Most multi-touch displays have some form of an input unit, and a display.
The display can be a front-screen projector, a rear-screen projector, or a flat panel (LCD or plasma). These are all well understood technologies, and your choice will be driven by cost and how it interacts with the input unit.
The input unit can have more implementations. When I write 'light' understand that it can be visible light or infrared.
- Rear-projected light, reflected by touch points, read by rear camera.
- Front-projected light, interrupted by touch points, read by rear camera.
- Front-projected light, reflected by the surface, interrupted by touch points, read by front camera.
- Side-mounted LEDs and photo-transistors, with row/column scanning.
- Touch-sensitive, capacitive grid.
That's just what I can think of off the top of my head. Any of the back-projection modes can use visible light; for front-projection you'll be better off using infrared.
For a more in-depth review, I suggest you do more research here:
If they use ordinary small "incandescent" filament bulbs you will be able to run them on DC just as well. For practical purposes AC and DC of equal RMS value will work the same.
IF you can access the bulbs you will be able to replace them in due course with LEDs of similar appearance and brightness - but probably not something you want to try unless essential.
The operating specifications are somewhat contradictory.
2.5, 3.5 or 6V is a VAST range, suggesting they may be 6V bulbs that will run on lower voltages if needed, but at lower brightness.
110VAC/35 lights is about 3 V/bulb. 110 VAC/50 ~= 2.2 V/bulb
6V, 1.2 Watts is at the high end of what you'd expect.
It sounds like these are nominally 6V bulbs.
But operation on 6VDC to start is not recommended.
SO
Starting with a single alkaline cell (1.5V nominal) would tell you something.
I'd expect a dim orange glimmer.
Then try to a alkalines in series. That's 3V nominal. Doing this with no series resistor is extremely unlikely to do any damage. Being very old there is a very small chance that it might but it's very unlikely. I'd expect an OK appearance - maybe not as bright as on some strings. From what you get you can decide what to do next.
Above 3V I'd start with a series resistor.
Start with a 22 ohm resistor and if not bright enough try 6V and 22 ohms.
If 4.5V 22 ohms is bright enough then something like 6V with 33 ohms may be similar.
Once you have a 6V pack looking OK you can adjust the resistor up or down to suit brightness. Er on the side of too dim if you want them to last.
I've suggested going to 6V as battery voltage will vary with time and using 6V and a resistor will keep the brightness more constant over the battery life.
For a high tech [tm] solution that helps protect the filaments a constant current driver may be used. Ask if of interest.
Resistors mentioned above can be half Watt or more. 1 Watt safer but probably not needed..
Constant current supply:
Am LM317 IC can be used to provide a simply built constant current feed.
A "problem" is that the circuit "uses up" a minimum of about 3Volt to operate. So, if you run it from 6V you can only get 3V out. Whether this is a problem depends on Vbulb when it is a bright as you want it. Ideally you'll want even more than 4 batteries :-(.
Here R1 is shown adjustable but you can use set resistors which are changed to suit. Maximum likely bulb power was given by 6V, 1.2 W = 200 mA lamp current (0.2 x 6V = 1.2W. )
Current source current = Vref/R or
Resistor = Vref/Icurrent_source.
Here V = 1.2V regulator Cref, I = 0.2A max.
So R = V/I = 1.2V/0.2A = 6 ohms.
So if you make R1 >= 6 ohms at all times, then Ilamp <= 200 mA.
Add extra resistance to R1 to get lower lamp current.
Connect B+ to Vin.
Iout to bulb top
Bulb bottom connects to battery -.
Best Answer
This "should" work.
2 transistors are driven by control IC.
Connections as shown shorts transistors.
That's about as nice a set of photos as I've seen with a question.
A welcome change.
Sharp, well lit, shows area of concern. Leads were initially not 100% clear but once my brain engaged (5am - up to shake off a nightmare :-) ) it was clear enough.
Why is the left hand common lead +ve and not -ve?
And even rougher :-).
Here is why the lh trace is +ve.
On the copper side image all diodes "point right" - diode cathodes right.
So, if AC goes to an Anode the Cathode is DC positive.
If AC goes to a cathode the anode is DC negative.
In this case the 1st and 3rd diodes from the top have the AC input connected to their Cathodes (rh side/ bar end in copper view) So their Anodes are DC positive and are joined together and have a trace running to PCB bottom left - which is thus DC positive.
A multimeter set to volts will confirm this.
Presumably (he asked again) there is an AC transformer and AC mains is NOT connected directly to the PCB.