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 25th March 2010, 04:53 PM #11 pdul   diyAudio Member   Join Date: Jan 2009 Location: Copenhagen You can use a capacitor as the current limiter instead of the resistor if you connect to the ac v before the rectifier and connect a diode in anti paralel with the led. I have seen it done, but unfortunatly i dont know the math. Maybe its worth looking into.
 25th March 2010, 05:07 PM #12 pdul   diyAudio Member   Join Date: Jan 2009 Location: Copenhagen My best guess is that the calculation is something like that of a 1st order high pass filter for a treble unit, so when you go below fn of the circuit the resistance will go up and since you have a fixed frequence in the supply it should be possiple to calculate something. Maybe some clever heads here can come up with the math
Mooly
diyAudio Moderator

Join Date: Sep 2007
Quote:
 Originally Posted by franke How will it affect that channel/board? Does it steal a little current or does it do more? I do not know much about electrical calculations ...
It draws current... it doesn't steal it from anything.
In real terms it's half of a quarter of nothing

If you are a perfectionist... and this will make Zero difference... I see that you have a 2K2 across each rail anyway. So on the rail you add the LED too (which can be either the plus or minus) remove the 2K2 on that rail (the LED supply one) to even things up. Easy

franke
diyAudio Member

Join Date: Apr 2008
Quote:
 Originally Posted by Mooly It draws current... it doesn't steal it from anything. In real terms it's half of a quarter of nothing If you are a perfectionist... and this will make Zero difference... I see that you have a 2K2 across each rail anyway. So on the rail you add the LED too (which can be either the plus or minus) remove the 2K2 on that rail (the LED supply one) to even things up. Easy
I think I'm beginning to understand. The current the LED will use is so small so it doesn't make a difference, unless you are a purist. There's lots of current going into the amp anyway.

thanks

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