Need Help for Voltage and Current circuit for High Current LED devices

[dhyaks]

Last week i get some High power LED (XML-T6 Series) emitter from my friend and i want to make some LED Driver circuit for these High Power LED.
Specs :
~3.7 - 4.2V
3A Max. Current

I still confusing about PWM circuit and because of limited component to make D.I.Y LED driver, i decide to make the driver from Adjustable regulator and Power transistor,

But the problem is : when i adjusting the regulator circuit without any load (LED) it can show 3,7V or 11,2V (for 3 Leds in series). but when i connected to load, voltage just drop too much and the current just show on 0.26A only and get HOT, can anybody give me a suggestion for my project?
The Source is from VLRA Battery specs 12V / 10A.

i try to think and collaborating of two circuit but i didn't sure it's can work well.
the diagram is like this :

i try to pick some Constant current Circuit from this:

and Voltage Regulator from discrete variable LM317 Voltage circuit.
Could someone give some opinion? better if it's not PWM / SMPS / Another Complex dimming circuit.

P.S : And i have 2pcs LED from Luminus that specs is 3,7V / 9A.
i want to drive it max from the circuit design that i have been think about.

 

[DerStrom8]

The LM317 adjustable voltage regulator can only source 1.5A, so chances are your LED is drawing much more current than the regulator can handle.

PWM is a much better option because it changes the average voltage by varying the width of the pulses sent to the LED. Linear regulators are extremely inefficient and I don't recommend them. Instead, you should be able to build a simple PWM circuit using a 555 timer and a few resistors and capacitors:



Replace the motor with your LED and you're set. Just make sure your MOSFET can handle >3A (>9A if you want to drive the LEDs you mentioned at full power).

 

[ronsimpson]

At full power the LED will get very hot. You need a heat sink on the LED.
Next
12V battery, 3V LED, 3Watts in the LED and 9 watts in the linear regulator. (big heat sink on the regulator)
12V battery, two 3V LEDs, 6W in the two LEDs and 6 watts in the regulator. (also big heat sink) (LEDs in series)
I was using 3V for the LED because the math is easy. Actually the LED voltage is 3.7 - 4.2V for 1 LED and 7.4 - 8.4V for 2 LEDs.

You regulator is only good for 1 to 1.5A. It is best for you to stay at a lower current level until you solve the "big ass heat sink problem".
**broken link removed**

 

[MikeMl]

I would not use either Strom's or Ron's circuit to drive a high power LEDs. Neither circuit is a constant current driver; both are essentially adjustable constant voltage sources, which is the wrong thing to drive a high power LED with...

If you start with a automobile battery, when engine is stopped, the battery voltage is <12.6V. With engine running, the voltage will be >14.4V.

With two LEDs in series, the worst case power dissipation in the current regulator will be (14.4V-7.4V)* 3A = 21W (highest input voltage, lowest forward drop on the two LEDs, highest current through the LEDs) . This requires a large heat sink (a good fraction of a square meter of metal). Good news is that you are in a huge car, mostly made of steel, so bolt the heat producing current regulator to the car body....

As pointed out, 3A LEDs must also be mounted onto a heat sink, otherwise you will destroy them...

 

[audioguru]

You show an LM317 with wrong value resistors driving a 15A 2N3055 power transistor so the output current should be very high.
The resistor from the output to the ADJ pin on the LM317 is supposed to be 120 ohms so that the output voltage does not rise when its load current is low. Then for an output of 20V the other resistor to ground must be 1.8k ohms. The 20V output of the LM317 drives the base of the 2N3055 and the emitter of the 2N3055 will be from 18V to 19.4V depending on the output current.
The current into your 4 ohm resistor should be about 4.6A and the output current from the LM317 should be about 130mA.

You do not show the input voltage so we cannot calculate how much heating will occur.

 

[DerStrom8]

You do not show the input voltage so we cannot calculate how much heating will occur.

Hi AG, the OP did mention what he is using for a power supply:

The Source is from VLRA Battery specs 12V / 10A.

 

[crutschow]

If you want to avoid the large power dissipation of a linear regulator and/or series resistor you can use a switching regulator in the constant current mode such as referenced here.

 

[MikeMl]

Here is a sim of the circuit that I linked the OP to on another forum (before it got shut down because it is "automotive").


Note that it does a pretty good job of holding the current through the LEDs constant as the battery discharges (X-axis of all the plots). What it fails at is keeping the current constant as a function of temperature. See the upper plot pane: Green is -20degC, Red is 0degC, Lt.Blue is 20degC and Dk.Blue is 40degC. The 30% current variation is due to the temperature dependance of the Vbe of Q1. You could use the circuit if you design for <2.3A, because even hot it will keep the LED current below its max 3A rating.

Look at the lower plot pane: It shows the power dissipation at the four temperatures in M1 (Green trace), one of the LEDs (Red Trace), and the resistor R1 (Lt.Blue Trace) at a function of battery voltage. Note that in the worst case, M1 dissipates 11.5W, so must be on a large heat sink. Note that the LED dissipates 8W, so it too must be on a heat sink. The emitter resistor R1 is dissipating 1.8W, so should be at least a 2W power resistor...

Even if you use a switching constant-current LED driver, the LED dissipation will not change; it will still have to mounted on a heat sink or you will destroy it...

 

[audioguru]

Hi AG, the OP did mention what he is using for a power supply: 12V.

Then why did his resistors try to make the output of the LM317 more than 20V?

 

[DerStrom8]

Then why did his resistors try to make the output of the LM317 more than 20V?

Don't ask me, ask him

 

[dhyaks]

Here is a sim of the circuit that I linked the OP to on another forum (before it got shut down because it is "automotive").


Note that it does a pretty good job of holding the current through the LEDs constant as the battery discharges (X-axis of all the plots). What it fails at is keeping the current constant as a function of temperature. See the upper plot pane: Green is -20degC, Red is 0degC, Lt.Blue is 20degC and Dk.Blue is 40degC. The 30% current variation is due to the temperature dependance of the Vbe of Q1. You could use the circuit if you design for <2.3A, because even hot it will keep the LED current below its max 3A rating.

Look at the lower plot pane: It shows the power dissipation at the four temperatures in M1 (Green trace), one of the LEDs (Red Trace), and the resistor R1 (Lt.Blue Trace) at a function of battery voltage. Note that in the worst case, M1 dissipates 11.5W, so must be on a large heat sink. Note that the LED dissipates 8W, so it too must be on a heat sink. The emitter resistor R1 is dissipating 1.8W, so should be at least a 2W power resistor...

Even if you use a switching constant-current LED driver, the LED dissipation will not change; it will still have to mounted on a heat sink or you will destroy it...

thanks Mr.Mike for suggestion, late night i just make that circuit and doing well (though i place a 12v/35W Automotive bulb).
check the voltage accros the Bulb there is 3,4V and the Current can steady in ~2A.
but is that circuit can operated well if i place 3 High Power in Series? (3,7V/3A per Led) because the voltage must be 10,2V ?
i try to change R2 (FET Bias) from 100k with 100k Potensiometer, try to tuning out and the stupid happen, i got the Potensio has blow up and fire. haha!
and FET + NPN Transistor has die. so i build up again. (My Bad).

Then why did his resistors try to make the output of the LM317 more than 20V?

no, i mean that VR / Pots just to make the Output voltage in nominal that i want. like if i just plug 1 LED so i tune it to 3,7V. if i plug 3 LED then i tune it to 11,2V.

 

[audioguru]

Sorry, your schematic shows 330 ohm and a 5k ohm fixed resistors, not a 5k pot.

Cree does not say your LEDs have a forward voltage of 3.7V. They say the typical forward voltage is 3.35V at 3A but might be less forward voltage.
Their current will be too high at 3.7V because it is off the end of their graph and is typically much higher than their maximum allowed current of 3A.
That is why LEDs need to be fed current, not voltage.
How will you cool your LEDs?

 

[dhyaks]

Sorry, your schematic shows 330 ohm and a 5k ohm fixed resistors, not a 5k pot.

Cree does not say your LEDs have a forward voltage of 3.7V. They say the typical forward voltage is 3.35V at 3A but might be less forward voltage.
Their current will be too high at 3.7V because it is off the end of their graph and is typically much higher than their maximum allowed current of 3A.
That is why LEDs need to be fed current, not voltage.
How will you cool your LEDs?

Up's sorry for my drawing, thanks for correction.
the LED is planted in a medium (with fan) heatsink. so i think the heat dissipation of LED can be handle by heatsink.
and yes, i will try to maximize the LED specs because i don't turn it on for a long time. maybe just 1-5 Minutes only.
if it's can work well then i want to make one for driving LED in motorcycle to fast and short time "beam" for signing the road, like grasstrack in the jungle

For All :
I found this site : PWM Constant Current Circuit

in the Site, said : "And now the fault-tolerant approach - dissipation through Q1 with a shorted load. Again, simply calculate the maximum supply power that Q1 will see. If the supply is unregulated and capable of large current flows in short time spans (e.g., batteries), again a fuse would be wise as a last-ditch protective measure against catastrophic failures by blowing the circuit open so it cannot try to short the supply. If the set current is 750mA, a 1A-1.5A fuse in series with this circuit would be a wise, and relatively inexpensive, addition - all the more so for portable projects using lithium batteries, given how these like to catch fire (and for nonrechargeable lithium primaries, explode) when shorted."

So can i change Mr.Mike circuit (R1) to Fuse rated 3A ? and what the formula to search the rating for R2 ?? still confuse

 

[MikeMl]

thanks Mr.Mike for suggestion, late night i just make that circuit and doing well (though i place a 12v/35W Automotive bulb).
check the voltage accros the Bulb there is 3,4V and the Current can steady in ~2A.
but is that circuit can operated well if i place 3 High Power in Series? (3,7V/3A per Led) because the voltage must be 10,2V ?
i try to change R2 (FET Bias) from 100k with 100k Potensiometer, try to tuning out and the stupid happen, i got the Potensio has blow up and fire. haha!
and FET + NPN Transistor has die. so i build up again. (My Bad).


no, i mean that VR / Pots just to make the Output voltage in nominal that i want. like if i just plug 1 LED so i tune it to 3,7V. if i plug 3 LED then i tune it to 11,2V.

To change the current through the LEDs, you do not change R2 (47K). Leave it alone... To change the current, you must change R1, the 0.33Ω resistor. The approximate LED current in the simulation is 0.7V/R1 = 0.7/0.33 = 2.12A.

Note that I deliberately set the current to ~2.2A; not 3A. This circuit is not precise enough to set the current to 3A, because if the transistor is different, or the resistor is out of spec, you will damage the LEDs.

I tried the sim with three LEDs instead of 2, and there is just barely enough compliance in the current source for the current (2.2A) to be maintained while the battery voltage is 11.2V or higher... That will reduce the heating in M1, but not in the LEDs...

Whenever you use a pot as a rheostat in place of a fixed resistor, you must put a min. value fixed resistor in series with the pot. This is to prevent a fire as you turn the pot down to zero Ω. The min value resistor will limit the current to a safe value. Without it, the current can high enough to blow up the pot, and the circuitry it is attached to, as you found out...

 

[dhyaks]

To change the current through the LEDs, you do not change R2 (47K). Leave it alone... To change the current, you must change R1, the 0.33Ω resistor. The approximate LED current in the simulation is 0.7V/R1 = 0.7/0.33 = 2.12A.

Note that I deliberately set the current to ~2.2A; not 3A. This circuit is not precise enough to set the current to 3A, because if the transistor is different, or the resistor is out of spec, you will damage the LEDs.

I tried the sim with three LEDs instead of 2, and there is just barely enough compliance in the current source for the current (2.2A) to be maintained while the battery voltage is 11.2V or higher... That will reduce the heating in M1, but not in the LEDs...

Whenever you use a pot as a rheostat in place of a fixed resistor, you must put a min. value fixed resistor in series with the pot. This is to prevent a fire as you turn the pot down to zero Ω. The min value resistor will limit the current to a safe value. Without it, the current can high enough to blow up the pot, and the circuitry it is attached to, as you found out...

Ah.. yes i understand to change current limit is recalculate the R1 Rating, but how you can say the R2 Rate is 47K ? is that any formula to calculate that? because when i try to turn the Pot and check the voltage across the LED, A Voltage is Increase along with i turning the Pot. from 3,4V to 6V before the pot begin fire. and how about using Fuse instead of Ceramic cement resistor?
Thank Mr.Mike

 

[audioguru]

You MUST NOT use a fuse to replace R1. The resistance of a fuse is almost a dead short and is not accurate.
The accurate resistance value of R1 together with the "thermometer" that is transistor Q2 sets the maximum current.

The value of R2 must not be too low or its current will be higher than the maximum allowed current in transistor Q2. The gate of a Mosfet draws no current so 47k is fine.
If R2 is 47k then with a 12.6V battery it has a voltage of 12.6V - 0.6V= 12.0V across it. Then Ohm's law says its maximum current is 12.0V/47k ohms= 0.25mA. If R2 is 3k ohms then its maximum current is about 4mA.

 

[MikeMl]

Ah.. yes i understand to change current limit is recalculate the R1 Rating, but how you can say the R2 Rate is 47K ? is that any formula to calculate that? because when i try to turn the Pot and check the voltage across the LED, A Voltage is Increase along with i turning the Pot. from 3,4V to 6V before the pot begin fire. and how about using Fuse instead of Ceramic cement resistor?
Thank Mr.Mike

R2 is just a pull up resistor that biases the gate of the NFET when the 2N3904 turns off. It has nothing to do with setting the current through the LEDs. This is the third time you have been told that. Do you not understand?

As Audioguru told you, you cannot substitute a fuse for a 0.33Ω resistor. The resistance of this resistor is the most critical parameter of this entire circuit. Change it, and the results will be predictable...

You came here to ask for help, but you obviously ignore what we are telling you.

 

[dhyaks]

R2 is just a pull up resistor that biases the gate of the NFET when the 2N3904 turns off. It has nothing to do with setting the current through the LEDs. This is the third time you have been told that. Do you not understand?

As Audioguru told you, you cannot substitute a fuse for a 0.33Ω resistor. The resistance of this resistor is the most critical parameter of this entire circuit. Change it, and the results will be predictable...

You came here to ask for help, but you obviously ignore what we are telling you.

soory i'm not to ignored your answer but just to ensure and understanding what your suggestion, thank you very much and i will try it. thank a lot..

 

[audioguru]

Have you looked at the datasheets for the transistor and the Mosfet to see which pin is which?

 

[ronsimpson]

**broken link removed**

Hopefully this link will work.
Power supply:
Vin 7-35 volts
Vout 1.25 to 30 (as long as vout is less than vin )
Current limit 0-3A. It states you need to add a heatsink at 3A level.

Set the voltage too high by a little. Set the current limit to 1A or something reasonable to start out at.

This switching power supply should generate much less heat than the linear regulator.

Really big version:
**broken link removed**