Replacing light bulbs with LED's for illuminating a radio's scale

[Richard9025]

Hi !
In my old big radio , the light bulbs that were supposed to light up the scale burned
So , I want to replace them with some LED's that can illuminate all of the radio scale .
The schematic showing how the light bulbs are connected :



BULB 1 , 2 and 3 are burned , the power led still works (35 years )
Can you give like ebay/amazon/tme links to the leds that can replace the burned bulbs , or at least specs ?
Thanks !

 

[Alec_t]

A problem with LEDs is that they are very directional compared to incandescent bulbs. Each bulb may therefore need to be replaced by several LEDs aimed in different directions to get enough light spread. Alternatively a single LED could be behind a diffuser; but this would reduce the brightness and require compensation by using greater LED current or a high efficiency type. Experimentation will be needed to get a reasonably uniform spread of light.
Impossible to give LED specs when we don't know the bulb specs or how the bulbs are positioned .

 

[safetywatto]

The equivalnce wattage ratigns on LED bulbs confuse the hell out of me

 

[Externet]

Bulbs 1,2,3 are supplied by AC. To use LEDs there, would be convenient to insert a 1N4001 diode in series to S4; and resistance values for R22,23,24 changed to 1KΩ/0.25W. Observe polarities on LEDs.

You will end with insufficient illumination spread for the dial after the surgery. Try 6V light bulbs of the same dimensions instead; they may last another 30 years. If you find 12V bulbs that fit, will be less bright with longer life.

 

[duke37]

Instead of a series diode, use two leds in inverse parallel, each will protect its butty from excess reverse voltage.
With 10V AC you could use two pairs in series.

 

[Richard9025]

More info :
First how long is the scale (30cm ruler also has inches)


Now the position of each bulb :
FIRST BULB


SECOND BULB



THIRD BULB



Now , these bulbs are covered with two metal pieces to reflect the light across all the scale
One metal piece :





How it should look like with all the bulbs working (a photo from the internet)



And how it looks after someone replaced the bulbs with leds (also from internet) :




There is not much info about the bulbs , there is more about the led's
How I made that drawing from above
There is a interconnection schematic in the service manual that shows where tha bulbs are connected
BEC = BULB
RETEA = POWER (power like from the wall wart , not power from a cb radio or amplifier or smth else)



What is in the orange rectangle from above is this :

The orange rectangle zoomed in


The leds that indicate what band it is (12 total green led's) are some MDE1531 , there is one red led for the fm stereo even if there are two fm bands (2 fm , 1 lw , 1 mw and 6 sw)
About the specs of the light bulbs inside . I will remove one and measure it
If I remove the metal pieces , I can fit some leds there
Now you know much more info

 

[Alec_t]

So, the dial is edge-lit. The LED effort shown looks to have 2 LEDs at each end of the dial. More LEDs would clearly give a better spread of light. If it were me I'd try probably 6 per end.

 

[Richard9025]

See if the following diagram is good


Also note that the radio scale is ALL GLASS , and on top of it is the "filter" showing the band , numbers , etc)

So these 2 metal pieces were reflecting the light into the glass , so that all the letters can be illuminated
On the photo number 7 on my previous post , he removed the metal pieces , because the blue leds dont equally light up the scale

Between the metal piece and the rest of the radio are 4 mm
If the drawing is above , what type of LED can I use , they must be under 4 mm , so some ultra-common 3mm leds . And , what voltage and miliampers rating should they be ? Dont forget the 10Ω/2w resistors .

 

[Externet]

Diagram is good. Color is your preference. Probably 'warm' white will go better with the age of receiver. There is also more efficient types that give more light with the same current.
Aim the LEDs beam spaced towards the edge of glass ends. Avoid under and over aiming the glass, or place masking.

Perhaps 1cm+ distant to avoid the concentrated spots and give more uniformity. You have to experiment. Adding a piece of diffuser as a matte-clear adhesive tape at the glass edges can help.
LEDs come in several beam angle flavors, you should investigate the data sheet for the ones you can source. That will affect the optimal distance choice from glass edge.
Luminosity will change with series resistors value, Aim for 10-15mA . Start with 1KΩ resistors and lower 100Ω steps without exceeding 75% of rated current, as they will be running on AC.

 

[Richard9025]

I'll use some white LEDs , 3mm 3.2V 12mA (3V-5mA 3,2V-12mA, 3,4V-20mA)

For DC is : The supply voltage minus The forward LED volts divided by the forward LED current and for AC , the result is divided by how many leds are in inverse parallel

So , for the following circuit :



R is 10v-3.2v divided by 6x 0.012A = 6,8:0,072 = 94Ω
So I can use 100Ω for R22 and R24 and for R23 , because there are only 2 leds , not 6 , a 270Ω resistor

See if the calculations are correct .
Thanks !

 

[duke37]

That diagram would work if all leds were identical. It would be better to put them into pairs so that one led does not have to take all the reverse voltage.

R = (10- 9.6) / 0.012 = 33
This however is too low to define the current with voltage or temperature variations.

The resistor will set the current and will be easy to calculate using DC, however, you have AC so there will be no current during part of the wave and in particular only half the wave will be active. I would select the resistor to give the correct brightness. Compare with 20mA DC.

Have you thought about glueing the leds directly to the edge of the glass to put most of the light into the scale?

 

[Alec_t]

If you were to use amber/yellow/green LEDs instead of white their Vf would be lower so you would have more 'headroom' to define their current better with a dropper resistor, assuming the LED arrays are as in post #8.

 

[Richard9025]

That diagram would work if all leds were identical

All the leds are identical

33

33 ohms ? kinda low

Have you thought about glueing the leds directly to the edge of the glass to put most of the light into the scale?

The scale is all glass , yes i thought of that , but first lets get the electrical part done
Just make a drawing on how to wire them and some specs if its incorrect what i did before,

 

[duke37]

The leds may be similar but they will not be identical and will be at different temperatures.

I would try two inverse parallel pairs in series and a single resistance.

At a rough guess (10-6.4)/0.0020 = 180
See how the brightness compares with 20mA on DC.

Use the circuit as in #10 but delete two diodes and cross connect each side as in #8.

 

[Externet]

AC reaches peaks (and valleys) of 1.41 times RMS = 14.1 Volts. for the 10 Volts AC.

Nothing wrong with asumming 10% extra if secondary of transformer puts 11V AC RMS instead.
That raises peaks to 15.5 V. Call it 16V for safe calculation.

The 16 V when used with three 3.33V Vf LEDs leaves 16-10=6V to use for the current calculation.

6V divided by 0.015 A = 400Ω for the schematic on post #10. Go for 390Ω 0.5W standard value.

Perhaps you like side looking LEDs----> http://www.mouser.com/Optoelectronics/LED-Emitters/Standard-LEDs-SMD/_/N-ec21j?P=1yp37eaZ1z0s07o

 

[duke37]

If you want to calculate properly, then work out the time of the peak of the sine wave above 6.4V. Integrate the voltage during this time to calculate the average voltage to use to calculate the resistance. Perhaps you went to school.
I would do it by trial and error.

 

[duke37]

I have had a brain storm and decided to do some calculus. 65 years is too long to leave it dormant.

Take a supply of 10V, peak of 14.1V.
Take drop of two leds totaling 6.4V.
Work out when the voltage is high enough to pass current. 0.476 to 2.67 radians.
Integrate sine wave between these limits. 14.1 * 1.78 = 25.13
Subtract the lower rectangle (6.4 * 2.2 ) from 25.13 = 10.96
Average height of active area = 10.96/2.2 = 4.98
Average height over one cycle = 4.98*2.2/3.14 = 1.74
V = IR
R = 1.74/0.02 = 87
So use a 91 or 100Ω resistor.
The current will be high over part of the cycle but will average about 20mA.

It would have been easier to get my computer working on LTspice

 

[kellys_eye]

Have you searched for modern MES light replacements?

They do exist and are simple screw-in replacement without all the hassle. Not exactly cheap (Farnell list them at around $10 each) but I suspect there will be cheap Chinese equivalents via the likes of eBay such as this one

http://www.ebay.co.uk/itm/2-x-LED-L...epid=0&hash=item46326cc149:g:pHgAAOSwk5FUvBm5

at two-for-$5 (ish)

 

[FuZZ1L0G1C]

Regarding the diffuser, I've been thinking of an idea, open to suggestions:
For camera flashguns, and on a larger scale, fluorescent light-fittings, you get a perspex diffuser.
These come in different thicknesses and pattern densities, from "moire-pattern" fine to bumpy.
As they don't block light, instead diffusing it, (maybe..) it will do the job.
LEDS: what about also strip-lighting top & bottom edges, as "LED" photo is dimmer on face.

 

[duke37]

You do not need a difuser, what you need is the light going into the edge of the glass. The light stays inside the glass due to total internal reflection When it encounters the print, the light can escape. Light bulbs must be placed close to the edge but leds will be better since the light source can be placed closer and glued into place.