"Power On" LED on a LM3875 Kit - diyAudio
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Old 1st December 2006, 12:27 PM   #1
Keyoke is offline Keyoke  United States
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Default "Power On" LED on a LM3875 Kit

Am I on crack here, or do I see a spot above R3 on the PSU in Peter's LM3875 kit for a LED?

I'm using a Avel torodial transformer @ 22+22 V secondaries, so that puts me at what, about 34V rectified at the PSU?

I'm assuming that's what that is.. it'd be awful nice if it was. If that's the case if I were to go with a 3.3v LED with the standard 20-25ma current you'd need a mighty large resistor (about 1.5K/2W) - and from what the online calcuators are telling me, that resistor is going to get mighty warm.

Is there a solution for this? I suppose I could add a small transformer and wire it in parallel with the Avel just to drive the LED, but that seems silly. There's gotta be a better way.

Course, if my understanding is incorrect and the board does NOT accomidate for a LED, well.. where would I add one I wonder? between V+/V- on the output side?
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Old 1st December 2006, 01:44 PM   #2
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Take about 10 turns of magnet wire around the toroid. Connect to LED. It's a very simple solution.
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Old 1st December 2006, 01:45 PM   #3
Nuuk is offline Nuuk  United Kingdom
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You have been found guilty of not reading the Decibel Dungeon Gainclone FAQ page !
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Old 1st December 2006, 01:48 PM   #4
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Oops... Nuuk watches all! That's where I learned the LED thing!

Your site is the best collection of GC info on the web.
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Old 1st December 2006, 01:54 PM   #5
Nordic is offline Nordic  South Africa
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As I'm reading this I'm bussy assembling a relay board and I also used that same concept to split off the 15V I need from a 30V supply.
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Old 1st December 2006, 01:59 PM   #6
Nuuk is offline Nuuk  United Kingdom
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We must have been posting at the same time Chipco!
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Old 1st December 2006, 02:23 PM   #7
cpemma is offline cpemma  United Kingdom
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Default Re: "Power On" LED on a LM3875 Kit

Quote:
Originally posted by Keyoke
...a 3.3v LED with the standard 20-25ma current
LEDs glow quite well on a tiny fraction of the "spec" current, which is quite often either a maximum or to give the quoted peak brightness (and a few hundred or more mCd is OTT for an indicator light).

5mA will increase the resistor (on 34V) to around 6.2k but reduce its wattage to under 1/4W.

http://led.linear1.org/how-is-led-br...ed-to-current/
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Old 2nd December 2006, 02:22 AM   #8
Keyoke is offline Keyoke  United States
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Quote:
Originally posted by chipco3434
Take about 10 turns of magnet wire around the toroid. Connect to LED. It's a very simple solution.

I didn't see this in the DD GC FAQ (how's that for abbreviating?).. 10 turns around the whole torroid?

Quote:
Originally posted by cpemma
LEDs glow quite well on a tiny fraction of the "spec" current, which is quite often either a maximum or to give the quoted peak brightness (and a few hundred or more mCd is OTT for an indicator light).

5mA will increase the resistor (on 34V) to around 6.2k but reduce its wattage to under 1/4W.
Does a 1/4W resistor generate less heat than a 2W one in this scenario? It seems an awfully steep drop from 34V -> ~3.0-3.3V.
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Old 2nd December 2006, 02:55 AM   #9
CarlosT is offline CarlosT  United States
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There are some LEDs that run on higher voltage if you're worried about the resistance. The other trick is to wire LEDs in series so that the effective voltage required goes up.
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Old 2nd December 2006, 07:12 AM   #10
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Quote:
Originally posted by Keyoke

10 turns around the whole torroid?

Through the hole in the transformer, not around its circumference.
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