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Old 25th March 2010, 03:53 PM   #11
pdul is offline pdul  Denmark
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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.
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Old 25th March 2010, 04:07 PM   #12
pdul is offline pdul  Denmark
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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
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Old 25th March 2010, 06:44 PM   #13
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by franke View Post
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
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Old 26th March 2010, 10:15 AM   #14
franke is offline franke  United Kingdom
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Quote:
Originally Posted by Mooly View Post
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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