Incandescents to LED Conversion and Problem

[jcjc]

Hello,
First, my knowledge and experience with electronics is very limited to say the least. So I have a vintage gaming machine made in 1975 that uses some old bayonet base lamps. Basically, the player shoots a ball into a numbered pocket and the corresponding light comes on. The light sockets are fairly corroded and so is the circuit board they're mounted to. I pulled the sockets off, cleaned the board and made a mount for an LED and it's resistor. I soldered the first one in and it worked fine. I made another and soldered it in and when the power is turned on, the LEDs come on by themselves but they are dim. So I tripped the switches controlling each one and they go from dim to their normal brightness as I expected. It seems there's some residual current in the circuit which may be normal but the LEDs are sensitive enough to "pick" it up. Also, in the past I had an engineer make some LEDs for this same type of conversion on a similar machine. His work perfectly and they have another component that I can't identify. From comparing pics online it looks like a small signal Schottky diode. I really dont't know what that is but was hoping someone here could comment both on the above problem and what the unknown (to me) component is and if is there to stop the LEDs from staying on. Here is a pic of the device that was made for me and works fine. These have a different mounting style and have different electrical specs or I would just place a few of these in the machine in question and see what happens. The component I'm trying to identify is the small cylinder on the left side of the base. Thanks for your time.


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I meant I'm trying to identify the component on the Right side of the base. Thanks.

 

[manishanand14]

It can be a zener diode.If you can desolder and remove the component from the PCB and take a closer pic revealing the part number.

 

[jcjc]

Thanks for your reply. I'll desolder it tomorrow and see if I can find a part number.

 

[jcjc]

Ok, I pulled the part numbers off which says "V" then under that is 4148. After searching online it seems to be a switching diode. So my question now is what is its purpose in this small LED assembly? Does it prevent the low level of current that apparently is always flowing from reaching the LEDs until they're actually switched on?
Thanks!

 

[manishanand14]

See the back side of the pcb and try to trace the circuit to know how the components are connected.I hope then you will get better idea.

 

[Inti Innovations]

I would venture a guess that the 1N4148 diode is there to prevent damage to the LED when inserted the wrong way round. 1N4148 has the reverse voltage of 20V and a typical white LED will have the reverse voltage of 5-10V.
The upshot of that arrangement is that whatever the voltage is in an off state it is reduced by 0.7V by the 1N4148 diode.

 

[chuckey]

To give a bulb a long life when its "flashed" on and off, they are operated at black heat, that is to say they have 70% of the working voltage put on them to keep the filaments hot(tish). I would suggest that you work out the circuit to see how this is done. It could be a resistor across the back of the switch. if you remove the resistor you may find that the black heat voltage goes. Could be that the switch changes the bulb voltage from , say 4 V to 6 V.
Frank

 

[jcjc]

To give a bulb a long life when its "flashed" on and off, they are operated at black heat, that is to say they have 70% of the working voltage put on them to keep the filaments hot(tish). I would suggest that you work out the circuit to see how this is done. It could be a resistor across the back of the switch. if you remove the resistor you may find that the black heat voltage goes. Could be that the switch changes the bulb voltage from , say 4 V to 6 V.
Frank

That's very interesting. The voltage is around 13 volts when switched on and around 4 volts when switched off. There's no resistors on the switches. This brings up another question and that is when I soldered the first and only LED in, it worked properly as in it was completely off when it was supposed to be. It was only when I attached the 2nd one that I experienced the unwanted illumination. With the power on I measured a few of the other locations on the board where the sockets used to be and they too show 4 volts.

 

[chuckey]

Well it still begs the question if the switch is really in series and is going open circuit, where is the voltage coming from? Only you can tell!!
Frank

 

[RF_Jim]

It was only when I attached the 2nd one that I experienced the unwanted illumination. With the power on I measured a few of the other locations on the board where the sockets used to be and they too show 4 volts.

Need to take a scope maybe and get an idea what the waveform really is; an average or even an RMS reading might be fooling us as to what is being applied ... be very careful with where the ground of the scope is attached too! I will not accept responsibility for damage. Maybe use a Fluke Scope-o-meter on battery for this measurement, and the scope (all rubber for safety) can the float like meters do ...

RF_Jim

 

[KeepItSimpleStupid]

I've done this numerous times. Use a resistor, a diode and a reverse biased Zenier diode. Since you have about 4V when off, probably a 3.3 V diode would work.

Your resistor would be <= (12-Vled-0.6-3.3)/ILed

The diode would protect the LED if inserted in an AC circuit.