LED for pointer lamp (Sansui 5050 and 6060)

[Analogdino]

I have two Sansui 5050's both with burned out pea (rice grain ?) filiament lamps in the tuner pointers. I replaced them with LED's.
Here's how:
First and obvious (but it has to be said), don't just connect in a miniature LED - it will burn out!

I used a HLMP-1340 LED (made by HP, 3-5 Systems Inc, Quality Techn and Chicago Miniature, perhaps others.) Most, if not all, similar substitutes will work. It's an ultrabright red (wavelength 635 nm), non diffused. It has a maximum forward rating is 30 mA with a drop of 2.2 (min), to 3.0 (max) VDC, but a PIV of only 5 VDC (all LED's have very low PIV ratings) We will run it well below this current.

The supply to these lamps is 6.3 VAC so the PIV could be 8.9 volts - too high.
To solve this we will connect a "back to front" diode across the LED, i.e. anode to cathode (see below.) Use any small, general purpose Si diode, such as an IN5392, or similar. Do NOT use a point contact detector diode (quite a different animal.)

Installation
1. Extract the dead lamp from the pointer carefully (don't sever or damage the tuning drive string that is over the top of the wire.)

2. Take a pair of very thin insulated flexible wires abut 9 inches long (I use 30 AWG flex, not solid, that we have in the lab) and solder one to each LED wire. Don't let the LED overheat, do it quickly and cool the joint with a wet sponge. Put about 2 inches of sleeving on, right up to the LED glass. Use the smallest sleeving that goes over the flexible wire and the soldered joint. Do not use thicker flex as it has to bend as the pointer moves and thicker wire can displace the pointer carriage on its track. The maximum size is what Sansui used - you may be able to salvage what was on the original lamp and just extend it for connection the other end.

3. Bend the LED wires to a right angle near (not at) the LED glass. You want the LED to go to the bottom of the hole in the plastic pointer and have the wires lie flat and level with the top of the pointer carriage. Easist way is to put the LED in, then bend the wires flat, then pull out the LED and set the right angle firmly with long-nosed pliers so that the bend in the wires and sleeving does not pull the pointer away from the vertical. Twist the wires together and cut them to the right length for free movement of the pointer to both ends of the scale, plus about 1 1/2 inches. Strip the wire ends carefully (only one chance!) Put the LED back in the pointer and dress the sleeving beween the tiny brass tabs; bend them in to hold. Lay the sleeved wire over the pointer carriage and tie it down to the center tiny dial string tab with a twist of thin copper wire. Keep the total height low, the box top is very close here. Test that there is no connection to ground from either thin wire with the DVM ohmmeter (the risk spot is the right angle bend of the LED connecting wires - they are very near the pointer carriage.

4. Connecting the PIV safety diode:
Connect the other diode across the stripped wire ends. When you have done this check with the diode check function of a DVM. When you test the pair together you should get a 0.6 VDC or so forward drop of the logic diode one way round and a 1.6 VDC or so forward drop of the LED the other way around. If you get 0.6 VDC and infinity, you have the diode the wrong way round - 50/50 chance! Reverse it. DO NOT omit this very important test!

5. The series resistor
Target around 12 mA for the LED "on" current (a 1/2 wave rectified AC.) Voltage to be dropped is 6.3v minus, say, 2v for the LED, or 4.3 volts, thus R = 4.3/12 Kohms = 358 ohms. But, the 5050 and 6060 models have an 18 ohm resistor in the pointer lamp cicrcuit, so we need a net 340 ohms. 330 ohms is a standard so use that, close enough. (The 7070's, 8080's and 9090's may be the same, but I've not seen the schematics.

6. Connections
Use the tags on the PCB under the white plastic lamp housing (any way around - it's AC but one side is grounded.) Connect one end of the resistor to either. Solder the diode (and flex) across the other end of the resistor and the other tag. Sleeve the resistor-to-diode bare wire. Dress the thin twisted flex to clear all obstacles as the pointer moves over the scale.

7. Test.
Switch on receiver. Pointer should glow red. If not, switch off immediately and trouble shoot.

I am working on a similar replacement for the stereo indicator. Will advise later.

Cheers,
Roger

[dgwojo]

Holy Chit, I'm glad I have 6.3v, grain-of-wheat lamps on hand, this sounds like a "labor of love"!! Dave.

[Inspiribomb]

Why the need for the diode? Wouldn't a resistor in series with the LED suffice?

[dgwojo]

Originally Posted by Inspiribomb

Why the need for the diode? Wouldn't a resistor in series with the LED suffice?

LED's require DC voltage!!

[EchoWars]

...another way of saying that LED's can't deal with reverse voltage like a regular diode can. A series diode along with the LED takes the Vrr punishment, and a resistor limits the forward current.

[Inspiribomb]

Ahhh, yet another thread that makes me realize just how little I know. Wish I woulda paid better attention in my electronics class, or at least retained some of it.

[Analogdino]

Originally Posted by EchoWars

...another way of saying that LED's can't deal with reverse voltage like a regular diode can. A series diode along with the LED takes the Vrr punishment, and a resistor limits the forward current.

Actually, it's a PARALLEL diode across the LED to shunt the LED's reverse voltage 1/2 cycle so that the LED sees only 0.6 volts PIV. The 330 ohm resistor limits both halves of the cycle, i.e. the forward current of both diodes.
I should have scanned and attached a schematic!
Now I want to do the same for the stereo indicator, but this is DC from an emitter follower (current amplifier) driven by the HA1196 stereo decoder chip. I'll post the solution later, when done...
BTW, I've done a similar lamp replacement on a classic Yamaha M4 amplifier where all the lamps had burned out.
Cheers,
Roger

[Analogdino]

Originally Posted by Inspiribomb

Why the need for the diode? Wouldn't a resistor in series with the LED suffice?

No, it does not block the reverse voltage (you need a diode.) But you still need the resistor it to limit the forward current.
Cheers,
Roger

[tdan]

Any particular reason you went with this led setup? The needle lamp on my 5000x died awhile back. The original according to my service manual was a 5v 60 ma. My local electronics shop said that the 5 volt lamps were no longer available. He said a 6 v 75 ma would work fine. They come with fine wire leads and are very easy to replace. 79 cents and 15 minutes later, as good as new!

[Analogdino]

Originally Posted by tdan

Any particular reason you went with this led setup? The needle lamp on my 5000x died awhile back. The original according to my service manual was a 5v 60 ma. My local electronics shop said that the 5 volt lamps were no longer available. He said a 6 v 75 ma would work fine. They come with fine wire leads and are very easy to replace. 79 cents and 15 minutes later, as good as new!

I like the fact that they are cold light, quite bright and "never" fail - just better technology than filament lamps. Also, I had a ready source of cheap miniature LED's and all the other bits were free, plus my previous experience with the Yamaha M4 lamp replacement (all the original switch lamps had burned out.)
Cheers,
Roger

[EchoWars]

A parallel diode would work, but if the diode opens then the LED dies. A series diode is generally how it's done.

6 'O one, 1/2 dozen the other.

[Analogdino]

Originally Posted by EchoWars

A parallel diode would work, but if the diode opens then the LED dies. A series diode is generally how it's done.

6 'O one, 1/2 dozen the other.

Series connection should work but it shares the PIV across the two diodes in an inverse ratio of their leakage currents at each of the shared reverse voltages - Hmm... hope that's clear! If the LED was being supplied by a significantly higher voltage than 6.3 VAC, one might speculate that the shared PIV across the LED could exceed its maximum PIV. Else, the LED's current near breakdown would likely start to go up and limit itself with the safety diode taking the rest of the PIV. This is getting a bit philosophical <g> and I've never studied LED's near PIV breakdown.
Parallel diode connection (if the series resistor is correctly chosen) does not allow the LED to get anywhere near its limits, its max PIV is only 0.6 VDC or so.
Cheers,
Roger

[pustelniakr]

EW is right. Series diode would work just as well, and, if the steering diode opened, no voltage would appear across the diode. In order for PIV to be a factor, you would need reverse current to damage the LED. The real benefit of the steering diode (series) is that the current in the negative direction is blocked, rather than simply wasted. This is not a factor for a single lamp, but when replacing more than 1 lamp, the savings can be significant. Also, you can cut the power dissipation, that the series resistor sees, in half, so it can be smaller.

All in all, I agree with EW...6 of one, half a dozen of another. The original concept expressed would work. As a matter of fact, its the method I recently shared for replacing incandescent meter lamps with white LEDs in another thread.

You could use back to back LEDs, if you have places to mount them.

Rich P

[Analogdino]

Originally Posted by SansuiSamRI

Instead of using a diode, would two opposing LEDs and a series drop resistor work or would that exceed the max PIV?

Yes, but the pointer only needs one and there's only space for one.
Cheers,
Roger