audioguru Posted March 8, 2006 Report Share Posted March 8, 2006 You shouldn't connect LEDs in parallel. Their forward voltages will be slightly different causing the one with the lowest voltage to hog all the current and maybe burn out. Each LED should have its own current-limiting resistor.In my 3V Ultra-bright LED Chaser project I use a Schmitt trigger oscillator from a 74HC14 to drive a transistor to blink its LEDs. You could make a small Schmitt trigger oscillator from a couple of transistors. Quote Link to comment Share on other sites More sharing options...
steev Posted March 9, 2006 Author Report Share Posted March 9, 2006 OOps i forgot.we r using 5 100Ohms resistors for each(5) LED seperately.to provide exact amount of current.but the problem is with the output supply of receiver circuit.its only working when we feed 3.6volts exactly.and when we connect led to this voltage they r burning out.(i fried up 10 led till now) :'(i know using lm317 we can get exactly 3volts(thanks to audio guru and ante).but problme is with heat.in other thread someone said that we can drop voltage by 0.6v using a diode.can i keep voltage exactly 3v from this diode method.or is there any seperate ckt for this (i need circuit as small as possible with no heat generation.) Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 9, 2006 Report Share Posted March 9, 2006 Please attach a schematic of your circuit. I don't understand how your LEDs can burn-out.A red LED has a forward voltage drop of about 1.8V. Most LEDs have a max current rating of 30mA.If you have a 100 ohm current-limiting resistor in series with it then 4.5V across the combination results in a current of 27mA. With only 3.6V across the combination of the 1.8V red LED and its series 100 ohm current-limiting resistor then the current is only 18mA. With 3V across the combination then the current is only 12mA.Anything that has continuous current in it and a voltage across it generates heat. Reduce the heat by reducing the current, extra voltage or both. Your choice of a 12V battery to power the low-voltage circuits wastes a lot of power. Quote Link to comment Share on other sites More sharing options...
steev Posted March 10, 2006 Author Report Share Posted March 10, 2006 ohh right.then i miscalculated the resistor value.problem here is i dono the forward voltage of the led iam using.i got them from a local electronics shop(shop guy said that he dono LED details).they r bright enough when i connect 2 AA cells in series.so can i consider its forward voltage as 1.8v.or there any specific method to measure the LED forward voltage Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 10, 2006 Report Share Posted March 10, 2006 ohh right.then i miscalculated the resistor value.problem here is i dono the forward voltage of the led iam using.they r bright enough when i connect 2 AA cells in series.so can i consider its forward voltage as 1.8v.or there any specific method to measure the LED forward voltageIf your LED has a forward voltage drop of 1.8V and you connect it to two AA cells in series (3V) without a current-limiting resistor then the extra 1.2V would create a current that is so high that the LED would instantly burn-out!You don't say the color of the LED that determines its voltage. My red LEDs are 1.8V, my blue LEDs are 3.1V and my white LEDs are 3.5V.You don't show a schematic nor say the voltage that is burning your LEDs when they each have a 100 ohm current-limiting resistor. Therefore maybe there is enough voltage to connect some LEDs in series. 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.