Switching to LED

Hi everyone,

I have an old JVC R-S33. There are three 12V 150mA lamps that light the tuner, attached is a small image cut from the schematic. I'm looking to replace all three lamps with LEDs. Only one of them is currently dead.

There are no sockets for the bulbs (they are held in place by a rubbery material that fits snugly over the glass, with the base sticking out completely). Wires just seem to run up into the base of each one, disappearing into a glob of dried adhesive. I'm hoping to cut them off, solder in sockets, and plug 12V LED bulbs into those. Will that work? Will I have to use resistors?

Sorry in advance if I'm being an idiot, but I'm totally new to this. Any help would be great. Thanks!

I also have a JVC R-S33. I recently ran across it in the free ads, and bought it used because it is identical to one of my first receivers that I bought new back around 1980 and sold a few years later. I like it alot.

A couple of thoughts about your proposed lamp retrofit.

1. I believe the incandescent lamps originally installed in this unit provide a diffused, non-focused light required to make the dial glow uniformly.

2. 12v LEDs are not that common. I believe most are rated around 3 volts. MOST are also more of a focused light, with little diffusion. If you can find 12v DIFFUSED LEDs, it might work okay. Otherwise, you'd have to reduce the voltage with resistance.

Might be just as easy to contact AKer dgwojo and try to obtain what's already there.

Hi Mark,

Thanks for your input. The LED vendor has told me that he'll be sending 12V LEDs. I didn't even know they were uncommon, so I guess I got lucky there. Hopefully they'll do the trick.

I bought green lights, hoping that might mitigate the diffusion problem a bit, and each of the three light bulbs is encased in that bluish rubber, illuminating that bluish strip of plastic that lines the back of the numbers on the dial (hopefully it won't turn out *too* blue). I wonder if those colored filters will provide even close to enough diffusion for the dial's glow to look uniform.

If it's not enough, maybe I'll try sanding or dotting the bulb tips with something opaque, or hell, I could even try mounting little reflectors in front of them. Resistors might be easier, but I'm such a n00b that I'm scared to try them unless I have to.

Once I cut out and replace the old 12V 150mA bulbs with sockets, and put 12V LEDs in those sockets, will I have set up any obvious electrical conflicts?

LED Retrofit

mist said:

Once I cut out and replace the old 12V 150mA bulbs with sockets, and put 12V LEDs in those sockets, will I have set up any obvious electrical conflicts?

Click to expand...

I am rapidly approaching my depth on this topic. Check this thread on a similar discussion, as it turns out, for the same receiver:

http://www.audiokarma.org/forums/showthread.php?t=30802

Chances are someone with more electronics experience will jump in on this topic.

Good luck with the retrofit. I will keep watching as I may have to do this same thing someday!

Thanks for the link, that thread did not turn up in my searching. It doesn't quite answer the question, but it is helpful in that it brings up the AC aspect, which I have considered but not yet addressed here.

I believe I read elsewhere that failing to filter the current might result in a rapid, flickering response of the LED. Anyone know what I could do about that? I suppose I should be asking about that as well as the question of whether or not I'll need a resistor to prevent shorts/killed LEDs.

You cannot drop in LEDs for incandescent bulb sockets. You must have supporting circuitry. The ones that are sold as replacements have this built in. It is move involved than it may look. Search through other threads here on the subject.

mist said:

I believe I read elsewhere that failing to filter the current might result in a rapid, flickering response of the LED. Anyone know what I could do about that? I suppose I should be asking about that as well as the question of whether or not I'll need a resistor to prevent shorts/killed LEDs.

Click to expand...

You're assuming that you'll see flickering of an LED at 60 times a second. I doubt that.

LED's aren't rated for voltage and current like an incandescent lamp. They are semiconductors, not resistive devices. The voltage 'rating' of an LED is the voltage that it takes to overcome the depletion layer of the P-N junction (otherwise known as the 'forward voltage', or 'Vf'). Don't sweat the deep meaning of that last bit of physics, just understand that once that Vf voltage is reached, the LED becomes a short in the circuit, and as such needs a current-limiting resistor.

Works like this: Say you have an LED which is 'rated' at 3V (2 to 3V is typical of modern high-brightness LED's). This means that when the LED is forward-biased, there will be 3V across it. First you decide how much current you want to run through the LED...most are rated for 20mA max, so let's say we want 15mA through it. If you have 12V for a source voltage, you have to calculate the current-limiting resistor. Since the LED itself is going to drop about 3V, subtract this from the supply, and use the remainder to calculate the resistor.
R=V/I
= (12-3)/.015
=9/.015
=600 ohms

So with this '3V' LED, a 600 ohm resistor will limit the current to about 15mA with a 12V source.

(Should also check that a standard 1/4W resistor will not be overtaxed. Power (P) = I² x R, = (.015)² x 600, = 135mW, or about 1/8 of a watt, so a 1/4W resistor is fine. Alternate calculation - P = V²/R, = 9²/600 [there will be 3V across the diode, and the rest of the supply voltage (9V) MUST be dropped across the resistor according to Kirchhoff's Voltage Law], = 81/600, =135mW. How about that? Ohm's Law works. )

If you are unsure about the Vf of your LED, you can measure it yourself. A 9V battery, a 1K resistor, and a multimeter will get the job done. Wire it up with some test leads and measure the voltage across the diode when conducting. Vf rises with current, but this method will get you well into the ballpark with a figure that is perfectly useful.

With an AC voltage source, things can get tricky. Diodes are supposed to allow current to flow in one direction, and oppose that flow when current is reversed (rectification). LED's are diodes, but they aren't designed for rectification duties. The standard max reverse voltage rating on LED's has been 5V for many years, and manufacturers don't condone their use as rectifiers, hence in most applications a rectifier is recommended to do the blocking of reverse current so the LED doesn't have to. But...another AK'er who sells LED assemblies has done some experimenting with some wide-dispersion LED's he has sourced, and found that they can take rather high peak reverse voltages despite of the manufacturers ultra-conservative rating. I do not know if all LED's are so conservatively rated and in my own applications it is usually easy enough to add a diode (but the additional voltage drop of the diode must be factored into the current calculation above).

All this is a moot point if the guy you're buying these from has done all the work for you by building an assembly with a limiting resistor and perhaps a protection diode. If you're simply buying a bare LED, you get to do this yourself. So...it depends on what you've bought, and it kinda sounds like you don't know.

Wow. Okay, so I think I followed that as well as I can without finding my old physics II book and a calculator. That said, you're right-- I don't know exactly what I bought. However, based on the particular vendor to whom you seem to be alluding, and what you've said about his products, and the fact that he told me he's sending 12V bulbs, I'm feeling pretty good about just soldering in some sockets and dropping in the new LEDs. Thanks very much for your reply, it was very helpful! :thmbsp:

If it's the whsh94a guy on Ebay it will be the easiest thing you ever did.

You can use LED's with any voltage, it's the current you have to limit. I have put an LED on a 12V source without a resistor just to see what happen and it exploded in my hand. A current limiting resistor is a must with any LED. The voltage associated with LED's is the forward voltage drop. Basically if you have a 3V LED, you'll need more than 3 volts before any current can flow and light the LED. This voltage is always subtracted from the source voltage when calculating current. Just do the basic Ohm's law math that Echowars posted and you can't go wrong.


One thing I have found is that LED's are very good at running at only a fraction of their rated current and still appearing as bright as the rated current. I have a phono pre that I built and with 180 microamps flowing through the LED it still lights just fine! The same goes for overdriving them, so I always like to run them a little low, so that they last for more than 10 years, lol.

So let me lay out a summary of what I think I understand:

I have 3 incandescent 12V 150mA bulbs that I want to replace. They are *not* in a series circuit (see schematic attachment in first post of thread). However, they are "grain of wheat" style bulbs that will have to be physically cut from the circuits. In their place, I will solder 3 bayonet sockets where the bulbs used to be.

I have ordered 3 custom green '12V' LED bulbs from a reputable AKer. These bulb assemblies are special in that they are high-dispersion, so they're less likely to cause an uneven lighting of the tuner dial that would probably result from typical LEDs placed at such intervals. They are green because a previous comment from a qualified individual posted in older thread tells me that's what I want for my receiver's green display. I will ultimately insert these LEDs into the bayonet sockets I solder into the line.

Here's where I things get murky for me.
These custom LEDs are 'rated' at 12V, meaning that each bulb will drop about 12V. To calculate the resistance I would need to keep the LED from exploding in my hand (see previous post), I will use Ohm's Law, as described above by EchoWars. This value will tell me what type of additional resistor I will need.

Resistance = (Source voltage - voltage dropped across LED) / reasonable amount of current for an LED.
|
V
R= (12V-12V) / .015A
= (0) / .015A
= 0

Hence, in this case, I believe I will not need to add an additional resistor to keep the bulb from totally freaking out on me. This will be either confirmed or smacked down in posts below. I am now suspecting that the mere introduction of a bayonet socket may add some amount of resistance to the line, and wondering if that current will be negligible, or a potential problem to overcome. Is that a dumb question?

Thanks again, everyone. This is very interesting and a huge help.

Doesn't work that way, and there's no such thing as a 12V LED.

Edit: I'll take that back...there may be such a thing as a 12V LED, but I've never heard of one and that isn't what you ordered (and all LED's require a limiting resistor). You need to ask questions from whoever you bought the LED's from, since they may be standard 3V LED's with a limiting resistor which is appropriate for a 12V voltage source. You could label such as assembly a '12V LED', but it is an inaccurate description.

I haven't heard of any led dropping that much voltage. If that's the case, by your calculations they won't light at all! I'll bet the voltage drop is between 1.5 and 3V. I don't think I've seen one with more than 3V drop. You can determine the forward voltage drop by connecting an led with a resistor and measuring the voltage drop across the led. I'd put a 1K resistor in line with it and measure the drop, then do the calculations. The bayonet won't matter. A few ohms when you are dealing with milliamps of current is negligible.

To EchoWars: Right, so I think I got that you were implying as much from your first post. That's why the '12V' was in single quotation marks, correct? In any event, if they are modified as such, does it mean that I will not need to use an additional resistor? And will resistance introduced by the socket and solder be negligible?

A useful link for calculating LED and resistor values. I have in the past successfully re-lamped an SX-1250 with LEDs and rectifier diodes.

http://led.linear1.org/1led.wiz

dew.

So then I'll pop in a 1K resistor just before the bayonet socket. Will that pretty much solve my dilemma, or do I still need more information about the LED assemblies (didn't use the word "assemblies" enough times in my previous summary post)? Honestly, I'm getting a little confused about what I'd even need to ask!

Edit: Errr... wait a second.. Okay, I'm totally lost.

mist said:

To EchoWars: Right, so I think I got that you were implying as much from your first post. That's why the '12V' was in single quotation marks, correct?

Click to expand...

I just picked a voltage...other than that, you've lost me here.

In any event, if they are modified as such, does it mean that I will not need to use an additional resistor?

Click to expand...

If they are modified to work properly with a 12V source, and you have 12V, why would you want to fool with something that is already set up to work with the voltage you have? Again, ask the seller WTH he's selling you.

And will resistance introduced by the socket and solder be negligible?

Click to expand...

What does your spidey-sense tell you? :scratch2:













(hint: the socket + solder resistance is likely substantially less than an ohm, but let's say it's some insane amount, like, 10 ohms. So now the series limiting resistance in my example a few posts ago has jumped from 600 ohms to 610 ohms. Recalculate the current through the LED. Did it change much?)

Okay, I have no idea what "spidey-sense" means. But that's okay. Until I read EchoWars's post, I also had no idea how much resistance would be imparted to a circuit by introducing a bayonet socket. So I'm glad I asked.

I've been told that what I'm getting is 12V, plug and play. I've been concerned about it because I need to add sockets, which as I understand, is not normally the case. I didn't want to annoy the seller too much with questions about the procedure because 1) he was already being pretty patient with me, and 2) I figured having the discussion in the public forum would possibly benefit others in the future, while giving me multiple sources of confirmation now. Markallen has already suggested that it will possibly benefit him (not to mention AgentWinchester).

So I'm just going to solder sockets in and install the LED assemblies with no worries. Hopefully I won't have many more questions about it, if it's getting annoying, please feel free to disengage and I really do appreciate the peace of mind everyone's help has given me thus far!

Yup - just plug the lamps into your new sockets - no resistors, calculations or other headscratching necessary.
DH

mist said:

Okay, I have no idea what "spidey-sense" means.

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No fan of Spiderman, eh? Tisk tisk...

Don, he never mentioned he was buying from you. If he had, I'd have told him not to sweat it.