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Recently, I started to design a LC (Liquid Crystal) shutter glass. The question is which component should I use. This is the LC shutter I might use: PolarSpeed®-S

However, there are a few more shutters to test so this shutter is not the only option. Anyway, we need fast switching response (below 100us) and to do that overdriving is essential.

Coming back to the question, this is the usual way to implement overdrive:

basic OVerdrive

you see, applying higher voltage for enough time of period then returning to nominal voltage level makes LC switching faster.

And this is the overdriving method of PolarSpeed®-S:

PolarSpeed®-S Overdrive method

PolarSpeed®-S uses two cells and this way it achieves below 50 uS rise and fall time.

So I was thinking there must be several ways to apply those voltage levels to LC cells.

1- An analog switch + Linear regulator(3 outputs)

schematic

simulate this circuit – Schematic created using CircuitLab

I couldn't find a deMUX in schematic so think those switches as 1:3 deMUX, everything is digital.

the other options are similar also:

2- Several Linear Transistor Driver

3- Several Voltage references + Swicthes

Or, can I:

4- Just use a few Op-Amps + resistors + switch

schematic

simulate this circuit

4- is the easiest and most flexible one because I already have 28V supply and all I will need are 3 Op-Amps, a voltage conterter to < -15V and switch. Also, this might be a good fit because I am allowed to change the gain of the opAmps so I can change the applied voltage to cells via changing resistors that way I can use the same pcb to test several shutters.

But I can not think of anything about pros and cons of those 4 options. So would you help me through this? What is the finest way for LCD overdriving? There may be other options I missed so if you know other options please share. Thanks per advance.

edit: a new idea by ChrisR is added below:

schematic

simulate this circuit

in the schematic GD(Gate driver) supplies are different in tme mean of overdrive purpose. GD1 supply is 14V - 0V, GD2 supply is 8V - 0V and GD3 is 14V - 8V. And just to simplify, diodes forward voltages are 1 V. So if I use this configuration, output will be like:

PWM and output

this I think, but I am not sure. Is this configuration safe to use?

edit(2): I decided to go with gate driver IX4428N and boost reg MIC2288YD5-TR.

schematic

simulate this circuit

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    \$\begingroup\$ It took me a while to realize that LC is meant to mean "liguid crystal", not inductor-capacitor. -1 for wasting people's time by saving yourself a few milliseconds of spelling out "liquid crystal" at least once. \$\endgroup\$ – Olin Lathrop Jan 16 '16 at 14:46
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    \$\begingroup\$ I apologize, I really didn't think that people may confuse LC with filter LC. Now I fixed the question, forgive me for your time. \$\endgroup\$ – Alper91 Jan 16 '16 at 15:03
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    \$\begingroup\$ @Alperözel - yes, if you have both 12V and 24 V available. Then put N-1 and N-12 in driver 1 and feed it 12V. Then put N-11 and N-13 in driver 2 and feed it 24V. You now have very simple logic to generate 0,12 or 24V on the shutter cells. If you use an ARM processor, the timers can be set to provide the correct PWM at the correct frame rate as well. \$\endgroup\$ – ChrisR Jan 25 '16 at 19:29
  • \$\begingroup\$ Thank you. I will do that way and let everyone know how it works... \$\endgroup\$ – Alper91 Jan 25 '16 at 19:49
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Try any Mosfet driver e.g. TC4420/29 from Microchip.

They switch in 25nS, will work at 4.5 to 20 or so volts, are small, cheap and designed for high speed switching of capacitive loads such as this shutter.

Some e.g. TC4427 are available in dual package with complimentary outputs in cases where you want a 10V swing over floating inputs (e.g. a bare LCD panel).

The specs, circuits and graphs all call for what is a digital drive. An analog solution such as an opamp will have difficulty generating the fast rise times (slew rates) you desire as they are not designed for fast switching.

For belt and braces, you can add a dual mosfet to the driver, but for he load you have, there is no need for extra circuitry, though these chips are at their limit near the 24 volt levels.

In order to achieve intermediate voltages, simply put a small capacitor across the output and feed the switch with a PWM. To boost current, add a dual FET to the output, or go to a 'high and low side' driver such as IRL2110.

The purpose of the 24 volt drive, from what I can see is to discharge the energy stored in the LC quickly, this can be done by reversing the 'sense' of the two outputs of a bridge configuration per the SE post below. You will not need the diodes referred to since your load appears to be capacitive.

To reverse sense, you would toggle the outputs i.e. whichever pin was high, would go low, and the pin that was lo would go high.

The beauty of the bridge configuration is that the load device sees an effective double supply voltage at the instant it is toggled. You could probably drive the entire device from a 12 volt source a the same speed as a 24 volt source without the danger of damaging it,since the effective 24V differntial is soon dissipated to the 12 volt capability of the supply.

To bring back to 12V, you would leave one driver driven low, and the other driven at 50%. 6V would be 25% etc. Measuring the voltage with an ADC would provide accurate control.

Please speak to the vendor / OEM for the shutter - I am sure they can advise on these SE solutions, or perhaps a better way.

schematic

simulate this circuit – Schematic created using CircuitLab

When Clk1 == Clk2 there are zero volts across shutter1 When CLK1 != Clk2 There are 12 volts across shutter1

at the instant you change Clk1 and Clk2 simultaneously, Shutter1 sees an effective -24V dropping to -12V (i.e. 12 volts in the opposite sense to what it was). This is effectively what happens in the graph on on page 7.

By changing both clocks simultaneously, then toggling one back to what it was, you can drop the shutter from 12V to 0V almost as fast as if you applied 24V per the manual. The speed of TC4420 is >> shutter speed (25nS vs 100us section 9 pg 7)

The low impedance of TC4420 is about 7ohms, if this is too large to switch the shutter fast enough, then add a dual low Rds Mosfet (e.g. SSD2025TF)to the output per the spec sheet for the TC4420.

I posted the question which you may find relevant

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  • \$\begingroup\$ Excuse me for my delay to respond, I had a very busy week. So I see, your point opamps are not good fits for the app, but, you saw the overdrive tchnique above, for a little period of time, I need to switch (lets say) 24V and then need to down 12V to do not harm the cell. For closing, for a short time of period -12V then pull up to GND. This is sorta what I will do. So, I will still need a swictch to apply different levels of voltage rgiht? \$\endgroup\$ – Alper91 Jan 23 '16 at 8:55
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    \$\begingroup\$ @Alperözel - Check with the OEM, do they have an appnote simulate the above circuit. From what I understand, these shutters act as capacitors. Better use High and Low side drivers in half bridge e.g. IR2110 with appropriate low Rds Mosfets to get the fastest switching. Get 12/24V either by PWM or separate drivers and power supplies. The point is these drivers are designed for fast switching, opamps for accurate slewing. \$\endgroup\$ – ChrisR Jan 25 '16 at 7:48
  • \$\begingroup\$ I think I will use IX4428N and up to 24 V adjustable boost Reg MIC2288YD5-TR. Each input will be controlled(each 4 pins of the shutter). Could you have a look at the last edit I will make in 5 mins? \$\endgroup\$ – Alper91 Jan 25 '16 at 18:29
  • \$\begingroup\$ Rather than fiddling with boost reg, have a look at Meanwell power supplies. They have a number of 12-24 volt booster modules, PSU with 5,12,24V outputs. I'm sure there are other manufacturers. digikey.com/product-search/en?keywords=meanwell \$\endgroup\$ – ChrisR Jan 25 '16 at 18:40
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    \$\begingroup\$ @Alperözel, no, just if space is not at a premium, it is quicker and less hassle to buy a working module :-) \$\endgroup\$ – ChrisR Jan 25 '16 at 20:03
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ad 1: it is completely wrong, you do always get the +5V if any of the switches is activated.

My proposal: if the intervals can be determined as an even number of base time, for example: t1 equals to (t4-t3) and (t2-t1), (t3-t2), (t5-t4), (t6-t5) are N * t1, then you could use a synchrnous counter made of flip-flops and use those analog swithes to switch +Vd, -Vd.

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  • \$\begingroup\$ I didn't quite understand. Where does +5V come from? Also, I can't use fpga, I need ultra low power consumption so I use ti series. \$\endgroup\$ – Alper91 Jan 16 '16 at 15:17
  • \$\begingroup\$ @Alperözel 5V comes from your schematics, from linear voltage regulator 7805 you depicted. What are ti series? If you started making ultra low power consumption with 7805 powered to 28V dc supply, then I have to tell you that the most power hungry FPGA consumes nothing compared to the power loss at 7805 regulator. \$\endgroup\$ – Marko Buršič Jan 16 '16 at 16:24
  • \$\begingroup\$ ohh sorry, it was just representation, I was trying to make a point. But you are right, LDO comsumes so much, so SMPS maybe? Ti MSP430 family is ultra low power I ll use MSP430FR5xx. \$\endgroup\$ – Alper91 Jan 17 '16 at 8:44

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