how to match a voltage source to 3w star led?

xchcui

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hello

i read on 3w star led data sheet that on If-700mA the min Vf is 3.4v and the max Vf is 4.9V.is it mean that i can directly(direct drive)the led with voltage source of 3.4v and increase it till it get 4.9V without any issue?
how much does it affect the bright of the led and the current?

thank in advance
 

AnAppleSnail

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Well, it's easiest to get a constant-current driver (LED driver) of appropriate current output and not have to worry about it.

The current through an LED changes greatly with small changes in current. Every LED has some forward voltage at the rated current, and there is variation. From there, adding voltage increases current a lot. Usually, a white LED driven at 4.7v will fail.
 

xchcui

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thanks for your reply.

so if i run the 3w star led with steady voltage source between 3.4vf and,lets say, 4.5vf(0.4v less than the max on the data sheet)without any resistor or other limiting circuit,it should be ok,should'nt it?

and by the way why in the data sheet they say max Vf(4.9v) and min Vf(3.4v) value for a certain If(700mA)?after all,as you mentioned, for every Vf on the led there is apecific rated current(If)?:thinking:
 

AnAppleSnail

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thanks for your reply.

so if i run the 3w star led with steady voltage source between 3.4vf and,lets say, 4.5vf(0.4v less than the max on the data sheet)without any resistor or other limiting circuit,it should be ok,should'nt it?

and by the way why in the data sheet they say max Vf(4.9v) and min Vf(3.4v) value for a certain If(700mA)?after all,as you mentioned, for every Vf on the led there is apecific rated current(If)?:thinking:

LEDs have individual variation. Voltage and current are related in an interesting way. Check out a Cree XP-E datasheet for a good example: Forward voltage vs. Current graph.

You'll probably find the 'ideal' voltage for this LED to be 3.7v. But if you buy and use a current-limiting LED driver, you won't have to worry about the voltage. It will be handled if you correctly feed the driver.
 

xchcui

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thanks AnAppleSnail.

you explain the necessity of current-limiting led driver for best results and i understand that.
but what about the data sheet question?why they give specification of 3.4Vf min. to 4.9Vf max. under the title"forward voltage,characterristics at If=700mA(Ta=25c degree)"if we know that there is a specific current that flow through the led for each specific voltage?
after all the current is not 700mA at 3.4vf and also at 3.5vf,3.7vf and etc.:confused:
 
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hellokitty[hk]

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It's already been said it's because individual led characteristics will vary and will even change, those are the maximum and minimum values that they will reach to.
 

AnAppleSnail

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At the rated current, every LED has some forward voltage. Call this "Vf." Increasing current increases the Vf of the LED by some amount. If it started with an awfully high Vf, its Vf will remain higher-than-average through the whole current range. If it began with an awfully low Vf, its Vf will remain lower than average.

So if you get an 'average' LED, you can easily calculate the exact voltage to get such-and-such current. But you don't have an average LED, you have a sample of LEDs. So whatever voltage you choose for your widgets will have varying current (And brightness, and heat) pass depending on the LEDs. In fact, it's difficult to set a voltage that will give maximum safe output for this reason: LEDs vary. So voltage control for an LED is not a good way to push it near maximum, because volts aren't precise. Current is.

If you're stuck with voltage control, use the nominal forward voltage and cross your fingers that your Vf variation isn't so great as to cause too-dim units or too-hot units... Or burned-out-in-a-tunnel units.
 

Steve K

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just as a comment to elaborate on AnAppleSnail's post, the key to using most semiconductor products, and perhaps most products used by engineers, is to carefully and completely read the datasheet for that product.

For instance, as I gaze upon the datasheet (version CLD-DS18 rev 12) for the Cree XP-E LED, I note that it lists on page 5 the forward voltage for the LED at various currents for each of the colors. For some of these currents, there is a typical Vf listed as well as a maximum Vf. The number for the max Vf is useful, since you can design your driver circuit knowing that this is the highest it will be under the given conditions. The more cynical/experienced engineers will ask "hey, what temperature is that data good for?", and as I look at the datasheet, it really doesn't say. I would assume that it only applies to 25C.

Looking elsewhere on page 5, I see that there is a spec for a temperature coefficient of voltage for the different colors. For white, it says there is a typical coefficient of -4mV/deg C. Therefore, if I expect my white LED to operate at 50C, I should expect a change of (50-25)*(-0.004) volts, or -0.1 volts. This puts the typical Vf at 50C at 1A equal to 3.4v instead of the 3.5v at 25C.

Also note on page 8 that there are graphs showing the typical change in Vf as the LED current changes. This is handy for estimating what Vf might be at 400mA, for example, or at least knowing how much the actual Vf will change as current changes.

The reader will also note that the datasheet doesn't list the max Vf for the higher current levels, and doesn't list any min Vf at all! For all you know, the Vf could drop to 2.5v for some random part. Would it?? Probably not, but Cree isn't willing to test all of their parts in order to guarantee it. If you asked the Cree applications engineer nicely, she might give you the values that they usually see, but that's nothing that they will stand behind. If it's important to the design of the driver, the engineer will test a number of samples, and possibly set up a test of the LEDs when they are received at the driver production facility.

So... that's my bit of advice for the day. :)
Read the datasheet carefully and completely. Learn what it means when things aren't included in the datasheet. Never "assume" that a specification will be what you want it to be just because it isn't in the datasheet.
I think Bob Pease did a nice article on "how to read a datasheet".... I might have to look for it among the columns he wrote for trade magazines.
 

xchcui

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so lets say that i choose(for example)a resistor of 2.6ohm to get 3.2vf with 700mA powered by 5vdc voltage source after calculation of(5v-3.2v)/0.7A=2.6ohm.how can i calculate the voltage and the current on the led and the voltage on the resistor when i increase or decrease the voltage source after i had already chosen the resistor value?lets say if i increase the voltage source to 5.3v.
 

AnAppleSnail

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so lets say that i choose(for example)a resistor of 2.6ohm to get 3.2vf with 700mA powered by 5vdc voltage source after calculation of(5v-3.2v)/0.7A=2.6ohm.how can i calculate the voltage and the current on the led and the voltage on the resistor when i increase or decrease the voltage source after i had already chosen the resistor value?lets say if i increase the voltage source to 5.3v. 5VDC in 2.6ohm R1 700mA current assumed 5.3VDC now Well, first let's recognize what you're assuming. Very few consumer parts supply 5VDC. Very few resistors are actually 2.6 ohm. And every LED has its Vf in some range. This is a strange setup. We aren't doing your homework for you, are we? So what is the real range of voltage for your PSU, over time and temperature (And line voltage?) What is the real value of that resistor (2 1-ohms and what, five three-ohm units in parallel?), with tolerance? What is the real Vf of the LED? What is the real current of the circuit? Here are four variables (Vin, R1, Vf, and I). Three of them make the fourth. It is easy for us to measure Vin, R1, and I in an assembled circuit. It requires some care to measure Vf, but can also be done outside the circuit. So let's assume three variables: Vin=5.0V, R1 = 2.6 ohms, and I = 700mA. Problemsolving strategies: We know some things and want others. Start with voltage: 5V = VR1 + Vf VR1 is voltage of R1, it's equal to current times R1: VR1 = 0.7A * 2.6 5 = (0.7*2.6)+Vf 3.18 = Vf So we get Vf of 3.18v. That isn't right, is it? One of these assumed numbers is wrong, then. Most LEDs from this product line will not pass 700 mA at 3.18v. The current is probably wrong. Now you do it at 5.3v. This is why voltage control for LEDs is an ugly beast.
 
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xchcui

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maybe i didn't ask the question right.i know how to calculate the voltage the resistance and ect. in the circuit,how to choose a resistor and how to measure the components with multimeter when it is working.the data that i gave was theoretically,only for example(for led 3vf-3.4vf).the question that i try to get an answer is-what happens after i have been chosen a resistor,after i did the all calculation and i have theoretical working circuit that i know the vf,the current,the resistance and etc. on each part of the circuit(and the circuit is working exactly as i plan on practical).and then,what will happen then,when i,suddenly, increase the voltage source from 5v to 5.3v,what then will happen to the led vf voltage?is it stay the same?is it increase?what will happen on the resistor?will it absorb all the change of the increasing voltage?how can i calculate the change of the values without measuring the components on the circuit while it is working?:banghead:
 

AnAppleSnail

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An increase in voltage will increase the current. This increases the forward voltage and the voltage drop of the resistor. Then the LED probably heats up and its forward voltage decreases sightly.

It's very hard to predict threse changes in great detail without getting a Vf/current curve for the exact LED sample you have. You can a assume several values to help. Start with an estimate for the LED forward voltage at several currents and interpolate the forward voltage and (R1 x current) to get the source voltage someplace on that interpolated curve.
 

AnAppleSnail

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An increase in voltage will increase the current. This increases the forward voltage and the voltage drop of the resistor. Then the LED probably heats up and its forward voltage decreases sightly.

It's very hard to predict threse changes in great detail without getting a Vf/current curve for the exact LED sample you have. You can a assume several values to help. Start with an estimate for the LED forward voltage at several currents and interpolate the forward voltage and (R1 x current) to get the source voltage someplace on that interpolated curve.
 

xchcui

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so i understand that it is very easy to calculate a resistor for the led but there isn't an easy and simple way to calculate,after that, these changes.The method that you gave me to find these changes is pretty long to find the value,but at least i can find it.
thank you very much for all the replies,you help me alot:twothumbs,now i can deal with the led while i aware to what i am doing by avoiding damage to the led,and better plan of the circuit.

by the way if i decide to use 3W 700mA constant current regulated led driver(or resistor) in series with some 3W star-led that mounted on a long piece of aluminium,how should i mount the led driver on the aluminium?(i believe the aluminium piece is going to be hot by coducting the heat through it,that has been caused by the leds,so i dont know how should i mount it without make any damage to it.)
 

AnAppleSnail

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That depends on the driver. Many drivers are built for flashlights, in which case it'll be a round, awkward package for your installation. I usually attach the driver firmly in a place it won't rattle around or be damaged. In many cases at low power (Probably dissipating less than a watt) it does not need a strong thermal connection. If it gets warm in your early testing, attach part of it to the aluminum in a way that doesn't conduct heat. There are products like thermal adhesive pads that work well with clamping to make thermal, non-electrical contacts.

The driver has similar heat tolerance to the LED. Running at very high temperatures for a long time affects its components more, though. Usually sticking it in the same chunk of metal as the LED works fine - That's how flashlights are built.
 
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xchcui

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AnAppleSnail,when you said:
Usually sticking it in the same chunk of metal as the LED works fine
how did you mean to stick it?with some kind of adhesive?
and when you said
If it gets warm in your early testing, attach part of it to the aluminum in a way that doesn't conduct heat.
do you mean that the driver need to cool-down by the air, and does not through the chunk aluminium?
 

Mr Happy

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so i understand that it is very easy to calculate a resistor for the led but there isn't an easy and simple way to calculate,after that, these changes

One is just as easy to calculate as the other. But what people are trying to tell you in this thread is that you don't know the actual Vf. The forward voltage drop varies from one sample to another, it varies with temperature and it varies with voltage. The data sheet gives you a range for a reason, because the exact value is variable and unknown.

However, let's take your calculation example. If we assume that the Vf is 3.7 V, and that we require 700 mA from a 5.0 V supply, then the calculation proceeds as follows:

1. Voltage drop over resistor = 5.0 - 3.7 = 1.3 V
2. Current through resistor = 700 mA = 0.7 A
3. Value of resistor = 1.3/0.7 = 1.86 ohms

Now let's suppose the supply voltage increases to 5.3 V. We can just work the calculation backwards:

1. Voltage drop over resistor = 5.3 - 3.7 = 1.6 V
2. Value of resistor = 1.86 ohms
3. Current through resistor = 1.6 / 1.86 = 0.86 A = 860 mA

It's almost exactly the same procedure one way or the other.

But I stress again, you don't really know the exact value of the Vf. That is why LEDs are usually operated with constant current drivers.
 

AnAppleSnail

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If the driver gets warm during operation, you can put it on a heatsink. 'Adhesive thermal pads' are good because they conduct heat fairly well, and are not conductive. They do require a clamping force to work well though. The driver can cool down inside most flashlights, with its "heat load" going into the flashlight and out into the environment. In sustained, extremely-hot operation (Painful to the hand), the driver components may have difficulties.
 

xchcui

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Mr Happy,as you mention,there is a problem with the calculation,because,like you said,when i increase the supply voltage the led vf increase(and doesn't stay constant) so the voltage drop on the resistor is less than 1.6v and the current less than 860mA.
but anyway i,now,understand why is so important to use a resistor in series with the led,and why is better,simple,safer and easy to use a constant current drive.
thank you very much,AnAppleSnail,for all your professional relpies and for all the members of the forum that help me to understand the issue.:thumbsup:
best regards:)
 
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