pier Posted April 13, 2007 Author Report Share Posted April 13, 2007 Hi AG Since i am weak in electronics i am getting confused . In the ckt the 3.9 Ohms is connected from o/p to adjust terminal so when the LED's are drawing 350mA current, then is the current flowing through the 3.9 Ohms resistor ? Or from where is it drawing current ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 13, 2007 Report Share Posted April 13, 2007 Yes, all the LED current flows through the 3.9 ohm resistor. When the current causes a voltage of 1.25V then the LM317 reduces the current by reducing the voltage to the load. Quote Link to comment Share on other sites More sharing options...
pier Posted April 20, 2007 Author Report Share Posted April 20, 2007 Hello AG Could these heat sink be sufficient for the regulator ? I first used PI49 and found the regulator over heating . Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 20, 2007 Report Share Posted April 20, 2007 Heatsinks are not selected for their looks. They are selected by calculating how much cooling you need, then selecting one with the correct spec's. The manufacturer will say how many degrees C the semiconductor's case temperature will rise per Watt of dissipation when it is bolted to the heatsink. It is assumed that no insulator is used and that thermal compound grease is used between the semiconductor and the heatsink. Quote Link to comment Share on other sites More sharing options...
pier Posted April 23, 2007 Author Report Share Posted April 23, 2007 Hi AG What should be the resistor value instead of 3.9 Ohms if i put only one LED ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 23, 2007 Report Share Posted April 23, 2007 What should be the resistor value instead of 3.9 Ohms if i put only one LED ?The 3.9 ohm resistor sets the regulated current to 325ma if there is only one of if there are many LEDs in series. With only one LED then a higher voltage will be across the LM317 causing it to heat more. Quote Link to comment Share on other sites More sharing options...
pier Posted April 23, 2007 Author Report Share Posted April 23, 2007 Hi AG So I would have to use a 6 volts stepdown 500mA transformer,bridge and a filter which would give an o/p voltage of 8.64 volts dc across the filter capacitor to reduce the voltage to lm317 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 23, 2007 Report Share Posted April 23, 2007 So I would have to use a 6 volts stepdown 500mA transformer,bridge and a filter which would give an o/p voltage of 8.64 volts dc across the filter capacitor to reduce the voltage to lm317 ?Almost. The rectifier bridge reduces the voltage to 7.1V. Then the max output voltage of the LM317 will be about 4.6V and the 3.9 ohm resistor reduces the max voltage for the LED to only 3.35V.Doesn't the LED need 3.9V? Quote Link to comment Share on other sites More sharing options...
pier Posted April 23, 2007 Author Report Share Posted April 23, 2007 Hello AG The LED forward voltage i measured is 3.48volts . So it wont be sufficient na ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 23, 2007 Report Share Posted April 23, 2007 There is not enough voltage if the LM317 is a weak one.If your mains voltage drops then the LED might dim more. Quote Link to comment Share on other sites More sharing options...
pier Posted April 24, 2007 Author Report Share Posted April 24, 2007 Hi AG Then I will have to use a 9 volts 500mA step down transformer. Will that give enough voltage for lm317 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 24, 2007 Report Share Posted April 24, 2007 A 9V transformer will provide plenty of DC voltage for the LM317 current regulator to drive a single 350mA LED. The LM317 will overheat without a small heatsink. Quote Link to comment Share on other sites More sharing options...
pier Posted April 24, 2007 Author Report Share Posted April 24, 2007 Hi AG I connected a 11 Ohms resistor as the load . I noticed that the o/p voltage of the lm317 regulator was 3.6volts dc when the input voltage to the regulator was varied from 7.82volts dc to 17volts dc . The lm317 is cooler then before now . Is this enough AG ??? as the forward voltage for the led is 3.48 volts dc ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 24, 2007 Report Share Posted April 24, 2007 3.6V in 11 ohms is a current of 327ma. If the forward voltage of your LED at 327mA is 3.48V then this current regulator is fine. Quote Link to comment Share on other sites More sharing options...
pier Posted April 27, 2007 Author Report Share Posted April 27, 2007 Hi AG I've seen some lens specially made for luxion 1 watt LED's at http://ledsupply.com/l2-op-025.php . What is your suggestion on this ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 27, 2007 Report Share Posted April 27, 2007 Who makes that lens? Does it work? Will it work well with a Luxeon LED?Philips probably make a suitable lens for their Luxeon LEDs. Quote Link to comment Share on other sites More sharing options...
pier Posted April 28, 2007 Author Report Share Posted April 28, 2007 Hello AG Philips has a certified a manufacturer for this http://www.polymer-optics.co.uk/Supa-Hex%20Range%20(Luxeon).pdf . Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 28, 2007 Report Share Posted April 28, 2007 Nice lenses. Quote Link to comment Share on other sites More sharing options...
pier Posted April 30, 2007 Author Report Share Posted April 30, 2007 Hi AG Assuming the current drawn by each LED is 350mA . I have connected a 500mA transformer for the whole circuit . Now to make 5 LED's work together, Will it be good to put them in series with a transformer of more o/p voltage? Because the i/p to the regulator is maximum 40 volts i guess !! I am not sure. Will it be safe ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 30, 2007 Report Share Posted April 30, 2007 5 LEDs in series need about 17.5V and the LM317 current regulator needs a minimum of 3.75V so a voltage more than 21.25V would be fine. Use an 18VAC transformer then the unregulated voltage will be 23.8VDC. Quote Link to comment Share on other sites More sharing options...
pier Posted May 1, 2007 Author Report Share Posted May 1, 2007 Hi AG What will be the input capacitor value after the bridge rectifier ?. For one LED the value i put was 1000uF when the transformer o/p voltage was 9 volts . Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 1, 2007 Report Share Posted May 1, 2007 I have this graph that shows the amplitude of peak-to-peak ripple voltage (at double the mains frequency) but it is for my 60Hz mains and is wrong for your 50Hz mains frequency.The ripple reduces the average DC voltage by half the amount of the p-p that is on the graph.For your 350mA then a 1000uF capacitor will have about a 2Vp-p ripple which is fine. Just add 1V to the minimum input voltage of the LM317 current regulator. Quote Link to comment Share on other sites More sharing options...
pier Posted May 1, 2007 Author Report Share Posted May 1, 2007 Hi AG So i can use 1000uF capacitor only !!!? Quote Link to comment Share on other sites More sharing options...
pier Posted May 2, 2007 Author Report Share Posted May 2, 2007 Hi AG Actually I made the circuit on bread board and by mistake i connected the capacitor(1000/25)in reverse but nothing happened to it . Was it supposed to blast ? But it was almost hot by the time i came to know my mistake and switched it off . Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 2, 2007 Report Share Posted May 2, 2007 A current of a few Amps or more will blow up an electrolytic capacitor that is connected backwards. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.