I'm having a little trouble grasping the basic concepts. I want to drive a XM-L2 U2 at max brightness (but not overdrive it) using a standard ac-dc wall plug. Can I do this with a wall plug rated at 5V 1A? I know I need a driver in between, but am not sure if a buck/boost affects voltage or amperage.
Led amperage basic question
- Thread starter meyerovb
- Start date
No.
And here is why I would not do it.
AC DC adapters voltage ratings are usually underrated by a large margin, up to twice their rated value. That 5V DC one you have, if you measure it's output may read like 7V or more. The XM-L2 wants to see like a 3 volt voltage across it and it is the voltage that will determine the current flowing through it. As it is most likely a non-regulated type, as soon as you connect it to the led, the 5v( or more ) will ramp up that current way past the 3A for a short time but just enough to probably fry it. I may be wrong but let's see what other CPF members think about. Good question though.
The buck/boost driver keeps the current in regulation by adjusting the voltage across the led (s). It affects both voltage and amperage ( current )
And here is why I would not do it.
AC DC adapters voltage ratings are usually underrated by a large margin, up to twice their rated value. That 5V DC one you have, if you measure it's output may read like 7V or more. The XM-L2 wants to see like a 3 volt voltage across it and it is the voltage that will determine the current flowing through it. As it is most likely a non-regulated type, as soon as you connect it to the led, the 5v( or more ) will ramp up that current way past the 3A for a short time but just enough to probably fry it. I may be wrong but let's see what other CPF members think about. Good question though.
The buck/boost driver keeps the current in regulation by adjusting the voltage across the led (s). It affects both voltage and amperage ( current )
Illum
Flashaholic
Theres two forms of AC/DC power supplies: the Wallwort "Linear" unregulated version or the SMPS "tabletop" regulated version. Wallworts are designed to be loaded roughly 50-70% at any given time [or else the heat resulted from the stepdown would melt the casing], by unregulated it means if no load is applied, the output voltage will read much higher than the rated output. 5V wallworts are pretty common nowadays and most are regulated [make sure it specifies 100-240V input]. Always make sure the power supply you are using is regulated, or else you will fry your driver the next time you plug in or turn on.
Driving an XML-2 at Max brightness would be driven by constant current at 3A, forward voltage would start at ~3.4V cold and slowly decrease as the emitter gets hot. This shift in forward voltage as a function to temperature is why its important to drive LEDs using a constant current driver and not a constant voltage driver. If current is not regulated, the LED will pull more current as it heats up, accelerating it to pull even more current and eventually die of thermal runaway.
With unregulated supplies especially, if its rated for 1A and you pull 3A, the transformer will saturate and funny things will happen, a number of which may result in an electrical fire. A wall plug thats rated at 5V 1A yields 5W total, to drive an XML at spec requires 3.4x3 = 10.2W, assuming 80% on your driver, you're looking at least 10.2/.8=~13W input. Unless you figure out a way to create energy from nothing I don't know how to help you. :)
Last time I tried driving an XML on 5V I had to use a 5V 3A regulated supply and a TI PTN78060W as the 3A driver. It worked okay... the output was a tad shy of 900 lumens under forced cooling and heatsink temp at 70F, really nothing to call home about. I never found why XML was so appealing. :thinking:
Driving an XML-2 at Max brightness would be driven by constant current at 3A, forward voltage would start at ~3.4V cold and slowly decrease as the emitter gets hot. This shift in forward voltage as a function to temperature is why its important to drive LEDs using a constant current driver and not a constant voltage driver. If current is not regulated, the LED will pull more current as it heats up, accelerating it to pull even more current and eventually die of thermal runaway.
With unregulated supplies especially, if its rated for 1A and you pull 3A, the transformer will saturate and funny things will happen, a number of which may result in an electrical fire. A wall plug thats rated at 5V 1A yields 5W total, to drive an XML at spec requires 3.4x3 = 10.2W, assuming 80% on your driver, you're looking at least 10.2/.8=~13W input. Unless you figure out a way to create energy from nothing I don't know how to help you. :)
Last time I tried driving an XML on 5V I had to use a 5V 3A regulated supply and a TI PTN78060W as the 3A driver. It worked okay... the output was a tad shy of 900 lumens under forced cooling and heatsink temp at 70F, really nothing to call home about. I never found why XML was so appealing. :thinking:
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Yoda4561
Flashlight Enthusiast
With any voltage source above around 3-3.2v you want a buck driver, not a boost driver. You use buck drivers when your voltage source is higher than the LED needs, you use a boost driver when it's lower. Leds work sort of like water valves, with the applied voltage being the valve handle that controls how much current can flow through it. Too much voltage and the LED will burn itself up trying to flow more current than it's capable of. While the wall wart may only be rated at 3 amps that just tells you how much it can deliver reliably, they will dump more short term at the expense of shortened lifetime. The XML may live long enough for the wall wart to become overloaded and suffer voltage drop (at 5v an XML will look like a dead short to the voltage source), but it's not doing either the LED or wall wart any favors.
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It is best that you look at the data sheet for an XML and look at the curve of voltage versus current. That will give you a bit of an understanding of the the XML responds to current/voltage. You need to realize though that this curve shifts by approximately +/- 250mV purely based on process variation. I.e. One XML could hit a target current at 3V, and one could hit it at 3.5V. You will not know till you get the part in your hand. To add to that, those values will change over the first 50 hours or so, and they change w.r.t. temperature. Forward voltage goes down as the temp goes up. So while it may hit a particular current at 3.2V when first turned on, you may only need 3.1V or less when it is hot.
For this reason, you need a constant current driver specifically for an LED, not a simple constant voltage power supply. You could start with a 12V, 1.5A power supply, then buy an appropriate LED buck driver that will hit the current you need. The LED driver automatically adjusts its output voltage so that the target current is maintained.
Semiman
Thanks, that's another question I had, what would be the difference between using, say, the 12V 1.5A psu and a 5V 4A psu (let's assume using a driver that can handle either, like a b3flex)? Could both power the led with to it's max brightness? Any advantages/disadvantages to either option?
argleargle
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First of all: I love this thread! :)
A lot of home made direct-drive LED stuff only works with the battery it was designed with because perhaps the designer didn't realize that the properties of the battery itself were limiting the voltage and current through the circuit. Perhaps the designer was attempting to come up with a cheaper solution by using direct-drive of the led (with or without a limit resistor?)
Example: With no calculations or math, using simple :poof: trial-and error, someone manages to come up with a driver-less direct-drive LED circuit that works when powered by "12 volts of specific button batteries." The tinker then hooks his circuit to a 12v car battery, thinking it "should work." Most likely, the circut blows. Why? The driver-less build relied upon resistance (reactance) of the LED and internal resistance of the battery stack to not explode. Different batteries have different capacities and each one has its own internal resistance. These things, of course, affect the circuit. Note: Button cells tend to supply extremely low peak current.
Related search terms: "internal battery resistance," "peak discharge amperage," "voltage sag," "current-limited regulator," "voltage-limited regulator."
The solution is a driver circuit of some type, although you can usually get away with tinkering with specific batteries and a limit resistor. This is also called "driverless" or "direct drive," and some implementations try to even leave out the limiting resistor. Warning: using different (higher current capable) batteries may well :poof: fry your home made experiment if you try this method.
Also: I completely agree with everyone else in this thread stating that each specific led might have different electrical properties than another one of the same type, even same maker.
If you've made a driverless circuit, the 5v psu with its 4a max might very well blow something that the 12v 1.5A psu will not. If the circuit design relies upon the PSU to *NOT* provide the full current that the electric circuit is "asking for," then it could definitely blow with other power supplies.
Another way to say it: A circuit that depends on "voltage sag" to not catch fire is definitely vulnerable to different battery types.
As always, someone jump in and correct me if I'm wrong please.
A lot of home made direct-drive LED stuff only works with the battery it was designed with because perhaps the designer didn't realize that the properties of the battery itself were limiting the voltage and current through the circuit. Perhaps the designer was attempting to come up with a cheaper solution by using direct-drive of the led (with or without a limit resistor?)
Example: With no calculations or math, using simple :poof: trial-and error, someone manages to come up with a driver-less direct-drive LED circuit that works when powered by "12 volts of specific button batteries." The tinker then hooks his circuit to a 12v car battery, thinking it "should work." Most likely, the circut blows. Why? The driver-less build relied upon resistance (reactance) of the LED and internal resistance of the battery stack to not explode. Different batteries have different capacities and each one has its own internal resistance. These things, of course, affect the circuit. Note: Button cells tend to supply extremely low peak current.
Related search terms: "internal battery resistance," "peak discharge amperage," "voltage sag," "current-limited regulator," "voltage-limited regulator."
The solution is a driver circuit of some type, although you can usually get away with tinkering with specific batteries and a limit resistor. This is also called "driverless" or "direct drive," and some implementations try to even leave out the limiting resistor. Warning: using different (higher current capable) batteries may well :poof: fry your home made experiment if you try this method.
Also: I completely agree with everyone else in this thread stating that each specific led might have different electrical properties than another one of the same type, even same maker.
If you've made a driverless circuit, the 5v psu with its 4a max might very well blow something that the 12v 1.5A psu will not. If the circuit design relies upon the PSU to *NOT* provide the full current that the electric circuit is "asking for," then it could definitely blow with other power supplies.
Another way to say it: A circuit that depends on "voltage sag" to not catch fire is definitely vulnerable to different battery types.
As always, someone jump in and correct me if I'm wrong please.
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The answer is a solid probably on the 5V, 4A supply. The required overhead of a B-Flex at 3A is 1.1V. The XML typical may be 3.4V, but it could be almost 4V, and if the 5V sags, you may not have enough oomph to get all the way there. You will probably be fine though. The 12V, 1.5A should be just fine.
ianfield
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A 3.3V switch mode PSU will be regulated within reasonable limits and might give you some light from the LED, which really wants about 3.4V when cold. Its probably going to be easier to find a wall wart with too much voltage than not enough, so a buck regulator is probably the way to go. Unfortunately - constant current output (essential to the life of your LED) is rather more tricky, but there are designs online if you're a patient searcher - a much easier solution is one of the many LED driver chips on the market.
If you stick with the 3.3V supply - you'll need a buck/boost converter (Sepic - Cuk etc) these are more complex and usually have 2 inductors + a hefty power transfer capacitor. Its only realy worth the extra expense for battery operation, where the fresh charge voltage starts off above 3.4 & end of life is somewhat lower than 3.4V
Steve K
Flashlight Enthusiast
Can you adjust the output of the driver to put out the desired 3 amps? If so, you should be in good shape. In fact, it's usually nice when the driver has a somewhat higher current capacity than you are using, as it tends to stress the parts less and provide a longer lifetime.
argleargle
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My understanding is that you'd be overdriving the crap out of it, the led doesn't "know" to only pull 3 amps. The driver "tells it what to do" and the power supply tells the driver "what it has to work with."
A "buck driver" limits the circuit down to a certain number. As the battery wears out, eventually the light "falls out of regulation." 4 amps to a single led would overdrive the leds I'm aware of, assuming your battery or power supply can actually provide the 4 amps. If the supply isn't up to the task, voltage and current will both sag beneath 4 amps and you'll have "undesired operation" either way you slice it.
Help any? Ask more questions if we are unclear. It's cheaper than an electrical fire.
If you used a 4 amp buck driver to drive a couple of leds in parallel, then you'd get away with it. Too much for single leds I'm aware of.
Perfect, thanks. What about if I use 3 1 amp bucks in parallel to drive a 3amp led, then dynamically switch the circuit to exclude 2 of them and drive a different 1amp led? Would that work? I understand each buck might have a +-5% margin on it's output, so it would increase the margin the more bucks I add, but just wanted to know if the theory is sound. Thanks.
ianfield
Newly Enlightened
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My apologies for confusing you.
Its important to make the point that you can't just use any old wall-wart to drive a LED. It must have higher voltage than the LED requires - and some provision *MUST* be made to regulate the current.
If your wall wart hes too low voltage, you can boost it with a flyback converter - if your wall wart has too high boltage you can drop it with a buck converter.
There are plenty of chips on the market that do specifically constant current for driving a LED - the manufacturers supply application notes with example circuits that make it almost as easy as a pre-packaged kit.
