DavidBanner Posted April 11, 2006 Report Share Posted April 11, 2006 I'd like double up on some of the leds connected to a doutx4 - i.e. put 2 leds where you would usually put 1.Will there be any issues with the extra current draw? Am I likely to fry something?? Quote Link to comment Share on other sites More sharing options...
doc Posted April 12, 2006 Report Share Posted April 12, 2006 Hi,I'm sure you mean the DOUT ....The max. current of the IC is 15-20 mA per Output. It depends slightly on the type of leds you are using. Normally nothing should happen if you'll connect two Leds on one output (They are getting a bit darker, b.t.w). If you want to make sure you won't fry anything, measure the current (between the IC Pin and the resistor). If the output is not more than 15 mA, you are safe ...greetsDoc... back in the ring ... /Baby break/ Quote Link to comment Share on other sites More sharing options...
DavidBanner Posted April 12, 2006 Author Report Share Posted April 12, 2006 yes, of course the dout - my bad - just corrected it...ta for the advice Quote Link to comment Share on other sites More sharing options...
Wilba Posted April 13, 2006 Report Share Posted April 13, 2006 Please read here: http://www.kpsec.freeuk.com/components/led.htmand scroll down to "Avoid connecting LEDs in parallel!"If you join them in parallel, you will need a resistor for each LED.You might want to put a bridge in the DOUT instead of a 220 Ohm resistor, connect each LED's anode to the DOUT output pin (single wire), and then put the two resistors between each LED's cathode and ground. Quote Link to comment Share on other sites More sharing options...
pay_c Posted April 13, 2006 Report Share Posted April 13, 2006 I really don´t know the big point at this problem. Just do the following:Instead of one LED:R_LED = (UV-U_LED)/I_LED = (5V-0,7V)/20mA = 215 Ohm = 220 Ohmput two LEDs in row (!) and lower the resistor on the DOUT module:R_LED = (UV-2*U_LED)/I_LED = (5V-1,4V)/20mA = 180 OhmThat´s it, the same for more LEDs (3 LEDs in row 150 Ohm and so on).Greetz! :) Quote Link to comment Share on other sites More sharing options...
Wilba Posted April 13, 2006 Report Share Posted April 13, 2006 You can drive two or more LEDs in series, but since we're talking about LEDs being driven from a DOUT, which outputs 5v, the combined voltage drop must be less than 5v. Blue and white LEDs will have a voltage drop of 2.75v or more, so you can't drive two in series with 5v.I was just suggesting the more generic solution, but pay_c is right - if possible, put them in series as the current draw is less than if you put them in parallel - i.e. you can drive two in series for the same current as just one... view it as exploiting the fact that you've got 5v x 20mA to work with. Quote Link to comment Share on other sites More sharing options...
pay_c Posted April 14, 2006 Report Share Posted April 14, 2006 @Wilba: Ooooops, you're right, I forgot the possibility for white/blue LEDs, sorry! In that case, David, the U_LED value is higher (red, green = 0,7 V, blue/white/pink = about 2 up to 3 volts (!), extra bright stuff partly even more than 3 volts).Greetz again! Quote Link to comment Share on other sites More sharing options...
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