audioguru Posted March 27, 2007 Report Share Posted March 27, 2007 The capacitors were still charged . Should i put and resistor across the capacitor bank so that they get discharged as soon as it is put off ? If so which would u suggest the value to be and its wattage ?5 pages ago, your circuit had 30uF for its supply filter. The max voltage is 370VDC so two 1/2W resistors in series should not arc. If their value each is 100k then they won't get too hot and they will be nearly completely discharged in (30uF x 100k) x 5= 15 seconds. Actually, the LEDs would discharge the capacitors quickly down to their voltage, then the resistors would continue the discharge. So depending on the total voltage of the LEDs, the discharge time is quicker than 15 seconds. Quote Link to comment Share on other sites More sharing options...
pier Posted March 30, 2007 Author Report Share Posted March 30, 2007 Hello AG I bought 2 new 1watt white led (lumileds), Its specifications are 3.5volts DC350 mA Since it is a costly one , As you suggested before i would like to use a isolation transformer(9 volts, 500mA), Bridge rectifier(w10m) a good filter(1000uF/16v) . For the 3.5 volts I dont know what to do . My idea was to use a 7805 with good heatsink and a voltage divider for 3.5volts DC and put the 2nos of 1watt LED's in series . Is this correct ?Or I have found a link http://www.national.com/webench/ledrefdesigns.do Should i use something from this Ag . Please Suggest me .Thanks Pier . Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 30, 2007 Report Share Posted March 30, 2007 How can you power two 3.5V LEDs in series (7V) with only 5V from a 7805?I didn't check the datasheet, but a 3.5V LED could be 3.0V, 3,5V or 4.0V. Some of them are 3.5V.Its absolute max current is probably 350mA. Do you want to operate it at its limit?You need two current regulators set to about 335mA each. Then you add their voltage plus the max LED voltage to determine the voltage that is needed.An LM317 makes a nice current regulator with just one resistor. Quote Link to comment Share on other sites More sharing options...
pier Posted March 31, 2007 Author Report Share Posted March 31, 2007 Hi AG, Sorry for the 7805, it was supposed to be 7808 . On the projects Q/A forum, the lm317 thread you have put up a circuit using 78xx series for current regulation . Shall i follow that ?One doubt : Assume when two leds(3.5v each) are connected in series the current(each led = 350mA) through both is same. So, is the total current drawn from 78xx series(to220) is the same as the flowing current(i.e, 350mA) ?... ??? Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 31, 2007 Report Share Posted March 31, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 3, 2007 Author Report Share Posted April 3, 2007 Hi AG How about this circuit, and the current regulator you suggested me . If correct what could be the value of r3 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 3, 2007 Report Share Posted April 3, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 4, 2007 Author Report Share Posted April 4, 2007 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) ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 4, 2007 Report Share Posted April 4, 2007 Yes, the LM317 will be a 320mA current regulator when the resistor is 3.9 ohms. Quote Link to comment Share on other sites More sharing options...
pier Posted April 4, 2007 Author Report Share Posted April 4, 2007 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 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 5, 2007 Report Share Posted April 5, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 5, 2007 Author Report Share Posted April 5, 2007 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. Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 5, 2007 Report Share Posted April 5, 2007 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: Quote Link to comment Share on other sites More sharing options...
pier Posted April 5, 2007 Author Report Share Posted April 5, 2007 Hi AG, So the value of the resistor is 1.2ohms ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 5, 2007 Report Share Posted April 5, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 5, 2007 Author Report Share Posted April 5, 2007 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 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 5, 2007 Report Share Posted April 5, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 6, 2007 Author Report Share Posted April 6, 2007 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 . Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 6, 2007 Report Share Posted April 6, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 9, 2007 Author Report Share Posted April 9, 2007 Hello AG The o/p of the lm317 is 15.45 volts(open voltage). Will it harm the led's . Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 9, 2007 Report Share Posted April 9, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 9, 2007 Author Report Share Posted April 9, 2007 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 ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 9, 2007 Report Share Posted April 9, 2007 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. Quote Link to comment Share on other sites More sharing options...
pier Posted April 12, 2007 Author Report Share Posted April 12, 2007 Hello AG One more doubt before connecting the load, should the 3.9 Ohms resistor also be 2 watt ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 12, 2007 Report Share Posted April 12, 2007 Hello AG One more doubt before connecting the load, should the 3.9 Ohms resistor also be 2 watt ?The 3.9 ohms resistor dissipates only (1.25V x 1.25V)/3,9= 400mW. A 1/2W resistor would be extremely hot, a 1W resistor will be fine. Quote Link to comment Share on other sites More sharing options...
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