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how to work with 3w leds??

docb

Feb 11, 2010
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I want a 3 watt LED, that is switched on/off by the voltage going to a tiny LED.
The 'trigger' is about 2 volts, so I was thinking about a MOSFET like an ndp6020p. (is there a better choice?)

My power source for the 3w LED would be a 9v wall wart, so I thought about a 7805 to get 5v. All sound ok so far?

Then to get the 5v down to what the LED wants, I thought about three diodes in series, just to drop to 2.9v.(.7x3v)

Can I just drive these 3 watt LEDs from that 2.9 volts, or is there more to it?
The LED specs say 3.2v @350ma.

Does this all sound sane? Thanks very much for any insight!!
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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LEDs are driven by a current source, not a voltage source. I'll try to find you the circuit for a a simple switchmode controller for high power LEDs.

This is a good solution, but I have one somewhere with a much lower parts count.

The attached image is the circuit referenced here and is somewhat similar to what my colleague uses (except I think his version is simpler!).

OK, here's another one. The problem with this one is that it has no +ve feedback and rather limited gain, and so if you have a slow rise time on your input voltage it can revert to beig a linear regulator with rather nasty (smoke producing) impact on the mosfet.
 

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docb

Feb 11, 2010
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I can't say I really understand what that means about being a driven by a current source, not a voltage source.

Right now I now have 2v tapped from an SMD LED, going to the base of a Q, which acts as the trigger to switch 5 volts to a bank of 100 small LEDs. It seems to work just fine.

Is what I want to do with one bigger 3 watt LED, really that different?
 

(*steve*)

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That circuit uses a resistor in series with the LED to limit the current. In your case, if you want to drive a 3W red LED from 9V, the resistor will be dissipating between 7 and 9 watts -- in other words you will be wasting 70% to 80% of the power you're putting in.

For low currents it's not a huge deal, but for a 3W LED it becomes significant. Those resistors will have to be BIG and they will get HOT.

Instructables generally shows you the cheapest and nastiest way of getting something done.

the REAL simplest project: if you get stuck, maybe just scrap the electronics and have a super-bright non-blinking light? all you need then is a switch and battery, and use maybe a 1-ohm resistor to save battery life.

That is a recipe to kill a high power LED really quickly.

You've come somewhere where you're likely to be told the right way. :)

You can still operate it from a small trigger voltage, but if you're running a single high power LED from a relatively high voltage source (9V is much greater than the LED voltage) you really should be looking at how best to regulate the current through it without wasting power.

this page has an appropriate circuit: http://www.kpsec.freeuk.com/trancirc.htm

Look at the "darlington pair" in the 4th section down. This is a suitable way to turn on your LED from a 2V signal. Place a 1k resistor in series with the base of the first transistor and it should work fine. In your case, a 6.8 ohm resistor in series with the diode is probably about right for a 3W device. Make sure the resistor is rated at about 10W and that the second transistor is capable of switching a couple of amps. It will probably need a small heatsink.
 

docb

Feb 11, 2010
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This what I'm using now for small LEDs and it seems to work fine for that. Where it says "low" it's hooked to + voltage on an SMD LED that acts as the trigger. Where is says "repeat for 24 leds", it's actually driving 100 with no problems at all.

The 2n5088 never gets warm at all, but I'm guessing it would with a 3watt LED, so I'd need a bigger one.

I have a few questions:

Overall is there a problem in this plan?

For the power in the diagram, I use a 7805 to regulate that 9v to 5v for the circuit.
Any issue with that? Or is 3 watts to much for the 7805? If 7805 is ok, How many LEDs can it drive? I have a 7805 driving 500 LEDs now and it doesn't get hot.

With the 3 watts I'd do the same, but is there something wrong with a couple of small diodes to drop the voltage from 5v down .7v twice to get down to 3.6 volts. Then the LED resistor could be quite small or even not needed, yes?

Why do I need a darlington pair with the bigger LEDs, if this is working with what I have? Is it just that one might assume the base voltage of Q might not be high enough to turn on a single transistor? In this case it seems to be.

Can you say why some circuits suggest a MOSFET? Is it because they use a lower "trigger" voltage, so you can skip having a darlington pair?

I guess I want to know why I can or can't I basically do exactly what I've done already but just beef up the transistor? And maybe power supply.

I see all these fancy LED drivers, but I don't see why I really need that..


 

(*steve*)

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This what I'm using now for small LEDs and it seems to work fine for that. Where it says "low" it's hooked to + voltage on an SMD LED that acts as the trigger.

Yeah, but major differences!

* The LEDs here run from 5V, you're talking about 9V
* The 2N5088 has a max current of 200mA a 3W LED requires well over an amp (depending on colour)
* Based on some assumptions, that circuit allows 300 mA to the LEDs and is in excess of the transistor's max current.

Where is says "repeat for 24 leds", it's actually driving 100 with no problems at all.

If you don't change the resistor, the total current will be far less (and will also probably limit the immediate risk of thermal runaway -- see below)

The 2n5088 never gets warm at all,

If that transistor doesn't get hot, it's because of one of the following:

* it's blown and is now a short or open circuit.
* it's only turned on for very brief periods
* it's not a 2N5088
* you're using a resistor > 3 ohms (perhaps 33?)

In any case, it's a terrible circuit. You're driving the LEDs in parallel (thus form what is essentially a voltage source). LEDs are not like light bulbs.

A light bulb starts with a relatively low resistance, and as current flows the wire heats up which increases its resistance. If 2 bulbs are put in parallel they will share the current because if one gets slightly more current it heats up more which reduces the current it can draw.

A LED may draw (say) 20 mA for a given voltage (let's call it 3.2 volts). But another LED form exactly the same batch may draw 21 mA at that same voltage, and another 19 mA. That doesn't sound like much of a difference, but the one drawing 21 mA will heat up slightly faster than the one drawing 20mA. The BIG difference is that the LED uses MORE current at a given voltage as it heats up. So the one drawing 21 mA starts to draw 22 mA -- that leaves less for the others which cool down a little and then require even less current. The additional current flows into the first LED which gets even warmer and draws more current, and so on...

Eventually one LED draws too much and fails. Then another LED takes up its place until it too fails (although it fails faster). Soon ALL the LEDs have failed. This is a major failure mode on LED Christmas lights that often have many LEDs in parallel.

If LEDs are placed in parallel they should each have their own current limiting resistor. If placed in series, each string requires a resistor.

but I'm guessing it would with a 3watt LED, so I'd need a bigger one.

Sure would.

I have a few questions:

Overall is there a problem in this plan?

Other than excessive power dissipation, for a single LED it could be made to work.

For the power in the diagram, I use a 7805 to regulate that 9v to 5v for the circuit.
Any issue with that? Or is 3 watts to much for the 7805? If 7805 is ok, How many LEDs can it drive?

The 7805 can supply between 100mA and 5A max depending on the type and pacage. I'll assume you have the fairly common TO-220 version which is typically rated at 1A. I'll also assume that your 9V source remains at about 9V under whatever load you are using. Oh yeah, I'll assume that your LEDs are white or blue.

You could probably drive a single 1W LED or more than 15 normal LEDs with that circuit and that regulator.

It would be able to supply almost enough current for a single 3W LED (although remember that the 7805 is not actually current limited to 1A) or up to 500 normal LEDs operating at 20mA

However your question is really nonsense. I could drive 500 LEDs at 20 mA, or 1 at 1A (it had better be a 3W LED), or 1000 at 10mA, or 1 million at 1 uA. It all depends on the current because, as I've said, LEDs are driven by a current source, not a voltage source.

I have a 7805 driving 500 LEDs now and it doesn't get hot.

This means that the average LED current (if we go by your circuit) is about 0.6mA (total of 300mA for all LEDs) and the power dissipated in the 7805 is about 1.2W. It should get warm.

With your circuit you can drive as many LEDS as you require, but as you add more LEDs they will get dimmer and dimmer...

Dissipating 1.2W, a 7805 will reach (without a heatsink) a case temperature 60 degrees C above ambient. In pretty much ALL ambient conditions (excluding Siberia in winter), this would be hot to the touch, and you wouldn't be able to keep your finger in it. It may be that your LEDs are not on for very long, or that your power supply is sagging under load, or that your LEDs are actually drawing far less current. Any of those might explain why your 7805 is not hot.

With the 3 watts I'd do the same, but is there something wrong with a couple of small diodes to drop the voltage from 5v down .7v twice to get down to 3.6 volts. Then the LED resistor could be quite small or even not needed, yes?

No, you need a current source, not a voltage source.

Whilst small LEDs are fairly tolerant (especially when lightly driven), high power LEDs need to have heatsinks and carefully controlled current if they are to be driven to maximum brightness (unless you don't care about their longevity).

The 7805 will dissipate 4W (and get pretty hot without a heatsink). The specs tell me it would reach a junction temperature of 260C above ambient -- which would cause it to self destruct if it didn't have an over temperature shutdown. Incidentally the case would reach a temperature of 240 above ambient which is easily enough to start a fire.

Presuming it has a heatsink, the diodes would drop about 1V each at 1A leaving less than the voltage required to drive the LED at full brightness, so the current would be less.

HOWEVER as the LED and diodes (and possibly the transistor) heat up, the current would increase and increase until either the regulator shut down, or the LED or Transistor started smoking. If you were really lucky, you might be able to destroy both the transistor and the LED and then the regulator.

It might make a good youtube video.

On the other hand it might just sit there working...

But either case is possible and it all depends on the tolerance of the various parts, the temperature, and a number of other random factors.

But please feel free to try.

On balance, I'd say that 2 diodes *might* be enough to prevent everything from going really pear shaped.

I also think you might be safe driving at 40 mph with bald tyres on ice. Hey, it might work, you might not slide. That intersection you slide through *may* be empty. (I don't drive that way)

Why do I need a darlington pair with the bigger LEDs, if this is working with what I have? Is it just that one might assume the base voltage of Q might not be high enough to turn on a single transistor? In this case it seems to be.

Because you'll be switching a heavier load and transistors that can handle the heavier load generally have lower gain. The lower gain translates into the need for a higher base current, and this can be provided by another transistor, typically connected as a darlington pair.

Can you say why some circuits suggest a MOSFET? Is it because they use a lower "trigger" voltage, so you can skip having a darlington pair?

No, they require a lower gate current, but at a higher voltage. With 2V, there are very few mosfets which will turn on sufficiently -- most require 5 to 15 volts.

I guess I want to know why I can or can't I basically do exactly what I've done already but just beef up the transistor? And maybe power supply.

You can, but it may not work, it may cause components to fail, or get very hot.


I see all these fancy LED drivers, but I don't see why I really need that..

It's because people with far more experience that you or I have determined how to get the best out of LEDs and the available power sources.

If excessive heat is not a problem, you can probably do it in a far easier way.

I live in a hot climate. Excessive heat increases my air-conditioning bill.

My parents travel in a mobile home. Excessive power dissipation reduces the life of their batteries (or increases fuel costs)

Some people think that efficiency is good (perhaps they pay for power).

I can see the advantage though if you live in a very cold climate and don't use a heat pump (which would generate heat more efficiently than your circuit)
 

docb

Feb 11, 2010
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First, thanks VERY much for the long reply.

I'm still lost of the current source vs voltage source thing, but Ill have to read up on it, as that concept still evades me.

I should maybe point out that all these are now used for flashing lights, so they are not on all the time. Maybe that's why it continues to work so far. :)

I will probably ask you more questions after I've digested all you have written.
 

Mitchekj

Jan 24, 2010
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Current source: supplies a constant current, or at least attempts to, regardless of the load. Voltage changes within a certain min/max window.
Voltage source: supplies a constant voltage, or at least tries to, regardless of the load. Current changes within a certain min/max window.

This is important for constant voltage sources with LEDs, particularly high powered ones, since the voltage drop across them lowers as heat rises. Thermal runaway can occur since the required drop (Vf) keeps going down, causing more current to flow, lowering Vf, ad nauseum.

Even if thermal runaway were not to occur, using a constant current supply will actually reduce power consumption as the LEDs heat up. Which is very handy.
 
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docb

Feb 11, 2010
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I see lm317 regulators can be set up as constant current source..

How about if my existing circuit used an lm317 to power the LED?



I'd set the regulator to output 3v.

(I'm guessing I need something bigger than a 2n5088)
 
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(*steve*)

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No, that wouldn't work. You have connected the 317 up incorrectly (but it's close!)

And yes, you would require a transistor capable of handling a higher current than the 2N5088.

Attached is something I came up with in about 30 seconds. My choice of transistor and resistor values may need tweaking.

The LM317 will need a heatsink.

This circuit (and any one using a constant current source) has the advantage of operating over a wide range of input voltages (albeit with lots of heat dissipated) and should be safe for the LED as long as the correct current is chosen.

There are probably far better PNP transistors than the 2N2955 -- let me check... Yeah, a TIP32C would be far better and the component values look reasonable for that -- it's also a pretty cheap part.
 

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(*steve*)

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This may be considered cheating a little, but it's a constant current supply for LEDs.

You could power this via the transistors as in my previous circuit, and connect the LED to its output.

It would draw less current and produce far less heat.

http://www.dealextreme.com/details.dx/sku.13557
 

docb

Feb 11, 2010
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How's this new schematic look?



Isn't 9v better so LM317 runs cooler?

If instead, I used the driver you linked to, is there a reason that I'd use this two transistor setup instead of my original circuit? (I'd change the 2n5088 to something bigger)

(I have that already built, so swapping out a transistor - and just using this driver you linked to - is a bit less to redo.)
 

docb

Feb 11, 2010
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How about this way:
(from the LM317 specs)



The 317 does the big work, and small transistor does the switching.

R1 stays at 240 and R2 sets the voltage I need for the LED, yes?
No reason not to use a 2n5088 instead, right?

Oh wait that would shut my power when I see 5v at base, yes? I'm getting confused.
 
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(*steve*)

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Isn't 9v better so LM317 runs cooler?

Yeah, way better. I must have grabbed that 12V from something else :)

If instead, I used the driver you linked to, is there a reason that I'd use this two transistor setup instead of my original circuit? (I'd change the 2n5088 to something bigger)

The issues is in providing enough base drive. A single transistor capable of switching 1A (preferably more since you don't want to run it near its limits) will have a lower gain (a TIP31A for example has a minimum gain of 10 with a collector current of 3A. That means we need to be able to supply at least 100mA of base current to allow 1A (It will actually require less, but always calculate based on the worst case). Your logic will not be able to give you the required 100mA.

An alternative is to use a darlington pair in a single package (like a BD681S) and use your original circuit (with the LM317 wired correctly). This provides a gain of 750 at 1.5A, so you'll need maybe 1.5mA of base current. Since the base has to be pulled up to at least 1.2V, the base resistor will be about 380 ohms. This device will have a Vce of 3V, so it will dissipate around 3W and will also require a heatsink.

(In contrast the TIP32C will have a Vce of around 1.2V and will probably not require a heatsink). The lower Vce also means the circuit will continue to operate from a lower voltage.
 

(*steve*)

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Oh wait that would shut my power when I see 5v at base, yes? I'm getting confused.

That circuit effectively creates a 1.2V regulator, I can't see it shuts it down. In any case, as when set up as a current regulator the LM317 is pretty much already regulating the lowest voltage it van (across a current sense resistor).

Another issue is that the 317 is floating above ground, well above your logic level. You could (I guess) use a 337, but I'm not sure it's a great option.
 

docb

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ok, so maybe this latest circuit (with D1 the other way around) is my best option.

I *might* replace the lm317 with one of those fancy LED drivers, but I assume I can simply wire the driver from collector of q1 to gnd, (and delete the lm317.)

 

(*steve*)

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Yes, that looks fine. The current required by the LED driver will be less than that required by the LED, so everything should run cooler.

The driver (if it is the one I suggested) has a bridge rectifier in the board which may be taken out of circuit for even higher efficiency -- but leave that for later.
 
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