ,brown, gold--which is apparantly 150 ohms ?

The instrument light picture has the one I need (bottom right -top picture)

I don't have room for linear voltage things.

 

The current regulator is a very nice one to use... Constant brightnes, reliable, be what I'd have used.



You calculate the ohm value as follows -

Vsupply - Vdrop over LEDS = Vrest

Vrest / current required = resistor value

If Vrest = 9V then

9V / 0.03A = 300R

That is not a standard value, the standard value you should use is 330R

so 9V / 330R = 27mA... close enough.

Colour code is orange orange brown gold (5% tolerance) and you can use a 1/4W or 1/2W, 1/2W will be better.

You can use 2 x 150R's in series as well, that will give you 300R



CDK's 1k resistor is good if you use the LED like a panel indicator... to see if the app is switched on or not.

Your 150R will be a bit hot on the LED's current, expect to replace the LED's about once a year or sooner if their current rating is 30mA and you run one off'n a 12V supply. You're driving 60mA with them...

 

Hi Fannie, Matt,

Voltage regulators will work nicely, except that they are resistive elements when they regulate, so they end up using just as much power (actually more) than a resistor. Their advantage is that they do what the name implies, they regulate, so even though they are resistive they are always the RIGHT resistor for the voltage applied. This is a good thing, for sure!

But we can do better! We can lose up to 95% of the power loss inherent with resistive approaches, AND have very good voltage regulation on top of that. But to understand why this is we have to make a little digression into how an inductor works.

Inductors are nothing more than coils of wire, usually wrapped on some sort of magnetic core. You can make a simple one by wrapping a coil of wire around a nail.

If you pass a current through an inductor this causes a magnetic field to form and grow because you just turned this current on. So, this growing magnetic field has to pass through the same coils of wire that produce it! This is an important thing to know, and I hope you will see why shortly, but for now just remember that this growing field is itself producing a current by cutting through the wires that produce it.

The question now is, what does this second current do?

Somebody by the name of Lenz observed this and found out the answer, thus getting the effect named after himself as "Lenz's Second Law of electricity," or something to that effect. Lenz's second law tells us that the magnetic field from any current will induce a second current with a second magnetic field that opposes the first. In the case of an inductor, that means that the current will build up slowly as it fights the self-induced current trying to go the opposite way.

So now we can ask a second question: "What happens if we suddenly remove the current source powering an inductor?"

And this is the most interesting answer: "The collapsing magnetic field induces a current flow in the inductor IN THE SAME DIRECTION AS THE INITIAL CURRENT THAT BUILT UP THE MAGNETIC FIELD! The inductor has stored the energy put into it in the form of a magnetic field, then when this energy source is removed it gives it back. It is, therefore, an energy storage device, and a damn efficient one, very near 100% efficient!

Here's how we use this effect. We put a voltage across an inductor, say 12 volts. The voltage builds up at some rate (dependent on the inductor). We keep an eye on this voltage and when it gets to whatever voltage we want, say 3 volts, we turn off the current source and the inductor's magnetic field collapses, thus pushing more current in the same direction. Now if we put a diode on this inductor between it and some load, like an LED, current from the inductor will power the LED at whatever voltage we choose, and we get there WITHOUT any series resistance.

The new switching regulators in an integrated circuit do all the heavy lifting to play this trick, and achieve efficiencies up to 95%. I'm at home now, and don't have any of my books here. But, in about 15 hours I should be able to gen up a circuit diagram for you all using a switching approach. I may also stick in a Pulse Width Modulator so you can control the brightness, but that will take me a little longer. Meanwhile, what LEDs would you like this circuit to power? By that I mean, how much current would you like to have at what voltage?

The wife is getting threatening, dinner is on the table, so see ya tomorrow!

BillyDoc

 

IMHO using a switching regulator for powering a LED is a bit like killing a mosquito with a laser guided missile.

Attached is a very simple circuit for a constant current source. The part may be labeled LM317C, TL317C or MC317, depending on the brand.
The resistor value determines the current: 22 ohms = 50 mA, 47 ohms = 25 mA, 100 ohms = 12 mA (approximately).
Instead of 1 LED you can wire 2, 3 or 4 pieces in series, the current will remain the same.

As BillyDoc pointed out correctly, the unused voltage drop across the circuit is a loss, but compared to any incandescent bulb it is neglectible.

 

Cree is just a manufacturer, they make all sorts of LEDs. Mine don't draw 300 milliamps, and the only heatsink they need is the board the are soldered to.

 

Hi Frosty,

Your color code would indeed indicate 150 ohms, with the gold band indicating 5% tolerance. The 540/560 in my post yesterday was an example of how to compute the needed resistor, but not necessarily applicable to your particular problem . . sorry about that. There are thousands of LEDs being made, so without more info I can't do much. What we need is some more data for your LEDs. If you can give me the type so I can look up the data, or the forward voltage and current then I'll be happy to calculate the resistor value for you.

BillyDoc

 

You are right about that, of course, but you can get some "economies of scale" if you use one switcher to power either lots of LEDs or some really powerful ones that use lots of current. Cree makes one white model that runs on 3.7 volts and sucks down an entire amp! I haven't seen it, but that is bound to be a lot of light. Over three and a half watts worth, actually. If you were to try and drop the voltage from your battery with a resistor to light one of these up you would be burning up roughly 75% of your power making heat in your resistor. Slightly more if you use a linear regulator.

You might think that putting several in series is a way to get around this, and sometimes that will work, but for it to truly work all your LEDs have to be very closely matched. If they aren't, one is going to have a slightly lower internal resistance than the others and could potentially cause problems. I've actually tried this approach and didn't have good luck with it. For reasons unknown to me, one would burn out and that would either short internally and cause the others to go, or open the circuit and all would simply cease to function. I was trying this with some really cheap LEDs so that may have been the problem.

Anyway, I really don't see using a switcher as a big problem and there are definitely advantages. My own plan is to have a separate wiring net just for LED lighting and power it from some central location. The design below can put out 5 amps, so that's a lot of LED light. Being a conservative sort of guy I designed it to only put out 2.8 amps, and at that power level it's 87% efficient. I picked this number because Cree makes those LEDs I mentioned above, and if you cut the power input down to 700 milliamps (which is what you get if you power them at 3.5 volts) they are bound to last longer than I will and the one switcher can coast along powering four of the things while dissipating only 0.39 watt it's own self. If it was a resistor doing this job and the battery was at 12 volts, then that resistor (or linear regulator) would be wasting almost 24 watts. Twenty four watts is actually a fair amount of heat if you have it in a small space.

Four Big Mother Cree LEDs kicking out almost ten watts of LED light is probably a lot more light than you want to live with in the cabin, though, unless you happen to be one of those "dudes" that never takes off his "shades." So, tomorrow I'll see about a dimmer. LEDs do not dim down well by just mucking with the input voltage, but they can be very effectively dimmed by turning them on and off rapidly and varying the "duty cycle" as you do. Which is to say, you might turn them on every ten milliseconds, then off briefly 9.5 milliseconds later and your perception would be of a steady bright light. This is because humans cannot perceive "flickering" that is happening at 100 cycles a second, which is what the ten millisecond cycle would produce. Now if I do the same thing and only turn the light on for one millisecond, then off for nine it's the same flickering frequency, but now only 10% of the light, and it will appear less intense. 90% less intense, actually, but the color of the light will be the same. This approach is called Pulse Width Modulation, if you are interested. Really intense lights like these Cree LEDs definitely need to be tamed down a bit. I guy could get a sunburn! Of course, modulating the light output also saves all that electricity, because the switcher will automatically cut the current back to compensate.

I think the diagram is self-explanatory, but if there are any questions just post them.

Opps! I see I forgot to label the diode and explain the switch. The diode is a "Schottky" type, rated for 5 amps and the unit is ON if the switch is OPEN or that input pin is just left unconnected.

BillyDoc

 

All this has gone over my head starting on post 3.

All I wanted was a resistance value and ho to do it. I understand I need more info on the LED but they are for sale loose in a box in the TV shop, they don"t even know what they are themselves . Yesterday I bought 2 x330's I will give then a shot today and judge by eye. They do get a bit warm with the 150,--warmer that I would have expected , I think Poken is right, I might be overdriving them.

 

Hi Frosty,

If you have an old potentiometer lying about use that to find the best resistance, then measure whatever it is with a multimeter -- if you have one. They shouldn't be getting hot. Make sure you start with too much resistance and gradually reduce it, not the other way around or the LED might flash brightly, once.

I got a "rush" project this morning and won't be back to the forum until Friday, so there will be some delay on that dimmer I promised. I want to build one and make sure theory matches practice before putting it up here.

BillyDoc

 

Hey I understood that post!! Yes I had thought about that ,I often try to get round adding sums up and do it mechanically.

I do have a couple of 1000 ohm pots. Im gonna do that now.

I bought some 330 yesterday but they were so small I lost them.

 

These LED seam to work with with same brightness with anything from 150's 330's and (900 with a potentiometer).

By the way some one said LED's dont respond well to potentiometers thats not true with a 150 resistor and 50K pot the LED dimmed from nothing to full perfectly.

The resistors driving one LEd gets too hot to touch, is that Ok

 

Frosty
Lets call this Lesson 4 in your electrical education.

The simple answer is NO. It is not OK At best it reduces life. At worst it will make a bad smell, smoke and could even start a fire.

The power in a resistor is:
Power = Vd^2/R

With your 12V system the situation could be something like this.
Supply voltage with charged battery of 14V. The voltage drop across the LED is 2V. Hence the drop across the resistor is 12V.

Not sure what value resistors you have ended up using but lets say orange, orange, brown. So the power is:
P = 12^2/330 = 144/330 = 0.43W

Now if you have a 1/2W resistor it will be OK. Should get warm but not too hot to touch. If you have 1/4W resistor it will be running too hot. The 1/2W are about 9mm long in the body and the 1/4W are about 6mm long.

Now if you ended up with 150ohm resistor then the power is almost 1W. The body needs to be something like 11mm long for this power dissipation level.

Rick W