white led

[audioguru2]

The datasheets for Limileds say to drive them with a current regulator.
If their total voltage is 7V or less (you must measure it because it could be 3V or up to 4V each) then a 7808 (measure it too) with a current-limiting resistor will feed them 335ma.

A 78xx as a current regulator gets very hot because it needs its own voltage plus voltage to make the current plus the output voltage required. An LM317 current regulator is cooler and needs less voltage.

 

[pier1]

Hi AG
        How about this circuit, and the current regulator you suggested me . If correct what could be the value of r3 ?

View attachment 40426

 

[audioguru2]

Hi Pier,
You have an LM317 voltage regulator and a two-transistors current regulator. You don't need the voltage regulator. The LM317 can be a current regulator by adding only a single resistor (1.25V/320mA= 3.9 ohms) to it as shown in its datasheet.

Two Lumileds might need a max voltage of about 8V, the LM317's current regulating resistor needs 1.25V and the LM317 needs a minimum of 2.5V for a total max voltage needed of 11.75V. Your 12V transformer has a peak voltage of 17V which is dropped to 15.4V by the rectifier bridge so there is plenty voltage.
The LM317 will dissipate about 4W max so it will need a heatsink.

 

[pier1]

Hi AG
        I could find an lm117 as a current regulator circuit in the datasheet (application notes) 317 is used as a voltage regulator and a reference with voltage regulator . Is this the same way i should do to the lm317 (picture) ?

View attachment 40427

 

[audioguru2]

Yes, the LM317 will be a 320mA current regulator when the resistor is 3.9 ohms.

 

[pier1]

Hi AG
        My multimeter measures 17.3 volts across the 100/25 capacitor. What will be the o/p of the LM317 ? Hows the circuit now ?

View attachment 40428

 

[audioguru2]

Hi Pier,
The output voltage of the LM317 current regulator is 1.25V higher than whatever is the total voltage for the two LEDs.

Your 390 ohms resistor reduces the LED current to only 1.25V/390= 3.2mA which is way too low. 3.9 ohms would regulate the current at 320mA to 330mA.
The 100uF capacitor should be 1000uF for 320mA without too much ripple.

Your 1uF capacitor is shorted and does nothing. A current regulator doesn't need an output capacitor.

 

[pier1]

Hi AG,

Your 390 ohms resistor reduces the LED current to only 1.25V/390= 3.2mA which is way too low. 3.9 ohms would regulate the current at 320mA to 330mA.

Should i take tho o/p at pin 2.

 

[audioguru2]

Should i take tho o/p at pin 2?

Then it will be a 1.25V voltage regulator, not a current regulator.

The datasheet shows how to make a current regulator and you had it almost correct before:View attachment 40429

 

[pier1]

Hi AG,
        So the value of the resistor is 1.2ohms ?

 

[audioguru2]

If you use 1.2 ohms then the current will be about 1A and your expensive LEDs will blow up!

Check the max current for your LEDs (350mA?) then calculate a suitable current-setting resistor. 3.9 ohms will give about 320mA.

 

[pier1]

Hi AG
        Thanks a lot for that . What will happen if the input voltage rise beyond 230 volts, as i measure 17.3 volts at 1000uF capacitor . And what is the calculation for finding out the o/p voltage of the lm317 AG ?

 

[audioguru2]

What will happen if the input voltage rise beyond 230 volts, as i measure 17.3 volts at 1000uF capacitor?

Then the LM317 gets warmer. The current is regulated so it remains the same.

What is the calculation for finding out the o/p voltage of the lm317?

We don't know what is the actual voltage of the LEDs because the voltage is a range of voltages and each LED is different. The LM317 needs about 2.5V to operate plus the current setting resistor needs 1.25V, plus the voltage of the LEDs.

 

[pier1]

Hi AG
        Which is the ideal resistor to be used in series with the LED to find out its exact forward voltage ? I have a 6volts battery as a source .

 

[audioguru2]

Hi Pier,
I don't know which LEDs you have. If they are the Lumileds Luxeon I then the max continuous current is 350mA if they have a good heatsink and their forward voltage is from 2.31V to 3.99V.

Use a 12 ohms resistor to limit the current while you measure their actual forward voltage.
Without a heatsink they will get very hot very quickly so keep the measurement time short.

View attachment 40430

View attachment 40431

 

[pier1]

Hello AG
            The o/p of the lm317 is 15.45 volts(open voltage). Will it harm the led's .

 

[audioguru2]

The LM317 with one resistor is a regulated current source, not a regulated voltage source. Its output voltage will adjust to whatever voltage is needed to produce its programmed current in the load.

 

[pier1]

Hello AG,
              Can I put a 25 ohms resistor in instead of the led's  to see the voltage across it ? If so should it be a 2watt ?

 

[audioguru2]

The LM317 with a 3.9 ohm current-regulating resistor will have a current of about 324mA. 324mA across 25 ohms is a voltage of 8.1V. The power in the 25 ohms resistor is 324mA squared x 25= 2.6W.

The LM317 needs to have an additional 1.25V across its 3.9 ohms resistor and an additional 2.5V at its input. Therefore the minimum input voltage to the LM317 is 8.1V + 1.25V + 2.5V= 11.85V.

If the input voltage of the LM317 is 17.3V and its output voltage is 8.1V then it will dissipate (17.3V - 8.1V) x 324mA= 3W and it will need a heatsink.

 

[pier1]

Hello AG
            One more doubt before connecting the load, should the 3.9 Ohms resistor also be 2 watt ?