nickagian Posted September 11, 2006 Report Share Posted September 11, 2006 Hi everybody!I want to design a project with some LEDs, something like a signboard and I need to use a current sink, which can drive up to approximately 500-600mA. I have thought of using a NPN transistor (for example 2N2222), but I do not know what the exact circuit should look like. I have simply connected the E to the negative supply (ground), the C to the negative lead of the LED (of course the + of the LED is connected through a resistor to the positive supply) and the B to ground, but it doesn`t work at all!Can anybody help me? Perhaps is there another solution, apart from using a transistor? (an IC maybe?) Quote Link to comment Share on other sites More sharing options...
ante Posted September 11, 2006 Report Share Posted September 11, 2006 Hi Nick,Why not use the LM317 or LM350? Quote Link to comment Share on other sites More sharing options...
nickagian Posted September 11, 2006 Author Report Share Posted September 11, 2006 Hi Nick,Why not use the LM317 or LM350?Well... I haven`t understand what you mean Ante... :-[ I`m sorry...How can the LM317 be used as a current sink?Actually I mean something like the following circuit (except that instead of 1 I want to use more LEDs), where the ON or OFF state of the LED can be changed by the voltage applied at the base of the transistor.But I obviously make a mistake because it simply doesn`t work at all! Quote Link to comment Share on other sites More sharing options...
ante Posted September 11, 2006 Report Share Posted September 11, 2006 OK Nick, I did misunderstand your question! :-[ I thought you needed regulated (limited) current for the LED’s! I can see now you are using a resistor for this. So this circuit does not work? What is trigging the base of the transistor? Quote Link to comment Share on other sites More sharing options...
nickagian Posted September 11, 2006 Author Report Share Posted September 11, 2006 OK Nick, I did misunderstand your question!It`s ok! Perhaps I didn`t express myself very clearly and I should have posted that picture too!Well I simply triggered it by connecting it to the ground or to the positive supply (later at my real circuit this will be connected to the output port of a PIC, but I guess it is the same thing!)I don`t know, I may have made a mistake... I will check it once more! Quote Link to comment Share on other sites More sharing options...
nickagian Posted September 11, 2006 Author Report Share Posted September 11, 2006 Thanks a lot Zeppelin! That resistor must be my mistake (I had thought of it, but couldn`t find the proper value). The guide is very helpful on that too!With a 5V supply you can't drive that many LEDs, tops 3, I guess-you need a higher supply voltage.I know that, I intend to use only two LEDs on series and organize all of them on doubles! Quote Link to comment Share on other sites More sharing options...
ante Posted September 11, 2006 Report Share Posted September 11, 2006 I believe it would work more reliable with only one LED in series with each transistor since the voltage is low. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 11, 2006 Report Share Posted September 11, 2006 Yeah, with only 5V for the supply, two 3.5V blue or white LEDs in series wouldn't be very bright. ;DMy red LEDs are only 1.8V and they would be fine with two in series. Quote Link to comment Share on other sites More sharing options...
nickagian Posted September 11, 2006 Author Report Share Posted September 11, 2006 Well I`ll use red LEDs, which -if I am not wrong- are 2.3V and I have tested two of them in series with a resistor of 47R. The current drawn was approximately 18mA and the brightness quite good! But I don`t doubt that alone it must have been much brighter! Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 11, 2006 Report Share Posted September 11, 2006 If you connect a single 2.3V LED in series with a 47 ohm resistor and across a 5V supply then the current will be 57.4mA and the LED (rated max is 30mA) will burn out. It will be very bright for a few milli-seconds. The resistor's value must be recalculated for just one. Quote Link to comment Share on other sites More sharing options...
nickagian Posted September 11, 2006 Author Report Share Posted September 11, 2006 Yes of course! ;D I didn`t mean that I have the intention to use a single LED with a 47 ohm resistor!! Thanks a lot anyway! Quote Link to comment Share on other sites More sharing options...
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