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Old 26th May 2012, 06:42 AM   #11
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Can you further elaborate on connecting the LED "across" +15 and -15
Three wires will go to your amp board... +15V and 0V and -15V.
If you understand that part of the hookup, mentally set it aside for the rest of this post.
If you don't understand it (or if I don't), state so in a reply.
The resistor and LED will connect across +15V and -15V like this: +15 ---/\/\/\--->|--- -15
Your "guess" sounds fine, and is probably how I would do it. When you've inserted the amplifier power wires into the supply board, the wire ends on the bottom side will create small posts that you can connect the LED wiring to.
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I'm going to stick with the standard IEC socket. Any recommendations for a smaller one.
No, they're pretty standard.
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No fuse needed.
Use a fuse!
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Old 26th May 2012, 07:14 AM   #12
benb is offline benb  United States
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There's a lot less smoke and fire in a fuse than in a transformer.
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Old 28th May 2012, 04:19 AM   #13
Mull3t is offline Mull3t  United States
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Ben, you have a good point.

sofaspud, this is exactly how I will do it then. I've also gotten a tip from a fellow diyaudio forum member to use two resistors, one on each side of the LED in order to take the load off one side of the LED. Wouldn't a zener diode or two help here too? I'm assuming that might be another way to skin a cat.
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Old 28th May 2012, 06:06 AM   #14
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"...Take the load off one side of the LED"? I haven't the slightest idea what that is supposed to mean. The fact that it wasn't posted here in the light should tell you something. There is no load on "one side of the LED," except the resistor i showed you in a previous post. And it's the only resistor you need. Other resistors and diodes in series would work to drop some voltage, but you definitely don't need them.
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Old 28th May 2012, 04:32 PM   #15
benb is offline benb  United States
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Hmm, there's this thing call Kirchoff's Current Law, that suggests to me it doesn't matter how many resistors are in a series circuit, nor what each value is as long as their resistance totals the same, the current through the circuit will be the same. Yes, you could use a zener diode to drop some voltage, but I don't see any advantage to that.

You can run the LED on a single regulator output, putting an extra 10mA load on it, or across the +15 and -15 regulated outputs putting 10mA on BOTH outputs (which appeals to the philosophical idea of symmetrical loading), or you could connect it to the INPUTS of the regulators, or even across the AC secondary (with a revesed diode across the LED to prevent too-high reverse voltage), which would make the LED flash at 50/60Hz.

Any of these will work, with an appropriately sized (resistance and power rating) series resistor. As "you can call me Bobby" says in his song while imitating some other pop-culture character, you doesn't have to call me Johnston.
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