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I have read this document: AN658 - LCD Fundamentals and the LCD Driver Module of 8-Bit PIC® Microcontrollers

I understand that the main concern is to maintain 0 VDC bias on every pixel when driving an LCD display.

I do not understand why there are multiple bias voltages required to drive backplanes (= common nodes.)

Why are such crazy, unintuitive waveforms required? enter image description here

Why I can't use just static biasing?
For example when I want to turn 1 pixel on:
1st frame would be BP0=1 (common), SEG0=0, the 2nd frame would be BP0=0 (common), SEG0=1. This would make 0 VDC bias, and the pixel would have the maximal contrast.

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    \$\begingroup\$ A specific difference charge voltage is necessary for a specific LCD video level. Erase charge sequence reduces memory effects. \$\endgroup\$ May 22, 2017 at 8:15
  • \$\begingroup\$ @TonyStewart.EEsince'75 could you be more explanatory? I don't understand your comment. If specific voltage is required, then why not to use that specific single voltage level (static biasing), instead of 3, 4, or 5 different levels? \$\endgroup\$ May 22, 2017 at 20:13
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    \$\begingroup\$ charge storage in LCD is critically dependent on voltage levels for white and black levels as well as other effects such as polarized viewing angle. Until you understand this, I cannot explain. \$\endgroup\$ May 23, 2017 at 17:54

3 Answers 3

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The voltages between BP and SEG for each segment need to fulfill two conditions:

  • The average voltage over the duration of the frame must be 0 (= no DC)
  • The RMS of the voltage must be high when we want the segment to be on and low when we want it to be off

So let's see how we can achieve that.

Let's assume we have a simple 4-segment display with two COM and two SEG connections, like this, taken from TI SLAA654A, page 6:

screenshot

With "V1" being 1 V and "V5" being 0 V, this Excel sheet calculates the RMS and DC offset for a full frame:

screenshot

The "Sum" cells are green when the DC is 0 and the "RMS" cells have a color scale depending on the RMS.

With formulas enabled, this is the result:

screenshot

Now let's try your proposal of using only 0 and 1 V:

screenshot

As you can see, the goal cannot be achieved. The segments BP0-SEG1 and BP1-SEG0 both have a DC offset and they are somewhat switched on.

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  • \$\begingroup\$ Thanks for very good explanation. I didn't know that there can be RMS voltage even on off segments. Is there any threshold for this RMS when segments are completely off? \$\endgroup\$ Jul 9, 2021 at 5:04
  • \$\begingroup\$ @Chupacabras Segments can be partially on, so there's no discrete threshold. \$\endgroup\$
    – AndreKR
    Jul 9, 2021 at 10:36
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Voltage at particular segment is a difference between voltage on two electrodes, lets call them row and column. There are 4 (four!!) cases:

  • Active segment in active row: |Vactivecol - Vactiverow| = Vmax.
  • Nonactive segment in active row |Vactiverow - Vnotcol| = 1/3 Vmax.
  • Nonactive segment in nonactive row while this segment column electrode is active |Vactivecol - Vnotrov| = 1/3 Vmax.
  • Nonactive segment in nonactive row |Vnotcol - Vnotrow| = 1/3 Vmax.
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Assuming you have your LCD with two commons (rows), C1 and C2, and two segments (columns) S1 and S2, which connect together in a matrix to give 4 displayable dots, C1S1, C1S2, C2S1, and C2S2. Assume you want to turn C1S1 on, while having the other three off, using only two voltage levels, call them 0 and 1. Obviously, you will have to apply 0 level to C1, and 1 to S1, otherwise C1S1 won't turn on. Obviously, you will have in this case to apply 1 on C2 (otherwise C2S1 will turn on too, as S1 is 1, and you don't want that).

By the same logic, you must apply 0 to S2 (because C1 is 0 and you don't want C1S2 to be on). Now, what does C2S2 dot does? Remark that its potential is now -1, as C2 has a 1 on it, while S2 has a 0, so it will turn on. That is why you can't use only two voltage levels in this case.

When playing with a LED matrix (or ROM memory by the way) you have the "advantage" that the current flows to the diodes in nodes in only one direction, to turn them on. For LCDs, only the voltage matter, not the direction, and the dots will turn on if you give them the right voltage potential, regardless of the direction.

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