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I just found a broken "laser projecting alarm clock". I took it apart, and am interested in only the laser projection arm. This consists of a laser diode (although I suspect that this is actually just a normal LED), a small LCD screen, and a plastic lens, which can be moved for focus. The LED has two wires. The LCD has 12 wires, which are connected more or less directly to the unit's microcontroller (a couple appear to have a single capacitor or resistor in between). The IC has an enamel blob and no markings. LCD wires are marked C1, C1, C2, and then S1 to S9. The board appears to be mostly running at about 3V (batteries are all 3V or 3.7V, and it has a 3V DC input).

The laser projection arm looks like a useful device. Since it can project time, then I assume it has at least a 3.5 digit 7-segment display (18:88), but the unpowered LCD just looks semi-transparent and green, and I can't see any markings. I would like to be able to use it for other projects, but I'm not sure how it works, nor how to start figuring it out. I have looked for data sheets of LCDs, but haven't found anything similar. I have a Arduino Leonardo, and was thinking of using it to figure out how the LCD is controlled.

So, a few of questions: Is the c1-3, s1-9 pinout common? Is it likely that the 5V from the Arduino could damage the (presumably) 3V LCD? What might be a good way to probe the functionality of the LCD?

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  • \$\begingroup\$ If it's laser, it isn't an LED. I'd say the laser either moves very quickly to show the time (you should see small motors or something) or the laser beam is filtered to show only the time. In that case, you need the filter instead of the laser. \$\endgroup\$ – user17592 Feb 3 '13 at 12:09
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    \$\begingroup\$ @CamilStaps: it says it's laser on the product advertisement, but I think that's just marketing. The diode looks like a normal LED, and it certainly doesn't move (glued in), nor does it have any moving mirrors or lenses. There is a filter - that's what I was calling the LCD. I haven't seen it in action, so I'm not entirely sure how the digits display. I assumed it's 7-segment, because it only has 12 wires, and I can't see how it could control more than 20-30 segments (3*9=27 which could control a 3.5 digit display), as it has no circuitry after the 12 wires. \$\endgroup\$ – naught101 Feb 3 '13 at 12:14
  • \$\begingroup\$ Right, thanks for the explanation. I thought you referred to the main LCD all the time. \$\endgroup\$ – user17592 Feb 3 '13 at 12:20
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    \$\begingroup\$ Look at the LCD with light at a glancing angle, and try with polarized glasses. Its electrodes are close to perfectly transparent so you need odd lighting to make them stand out better. If you can rig the Arduino to generate 50Hz complementary square waves you could connect one end to a C and the other to an S and see which segment lights up (goes clear). Use a resistive divider to step the Arduino outputs down to 3V each leg. \$\endgroup\$ – Brian Drummond Feb 3 '13 at 12:21
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    \$\begingroup\$ If the light emitting part flickers on a tiny bit at 5 Volts, it could potentially be a diode laser. A photograph of the part would help. \$\endgroup\$ – Anindo Ghosh Feb 3 '13 at 12:58
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Based upon the labeling, I would guess that the LCD glass has a total of 27 segments which are multiplexed in a 3x9 arrangement. The easiest way to produce a map is probably to temporarily wire all 12 LCD wires to CPU output pins, and then write a program that will alternately write (011111111111) and (100000000000) to the display (at 1kHz or so) until you've observed what segments are connected to the first pin, then (101111111111) and (010000000000) until you've mapped out what segments are connected to the second pin, etc. When driving the display, you must make certain that amount of time the outputs spend in any state is about the same as the amount of time they spend in the all-bits-complemented state; failure to do this may cause temporary damage to the display within seconds [meaning the display will look "odd" for awhile, but may eventually return to normal], and destroy it within minutes.

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  • \$\begingroup\$ Hrm. If we assume that the c1-3 and s1-9 are sets of power and ground wires, then wouldn't it make more sense to do (100000000000) and (000111111111) etc? and then I could probably just leave the first set on 3 times as long to a balance the charge, or something, since each c pin is likely to correspond to ~3 s pins, right? \$\endgroup\$ – naught101 Feb 6 '13 at 1:37
  • \$\begingroup\$ If the first three wires are common wires, then driving the display alternately with the two patterns shown would cause the 18 segments attached to commons 2 and three to alternate between one polarity of DC and zero volts. Not good. You need to balance each combination of twelve voltages with the opposite combination of all twelve voltages. \$\endgroup\$ – supercat Feb 6 '13 at 5:14
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The small LCD glass that you see acts as a light shutter and works in a mode for light passing through the glass. The segments of the display are controlled so that when activated they allow the light to pass through. When deactivated they do not allow the light to pass through.

I strongly suggest that the display has three seven segment sections that are designed to be driven in a multiplex manner. The lines C1, C2 and C3 are the common connection for each of the multiplex sections and S1 through S7 are used to select the individual segments. One of the common connections are also used for the leading "1" digit and the center ":" character. These are individually controlled by the corresponding S8 and S9 segment selection lines.

I can make no call on the LED versus LASER discussion since I have not seen your particular unit. It would seen unlikely that a laser diode would have a wide enough beam width to light up the whole back of the LCD glass.

* Update * I took a look at the clock product link that you provided and it looks like the adjustable projector arm on the side includes optics in addition to a very small LCD screen. With this setup it is quite possible than a super bright LED could illuminate the back side of the LCD glass.

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