Dial Light Bulbs for old Lafayette Stereo

Den tirsdag den 1. juli 2014 02.41.14 UTC+2 skrev snipped-for-privacy@home.com:

-Lasse

While you have them out why not convert them to LED's and not worry about expensive filament lamps ??

Bayonet style Or leaded (Grain of Wheat)?

Cheers

I was thinking about that. But First off, this is AC voltage, so I'd have to add a rectifier diode and a filter cap. Then, how do I determine the ohm rating for the resistor to use in series with the LED?. I assume one diode would cut the voltage down to around 4 or

5VDC. (One lead from the power transformer is grounded as well as each light).

$1 each quantity one, $0.85 each quantity ten at All Electronics.

Standard disclaimers apply: I don't get money or other consideration from any companies mentioned.

Matt Roberds

Try All Electronics. I've been buying from them for over 30 years and have been happy with their products and service. Their shipping is not cheap, but if you are in the Los Angeles, CA area you can visit their store.

See:

Regards,

Dave

You don't need a filter cap, just use a zener with a breakdown voltage less than the LEDs reverse breakdown in parallel with it, then resistor to ground. Then the LED is protected from reverse bias, and overvoltage.

OK, if you want to go LED, you do not need filtering. You should have a blo cking diode because LEDs do not like reverse volatage so much. Then you nee d to get thew current right.

As for the channel imbalance, adfjusting is not going to do it and I suggge st you do not try to adjust it. You got bias and offset, it is highly unlik ely there is a gain control. Even old ass tube amps didn't have them. What you have is bad coupling caps.

Don't mess with any comntrols. I just found a print on the thing and it has one of those all f***ed up circuits. This is not a normal amplifier.

Click where it says "get manual".

Anyway, the first thing you have to do is make sure the balance is off on a ll sources. You probably ponly listened to the FM so far. Try the AAM, and try AUX. Feed it from your computer and get something on youtube or whateve r. If the imbalance is the same on all sources, look to the power amp. Whja t you really need is an ESR meter to check the caps, or you could probably replace them all in the power amp circuit. I see the unit does not use regu alt tone controls, the probalem could be there.

Actually, the most likely problem is C401, 402 or 404. I wouild probably re place them all. There are L and R of each.

This thing has a funny power amp circuit, of a type I REALLY do not like to work on. It is imperfect and can be a bit quirky, so don't blow it. Makes a big mess. Actually the design looks like something Magnavox would do. Do not mess with the adjustments unless you have to change transistors. Take m y word for it.

Looking at the print, I am kinda impressed by the FM tuner section. Sometim es these old recievers can pick up a stick in the mud in china. That looks like one of them. Three tuned RFs. Using FETs. Though the IF strip does not use ceramic filters, it looks quite well designed. I can instruct you how to align it properly if you have a scope. You can just use an off air signa l, but hooking up the scop properly is the key to doing it right.

There is also some discretion in this, if you really do have to align it, a nd this is not all that unlikely, you can choose between having a narrower IF bandwidth and better sensitivity, or a wider bandwidth which will give y ou lower distortion, IF you got enough signal.

Thre unit also should be tweaked in the M?PX section becasue it is not PLL. I can tell you how to tweak that, but you should stay away from the pots. the main thing to adjust there is T207, which they call A34, but that is a Sams thing I think.

Whatever, this old vintage stuff is cool. The amp is not really all that ba d, it looks worse because of Sams' way of drawing it. It's not like those S onys that have no PNP transistors. There were a few others like that and I do not like any of them. this is not that though.

It says it has a 70 volt supply to the outputs. That means it clips at 76 w atts a channel - woith no load. you can expect to lose at least 10 % with e ight ohm speakers which will leave you with about 60 watts a side. The rati ng is much lower than that because they give that distortion spec, it is pr obably rated 35 a channel if that. But it will give you 50 or 60, bu not wi thin their distortion spec. This is common of older equipment.

the technical reason is that they couldn't keep it linear all the way to th e power rails. Now they can. A 100 WPC amp today is juyst about that, becau se the circuitry as well as the components have improved. Fact is though, i f it is rated 30 WPC, and puts out double that, it might be hitting 1 or 2 % distortion. Of course most speakers are much higher than that at those le vels anyway.

Not a bad piece. I wouldn't trade you my Phase Linear fot it, but I would t rade you a 200 watt (stereo only, no surrond, which is good) Pioneer for it .

Anyway, looking at the print, where I see the thing fall down the most is i n the phono preamp. Of course that is not a big deal for alot of people the se day. You could also get an outborad preamp but really, fukit.

Have fun with it. It'll jam. I think it's not just old enough to drink, I t hink it is old enough for a military pension.

The 8VAC, rectified & filtered will provide about 10.7 V DC. White LEDs drop about 3.2 volts, leaving about 7.5 volts to be dropped by the resistor. The LED will have max current rating. See below for a typical white LED:

You don't want to run it at that rating - you should de-rate it to be gentle on the LED. (It will still be plenty bright even at half the max current.) So, divide the max by 2 to get the current that will flow. Then use the formula R = E/I where E is 7.5 volts and I is half the LED max current rating. Example: a 3.2 volt 30 mA white LED connected to 10.7 V DC R=E/I so R = (10.7 - 3.2)/.015 = 7.5/.015 = 500 ohms. A close standard value is 470 ohms. Using 470 ohms, the LED current will be 7.5/470 or about 15.9 mA.

• 10.7V ---[LED]---[470 ohms]--- gnd

You can wire two or three white LEDs in series. That changes the resistor value, because each LED drops about 3.2 volts. With three in series, the total LED voltage drop would be about 9.6 volts, meaning you want to drop about 1.1 volts across the resistor. So R = 1.1/.015 or about 73 ohms. With two LEDs in series, the LED Vdrop is about 6.4, so you want to drop about 4.3 volts across the resistor. R = 4.3/.015 or about 286 ohms.

For multiple LEDs (assuming 3.2 volts for each LED) rated at a maximum LED current of 30 mA you can do this:

• 10.7V ---+---[LED]---[470 ohms]------------------+--- gnd | | +---[LED]---[LED]---[270 ohms]----------+ | | +---[LED]---[LED]---[LED]---[75 ohms]---+

You can add as many 1, 2 or 3 LED strings as you like, up to

16 strings, total, which would draw about .015 mA each for a total draw of about 240 mA.

You can also dim the LEDs by increasing the resistor values.

Ed

--
Not true. :-) 

Since it's the current through the LED which heats it up, and the 
reverse voltage across it which pokes holes through its depletion 
region, you can do this and kill three birds with one stone: 

.        +--[]--+ 
. R||E              | 
. I||C---+---[Rs]---+  
.        | 
.       GND 


1. The resistor will limit the current through each LED as it takes 
   its turn being ON. 

2. The Vf of one LED is the Vr of the other, making it impossible to 
   overstress the OFF LED. 

3. Keeping the LEDs close together will double the apparent ripple 
   frequency, making the ripple undetectable and eliminating the 
   need for a DC supply.  


>Then, how do I determine the ohm rating for the resistor to use in series with the 
>LED?.