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I am reverse engineering a LCD (this is a raw bit of glass, i.e. no controller) and trying to understand the signals... there are 16 lines of which 4 are backplanes and 12 are segment lines.

The backplanes are driven, in a 60 Hz repeating 8-step sequence, at 4 levels... 0, 1v12, 2v24 and 3v36.

I took the output from one segment line and subtracted the individual backplane signal to get 4 different signals (effectively the demultiplexed signal)..

The average voltage for these signals is zero (i.e. no DC bias), the segment that is illuminated has a signal with a RMS of 3v, the segment off signals have RMS 1v.

Why does a segment off signal have an amplitude, could it not just be fed with zero or is that impossible in an LCD multiplexing scheme?

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To allow continued operation without damage, every segment on an LCD panel must have an average of zero volts DC across it. Further, if one or more segments is active and one or more rows is active, there will be four kinds of segments. Assume the active row is driven with voltage R and the inactive rows with voltage I. The active segments are driven with voltage A and the inactive ones with voltage B. There will be four kinds of segments, driven with the following voltages

A-R: Active segment on active row
B-R: Inactive segment on active row
A-I: Active segment on inactive row
B-I: Inactive segment on inactive row

Ideally, one would like to have A-R be a nice big voltage while the other three differential voltages are zero, but that is mathematically impossible. Since the amount to which a display segment is darkened is a function of the RMS voltage (which doesn't care about polarity), having A be one rail, R and B be the opposite rail, and I be mid-rail (swap the R/B and A each frame to avoid DC offset), will mean that |A-R| will be full voltage, |B-R| will be zero, and both |A-I| and |B-I| will be half voltage. Note that |A-I| and |B-I| are equal--that means that the darkness of a segment on an inactive row will not be affected by whether the segment on the active row is active or not; if you see LCDs with "streaks" on the columns, it generally means that for some reason those two voltages are not quite equal.

If one has a 3:1 multiplex display with the above drive method, then active segments will spend 1/3 of the time with full voltage, and 2/3 of the time with 1/2 voltage. Their RMS drive will thus be sqrt(1/3 + 2/3*1/4) = sqrt(1/2). Inactive segments will spend 1/3 of the time with zero voltage and 2/3 with 1/2 voltage, so their RMS drive will be sqrt(2/3*1/4) = sqrt(1/6). The ratio of RMS voltage will thus be sqrt(3):1.

One may improve things slightly if on half the frames (reverse polarity on the other half) one sets R to be VDD, I to be 1/3 VDD, A to be zero, and B to be 2/3 VDD. In that case, |A-R| will be VDD, |B-R| will be 1/3 VDD, and both |A-I| and |B-I| will also be 1/3 VDD. Active segments will spend 1/3 of the time with full voltage and 2/3 with 1/3 voltage, for an RMS drive of sqrt(1/3 + 2/3*1/9) = sqrt(11/27). Inactive segments will spend their entire time at 1/3 voltage, for an rms drive of sqrt(1/9) [i.e. sqrt(3/27)]. The ratio of RMS voltage will thus be sqrt(11/3):1.

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  • \$\begingroup\$ Wow! That is a great answer! Thanks for spending the time to write that out. \$\endgroup\$ – NivagSwerdna Feb 10 '13 at 13:20
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Your last guess is correct, it isn't possible to have zero volts across inactive segments in a multiplexed LCD.

The multiplexing takes advantage of the nonlinearity of the liquid crystal response to voltage. As long as the AC voltage remains below a certain threshold, the segment remains inactive, and if the voltage is pushed above that threshold, it becomes active.

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  • \$\begingroup\$ For normal LCD materials, RMS is what's relevant, rather than peak-to-peak voltage. Average DC voltage must be zero, but RMS voltage is what determins contrast. It's possible some super-twist materials may have different behavior, though I have no reason to believe that to be the case. \$\endgroup\$ – supercat Feb 10 '13 at 18:41
  • \$\begingroup\$ @supercat: OK, I'll take your word for it. \$\endgroup\$ – Dave Tweed Feb 10 '13 at 23:21

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