Patent Publication Number: US-8111228-B2

Title: Method and device to optimize power consumption in liquid crystal display

Description:
FIELD OF INVENTION 
     Liquid Crystal Displays (LCDs) consume less power as compared to other display devices. They are also flat panel devices with negligible depth. LCDs can be operated with low voltage power sources and hence are extensively used in portable products. Although liquid crystal displays consume less power, it is desirable to reduce the power consumption further so that the frequency of replacing or charging the cells in portable equipment is reduced. We have used the multi-step waveform profile to reduce power consumption in liquid crystal displays. 
     BACKGROUND OF INVENTION 
     Marks [1] has shown that the power consumption of non-multiplexed displays can be reduced to about 50% when the charge across the pixels is discharged by shorting the two electrodes for a short time interval before charging the pixels to a voltage with opposite polarity. He has also analyzed and estimated the power consumption in multiplexed LCDs [2], based on the model of the matrix display shown in  FIG. 1 . Each pixel in the LCD is represented as a capacitor. One electrode of the pixel is connected to a row address line and the other electrode is connected to a column address line. Capacitance of pixel depends on the state of the pixel since the effective dielectric constant is determined by the orientation of the liquid crystal molecules. The capacitance of the ON pixel C on  is at least twice the capacitance of the OFF pixel C off  in nematic liquid crystal displays because the dielectric constant of the rod-like liquid crystal molecules is higher when measured parallel to its long axis as compared to the other two perpendicular directions. 
     Display drivers are used to apply the waveforms to the rows and columns of the matrix display. Power consumption of the panel is the power dissipated in the resistors while charging and discharging the pixels to voltages as dictated by the addressing technique. Marks [2] has estimated the power consumption in a matrix display driven by the conventional line-by-line addressing when the worst-case pattern that consists of alternate ON and OFF pixels is displayed. He has shown that the power consumed by the multiplexed display is proportional to N 2 M. Here, N is the number of lines multiplexed and M is the number of columns in the matrix display. This analysis is restricted to just one polarity inversion per frame. Frequent polarity reversal is introduced in the addressing waveforms to improve the brightness uniformity of the display. It induces transitions in places where there were no transitions and suppresses transitions in some other places. Polarity of the addressing waveforms is changed after scanning few address lines in most of the passive matrix LCDs. We have extended the analysis of power consumption in the line-by-line addressing technique by including polarity inversion as an additional parameter. 
     Power is dissipated in the drive circuit when pixels in the passive matrix displays are charged and discharged. Substituting the select pulses in the scanning waveforms with multi-step waveforms will reduce the power dissipation. The rows in the matrix displays are selected with a pulse because they are easy to generate. 
     Analysis of Line-by-Line Addressing with Multiple Polarity Inversions in a Frame 
     Let V r  and V c  be the amplitudes of the row and column voltages. Let n p  be the number of polarity inversions in a frame (Marks had assumed n p =1 in his analysis) and f be the frame frequency of the line-by-line addressing. Let C on  and C off  be the capacitance of the pixels in ON and OFF states respectively. Table I gives the voltage transition across the pixels based on the two neighboring pixels in a column when the row (i) is unselected and the row (i+1) is selected. These transitions depend on the state of the pixels in rows (i) and (i+1) as well as the polarity inversion. The voltage transitions when a polarity inversion is introduced are shown within the parentheses. 
     
       
         
           
               
             
               
                 TABLE I 
               
             
            
               
                   
               
               
                 VOLTAGE TRANSITIONS ACROSS PIXELS 
               
               
                 IN LINE-BY-LINE ADDRESSING 
               
            
           
           
               
               
               
            
               
                   
                 State of the 
                 Voltage swing across 
               
               
                   
                 pixel in 
                 the pixels in 
               
            
           
           
               
               
               
               
               
            
               
                 Row 
                 row 
                   
                 row 
                 other 
               
               
                 (i) 
                 (i + 1) 
                 row (i) 
                 (i + 1) 
                 rows 
               
               
                   
               
               
                 ON 
                 ON 
                 V r   
                 V r   
                 0 
               
               
                   
                   
                 (V r  + 2 V c ) 
                 (V r  + 2 V c ) 
                 (2 V c ) 
               
               
                 ON 
                 OFF 
                 V r  + 2 V c   
                 V r  − 2 V c   
                 2 V c   
               
               
                   
                   
                 (V r ) 
                 (V r ) 
                 (0) 
               
               
                 OFF 
                 ON 
                 V r  − 2 V c   
                 V r  + 2 V c   
                 2 V c   
               
               
                   
                   
                 (V r ) 
                 (V r ) 
                 (0) 
               
               
                 OFF 
                 OFF 
                 V r   
                 V r   
                 0 
               
               
                   
                   
                 (V r  − 2 V c ) 
                 (V r  − 2 V c ) 
                 (2 V c ) 
               
               
                   
               
            
           
         
       
     
     Case 1: Power consumed by a blank screen when all the pixels are OFF. Power consumed in a column during a transition i.e., when the row (i+1) is selected and polarity of the voltages applied to the two rows remains unchanged is as follows. 
     
       
         
           
             
               
                 
                   
                     P 
                     
                       tran 
                       . 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             C 
                             off 
                           
                           ⁢ 
                           
                             V 
                             r 
                             2 
                           
                         
                         2 
                       
                       + 
                       
                         
                           
                             C 
                             off 
                           
                           ⁢ 
                           
                             V 
                             r 
                             2 
                           
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             N 
                             - 
                             2 
                           
                           ) 
                         
                         ⁢ 
                         
                           
                             C 
                             off 
                           
                           ⁡ 
                           
                             ( 
                             0 
                             ) 
                           
                         
                       
                     
                     = 
                     
                       
                         C 
                         off 
                       
                       ⁢ 
                       
                         V 
                         r 
                         
                           2 
                           ⁢ 
                           
                               
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
     The first term corresponds to the power dissipated while discharging the pixel in row (i) from V r −V c  to −V c  and the second term corresponds to the charging the pixels in row (i+1) from −V c  to V r −V c  while the third term corresponds to the rest of the (N−2) pixels in a column without any change in the voltage across them. Similarly, the power dissipated when the polarity of the select voltage changes is given in (2). 
     
       
         
           
             
               
                 
                   
                     P 
                     
                       tran 
                       . 
                     
                     ′ 
                   
                   = 
                   
                     
                       
                         
                           
                             C 
                             off 
                           
                           ⁡ 
                           
                             ( 
                             
                               
                                 V 
                                 r 
                               
                               - 
                               
                                 2 
                                 ⁢ 
                                 
                                   V 
                                   c 
                                 
                               
                             
                             ) 
                           
                         
                         2 
                       
                       2 
                     
                     + 
                     
                       
                         
                           
                             C 
                             off 
                           
                           ⁡ 
                           
                             ( 
                             
                               
                                 V 
                                 r 
                               
                               - 
                               
                                 2 
                                 ⁢ 
                                 
                                   V 
                                   c 
                                 
                               
                             
                             ) 
                           
                         
                         2 
                       
                       2 
                     
                     + 
                     
                       
                         ( 
                         
                           N 
                           - 
                           2 
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           
                             
                               C 
                               off 
                             
                             ⁡ 
                             
                               ( 
                               
                                 2 
                                 ⁢ 
                                 
                                   V 
                                   c 
                                   
                                       
                                   
                                 
                               
                               ) 
                             
                           
                           2 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
     Power consumption in a column during a frame is given by
 
 P   column (frame)=( N−n   p ) P   tran.   +n   p   P′   tran.   (3)
 
     Power consumed by the whole display panel is obtained by multiplying (3) by M, the number of columns in the display and f, the frame frequency as shown in the following equation.
 
 P   ALL     —     OFF   =MC   off   V   c   2 ( N   2   +n   p (2 N− 4√{square root over ( N )})) f   (4)
 
     Case 2: Power consumed by a blank screen, when all the pixels are ON is given in the following expression.
 
 P   ALL     —     ON   =MC   on   V   c   2 ( N   2   +n   p (2 N+ 4√{square root over ( N )})) f   (5)
 
     Case 3: Power consumed when a checkerboard pattern is displayed is given in (6). Here, the number of pixels in ON and OFF states are equal and the neighboring pixels in the vertical as well as the horizontal direction are in the opposite states. 
     
       
         
           
             
               
                 
                   
                     P 
                     
                       ON 
                       ⁢ 
                       _ 
                       ⁢ 
                       OFF 
                     
                   
                   = 
                   
                     
                       
                         
                           MV 
                           c 
                           2 
                         
                         2 
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             
                               
                                 
                                   
                                     C 
                                     on 
                                   
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         3 
                                         ⁢ 
                                         
                                           N 
                                           2 
                                         
                                       
                                       + 
                                       
                                         4 
                                         ⁢ 
                                         N 
                                         ⁢ 
                                         
                                           N 
                                         
                                       
                                       - 
                                       
                                         
                                           n 
                                           p 
                                         
                                         ⁡ 
                                         
                                           ( 
                                           
                                             
                                               2 
                                               ⁢ 
                                               N 
                                             
                                             + 
                                             
                                               4 
                                               ⁢ 
                                               
                                                 N 
                                               
                                             
                                           
                                           ) 
                                         
                                       
                                     
                                     ) 
                                   
                                 
                                 + 
                               
                             
                           
                           
                             
                               
                                 
                                   C 
                                   off 
                                 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     
                                       3 
                                       ⁢ 
                                       
                                         N 
                                         2 
                                       
                                     
                                     - 
                                     
                                       4 
                                       ⁢ 
                                       N 
                                       ⁢ 
                                       
                                         N 
                                       
                                     
                                     - 
                                     
                                       
                                         n 
                                         p 
                                       
                                       ⁡ 
                                       
                                         ( 
                                         
                                           
                                             2 
                                             ⁢ 
                                             N 
                                           
                                           - 
                                           
                                             4 
                                             ⁢ 
                                             
                                               N 
                                             
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                         ] 
                       
                     
                     ⁢ 
                     f 
                   
                 
               
               
                 
                   ( 
                   6 
                   ) 
                 
               
             
           
         
       
     
     We have also introduced duty cycle in the pulses of the line-by-line addressing technique. Power consumption after the inclusion of duty cycle is analyzed and compared in the next section. 
     REFERENCES 
     
         
         [1] Burton. W. Marks, “Power reduction in liquid crystal display modules”,  IEEE Trans. Electron Devices , Vol. ED-29, No. 12, pp. 1884-1886, 1982. 
         [2] Burton. W. Marks, “Power consumption of multiplexed liquid crystal displays”,  IEEE Trans. Electron Devices , Vol. ED-29, No. 8, pp. 1218-1222, 1982. 
         [3] T. N. Ruckmongathan, M. Govind and G. Deepak, “Reducing power consumption in liquid crystal displays”  IEEE Trans. Electron Devices , submitted for publication. 
         [4] T. N. Ruckmongathan, Techniques for Reducing the Hardware Complexity and the Power Consumption of Drive Electronics, Proceedings of the Asian Symposium on Information Display (ASID&#39;06), Oct. 8-12, 2006, pp 115-120. 
       
    
    
    
     
       BRIEF DESCRIPTION OF ACCOMPANYING DRAWINGS 
         FIG. 1 : shows a model of matrix display used for computing power consumption. 
         FIG. 2 : shows plots of normalized power consumption in matrix displays driven by line-by-line addressing with duty cycle control and the conventional line-by line addressing without duty cycle control and n p  polarity inversions in a frame. 
         FIG. 3 : shows plot of normalized power consumption as a function of n p , the number of polarity inversions in a frame (N=100). These plots are normalized to power consumed when the screen is blank and n p  is one. 
         FIG. 4 : shows a multi-step voltage profile that will replace a pulse of amplitude Vp in the addressing techniques. 
         FIG. 5 : shows plot of the percentage increase in maximum amplitude of the multi-step voltage profile for several values of ‘s’, the number of steps in a multi-step voltage profile when T 0 =T s , T f ≧2T s  and T s ≧0.01T. 
         FIG. 6 : shows plot of ratio of power consumption for several values of s, the number of steps in the multi-step voltage profile. Power consumption is normalized to that of a pulse having a duty cycle (T 0 =T s ) with T f ≧2T s  and T s ≧0.01T. 
         FIG. 7 : shows typical waveforms when multi-steps are introduced in line-by-line addressing 
         FIG. 8 : shows typical waveforms when multi-steps are introduced in static drive 
         FIG. 9 : Multi-step waveform profile with RC distortion and s=4. Ideal waveform is shown in dotted lines and it is held at zero during T L    
         FIG. 10 : Power dissipation (normalized to that of a single pulse) vs. normalized time constant for different values of S; T L =0.1T and T f =2T s    
         FIG. 11 : Peak voltage (normalized to that of a single pulse) vs. normalized time constant for different values of S; T L =0.1T and T f =2T s    
         FIG. 12 : Magnitude of voltage to be applied during T L  (normalized to the corrected peak voltage) as a function of normalized time constant; T L =0.1T and T f =2T s    
         FIG. 13 : Shows a system block diagram with two Voltage Level Generators (VLG) to provide voltages to LCD drivers along with multiplexer. 
     
    
    
     OBJECTS OF THE INVENTION 
     The primary objective of the invention is to develop a method to optimize power and improve brightness uniformity of pixels consumption in Liquid Crystal Display. 
     Another objective of the invention is applying multi-step waveform for selecting pre-determined address lines. 
     Still another objective of the present invention is maintaining ratio of step-width (T s ) and pulse width (T) between 0.02 to 0.25. 
     Still another objective of the present invention is making final step duration (T f ) greater than or equal to twice the step width (T s ) to optimize the supply voltage of the driver circuit of the Liquid Crystal Display. 
     STATEMENT OF INVENTION 
     The present invention is related to a method to optimize power consumption and improve brightness uniformity of pixels in Liquid Crystal Display, wherein said method comprises steps of: applying a multi-step waveform for selecting pre-determined address lines of the LCD, wherein said multi-step waveform comprises ascending steps and descending steps maintaining ratio of step-width (T s ) and pulse width (T) between 0.02 to 0.25; making top most step duration (T f ) greater than or equal to twice the step width (T s ); and applying a negative voltage to make voltage across the pixels zero at the end of multi-step time interval to optimize power supply voltage in Liquid Crystal Display; a device to optimize power consumption and improve brightness uniformity of pixels in Liquid Crystal Display, wherein said device comprises: voltage level generator (VLG) to provide voltages to the drivers, and analog multiplexer (number of steps: 1) with variable resistor (Rs) connected to the VLG wherein Rs is dictated by multi-step waveform to determine the voltage applied to the LCD; 
     DETAILED DESCRIPTION OF THE INVENTION 
     The primary embodiment of the invention is a method to optimize (reduce) power consumption and improve brightness uniformity of pixels in Liquid Crystal Display, wherein said method comprises steps of applying a multi-step waveform for selecting pre-determined address lines of the LCD, wherein said multi-step waveform comprises ascending steps and descending steps; maintaining ratio of step-width (T s ) and pulse width (T) between 0.02 to 0.25; making top most step duration (T f ) greater than or equal to twice the step width (T s ); and applying a negative voltage to make voltage across the pixels zero at the end of multi-step time interval to optimize power supply voltage in Liquid Crystal Display. 
     In yet another embodiment of the present invention the number of steps is ranging from 2 to 16. 
     In still another embodiment of the present invention, the step-width (Ts) is not same for all steps. 
     In still another embodiment of the present invention applying a voltage of opposite polarity to that of peak voltage as the last step (T L ) of multi-step waveform to bring the multi-step waveform to zero at the end of the period ‘T’. 
     In still another embodiment of the present invention, amplitude of the voltage is less than the peak voltage of the multi-step waveform profile. 
     In still another embodiment of the present invention, amplitude of the multi-step waveform is varied to increase energy delivered to the pixels. 
     In still another embodiment of the present invention, amplitude of first step of the multi-step waveform is increased. 
     In still another embodiment of the present invention, amplitude of the entire steps is increased uniformly. 
     In still another embodiment of the present invention, amplitude of topmost step is increased. 
     Another main embodiment of the present invention is a device to optimize power consumption and improve brightness uniformity of pixels in Liquid Crystal Display, wherein said device comprises: voltage level generator (VLG) to provide voltages to the drivers, and analog multiplexer (number of steps: 1) with variable resistor (Rs) connected to the VLG wherein Rs is dictated by multi-step waveform to determine the voltage applied to the LCD as shown in the  FIG. 13 . 
     In yet another embodiment of the present invention, the VLG is generated by multiplexing 2 VLG&#39;s with a selection bit as shown in the  FIG. 13 . 
     In still another embodiment of the present invention the multiplexer that are common to drivers reduces number of voltages selected inside multi-stage drivers. 
     Line-by-Line Addressing with Duty Cycle Control 
     Power consumed by the display panel depends on the number of transitions in the voltage across the pixels and the magnitude of these transitions. Number of transitions in turn depends on the image being displayed and the addressing technique. Number of transitions in the addressing waveform can be made independent of the image if the voltage in the row and column waveforms is chosen to be the same for a fraction (T 0 ) of the row select time T. This introduces transitions in both row and column waveforms and the voltage across the pixel is zero during the interval T 0 . Amplitude of the row and column waveforms has to be increased by a factor 
               (     T     T   -     T   0         )           
to ensure that the rms voltage across pixel is same as that of the conventional line-by-line addressing technique. This technique will be referred to as line-by-line addressing with duty cycle control. Introduction of duty cycle has the advantage of good brightness uniformity of pixels [4]. Although the number of transitions is the same across all pixels, power consumption depends on the number of pixels in the ON and OFF states because the capacitance of the pixel depends on its state. Power consumption of the multiplexed display driven by line-by-line addressing when 50% of the pixels are driven to ON state is given in (7).
 
     
       
         
           
             
               
                 
                   
                     P 
                     
                       line 
                       ⁢ 
                       
                         - 
                       
                       ⁢ 
                       by 
                       ⁢ 
                       
                         - 
                       
                       ⁢ 
                       line 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       with 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       duty 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       cycle 
                     
                   
                   = 
                   
                     
                       MV 
                       c 
                       2 
                     
                     ⁢ 
                     
                       X 
                       ( 
                       
                         T 
                         
                           T 
                           - 
                           
                             T 
                             0 
                           
                         
                       
                       ) 
                     
                     ⁢ 
                     f 
                   
                 
               
               
                 
                   ( 
                   7 
                   ) 
                 
               
             
           
         
       
     
     Wherein
 
 X=└C   on ( N   2   +N√{square root over (N)} )+ C   off ( N   2   −N√{square root over (N)} )┘  (8)
 
     Power consumption (as a function of N) of a display driven by the line-by-line addressing with duty cycle is compared with that of a display driven by conventional line-by-line addressing in  FIG. 2 . This comparison is done when the three specific patterns (case I to III) are displayed. These plots are based on the assumption that C on =2C off  and T 0 =0.05T. Least amount of power is consumed when all the pixels in the display are OFF and the number of polarity reversals in a frame (n p ) is one. Hence, the power consumption is normalized to the minimum power in these plots. 
     Normalized power consumption as a function n p , the number of polarity inversions in a frame (when N, the number of lines multiplexed is 100) is plotted in  FIG. 3 . Power consumption while displaying blank patterns increases with n p  because the number of transitions in waveforms increases with the n p . In case of the checkerboard pattern, the power consumption decreases with n p  due to a decrease in number of transitions with increase in n p . Power consumption of the line-by-line addressing technique with duty cycle control is also shown in the  FIG. 2 . Power consumption of the line-by-line addressing with duty cycle control also depends on the image because the capacitances of the ON and OFF pixels are not equal although, the number of transitions in the addressing waveforms are equal. Introduction of duty cycle improves the brightness uniformity of pixels in the display because the waveforms across all the pixels are distorted to the same extent (since number of transitions are equal) and the reduction in RMS voltage across the pixels can be compensated by just increasing the peak amplitude of the pulses. We have also considered some typical images of size 100×100 pixels and we have estimated the power consumption when these images are displayed. The results are summarized in Table II. 
     
       
         
           
               
             
               
                 TABLE II 
               
             
            
               
                   
               
               
                 NORMALIZED POWER CONSUMPTION FOR SOME TEST IMAGES 
               
            
           
           
               
               
               
            
               
                   
                 Line-by-line 
                   
               
               
                   
                 addressing 
                 Line-by-line addressing with n p   
               
               
                   
                 with duty 
                 polarity reversals per frame 
               
               
                   
                 cycle control 
                 (without duty cycle) 
               
            
           
           
               
               
               
               
               
               
            
               
                 Image 
                 (T 0  = 0.05 T) 
                 n p  = 1 
                 n p  = 25 
                 n p  = 50 
                 n p  = 100 
               
               
                   
               
               
                 
                   
                 
                 3.3917 
                 2.3911 
                 2.7849 
                 3.2277 
                 4.0621 
               
               
                 
                   
                 
                 3.8911 
                 2.2777 
                 2.9791 
                 3.7014 
                 5.1211 
               
               
                 
                   
                 
                 2.9960 
                 1.5913 
                 2.2226 
                 2.8566 
                 4.1311 
               
               
                 
                   
                 
                 3.8370 
                 2.2219 
                 2.9283 
                 3.6480 
                 5.1037 
               
               
                 
                   
                 
                 3.3553 
                 1.8255 
                 2.4867 
                 3.2064 
                 4.5881 
               
               
                   
               
            
           
         
       
     
     Power consumptions shown in the Table II are normalized to that of the blank pattern with all the pixels in OFF state and having just one polarity reversal per frame. It is evident from the table that the power consumption of line-by-line addressing when the duty cycle control is introduced is about the same as that of line-line addressing wherein the polarity is reversed after scanning every two-address line. Analysis presented in this section implies that the introduction of duty cycle will not reduce the power consumption in multiplexed matrix LCDs although it is quite effective in reducing the power consumption in non-multiplexed displays. Principle of a new technique to reduce power consumption in multiplexed as well as non-multiplexed displays is presented next. 
     Principle of Reducing Power Consumption 
     It is well known that the power dissipated in the resistor while charging or discharging a capacitor in a RC circuit using a single step of amplitude V p  is given by 
                 CV   p   2     2     .         
Hence, the total power consumed during a charge-discharge cycle is CV p   2 . Now, if the capacitor is charged and discharged using two steps, each of amplitude
 
               V   p     2         
then the power consumption will be
 
               CV   p   2     2         
i.e. 50% of the power dissipated as compared to a single pulse of amplitude V p . Similarly, the power consumption can be reduced by a factor ‘s’ by introducing ‘s’ steps to charge a capacitor to a voltage V p  and using an equal number of steps to discharge it to ground potential.
 
     We propose to use the multi-step voltage profile shown in  FIG. 4  to reduce the power consumption in multiplexed as well as non-multiplexed displays. It has (s−1) ascending and descending steps of equal duration T s  while the final step with a maximum voltage of V lasts for the duration T f . Amplitude of the multi-step voltage profile is zero during the period T o . Step size of the ascending as well as descending steps are equal (V/s) and the total period is T=2(s−1)T s +T f +T 0 . This multi-step profile reduces to the single pulse of the conventional addressing technique when s=1 and T 0 =0 as shown in the  FIG. 4  using dashed lines. LCDs are slow responding devices and their response times are usually in milliseconds. The response depends on the energy delivered to the pixel and the actual wave shape is not important as long as the period of the waveform is small as compared to the response times. Hence, the RMS voltage across the pixel decides the state of the pixel. Peak amplitude (V) of the multi-step voltage profile that will deliver the same energy as a pulse of amplitude V p  and of duration T can be obtained equating the RMS voltage of the multi-step voltage profile to that of a pulse as shown in (9). 
     
       
         
           
             
               
                 
                   
                     
                       
                         1 
                         T 
                       
                       ⁢ 
                       
                         { 
                         
                           
                             2 
                             ⁢ 
                             
                               
                                 ∑ 
                                 
                                   i 
                                   = 
                                   1 
                                 
                                 
                                   s 
                                   - 
                                   1 
                                 
                               
                               ⁢ 
                               
                                 
                                   
                                     ( 
                                     
                                       
                                         V 
                                         s 
                                       
                                       · 
                                       ⅈ 
                                     
                                     ) 
                                   
                                   2 
                                 
                                 ⁢ 
                                 
                                   T 
                                   s 
                                 
                               
                             
                           
                           + 
                           
                             
                               V 
                               2 
                             
                             ⁢ 
                             
                               T 
                               f 
                             
                           
                         
                         } 
                       
                     
                   
                   = 
                   
                     V 
                     p 
                   
                 
               
               
                 
                   ( 
                   9 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         
                           V 
                           2 
                         
                         T 
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             
                               ( 
                               
                                 
                                   
                                     ( 
                                     
                                       s 
                                       - 
                                       1 
                                     
                                     ) 
                                   
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         2 
                                         ⁢ 
                                         s 
                                       
                                       - 
                                       1 
                                     
                                     ) 
                                   
                                 
                                 
                                   3 
                                   ⁢ 
                                   s 
                                 
                               
                               ) 
                             
                             ⁢ 
                             
                               T 
                               s 
                             
                           
                           + 
                           
                             ( 
                             
                               T 
                               - 
                               
                                 T 
                                 0 
                               
                               - 
                               
                                 2 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     s 
                                     - 
                                     1 
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   T 
                                   s 
                                 
                               
                             
                             ) 
                           
                         
                         ] 
                       
                     
                   
                   = 
                   
                     V 
                     p 
                   
                 
               
               
                 
                   ( 
                   10 
                   ) 
                 
               
             
           
         
       
     
     Hence the maximum amplitude of the multi-step profile is 
     
       
         
           
             
               
                 
                   V 
                   = 
                   
                     
                       
                         
                           T 
                           
                             T 
                             - 
                             
                               T 
                               0 
                             
                           
                         
                         ⁢ 
                         
                           ( 
                           
                             
                               3 
                               ⁢ 
                               
                                 s 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     T 
                                     - 
                                     
                                       T 
                                       0 
                                     
                                   
                                   ) 
                                 
                               
                             
                             
                               
                                 3 
                                 ⁢ 
                                 
                                   s 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       T 
                                       - 
                                       
                                         T 
                                         0 
                                       
                                     
                                     ) 
                                   
                                 
                               
                               - 
                               
                                 
                                   ( 
                                   
                                     
                                       4 
                                       ⁢ 
                                       s 
                                     
                                     + 
                                     1 
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     s 
                                     - 
                                     1 
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   T 
                                   s 
                                 
                               
                             
                           
                           ) 
                         
                       
                     
                     · 
                     
                       V 
                       p 
                     
                   
                 
               
               
                 
                   ( 
                   11 
                   ) 
                 
               
             
           
         
       
     
       FIG. 5  shows the peak amplitude V of the multi-step voltage profile as a function of the duration T s , normalized to T for several values of ‘s’, the number of steps. The peak voltage increases with T s  and it is preferable to choose a small value for T s . The peak voltage decreases and approaches the magnitude of a single pulse in the conventional line-by-line addressing as T f , the duration of the maximum voltage is increased. The power dissipated in the resistor while charging and discharging a pixel (capacitor) by applying the waveform shown in  FIG. 4  is given in the following expression. 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           P 
                           
                             multi 
                             ⁢ 
                             
                               - 
                             
                             ⁢ 
                             step 
                           
                         
                         = 
                         
                           
                             sC 
                             ( 
                             
                               V 
                               s 
                             
                             ) 
                           
                           2 
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           C 
                           ⁢ 
                           
                             T 
                             
                               T 
                               - 
                               
                                 T 
                                 0 
                               
                             
                           
                           ⁢ 
                           
                             ( 
                             
                               
                                 3 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     T 
                                     - 
                                     
                                       T 
                                       0 
                                     
                                   
                                   ) 
                                 
                               
                               
                                 
                                   3 
                                   ⁢ 
                                   
                                     s 
                                     ⁡ 
                                     
                                       ( 
                                       
                                         T 
                                         - 
                                         
                                           T 
                                           0 
                                         
                                       
                                       ) 
                                     
                                   
                                 
                                 - 
                                 
                                   
                                     ( 
                                     
                                       
                                         4 
                                         ⁢ 
                                         s 
                                       
                                       + 
                                       1 
                                     
                                     ) 
                                   
                                   ⁢ 
                                   
                                     ( 
                                     
                                       s 
                                       - 
                                       1 
                                     
                                     ) 
                                   
                                   ⁢ 
                                   
                                     T 
                                     s 
                                   
                                 
                               
                             
                             ) 
                           
                           ⁢ 
                           
                             V 
                             p 
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   12 
                   ) 
                 
               
             
           
         
       
     
     Power consumed when a pixel is charged and discharged with a pulse of duration (T−T 0 ) is given in (13). 
     
       
         
           
             
               
                 
                   
                     P 
                     pulse 
                   
                   = 
                   
                     
                       C 
                       ( 
                       
                         T 
                         
                           T 
                           - 
                           
                             T 
                             0 
                           
                         
                       
                       ) 
                     
                     ⁢ 
                     
                       V 
                       p 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   13 
                   ) 
                 
               
             
           
         
       
     
     Reduction in power consumption while using the multi-step voltage profile as compared to that of a pulse is given in the following expression. 
     
       
         
           
             
               
                 
                   
                     
                       P 
                       
                         multi 
                         ⁢ 
                         
                           - 
                         
                         ⁢ 
                         step 
                       
                     
                     
                       P 
                       pulse 
                     
                   
                   = 
                   
                     ( 
                     
                       
                         3 
                         ⁢ 
                         
                           ( 
                           
                             T 
                             - 
                             
                               T 
                               0 
                             
                           
                           ) 
                         
                       
                       
                         
                           3 
                           ⁢ 
                           
                             s 
                             ⁡ 
                             
                               ( 
                               
                                 T 
                                 - 
                                 
                                   T 
                                   0 
                                 
                               
                               ) 
                             
                           
                         
                         - 
                         
                           
                             ( 
                             
                               
                                 4 
                                 ⁢ 
                                 s 
                               
                               + 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             ( 
                             
                               s 
                               - 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             T 
                             s 
                           
                         
                       
                     
                     ) 
                   
                 
               
               
                 
                   ( 
                   14 
                   ) 
                 
               
             
           
         
       
     
     This ratio of power consumption as a function the step width T s  normalized to the select time T is shown in  FIG. 6  for several values of ‘s’, the number of steps in the multi-step voltage profile. 
     Power consumption decreases with the increase in the number of steps (s). A good reduction in power consumption can be achieved with just two steps as long as the duration T s  is small. A large value of T s  decreases T f  (the duration of the maximum voltage V) while increasing the amplitude of V and this is not favorable for reducing the power consumption. A line-by-line addressing technique incorporating the multi-step voltage profile is proposed in the following section to reduce the power consumption in passive matrix LCDs. 
     Line by Line Addressing with Multi-Step Waveforms 
     Let us consider the conventional line-by-line addressing technique and replace the pulses in the row and column waveforms with the multi-step profiles as shown in  FIG. 7 . Let V x  and V y  be the maximum amplitudes of row and column voltages. The RMS voltage across a pixel is as follows. 
     
       
         
           
             
               
                 
                   
                     V 
                     RMS 
                   
                   = 
                   
                     
                       
                         1 
                         NT 
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             E 
                             select 
                           
                           + 
                           
                             
                               ( 
                               
                                 N 
                                 - 
                                 1 
                               
                               ) 
                             
                             ⁢ 
                             
                               E 
                               
                                 non 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 select 
                               
                             
                           
                         
                         ] 
                       
                     
                   
                 
               
               
                 
                   ( 
                   15 
                   ) 
                 
               
             
           
         
       
     
     Here, E select  is the energy delivered to a pixel during the select interval T. Energy delivered to the pixel during the rest of the (N−1) row select intervals when the other rows in the matrix are selected is given by E non select . 
     
       
         
           
             
               
                 
                   
                     E 
                     select 
                   
                   = 
                   
                     
                       2 
                       ⁢ 
                       
                         
                           ∑ 
                           
                             i 
                             = 
                             1 
                           
                           
                             s 
                             - 
                             1 
                           
                         
                         ⁢ 
                         
                           
                             
                               ( 
                               
                                 
                                   
                                     ( 
                                     
                                       
                                         V 
                                         x 
                                       
                                       ± 
                                       
                                         V 
                                         y 
                                       
                                     
                                     ) 
                                   
                                   s 
                                 
                                 ⁢ 
                                 i 
                               
                               ) 
                             
                             2 
                           
                           ⁢ 
                           
                             T 
                             s 
                           
                         
                       
                     
                     + 
                     
                       
                         
                           ( 
                           
                             
                               V 
                               x 
                             
                             ± 
                             
                               V 
                               y 
                             
                           
                           ) 
                         
                         2 
                       
                       ⁢ 
                       
                         T 
                         f 
                       
                     
                   
                 
               
               
                 
                   ( 
                   16 
                   ) 
                 
               
             
           
         
       
     
     The instantaneous voltage across an ON pixel is (V x +V y ), while it is (V x −V y ) across an OFF pixel. This is shown by the symbol ‘±’ in (16). 
     
       
         
           
             
               
                 
                   
                       
                   
                   ⁢ 
                   
                     
                       E 
                       
                         non 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         select 
                       
                     
                     = 
                     
                       
                         2 
                         ⁢ 
                         
                           
                             ∑ 
                             
                               i 
                               = 
                               1 
                             
                             
                               s 
                               - 
                               1 
                             
                           
                           ⁢ 
                           
                             
                               
                                 ( 
                                 
                                   
                                     
                                       V 
                                       y 
                                     
                                     s 
                                   
                                   ⁢ 
                                   i 
                                 
                                 ) 
                               
                               2 
                             
                             ⁢ 
                             
                               T 
                               s 
                             
                           
                         
                       
                       + 
                       
                         
                           V 
                           y 
                           2 
                         
                         · 
                         
                           T 
                           f 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   17 
                   ) 
                 
               
             
             
               
                 
                   
                     V 
                     RMS 
                   
                   = 
                   
                     
                       
                         ( 
                         
                           
                             
                               3 
                               ⁢ 
                               
                                 s 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     T 
                                     - 
                                     
                                       T 
                                       0 
                                     
                                   
                                   ) 
                                 
                               
                             
                             - 
                             
                               
                                 ( 
                                 
                                   
                                     4 
                                     ⁢ 
                                     s 
                                   
                                   + 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 ( 
                                 
                                   s 
                                   - 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 T 
                                 s 
                               
                             
                           
                           
                             3 
                             ⁢ 
                             sT 
                           
                         
                         ) 
                       
                       · 
                       
                         ( 
                         
                           
                             
                               
                                 V 
                                 x 
                                 2 
                               
                               ± 
                               
                                 2 
                                 ⁢ 
                                 
                                   V 
                                   x 
                                 
                                 ⁢ 
                                 
                                   V 
                                   y 
                                 
                               
                             
                             + 
                             
                               NV 
                               y 
                               2 
                             
                           
                           N 
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   18 
                   ) 
                 
               
             
           
         
       
     
     It can be shown that the selection ratio 
             (         V   ON     ⁡     (   RMS   )           V   OFF     ⁡     (   RMS   )         )         
will be a maximum when the condition V x =√{square root over (N)}V y  is satisfied.
 
     Expression for the RMS voltage across a pixel in a display driven by the conventional line-by-line addressing technique is 
     
       
         
           
             
               
                 
                   
                     V 
                     
                       RMS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       conventional 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             V 
                             r 
                             2 
                           
                           ± 
                           
                             2 
                             ⁢ 
                             
                               V 
                               r 
                             
                             ⁢ 
                             
                               V 
                               c 
                             
                           
                         
                         + 
                         
                           NV 
                           y 
                           2 
                         
                       
                       N 
                     
                   
                 
               
               
                 
                   ( 
                   19 
                   ) 
                 
               
             
           
         
       
     
     It is similar to (18) expect for the first term in (18). We can show that the maximum amplitude of the step voltage profiles are related to that of the conventional line-by-line addressing as shown in (20) and (21) 
     
       
         
           
             
               
                 
                   
                     V 
                     x 
                   
                   = 
                   
                     
                       
                         ( 
                         
                           
                             3 
                             ⁢ 
                             sT 
                           
                           
                             
                               3 
                               ⁢ 
                               
                                 s 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     T 
                                     - 
                                     
                                       T 
                                       0 
                                     
                                   
                                   ) 
                                 
                               
                             
                             - 
                             
                               
                                 ( 
                                 
                                   
                                     4 
                                     ⁢ 
                                     s 
                                   
                                   + 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 ( 
                                 
                                   s 
                                   - 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 T 
                                 s 
                               
                             
                           
                         
                         ) 
                       
                     
                     · 
                     
                       V 
                       r 
                     
                   
                 
               
               
                 
                   ( 
                   20 
                   ) 
                 
               
             
             
               
                 
                   
                     V 
                     y 
                   
                   = 
                   
                     
                       
                         ( 
                         
                           
                             3 
                             ⁢ 
                             sT 
                           
                           
                             
                               3 
                               ⁢ 
                               
                                 s 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     T 
                                     - 
                                     
                                       T 
                                       0 
                                     
                                   
                                   ) 
                                 
                               
                             
                             - 
                             
                               
                                 ( 
                                 
                                   
                                     4 
                                     ⁢ 
                                     s 
                                   
                                   + 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 ( 
                                 
                                   s 
                                   - 
                                   1 
                                 
                                 ) 
                               
                               ⁢ 
                               
                                 T 
                                 s 
                               
                             
                           
                         
                         ) 
                       
                     
                     · 
                     
                       V 
                       c 
                     
                   
                 
               
               
                 
                   ( 
                   21 
                   ) 
                 
               
             
           
         
       
     
     The multiplying factor is the same for both row and column waveforms and is similar to that in (11) of section V. Number of transitions across the pixels is independent of the image when the duty cycle control is introduced with a finite T 0 . Here, the power consumption depends just on the number of ON and OFF pixels in the image because the capacitance of the ON and OFF pixels are not equal. It does not depend on the number of polarity inversions or the data sequences involved in forming the image. Expressions for the power consumed by ON and OFF pixels are given in (22) and (23) respectively. 
     
       
         
           
             
               
                 
                   
                     P 
                     
                       ON 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       pixel 
                     
                   
                   = 
                   
                     
                       
                         sC 
                         ON 
                       
                       NT 
                     
                     ⁡ 
                     
                       [ 
                       
                         
                           
                             ( 
                             
                               
                                 
                                   V 
                                   x 
                                 
                                 + 
                                 
                                   V 
                                   y 
                                 
                               
                               s 
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             ( 
                             
                               N 
                               - 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             
                               ( 
                               
                                 
                                   V 
                                   y 
                                 
                                 s 
                               
                               ) 
                             
                             2 
                           
                         
                       
                       ] 
                     
                   
                 
               
               
                 
                   ( 
                   22 
                   ) 
                 
               
             
             
               
                 
                   
                     P 
                     
                       OFF 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       pixel 
                     
                   
                   = 
                   
                     
                       
                         sC 
                         OFF 
                       
                       NT 
                     
                     ⁡ 
                     
                       [ 
                       
                         
                           
                             ( 
                             
                               
                                 
                                   V 
                                   x 
                                 
                                 - 
                                 
                                   V 
                                   y 
                                 
                               
                               s 
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             ( 
                             
                               N 
                               - 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             
                               ( 
                               
                                 
                                   V 
                                   y 
                                 
                                 s 
                               
                               ) 
                             
                             2 
                           
                         
                       
                       ] 
                     
                   
                 
               
               
                 
                   ( 
                   23 
                   ) 
                 
               
             
           
         
       
     
     Power consumption of the display is given in (24). Here, x j  is the number of ON pixels in the display with N rows and M columns.
 
 P=x   j   P   ON pixel +( N·M−x   j ) P   OFF Pixel   (24)
 
     Ratio of power consumption of multi-step line-by-line addressing to that of conventional line-by-line addressing with duty cycle control is given by 
     
       
         
           
             
               
                 
                   
                     
                       P 
                       
                         multi 
                         ⁢ 
                         
                           - 
                         
                         ⁢ 
                         step 
                       
                     
                     
                       P 
                       pulse 
                     
                   
                   = 
                   
                     ( 
                     
                       
                         3 
                         ⁢ 
                         
                           ( 
                           
                             T 
                             - 
                             
                               T 
                               0 
                             
                           
                           ) 
                         
                       
                       
                         
                           3 
                           ⁢ 
                           
                             s 
                             ⁡ 
                             
                               ( 
                               
                                 T 
                                 - 
                                 
                                   T 
                                   0 
                                 
                               
                               ) 
                             
                           
                         
                         - 
                         
                           
                             ( 
                             
                               
                                 4 
                                 ⁢ 
                                 s 
                               
                               + 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             ( 
                             
                               s 
                               - 
                               1 
                             
                             ) 
                           
                           ⁢ 
                           
                             T 
                             S 
                           
                         
                       
                     
                     ) 
                   
                 
               
               
                 
                   ( 
                   25 
                   ) 
                 
               
             
           
         
       
     
     This factor is same as that in (14) and hence comparison of the power consumption shown in  FIG. 6  holds good for the line-by-line addressing with multi-step voltage profile. 
     Static Drive with Multi-Step Waveforms 
     A non-multiplexed display can be treated as a special case of a matrix display having just one row (N=1). Typical multi-step waveforms for a non-multiplexed display are shown in  FIG. 8 . 
     Analysis with Distortion in the Multistep Waveform: 
     The analysis presented in [3] is based on the assumption that the time constant (τ=RC) of the drive circuit is small as compared to the step width (T s ). It depends on the number of steps (s) in the multi-steps, frame refresh frequency (f), T f  duration of the voltage V, and the number of lines that are multiplexed (N) in the display. The time constant (τ) may become comparable to the step width (T s ) as the select time (T=T f +(2·s−1)T s ) decreases with increasing N and s. Then the decrease in the rms voltage across the pixel due to the distortion in the addressing waveforms will no longer be negligible. Effects of RC distortion on the rms voltage, power dissipation and the peak voltage in multi-step and conventional pulse based waveforms are presented in the following section. 
     I. Analysis of Multi-Step Waveforms with Distortion 
     Let us consider a multi-step waveform profile with RC distortion as the select waveform. It has a period (T) as shown in  FIG. 9 . 
     Each ascending and descending step has a duration T s  while the peak voltage is applied during T f . The ideal waveform without distortion is shown using dotted lines. The waveform profile in  FIG. 9  may be split into four distinct intervals as given below. 
     i) Ascending steps of duration T s . 
     ii) Flat region (at the top) of duration T f  when the voltage is V). 
     iii) Descending steps of duration T s . 
     iv) Duration T L , when the voltage applied to the pixel is zero to ensure full discharge of voltage across the pixel. 
     Piecewise expression for the voltage across a pixel during an ascending step is given by: 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       pixel 
                     
                     ⁡ 
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       k 
                       ⁢ 
                       
                         V 
                         s 
                       
                     
                     - 
                     
                       
                         V 
                         s 
                       
                       ⁢ 
                       
                         ( 
                         
                           
                             ∑ 
                             
                               j 
                               = 
                               0 
                             
                             
                               k 
                               - 
                               1 
                             
                           
                           ⁢ 
                           
                             ⅇ 
                             
                               - 
                               
                                 
                                   j 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     T 
                                     s 
                                   
                                 
                                 τ 
                               
                             
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         ⅇ 
                         
                           - 
                           
                             
                               t 
                               - 
                               
                                 
                                   T 
                                   s 
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     k 
                                     - 
                                     1 
                                   
                                   ) 
                                 
                               
                             
                             τ 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
     This expression is valid for the k th  (1≦k≦s−1) ascending step in the interval (k−1)T s ≦t≦kT s . Similarly the expression for descending steps is as follows: 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       pixel 
                     
                     ⁡ 
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       V 
                       s 
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           ( 
                           
                             s 
                             - 
                             k 
                           
                           ) 
                         
                         + 
                         
                           
                             c 
                             k 
                           
                           ⁢ 
                           
                             ⅇ 
                             
                               - 
                               
                                 
                                   t 
                                   - 
                                   
                                     
                                       T 
                                       . 
                                     
                                     f 
                                   
                                   - 
                                   
                                     
                                       T 
                                       s 
                                     
                                     ⁡ 
                                     
                                       ( 
                                       
                                         s 
                                         + 
                                         k 
                                         - 
                                         2 
                                       
                                       ) 
                                     
                                   
                                 
                                 τ 
                               
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
             
               
                 
                   
                     Wherein 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       c 
                       k 
                     
                   
                   = 
                   
                     [ 
                     
                       
                         ( 
                         
                           
                             ∑ 
                             
                               j 
                               = 
                               0 
                             
                             
                               k 
                               - 
                               1 
                             
                           
                           ⁢ 
                           
                             ⅇ 
                             
                               - 
                               
                                   
                               
                               ⁢ 
                               
                                 
                                   j 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     T 
                                     s 
                                   
                                 
                                 τ 
                               
                             
                           
                         
                         ) 
                       
                       - 
                       
                         
                           ( 
                           
                             
                               ∑ 
                               
                                 j 
                                 = 
                                 0 
                               
                               
                                 s 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               ⅇ 
                               
                                 - 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   
                                     j 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       T 
                                       s 
                                     
                                   
                                   τ 
                                 
                               
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           ⅇ 
                           
                             - 
                             
                               
                                 
                                   
                                     T 
                                     s 
                                   
                                   ⁡ 
                                   
                                     ( 
                                     
                                       k 
                                       - 
                                       1 
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   T 
                                   f 
                                 
                               
                               τ 
                             
                           
                         
                       
                     
                     ] 
                   
                 
               
               
                 
                   ( 
                   5 
                   ) 
                 
               
             
           
         
       
     
     The expression in (27) is valid for the k th  (1≦k≦s−1) descending step when the pixel is discharged to a lower voltage during the interval ((s+k−2)T s +T f )≦t≦((s+k−1)T s +T f ). When the time constant is large, the voltage across the pixel (capacitor) may not discharge completely during T L . The residual voltage across the pixel at the end of each select time will change the waveform across the pixel in successive time intervals. As a result, the rms voltage (across the pixel) will depend on the sequence of voltages in the addressing waveform, which in turn depends on the image that is displayed. Pixels that are driven to the same state in different columns in the matrix display will have different sequence of voltages and hence the rms voltage across them will not be equal. It results in poor brightness uniformity of pixels in the display. In order to maintain good brightness uniformity and avoid cross talk, we propose to apply a negative voltage (β) during T L  as shown in the  FIG. 9 , that will force the pixels to discharge completely. The negative voltage (β) is given in the following expression: 
     
       
         
           
             
               
                 
                   β 
                   = 
                   
                     
                       ( 
                       
                         V 
                         s 
                       
                       ) 
                     
                     ⁢ 
                     
                       
                         ( 
                         
                           1 
                           
                             1 
                             - 
                             
                               ⅇ 
                               
                                 
                                   T 
                                   0 
                                 
                                 τ 
                               
                             
                           
                         
                         ) 
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             ( 
                             
                               
                                 ∑ 
                                 
                                   j 
                                   = 
                                   0 
                                 
                                 
                                   s 
                                   - 
                                   1 
                                 
                               
                               ⁢ 
                               
                                 ⅇ 
                                 
                                   - 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     
                                       j 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         T 
                                         s 
                                       
                                     
                                     τ 
                                   
                                 
                               
                             
                             ) 
                           
                           - 
                           
                             
                               ( 
                               
                                 
                                   ∑ 
                                   
                                     j 
                                     = 
                                     0 
                                   
                                   
                                     s 
                                     - 
                                     1 
                                   
                                 
                                 ⁢ 
                                 
                                   ⅇ 
                                   
                                     - 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       
                                         j 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           T 
                                           s 
                                         
                                       
                                       τ 
                                     
                                   
                                 
                               
                               ) 
                             
                             ⁢ 
                             
                               ⅇ 
                               
                                 - 
                                 
                                   
                                     
                                       
                                         T 
                                         s 
                                       
                                       ⁡ 
                                       
                                         ( 
                                         
                                           k 
                                           - 
                                           1 
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       T 
                                       f 
                                     
                                   
                                   τ 
                                 
                               
                             
                           
                         
                         ] 
                       
                     
                   
                 
               
               
                 
                   ( 
                   6 
                   ) 
                 
               
             
           
         
       
     
     Voltage across the pixel during subsequent time intervals will not depend on the voltage across the pixel during the select time because the voltage across the pixel is fully discharged within (T) by applying a voltage of opposite polarity as shown in the  FIG. 9 , even when the time constant is large. The analysis is valid irrespective of the time constant and such a voltage profile can substitute a pulse in any addressing scheme. Energy delivered to a pixel during the select time T is computed as follows: 
     
       
         
           
             
               
                 
                   
                     E 
                     
                       RC 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       distortion 
                     
                   
                   = 
                   
                     [ 
                     
                       
                         E 
                         s 
                       
                       + 
                       
                         E 
                         s 
                         ′ 
                       
                       + 
                       
                         ( 
                         
                           
                             ∑ 
                             
                               k 
                               = 
                               1 
                             
                             
                               s 
                               - 
                               1 
                             
                           
                           ⁢ 
                           
                             E 
                             k 
                           
                         
                         ) 
                       
                       + 
                       
                         ( 
                         
                           
                             ∑ 
                             
                               k 
                               = 
                               1 
                             
                             
                               s 
                               - 
                               1 
                             
                           
                           ⁢ 
                           
                             E 
                             k 
                             ′ 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                 
               
               
                 
                   ( 
                   30 
                   ) 
                 
               
             
           
         
       
     
     Wherein E k  (or E′ k ); energy delivered to the pixel, is obtained by squaring and integrating the instantaneous voltage across the pixel i.e. V pixel (t) over the duration of the k th  (1≦k≦s−1) ascending (or descending) step i.e. 
     
       
         
           
             
               
                 
                   
                     E 
                     k 
                   
                   = 
                   
                     
                       ∫ 
                       
                         
                           ( 
                           
                             k 
                             - 
                             1 
                           
                           ) 
                         
                         ⁢ 
                         
                           T 
                           s 
                         
                       
                       
                         kT 
                         s 
                       
                     
                     ⁢ 
                     
                       
                         
                           
                             ( 
                             
                               
                                 V 
                                 pixel 
                               
                               ⁡ 
                               
                                 ( 
                                 t 
                                 ) 
                               
                             
                             ) 
                           
                           2 
                         
                         R 
                       
                       ⁢ 
                       
                         ⅆ 
                         t 
                       
                     
                   
                 
               
               
                 
                   ( 
                   31 
                   ) 
                 
               
             
           
         
       
     
     E s  and E′ s  correspond to the energy delivered during T f  and T L  respectively. The power dissipation over a row select time expressed as 
     
       
         
           
             
               
                 
                   
                     P 
                     ⁡ 
                     
                       ( 
                       
                         RC 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         distortion 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       1 
                       T 
                     
                     ⁡ 
                     
                       [ 
                       
                         
                           P 
                           s 
                         
                         + 
                         
                           P 
                           s 
                           ′ 
                         
                         + 
                         
                           ( 
                           
                             
                               ∑ 
                               
                                 k 
                                 = 
                                 1 
                               
                               
                                 s 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               P 
                               k 
                             
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               ∑ 
                               
                                 k 
                                 = 
                                 1 
                               
                               
                                 s 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               P 
                               k 
                               ′ 
                             
                           
                           ) 
                         
                       
                       ] 
                     
                   
                 
               
               
                 
                   ( 
                   32 
                   ) 
                 
               
             
           
         
       
     
     Here, P k  and P′ k  represent the power dissipated during the k th  ascending and descending steps respectively. Whereas P s  and P′ s  correspond to the durations T f  and T L  respectively. 
     The energy delivered is less than that of the ideal waveform profile due to the distortions in the waveform and the introduction of negative voltage (β). In order to achieve the same effect as that of a single pulse the energy has to be same because LCDs are rms responding devices. The overall decrease in the energy delivered to the pixel can be increased by increasing the amplitude using one of the three possible correction methods outlined here: 
     1) Amplitude of the first step is increased. 
     2) Amplitude of all the steps are uniformly increased. 
     3) Amplitude of the topmost i.e. s th  step is increased. 
     Increase in peak amplitude for scheme 2 can be analytically obtained as follows. If 
             γ   =       E   ideal       E     RC   ⁢     -     ⁢   distortion               
then the excitation voltages of all the steps have to be increased by a factor √{square root over (γ)} to obtain the energy of the ideal waveforms even when the waveforms are distorted. Power dissipation also increases by a factor γ i.e., the above expression is multiplied by this factor. It is not possible to estimate the increase in peak amplitude and power dissipation for cases 1 and 3 analytically. However, they can be estimated numerically. The three methods of correcting the energy in the select profile may be compared using the following criteria:
 
     1. Increase in the peak voltage. 
     2. Reduction in power dissipation. 
     We found the increase in the amplitude of the peak voltage is the least when the amplitude of the first step alone is corrected (case 1); and it also has the lowest power dissipation over a range of time constants. 
     Application of a negative voltage to bring the voltage across the pixel to zero at the end of the select time interval (without affecting the energy delivered during the select time) ensures that the power dissipation due to one pixel does not depend on the states of the neighboring pixels just as the introduction of a short duration of zero voltage across the pixel in [3]. We have included these corrections in further analysis and comparison of power dissipation and supply voltage in this paper. 
     Power dissipation of the multi-step waveform as compared to the power dissipation of a single pulse (having the same duty cycle) is plotted as a function of the RC time constant (τ) in  FIG. 10 . The durations T L  and T f  are chosen to be 0.1T and 2T s  respectively. The power dissipation decreases significantly with increase in s when the time constant is small. Saving in power achieved by increasing s is small when the time constant is large. About 30% reduction in power dissipation can be achieved with just three steps when τ=0.2T. The multi-step profile tends towards a triangular waveform when the number of steps is greater than 3 and T f =2T s  as discussed in section IV. Power dissipation of a triangular waveform is plotted in  FIG. 10  for comparison. The triangular waveform has the lowest power dissipation, especially for small values of τ. However, it is achieved with a 70% increase in the peak voltage.  FIG. 11  compares the supply voltage of the multi-step waveform with that of a single pulse for various values of τ and s. It is evident from the plots in  FIGS. 10 and 11  that at lower values of time constant τ, a large reduction in power can be achieved with higher values of s and consequently a higher supply voltage. However, it may be adequate to choose a small value of s because it is possible to achieve reasonable reduction in power with a moderate increase in supply voltage when τ is large. 
     Amplitude of the voltage (β), applied during T L  to completely discharge the pixel within the select time is plotted in  FIG. 12  as a function of the time constant τ. It is normalized to the peak voltage for the corresponding number of steps (s). Amplitude of β is comparable to the peak voltage when τ is large. However, most of the addressing techniques, especially multi-line techniques have both positive and negative voltages and hence the supply voltage of the driver circuit will not double in such cases. 
     As the number of steps is increased, the multi-step waveforms tend towards triangular or trapezoidal profiles [4]. They have a gradual change in amplitude and do not have any abrupt changes, which contribute to higher power dissipation while charging and discharging pixels. 
     CONCLUSION 
     The fundamental nature of the idea of substituting multi-step voltage profiles in place of a single voltage of a pulse to reduce power consumption ensures that it will hold good in combination with any other addressing technique suitable for driving the RMS responding matrix LCDs. Scaling of amplitude of the row and column waveforms and the reduction in power consumption etc., will be the same as the outcome of the analysis presented in the section V. The multi-step voltage profile reduces to a triangular waveform when s→∞. Application of such triangular waveforms to drive the matrix display is outside the scope of this paper and it will be presented elsewhere.