I recently bought a (circa mid-1990s) NAD 502 CD Player at an auction. It plays CDs just fine, but the LED display is completely dead. As it turns out the display is backlit by two incandescent light bulbs, so the dead display is obviously a common problem with these players. Incandescent bulbs would last ten years at most. One blogger says he swapped out the old bulbs with LEDs, and that's what I'd like to do. Here's how the blogger described his repair:

The lamps are supplied from the 0V rail with current going into the -15V rail from the 7915 regulator, so there are two sets of 0V pads and two sets of -15V pads. I wired a 1k resistor in series with two LEDs in series to give an adequate brightness, and this draws less current from the regulator than the capsule lamps.

http://blog.martincowen.me.uk/repairing-a-nad-502-cd-player.html

I'm not sure exactly what these words mean, but I'm assuming I'll remove one of the bulbs and leave the other dead bulb connected? Then use those two wires of the disconnected bulb to power the two LEDs and 1k resistor wired in series? Here's a photo of the two bulbs in question. Sorry if this question is very basic. But I want to have a better understanding before I attempt this repair.

 

Yes, that is what he means.

0V -> resistor -> +led1- -> +led2- -> -15V

The + is the anode of the led and - is the cathode.

 

Service manual here. https://elektrotanya.com/nad_502_sm.pdf/download.html.

Yes put the two leds in series with the 1K resistor across the -15V and ground , ( anodes to ground) as you can see the bulbs are in parallel across the supply. (You can alter the brightness by changing the resistor value, lower brightness higher resistance and vice versa)

 

Yes, that is what he means.

0V -> resistor -> +led1- -> +led2- -> -15V

The + is the anode of the led and - is the cathode.

Just to clarify, I could snip off one of the two bulbs and use those leads to wire the resistor+led1+led2? Or remove both older bulbs?

 

Just to clarify, I could snip off one of the two bulbs and use those leads to wire the resistor+led1+led2? Or remove both older bulbs?

Either way is fine.
The dead bulb does nothing so it makes no difference if it's left in or removed.

 

Either way is fine.
The dead bulb does nothing so it makes no difference if it's left in or removed.

Sorry to be so pedantic here. If I did snip off both original bulbs, I would connect the positive from one of the old leads to the negative of the other lead, then use the remaining two leads to power the new resistor+led1+led2 circuit? If you were going to do it, would you just snip off one or both of the two original bulbs? Something about leaving the old bulb in place sort of bothers me for some reason. LOL. Thanks for your help.

 

I would connect the positive from one of the old leads to the negative of the other lead

If this results in directly connecting together two leads that used to feed one bulb, that would create a short and cause thr power supply to fail. Either leave one dead bulb in place or leave its leads unconnected. Use the other bulb leads to power a pair of LEDs. You could power four LEDs, two from each of the original bulb leads.

 

If this results in directly connecting together two leads that used to feed one bulb, that would create a short and cause thr power supply to fail. Either leave one dead bulb in place or leave its leads unconnected. Use the other bulb leads to power a pair of LEDs. You could power four LEDs, two from each of the original bulb leads.

Okay, thanks. I like this idea of two separate resistor+led1+led2 circuits. I believe I'll go this way. Thanks again.

 

If you look at the board, you will probably see that the + and - connections from both bulbs go to the same place. You are not gaining anything by using both. In fact you could put four in series in place of one bulb, and use half the power of putting two in place of each bulb.

 

If you look at the board, you will probably see that the + and - connections from both bulbs go to the same place. You are not gaining anything by using both. In fact you could put four in series in place of one bulb, and use half the power of putting two in place of each bulb.

Exactly! I was lazy and didn’t perform a power inventory. 15V is plenty to supply the Vf of four LEDs with enough headroom for a current limiting resistor. As long as the TS is not tempted to connect ANYTHING to where the second incandescent bulb leads were connected.

 

Exactly! I was lazy and didn’t perform a power inventory. 15V is plenty to supply the Vf of four LEDs with enough headroom for a current limiting resistor. As long as the TS is not tempted to connect ANYTHING to where the second incandescent bulb leads were connected.

This conversation has been very helpful. I think what I'll do is use a 470 ohm resistor and 3 leds connected to one side. I'll leave the second bulb alone. I've never dealt with negative voltage before, but I gather the current flows from 0V to the -15 volt rail. Just have to remember that when I get to the soldering stage. Anyway, thanks again for all the help.

 

At this point you also have the option of selecting a different colorfor the illumination, So one thing to think about. AND DO NOT connect any of the snipped off bulb leads to any other, OR to anything else. The damage to the power supply will be a real inconvenience.

 

At this point you also have the option of selecting a different colorfor the illumination, So one thing to think about. AND DO NOT connect any of the snipped off bulb leads to any other, OR to anything else. The damage to the power supply will be a real inconvenience.

I opened up the NAD 502 today and as it turns out, after all the talk about snipping off the bulb, there wasn't enough of a lead to solder to my led assembly. As such I had to remove the entire board so as I could get to the underside, and from there I had to improvise by running wires through some holes to the topside. I ended up using 2 leds + 470 ohm resistor because the space under the plastic cover was so tight. I also had to take a bit of a risk, using an led+resistor to figure out which lead was positive and which was negative. I'm sure some of the more experienced users here are rolling your eyes at this point, but all's well that ends well!

The light from the 2 leds is bright enough if perhaps a little uneven, though I think that's how this thing was designed. If I ever go back in there again, I'll create some kind of led assembly with a diffuser to get a more even light. Anyway, thanks everyone for your help!

 

A very simple and cheap way to diffuse LED light is with a bit od sandpaper, just sand the rounded portion a bit and the light is diffused. Difficult to UNDO, but probably you will be happy with it.