High power led dimming

Discussion in 'General Electronics Chat' started by ktnch, Jan 19, 2011.

  1. ktnch

    Thread Starter New Member

    Jan 19, 2011
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    Hi newbie here be gentle, Got a 32-36v psu with 1400ma regulated current also bought a 50w 32-36v led 1700ma max. Now these are sold as a stand alone working pair, what it the simplest way to get the led dimming, I am about to buy an Arduino as its a project for a reef lighting system. We need to get a rtc and a touch screen controller for the arduino as well.

    Could i use the pwm feed from the arduino to feed a tc1427 then 2 mosfets IRL13705N Slight overkill as the mosfets are rated at 52amp 55v but comes on an arduino board

    The build will be 6 x 50w leds ran of one dimming channel then 10 royal blues ran of another pwm channel and finally violet leds running from a third pwm channel

    At first i was going to have a common psu supplying a 30amp 36v pwm speed control but the constant current drivers for the leds would have been far to much so the cheapest way is one psu per led

    Thank you

    Kev
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Sure, you can use the PWM output. If the base PWM frequency is over 200 Hz, you shouldn't see any flicker.
     
  3. ktnch

    Thread Starter New Member

    Jan 19, 2011
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    Thank you for that, I had started a thread on the marine fish forum i am on and got told it could not be done, I kept trying to tell them that the mosfets are not there to change the current or voltage but simply as a switch the psu i have for the led is regulated under the max of the led, I got told something about feedback loops on the pwm circuit any ideas ?

    infact this is what i got told can you shed any light on it

    Think of a light bulb. If you set the voltage across it, current will rise/fall according to the voltage. Less volts, more current. This is because the filament has an electrical resistance that increases with the filaments temperature. The filaments resistance increases with temperature, so acting to regulate current, due to power being dissipated by the filament heating up. This means if you change the current or voltage across the filament, the other will change as resistance will change according the the change you've made to voltage or current.

    Basically, most electrical devices obay the law V=IR; Voltage is equal to current timesed by resistance.

    Ok, so, basic electronics that you probably already knew out of the way, we'll now get onto the intermediate electronics that applies to LED's.

    LED's do not have a resistance when a forward voltage is applied. With a negative voltage being applied, the resistance is near infinity. What's the implication of this?

    Well, infinite resistance means no current flow, so no power is dissapated - effectively you have a open (off) circuit

    However, apply a nominal forward voltage

    V = IR
    5v = I (infinitely small value, which approximates to)0

    Rearage to resolve I

    I = V/R
    I = 5/0

    Anything divided by zero equals infinity. Therefore, for any nominal forward voltage, an LED will try to draw an infinitely large amount of current.

    Most PSU's and MOSFETS do not regulate current, only voltage. So your applying a nominal forward voltage to the LED and allowing it to set it's own current. This is theoretically always going to be an infinite current.

    P = IV; Power equals current times by voltage

    so, using the above current draw example

    P = infinity times 5
    P = infinity.

    OK, that's a sterile theoretical model. Lets make things a bit more complex and hence closer to reality. You always have some inefficiency in a circuit. Let say 2% for arguments sake. Inefficiency is usually caused by heat as resistance from components. LED's don't have perfectly zero resistance, but they don;t have a resistance that's easy to detect. Likewise wire, the MOSFET and the PSU. Hence, all these will create heat with power going though them. With infinite heat, the components will reach infinitely high temperatures.

    Basically, with no current regulation, an LED will draw as much current as it can get. This will lead to a component failure by overheating, as currents far higher than any of your components are rated for will flow.

    The electrical inefficiency of the circuit will allow no regulation of current at very low voltages, as this component electrical resistance acts as a small current regulator. Per-say, at low voltages of about 0.25V and below you can get away without current regulation. However, if you want to be able to see the light your LED's are emitting, you will have to regulate current.

    All the devices you've linked to so far regulate voltage only, allowing whatever they are connected to to set the current draw. This is where your problem lies, and why your barking up the wrong tree. [​IMG] When working with high power LED's you need to focus on devices that will supply a regulated current.

    Given a regulated current, LED's will set a nominal voltage, according to their power output. So, given a 5W LED at 1,000mA;

    P = IV
    5 = 1V

    Resolve V

    V = 5/1
    V = 5

    Similarly, you can work out voltage draw for a given lower current knowing what current is and working it out as a proportion of power. Take the above 5W LED and drive it at 500mA;

    P is proportional to I

    I is half the rated, so P will be half the rated

    2.5 = 2.5V
    V = 2.5/2.5
    V = 1V

    Basically, LED's obay the P=IV law, but not the V=IR law, as they have no resistance for practical or theoretical applications near full load [​IMG]

    Hope this clarified somewhat for you.
     
  4. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    That's almost completely nonsense, with a couple of half grains of truth. If the supply you have is designed for the LEDs then you shouldn't have to worry about anything apart from good heatsinking and airflow around the LEDs.
    There's plenty of good info out there if you're interested, a good source is datasheets for LED driver chips.
     
  5. tom66

    Senior Member

    May 9, 2009
    2,613
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    There is one concern I could have though. If you turn the load off while it is regulating current through a MOSFET, it may try to output the highest voltage and when the fet turns back on the LED would be hit with a spike of current. Maybe the supply will automatically limit voltage, or maybe the spike won't prove an issue for the LED.
     
  6. Kirtho

    New Member

    Jan 18, 2011
    6
    0
    LEDs have a breakover voltage that varies depending on type. Once that breakover voltage is reached, they will draw as much current as you supply. You have to have a resistor in series with the LED to limit the current.

    I just hooked up a high brightness white LED to a 9 volt battery with a 150 volt resistor to limit the current, using alligator clips. This is the easy way to find out a few things. Here are my measurements:

    Battery: 8.05 volts (excluding voltage across multimeter used for measuring current)
    LED: 2.78 volts
    Resistor: 5.27 volts
    Current measured: 35.4 milliamps

    The LED acts like a zener diode and maintains a fairly constant voltage. The current is equal to the voltage across the resistor divided by its resistance in ohms. 5.27 volts divided by 150 ohms equals 35.1, so you know the resistor I used is pretty close to its nominal value.

    Whatever you use for a supply, pulse-width modulated or not, you have to have some kind of current limiting element. The wattage of that resistor has to be chosen to withstand the current that would be flowing through it if you simply wired it to the power supply. In my projects I actually feel the resistor. It's doing pretty good when a quarter watt of heat doesn't seem to make it particularly warm, I'm using a half-watt resistor.

    This is also inefficient. Almost twice as much power is being wasted by the resistor as is being used by the LED.

    Ktnch has a driver board that is made for the LED. Now he wants to use the Arduino board to control the LED instead. When you do that, Ktnch, you are going to need a current-limiting resistor. It must be chosen to limit the LED's current to its maximum, and you probably won't be able to see the difference if you feed 1200 ma or so to an LED that can pull 1700 ma. You have to be aware of heat and when electronics engineers measure heat they measure it in watts. 1700 ma times 3.2 volts equals 5.4 watts of heat.
     
  7. ktnch

    Thread Starter New Member

    Jan 19, 2011
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    " Ktnch has a driver board that is made for the LED. Now he wants to use the Arduino board to control the LED instead. When you do that, Ktnch, you are going to need a current-limiting resistor. It must be chosen to limit the LED's current to its maximum, and you probably won't be able to see the difference if you feed 1200 ma or so to an LED that can pull 1700 ma. You have to be aware of heat and when electronics engineers measure heat they measure it in watts. 1700 ma times 3.2 volts equals 5.4 watts of heat" heat they measure it in watts. 1700 ma times 3.2 volts equals 5.4 watts of heat.

    Where did the 3.2 volts come from ?

    No i have no led driver board i have a psu that is designed for the led having the current regulated at 300ma under the leds max ma. I am using the Arduino to control the mosfet by pwm not to regulate current or voltage but to turn the led on and off like a simple light switch so the mosfet circuit will not require any regulation
     
    Last edited: Jan 20, 2011
  8. ktnch

    Thread Starter New Member

    Jan 19, 2011
    22
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    Spike of current, the current or flow can not be over 1400ma which is the max the regulated psu will provide can it ?

     
  9. tom66

    Senior Member

    May 9, 2009
    2,613
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    In theory. However, the supply will have a small capacitor on the output. When the MOSFET goes open circuit, no current will flow. The feedback action will cause the supply to hit the maximum output as it tries to increase the current. When the LED is reconnected, the voltage does not drop instantly (as the capacitor stores a voltage - technically a charge, but think of it as a voltage for now), and the spike of current flows through the LED. Potentially, the extra heat caused by this could damage the LED, especially at higher frequencies. In some cases, the load dump could be 10A+. One way to protect against this would be a series limiting resistor (say limiting current to twice the maximum), but this would be troublesome for high power LEDs as it may waste a lot of heat.
     
    Last edited: Jan 20, 2011
  10. ktnch

    Thread Starter New Member

    Jan 19, 2011
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    It will be interesting to find out, ordered the 50w led today and a 90 deg lens with the led regulated psu
     
  11. Kirtho

    New Member

    Jan 18, 2011
    6
    0
    I made a mistake. You said 32 volts and I automatically thought 3.2 volts.

    By "driver board" I meant the PSU.

     
  12. Kirtho

    New Member

    Jan 18, 2011
    6
    0
    I think they use a charge pump arrangement that uses feedback from the current-limiting resistor. I've read specs before that show this. This way a 0.1 ohm resistor can do that I squared R thing and dissipate less than half a watt for 1400 mA. With the capacitor across the LED the LED will never receive more than the amount of current that is fed to both. I would do it that way just because of the amount of radio frequency interference that would be generated by running a long PWM line. This is more complex but it is well worth learning how to do. This way a filtered current can be delivered to the LED. What you have when you are done is a digitally controlled constant current source.

    This might be worth it for Ktnch to learn how to do.

     
  13. ktnch

    Thread Starter New Member

    Jan 19, 2011
    22
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    Ah had me wondering lol

    So led max flow 1700ma and 36v so thats 61.2 watts of heating is that correct ?

    If we have problems with the pwm and current surging i can simply add a constant led driver, Found one for under £10 that the voltage and current is adjustable by on board pots

    Ah was going to say this one but is max 30v

    The company who i am buying the led and driver off is making a led pwm driver dimmer rated for 20-100watts this will be adjustable inside the unit might have to hang fire for them to sort this
     
    Last edited: Jan 20, 2011
  14. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    That's not a charge pump, that's a boost converter, or a buck converter.
     
  15. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    Yes but in practice the response time will not be zero and this will mean on each switching off to on transition the LED will be hit with a surge of current. Think about when the terminals are disconnected: the controller will compensate by increasing voltage, but the voltage does not change instantaneously.
     
  16. ktnch

    Thread Starter New Member

    Jan 19, 2011
    22
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    Its a led dc-dc driver isnt it it converts a non regulated supply for a led to regulated voltage and current

    The specs for the above item although i know it will not power my 36v led because of the voltage and psu. If i could find one like this with a higher voltage and with pwm it would be spot on


    LED Power Driver Constant Voltage / Current Adjustable CC CV
    Specification:

    * Module Properties: Non-isolated step-down constant current, constant voltage module (CC CV) charging module.
    * Input Voltage: DC 5~30 V
    * Output Voltage: DC 1.25~26 V ( adjustable, O/p Voltage < I/p Voltage by 3V )
    * Output Voltage < Input Voltage: 3V
    * Output Current: rated 2A, the largest 2.6A
    * Constant Current Range: 0.2~2A (adjustable)
    * Efficiency: 86%
    * Output Ripple: 0.05V
    * Potentiometer adjustment direction: clockwise (increase), counterclockwise (decrease)
    * No-load Current: Typical 10mA
    * Load Regulation: ± 1%
    * Voltage Regulation rate: ± 0.5%
    * Dynamic response speed: 5% 200uS
    * Output Short-circuit Protection : Yes, Constant Current
    * Input Reverse Protection: None


    similar to this item
     
    Last edited: Jan 20, 2011
  17. Kirtho

    New Member

    Jan 18, 2011
    6
    0
    I don't know how efficient the LED is. If more energy leaves the die as radiation, less energy heats the die. If it's fifty percent energy efficient, you have only 32 watts to contend with, and so on.

    If you want to be able to dim the lights from the outside, you will either have to use a different controller or remove the potentiometer that sets the constant current and run wires to one that you can operate by hand. Doing six in tandem gets complicated. I can design something like this from the ground up. Do you really want software control or just a way to make it work?

     
  18. ktnch

    Thread Starter New Member

    Jan 19, 2011
    22
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    Going to try the regulated psu and led with the arduino using pwm to control the tc1427 and them controlling the IRL137O5N if it doesnt work then will look at other alternatives i could always go back to the original plan and use a 48v psu with around 14 amps and use dimming buckpucs to control the leds, Just a lot more expensive that way
     
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