Constant Current for LED

[kd7eir]

I am a bit confused.  I understand (I think) that LED's are current driven devices, and that the forward voltage across them is merely a function of their construction, not a voltage that is required to drive them.

I need to drive several 3 watt LED's that need 700ma, and have a forward voltage of 3-4V.  Given this, I designed a constant current power supply that is providing 500ma, and is fed with a 5V input to test one of these LED's.

My confusion - Do I need to feed the power supply with 3-4V x the number of LED's, or can I provide any input voltage as long as I limit the current to 700ma or less?

My goal is to have as little heat developed by the power supply as possible.

Thank you

[IanB]

My confusion - Do I need to feed the power supply with 3-4V x the number of LED's, or can I provide any input voltage as long as I limit the current to 700ma or less?

The LED will "need" the 3-4 V foward voltage to drive the current through it. You can easily observe this if you measure the voltage between the LED terminals when the LED is running.

Your LED driver must be able to provide at least as much voltage as the LED string needs. Any supply voltage higher than what the LEDs need must be dropped or absorbed by the driver.

How efficient this is depends on the design of the driver. A linear driver (rare) will absorb the extra voltage as heat. A switching driver (more common) will only dissipate as much heat as its operating efficiency determines.

[kd7eir]

Thank you IanB, that clears up most of my confusion.

If I have 10 of these LED's in a string, the driver needs to provide 30V to the string, which will be dropped as ~ 3V per LED.  Is that correct?

[MikeK]

Interestingly enough a recent issue of Elektor had a article on constant current sources for a string of LEDs.  In fact, I posted a question here about one of the quiz questions.  "Electronics for Starters", April 2012 issue.

Yes, you need the voltage per LED x the number of LEDs.

[Rerouter]

obviosly the closer you feed the thing to dropout voltage the less power will be dissipated by your parts, however you should post a schemaitc or even a crude drawing of your design, so we can have a look at other factors that might not be common to all approaches,

here is one i am familiar with Falstads Constant Current

for a single LED this will start driving from 5V to the thermal limit of the transistors, for a 4.1 volt led, 9V for 2, and so on,

[Rerouter]

though being how its only 2 transistors and 2 resistors, i myself would probably be game to just build separate constant current drives per LED,
as it only the smallest amount more wattage on the parallel configuration (

6.1W for 2 in series being powered off 8.9V (660mA)
6.1W for 2 separate drivers in parallel off 4.8V (630mA)
that is with them both using the 220 ohm bias, by increasing it, and lowering your drive a little you can get it running slightly more efficiently, but it is a trade off for brightness,
equally increasing the gain of your upper transistor and lowering the gain of your lower can help get it a little brighter per mW input, but its a tradeoff to just buying bulk transistors,

[kd7eir]

I have attached a schematic of the circuit that I am using.

[amyk]

The important thing to note is that LEDs are diodes so they have the same exponential IV curve like any other diode. They are designed to be operated at a specific current, and due to this IV curve a small change in voltage / temperature can cause a large change in current, which is why CC supply is needed.

[MikeK]

If 700mA is passing through those LEDs, then 700mA is also passing through that 1 ohm resistor, which means it dissipates 0.7 watts.  You'll need at least a 1W resistor

[BravoV]

As you said to minimize heat, this simple transistor base won't cut it cause the high current sense resistor requirement, I think you should start using op-amp to have much smaller sense resistor.

[kd7eir]

The 1 ohm resistor is now a 5 watt cement resistor, I realized that mistake the first time I used this.  I will look into using an op-amp for the circuit, thank you.  Back to the books...