Luxman MB3045 pilot lamp

[acorkos]

I need to replace the neon pilot lamp on this amplifier....there is 129VDC across the bulb connections. I can't find a suitable replacement....can you recommend a way to lower the voltage to the lamp to around 12V to work with a replacement LED that I found to fit the panel hole? Thanks!

[Enzo]

You don;t understand the circuit quite. If the bulb isn't lighting, it is an open circuit. This circuit has the lamp and a 150k resistor in series across 105v. The reason you read 129v, other than that it ought to be 105v, is because no voltage drops across the resistor. If any current was flowing, there would be a corresponding voltage drop. Neon bulbs need a minimum voltage to even fire at all. The NE2 bulb, probably as common a neon as it gets, can need as much as 90v to fire on DC. SO your voltage sounds good anyway. Why not just get a new neon bulb? I think Radio Shack might even still sell NE2 bulbs.

But if you want to use a LED, it isn't about making 12v. Unless you have a whole LED indicator assembly. A plain old LED by itself needs to have its current controlled, not voltage. A typical LED will just naturally have about a 1.2v drop across it. This varies with color and other factors though, so somewhere in the volt or two range. If we set the target current as 10ma, which ought to be plenty, and we know the supply voltage, 105v or 129v, pick one. We then calculate a resistor to limit current to 10ma. Normally we subtract the LED's own voltage from the supply, so 105 - 2 = 103. R = V/I so R = 103/.01 = 10,300 = 10k resistor. If you connected the LED in place of the neon, that 150k resistor would hold current to less than 1ma. The LED would probably light, but not very brightly.


If you have an indicator assembly, that just means an LED and resistor in one package. SO if it needed 12v, assuming about 10ma again, 105v - 12v = 93v R = V/I = 93/.01 = 9300 ohms. As far as I am concerned that same 10k resistor ought to work fine. Chances are such an indicator has a 1k resistor in it, which adds up with our example.

[acorkos]

...We then calculate a resistor to limit current to 10ma. Normally we subtract the LED's own voltage from the supply, so 105 - 2 = 103. R = V/I so R = 103/.01 = 10,300 = 10k resistor. If you connected the LED in place of the neon, that 150k resistor would hold current to less than 1ma. The LED would probably light, but not very brightly.

thanks for that explanation. do you mean removing the 150K resistor and installing the 10K resistor in its place?

[Jazz P Bass]

What general size is the lamp.
Radio Shack sells a narrow diameter, 5/16" 120V, 1/3 W neon lamp.
Red or green. A pack of two. 120VAC Neon Lamp Assembly (2-Pack) : Lamp Assemblies | RadioShack.com
Or Google 120V/1/3 w neon lamp.
Solico is the major manufacturer of neon lamps.

[Enzo]

That is an indicator assembly - a bulb and resistor. If you use one of those, you need to remove the 150k resistor. Apparently radio Shack no longer sells bare neon bulbs.