Patent Publication Number: US-8120442-B2

Title: Semiconductor device

Description:
CROSS-REFERENCE TO RELATED APPLICATION 
     The present invention claims priority from Korean patent application number 10-2008-0052753, filed on Jun. 4, 2008, which is incorporated by reference in its entirety. 
     BACKGROUND OF THE INVENTION 
     The present invention relates to a semiconductor design technology; and more particularly, to a technology of removing crosstalk that may occur between transmission lines of the semiconductor device. 
     In semiconductor devices, when a signal is transmitted via a transmission line, the signal may be affected by electromagnetic fields, which will be discussed in detail below. 
     When a dielectric is between metals, there is capacitance between them. In general, a transmission line is made of metal and a dielectric is arranged between transmission lines, and thus, capacitance may exist between them. It is known that even air is a dielectric whose relative dielectric constant is ‘1’. Therefore, even when transmission lines are arranged in the air, there may also be capacitance between them. When an alternate current of high frequency band flows onto a transmission line, electrical energy interference occurs due to the influence of capacitance between transmission lines, i.e., mutual capacitance, as the current goes toward a higher frequency band. This also affects the characteristic impedance value of transmission line, and so on. In addition, as an alternate current flows onto a transmission line, a magnetic field is induced on the transmission line, which produces mutual inductance that affects a magnetic field of another transmission line. This mutual inductance causes magnetic energy interference that affects the inductance value of each transmission line, etc, which in turn affects the characteristic impedance value of each transmission line. 
     Meanwhile, coupling is a phenomenon in which alternate energies such as an electric field and a magnetic field are mutually transferred between separated spaces or transmission lines. As metals are closed to each other, this phenomenon gives rise to interference with a signal and acts as an undesired parasitic effect. This unnecessary coupling is often called “crosstalk” in view of ElectricMagnetic Interference (EMI). In the invention, therefore, unnecessary mutual interference caused by coupling is defined as crosstalk, which will be set forth below. 
     As noted earlier, electromagnetic interference that occurs between adjacent transmission lines is referred to as “crosstalk”. This phenomenon arises due to both mutual inductance and mutual capacitance. Such mutual inductance and mutual capacitance have influence on the total inductance and the total capacitance of transmission line. The following is an explanation about the influence of mutual inductance and mutual capacitance in each mode, in an ODD mode and EVEN mode analyze in a coupled transmission line theory. 
       FIG. 1  shows an equivalent circuit model of a coupled transmission line, which includes an equivalent circuit  100  of inductance and capacitance, an inductance equivalent circuit  110  and a capacitance equivalent circuit  120 . 
     Signal modes between adjacent transmission lines can be largely classified into an ODD mode and an EVEN mode. When there are two transmission lines, the ODD mode is when signals having a 180-degree phase difference, but the same amplitude, are applied to the two transmission lines, respectively. First, regarding inductance, a voltage is generated by inductive coupling and currents I 1  and I 2  flowing onto two transmission lines in the inductance equivalent circuit  110  have the same amplitude but opposite directions to each other. Assuming that the self-inductance L 11 =L 22 =L 0  and the mutual inductance L 12 =L M , V 1  and V 2  of the inductance equivalent circuit  110  may be expressed as: 
     
       
         
           
             
               
                 
                   
                     V 
                     1 
                   
                   = 
                   
                     
                       
                         L 
                         0 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                     + 
                     
                       
                         L 
                         M 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     1 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     V 
                     2 
                   
                   = 
                   
                     
                       
                         L 
                         0 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                     + 
                     
                       
                         L 
                         M 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     2 
                     ) 
                   
                 
               
             
           
         
       
     
     In the ODD mode, I 1 =−I 2  and V 1 =−V 2 , and thus, these equations may be represented as follows: 
     
       
         
           
             
               
                 
                   
                     V 
                     1 
                   
                   = 
                   
                     
                       
                         
                           L 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               I 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           L 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 - 
                                 
                                   I 
                                   1 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             L 
                             0 
                           
                           - 
                           
                             L 
                             M 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     3 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     V 
                     2 
                   
                   = 
                   
                     
                       
                         
                           L 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               I 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           L 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 - 
                                 
                                   I 
                                   2 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             L 
                             0 
                           
                           - 
                           
                             L 
                             M 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     4 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     L 
                     ODD 
                   
                   = 
                   
                     
                       L 
                       11 
                     
                     - 
                     
                       L 
                       M 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     5 
                     ) 
                   
                 
               
             
           
         
       
     
     As can be seen from Eq. (5) above, the total inductance L ODD  in the ODD mode becomes smaller than the self-inductance L 11  by the mutual inductance L M . 
     Similarly, relating to capacitance, assuming that in the capacitance equivalent circuit  120  the self-capacitance C 1G =C 2G =C 0  and the mutual capacitance C 12 =C M , I 1  and I 2  of the capacitance equivalent circuit  120  may be expressed as: 
     
       
         
           
             
               
                 
                   
                     I 
                     1 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   1 
                                 
                                 - 
                                 
                                   V 
                                   2 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               C 
                               0 
                             
                             - 
                             
                               C 
                               M 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       - 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     6 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     I 
                     2 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   2 
                                 
                                 - 
                                 
                                   V 
                                   1 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               C 
                               0 
                             
                             - 
                             
                               C 
                               M 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       - 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     7 
                     ) 
                   
                 
               
             
           
         
       
     
     In the ODD mode, I 1 =−I 2  and V 1 =−V 2 , and thus, these equations may be defined as: 
     
       
         
           
             
               
                 
                   
                     I 
                     1 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   1 
                                 
                                 - 
                                 
                                   ( 
                                   
                                     - 
                                     
                                       V 
                                       1 
                                     
                                   
                                   ) 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             C 
                             0 
                           
                           + 
                           
                             2 
                             ⁢ 
                             
                               C 
                               M 
                             
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             V 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     8 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     I 
                     2 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   2 
                                 
                                 - 
                                 
                                   ( 
                                   
                                     - 
                                     
                                       V 
                                       2 
                                     
                                   
                                   ) 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             C 
                             0 
                           
                           + 
                           
                             2 
                             ⁢ 
                             
                               C 
                               M 
                             
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             V 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     9 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                           C 
                           ODD 
                         
                         = 
                         
                           
                             
                               C 
                               
                                 1 
                                 ⁢ 
                                 G 
                               
                             
                             + 
                             
                               2 
                               ⁢ 
                               
                                 C 
                                 M 
                               
                             
                           
                           = 
                           
                             
                               C 
                               11 
                             
                             + 
                             
                               C 
                               M 
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         
                           C 
                           11 
                         
                         = 
                         
                           
                             C 
                             
                               1 
                               ⁢ 
                               G 
                             
                           
                           + 
                           
                             C 
                             M 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     10 
                     ) 
                   
                 
               
             
           
         
       
     
     As can be seen from Eq. (10) above, the total capacitance C ODD  in the ODD mode becomes greater than the self-capacitance C 1G  by 2C M . 
     Based on the total inductance L ODD  and the total capacitance C ODD  in Eqs. (5) and (10) above, Z ODD  and TD ODD  may be defined as: 
     
       
         
           
             
               
                 
                   
                     Z 
                     ODD 
                   
                   = 
                   
                     
                       
                         
                           L 
                           ODD 
                         
                         
                           C 
                           ODD 
                         
                       
                     
                     = 
                     
                       
                         
                           
                             L 
                             11 
                           
                           - 
                           
                             L 
                             12 
                           
                         
                         
                           
                             C 
                             11 
                           
                           + 
                           
                             C 
                             12 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     11 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     TD 
                     ODD 
                   
                   = 
                   
                     
                       
                         
                           L 
                           ODD 
                         
                         ⁢ 
                         
                           C 
                           ODD 
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               L 
                               11 
                             
                             - 
                             
                               L 
                               12 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           ( 
                           
                             
                               C 
                               11 
                             
                             + 
                             
                               C 
                               12 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     12 
                     ) 
                   
                 
               
             
           
         
       
     
     Meanwhile, the EVEN mode is when signals having the same phase and the same amplitude are applied to the two transmission lines, respectively. First, as for inductance, a voltage is generated by an inductive coupling and currents I 1  and I 2  flowing onto the two transmission lines in the inductance equivalent circuit  110  have the same amplitude and the same direction. Assuming that the self-inductance L 11 =L 22 =L 0  and the mutual inductance L 12 =L M , V 1  and V 2  of the inductance equivalent circuit  110  may be expressed as in Eqs. (1) and (2) above. And, in the EVEN mode I 1 =I 2  and V 1 =V 2 , and thus, these equations may be represented as: 
     
       
         
           
             
               
                 
                   
                     V 
                     1 
                   
                   = 
                   
                     
                       
                         
                           L 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               I 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           L 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 I 
                                 1 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             L 
                             0 
                           
                           + 
                           
                             L 
                             M 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     13 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     V 
                     2 
                   
                   = 
                   
                     
                       
                         
                           L 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               I 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           L 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 I 
                                 2 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             L 
                             0 
                           
                           + 
                           
                             L 
                             M 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             I 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     14 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     L 
                     EVEN 
                   
                   = 
                   
                     
                       L 
                       11 
                     
                     + 
                     
                       L 
                       M 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     15 
                     ) 
                   
                 
               
             
           
         
       
     
     As can be seen from Eq. (15), the total inductance L EVEN  in the EVEN mode becomes the self-inductance L 11  plus the mutual inductance L M . 
     Similarly, assuming that in the capacitance equivalent circuit  120  the capacitance may be expressed as Eqs. (6) and (7) above, and in the EVEN mode I 1 =I 2  and V 1 =V 2 , and thus, I 1  and I 2  may be represented again as: 
     
       
         
           
             
               
                 
                   
                     I 
                     1 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               1 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   1 
                                 
                                 - 
                                 
                                   V 
                                   1 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         C 
                         0 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             V 
                             1 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     16 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     I 
                     2 
                   
                   = 
                   
                     
                       
                         
                           C 
                           0 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               V 
                               2 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           C 
                           M 
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               ( 
                               
                                 
                                   V 
                                   2 
                                 
                                 - 
                                 
                                   V 
                                   2 
                                 
                               
                               ) 
                             
                           
                           
                             ⅆ 
                             t 
                           
                         
                       
                     
                     = 
                     
                       
                         C 
                         0 
                       
                       ⁢ 
                       
                         
                           ⅆ 
                           
                             V 
                             2 
                           
                         
                         
                           ⅆ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     17 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                           C 
                           EVEN 
                         
                         = 
                         
                           
                             C 
                             
                               1 
                               ⁢ 
                               G 
                             
                           
                           = 
                           
                             
                               C 
                               11 
                             
                             - 
                             
                               C 
                               M 
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         
                           C 
                           11 
                         
                         = 
                         
                           
                             C 
                             
                               1 
                               ⁢ 
                               G 
                             
                           
                           + 
                           
                             C 
                             M 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     18 
                     ) 
                   
                 
               
             
           
         
       
     
     Accordingly, the total capacitance C EVEN  in the EVEN mode becomes equal to the self-capacitance C 1G , as shown in Eq. (18). Using the total inductance L EVEN  and the total capacitance C EVEN  in Eqs. (15) and (18) above, Z EVEN  and TD EVEN  may be defined as: 
     
       
         
           
             
               
                 
                   
                     Z 
                     EVEN 
                   
                   = 
                   
                     
                       
                         
                           L 
                           EVEN 
                         
                         
                           C 
                           EVEN 
                         
                       
                     
                     = 
                     
                       
                         
                           
                             L 
                             11 
                           
                           + 
                           
                             L 
                             12 
                           
                         
                         
                           
                             C 
                             11 
                           
                           - 
                           
                             C 
                             12 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     19 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     TD 
                     EVEN 
                   
                   = 
                   
                     
                       
                         
                           L 
                           EVEN 
                         
                         ⁢ 
                         
                           C 
                           EVEN 
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               L 
                               11 
                             
                             + 
                             
                               L 
                               12 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           ( 
                           
                             
                               C 
                               11 
                             
                             - 
                             
                               C 
                               12 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     20 
                     ) 
                   
                 
               
             
           
         
       
     
     As mentioned above, the characteristic impedance of transmission line varies depending on adjacent transmission lines and signal modes (signal transmission modes) due to the influence of coupling between the adjacent transmission lines, which gives rise to a difference in the transmission rates of signals. In result, such difference in the transmission rates acts as a factor that impairs a timing margin. 
       FIG. 2  is a view showing several types of crosstalk generated depending on signal transmission modes. 
     Referring to  FIG. 2 , signal modes between adjacent transmission lines can be largely classified into an EVEN mode  310  and an ODD mode  320 . When there are two transmission lines, the EVEN mode  310  is when signals having the same phase and the same amplitude are applied to the two transmission lines, respectively. On the other hand, the ODD mode  320  is when signals having a 180-degree phase difference but the same amplitude are applied to the two transmission lines, respectively. In addition, when there is a variation in only a signal on any one of the two transmission lines, it is regarded that the EVEN mode  310  and the ODD mode  320  overlap each other, which is called an overlap mode  300 . 
     Now, what effect signal modes between two transmission lines would have on transmission signals on those transmission lines will be described with reference to  FIG. 2 . 
     First, referring to an EVEN mode model  311 , when the EVEN mode is set between two transmission lines, it can be seen, from the signal shape  312  of the EVEN mode, the fact that the transmission rate of signal is the slowest. Further, referring to an ODD mode model  321 , when the ODD mode is set between two transmission lines, it can be found, from the signal shape  322  of the ODD mode, the fact that the transmission rate of signal is the fastest. Lastly, referring to an overlap mode model  301 , when the ODD mode is built up between two transmission lines, it can be found, from the signal shape  302  of the overlap mode, the fact that the signal shapes  312  and  322  of the EVEN mode and the ODD mode overlap each other. 
     As discussed above, there occurs a difference in the transmission rates of signals, i.e., skew, depending on signal modes between adjacent transmission lines. Among various methods developed to compensate such skew, the most frequently used method is a Delayed Locked Loop (DLL). However, this type of devices compensates such skew by synchronizing with a reference signal such as a clock, without having to consider which type of crosstalk occurred at all. Therefore, such devices require numerous control signals therein, thus making their circuits complicated and wider in area, together with much power consumption. 
     SUMMARY OF THE INVENTION 
     Embodiments of the present invention are directed to providing a semiconductor device capable of reducing skew caused by crosstalk. 
     In accordance with an aspect of the invention, a semiconductor device includes a plurality of transmission lines for conveying signals, a plurality of transition detectors, each checking whether a transmission signal on each of the plurality of transmission lines is transited, and if so, detecting its transition shape, a signal mode determining unit for determining signal transmission modes between adjacent transmission lines out of the plurality of transmission lines in response to output signals from the plurality of transition detectors and a plurality of delay units, each being coupled to each of the plurality of transmission lines for adjusting a transmission delay of the transmission signal depending on a corresponding output signal from the signal mode determining unit. 
     In accordance with another aspect of the invention, a semiconductor device includes a plurality of transmission lines for conveying signals, a plurality of transition detectors, each checking whether a transmission signal on each of the plurality of transmission lines is transited, and if so, detecting its transition shape, a signal mode determining unit for determining signal transmission modes between adjacent transmission lines out of the plurality of transmission lines in response to output signals from the plurality of transition detectors, a plurality of clock delay units, each delaying a clock signal being applied in response to a corresponding output signal from the signal mode determining unit and a plurality of latch units, each being coupled to each of the plurality of transmission lines for latching the transition signal in response to an output signal from a corresponding one of the plurality of clock delay units. 
     As discussed above, the prior art device employs complicated circuits such as a DLL circuit in order to remove skew caused by crosstalk. On the contrary, the invention detects the type of transition of transmission signal on each transmission line and judges a signal transmission mode between adjacent transmission lines based on the detected signal and adjusts a delay of transmission signal depending on the signal transmission mode. That is, the invention adjusts a delay difference of transmission signals based on such a delay difference by signal transmission modes, thereby removing skew. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  is an equivalent circuit model of a coupled transmission line. 
         FIG. 2  is a view showing the forms of crosstalk generated depending on signal transmission modes. 
         FIG. 3A  is a block diagram of a semiconductor device in accordance with one embodiment of the present invention. 
         FIG. 3B  is a block diagram of a semiconductor device in accordance with another embodiment of the invention. 
         FIG. 4  is a block diagram of a semiconductor device in accordance with still another embodiment of the invention. 
         FIG. 5  is a block diagram of a semiconductor device in accordance with a further embodiment of the invention. 
         FIG. 6  is circuit diagram illustrating an example of the signal transition detector in accordance with the invention. 
         FIG. 7  is a circuit diagram illustrating another example of the signal mode determining unit in accordance with the invention. 
         FIG. 8  is a circuit diagram illustrating an example of the delay unit in accordance with the invention. 
         FIG. 9  is a circuit diagram illustrating another example of the delay unit in accordance with the invention. 
     
    
    
     DESCRIPTION OF SPECIFIC EMBODIMENTS 
     Hereinafter, preferred embodiments of the present invention will be described in detail with reference to the accompanying drawings, so that the invention can readily be practiced by those skilled in the art to which the invention pertains. 
     Typically, in logic circuits, a digital logic signal is identified as a logic high level or a logic low level, which is expressed as ‘1’ or ‘0’, respectively. In addition, it may further have a high impedance (HI-Z) state, etc., as further discussed below. 
       FIG. 3A  is a block diagram showing the configuration of a semiconductor device in accordance with one embodiment of the invention. Referring to  FIG. 3A , the semiconductor device of this embodiment includes a first transmission unit having first transmission line  4 A, first signal transition detector  400  and first delay unit  430 , and a second transmission unit includes second transmission line  4 B, second signal transition detector  410  and second delay unit  440 . The first and second transmission lines  4 A and  4 B are provided for conveying signals. The first and second signal transition detectors  400  and  410  detect transition shapes of transmission signals S 1  and S 2  loaded on the first and the second transmission lines  4 A and  4 B, respectively. A signal mode determining unit  420  determines signal transmission modes between the first and the second transmission lines  4 A and  4 B adjacent to each other in response to output signals S 1   t  and S 2   t  from the first and the second signal transition detectors  400  and  410 . The first and second delay units  430  and  440  are coupled to each of the first and the second transmission lines  4 A and  4 B adjust transmission delays of the transmission signals S 1  and S 2  depending on output signals D 1  and D 2  from the signal mode determining unit  420 , respectively. 
     Now, an operation of the semiconductor device having the configuration as above will be described in detail. First, each of the first and the second signal transition detectors  400  and  410  checks whether the transmission signal S 1  or S 2  on the first or the second transmission line is transited, and if so detects its transition shape with time, to output a check and detection result S 1   t  or S 2   t . The signal mode determining unit  420  receives an output signal from each of the first and second signal transition detectors  400  and  410  and determines whether signal transmission modes established between the first and second transmission lines  4 A and  4 B are an EVEN mode, an ODD mode, or an overlap mode. Each of the first and the second primary delay units  430  and  440  delays each of the transmission signals S 1  and S 2  on each of the transmission lines depending on output signals D 1  and D 2  of the signal mode determining unit  420 . That is, each of the first and the second delay units  430  and  440  adjusts a delay of each of the transmission signals S 1  and S 2  depending on the signal transmission modes between the two transmission lines  4 A and  4 B, thereby compensating skew caused by signal modes. 
       FIG. 3B  is a block diagram showing the configuration of a semiconductor device in accordance with another embodiment of the invention. 
     Referring to  FIG. 3B , the semiconductor device of this embodiment includes a first transmission unit having a first transmission line  4 A, first signal transition detector  400 , first clock delay unit  450 A and first latch unit  430 A, and a second transmission unit having second transmission line  4 B, second signal transition detector  410 , second clock delay unit  460 A and second latch unit  440 A. The first and second transmission lines  4 A and  4 B for convey signals; the first and second signal transition detectors  400  and  410  detect transition shapes of transmission signals S 1  and S 2  on the first and the second transmission lines  4 A and  4 B, respectively; a signal mode determining unit  420  determine signal transmission modes between the first and the second transmission lines  4 A and  4 B adjacent to each other in response to output signals S 1   t  and S 2   t  from the first and the second signal transition detectors  400  and  410 ; the first and second clock delay units  450 A and  460 A delay clock signals CLK being applied thereto in response to output signals D 1  and D 2  from the signal mode determining unit  420 , respectively; and first and second latches  430 A and  440 A coupled to each of the transmission lines latch the transmission signals in response to output signals from the clock delay units  450 A and  460 A, respectively. 
     Now, an operation of the semiconductor device having the configuration as above will be described in detail. First, each of the first and the second signal transition detectors  400  and  410  checks whether the transmission signal S 1  or S 2  on the first or the second transmission line is transited, and detects its transition shape with time if so, to a output check and detection result S 1   t  or S 2   t , respectively. The signal mode determining unit  420  receives an output signal from each of the first and the second signal transition detectors  400  and  410  and determines whether signal transmission modes established between the first and the second transmission lines  4 A and  4 B are an EVEN mode, an ODD mode, or an overlap mode. Each of the first and the second clock delay units  450 A and  460 A delays the clock signal CLK being applied thereto in response to a corresponding one of output signals D 1  and D 2  from the signal mode determining unit  420  by a predetermined time period to output a delayed clock CLK_D. Each of the first and the second latches  430 A and  440 A coupled to each of the transmission lines latches each of the transmission signals in response to an output signal from each of the clock delay units. That is, by adjusting an input time of the clock signal CLK to control a point of time the transmission signal on each of the transmission lines is to be latched, skew generated by signal modes can be compensated. Although the single clock signal CLK has been used by way of example, plural clocks may be used where necessary. 
       FIG. 4  is a block diagram showing the configuration of a semiconductor device in accordance with still another embodiment of the invention. Referring to  FIG. 4 , the semiconductor device of this embodiment includes transmission lines  5 A,  5 B,  5 C, and  5 D for conveying signals; signal transition detectors  500  to  503  for detecting transition shapes of transmission signals S 1 , S 2 , S 3 , and S 4  on the transmission lines  5 A,  5 B,  5 C, and  5 D, respectively; signal mode determining units  510  to  513  for determining signal transmission modes between the adjacent transmission lines in response to output signals S 1   t , S 2   t , S 3   t , and S 4   t  from the signal transition detectors  500  to  503 , respectively; and delay units  520  to  523  coupled to each of the transmission lines for adjusting transmission delays of the transmission signals S 1 , S 2 , S 3 , and S 4  depending on output signals D 1 , D 2 , D 3 , and D 4  from their respective adjacent mode determining units, respectively. 
     An operation of the semiconductor device configured as above is substantially identical to that of the embodiment in  FIG. 3A . However, in contrast to the embodiment of  FIG. 3A , here the invention is applied to adjacent transmission lines out of plural transmission lines arranged in the semiconductor device. Such an operation of the embodiment of  FIG. 4  will be described below. 
     First, each of the signal transition detectors  500  to  503  checks whether the transmission signal S 1 , S 2 , S 3 , or S 4  on its corresponding transmission line is transited, and detects its transition shape with time if so, to an output check and detection result S 1   t , S 2   t , S 3   t , or S 4   t . Each of the signal mode determining units  510  to  513  determines a signal transmission mode between the adjacent transmission lines in response to an output signal S 1   t , S 2   t , S 3   t , or S 4   t  from each of the signal transition detectors  500  to  503 . Each of the delay units  520  to  523  delays a transmission delay of the transmission signal S 1 , S 2 , S 3 , or S 4  in response to an output signal D 1 , D 2 , D 3 , or D 4  from each of the plural signal mode determining units  420 . That is, each of the delay units  520  to  523  adjusts a delay of the transmission signal S 1 , S 2 , S 3 , or S 4  depending on signal transmission modes between adjacent transmission lines out of the signal transmission lines  5 A,  5 B,  5 C, and  5 D, thereby compensating skew generated by signal modes. 
       FIG. 5  is a block diagram showing the configuration of a semiconductor device in accordance with still another embodiment of the invention. Referring to  FIG. 5 , the semiconductor device of this embodiment includes transmission lines  6 A,  6 B,  6 C, and  6 D for conveying signals; signal transition detectors  600  to  603  for detecting transition shapes of signals S 1 , S 2 , S 3 , and S 4  on the transmission lines  5 A,  5 B,  5 C, and  5 D, respectively; a signal mode determining unit  610  for determining signal transmission modes between the adjacent transmission lines in response to output signals S 1   t , S 2   t , S 3   t , and S 4   t  from the signal transition detectors  600  to  603 ; and delay units  620  to  623  coupled to each of the transmission lines for adjusting transmission delays of the signals S 1 , S 2 , S 3 , and S 4  depending on output signals D 1 , D 2 , D 3 , and D 4  from the signal mode determining unit  610 , respectively. 
     Now, an operation of the semiconductor device as configured above will be described below. 
     First, each of the signal transition detectors  600  to  603  checks whether the transmission signal S 1 , S 2 , S 3 , or S 4  on its corresponding transmission line is transited, and detects its transition shape with time if so, to output a check and detection result S 1   t , S 2   t , S 3   t , or S 4   t . The signal mode determining unit  610  determines signal transmission modes between adjacent transmission lines in response to an output signal S 1   t , S 2   t , S 3   t , or S 4   t  from each of the signal transition detectors  600  to  603 . Each of the delay units  620  to  623  adjusts a transmission delay of the transmission signal S 1 , S 2 , S 3 , or S 4  in response to an output signal D 1 , D 2 , D 3 , or D 4  from the signal mode determining unit  620 . That is, each of the delay units  520  to  523  adjusts a delay of the transmission signal S 1 , S 2 , S 3 , or S 4  by taking account of all signal transmission modes between adjacent transmission lines together out of the signal transmission lines  5 A,  5 B,  5 C, and  5 D, thereby compensating skew generated by signal modes. This embodiment adjusts a transmission delay of each transmission signal by determining all signal modes between adjacent transmission lines together, differently from determining only individual signal transmission modes between adjacent transmission lines in the embodiment in  FIG. 3A . 
     In the embodiments illustrated in  FIGS. 3A to 5 , the semiconductor device may include plural auxiliary delay units which are connected to each transmission line before the delay unit or latch unit to delay transmission signals for a predetermined time period. That is to say, a delay control signal or a delayed clock signal may not reach at a desired point of time if there is a large operation delay time in the signal transition detector and the signal mode determining unit. In order to remove this problem, the auxiliary delay units may delay the transmission signals to some degrees in advance. 
       FIG. 6  is circuit diagram illustrating an example of the signal transition detector in accordance with the invention. 
     Referring to  FIG. 6 , the signal transition detector includes a delay element  700  for delaying a first transmission signal S 1  on a transmission line, and a signal comparator  710  for comparing an output signal S 1   t-1  from the delay element  700  with a second transmission signal S 1   t  on the transmission line. The signal comparator  710  is composed of an exclusive OR gate XOR for comparing the two signals. 
     An operation of the signal transition detector having the configuration as above will now be described in detail. 
     First, the delay element  700  delays the first transmission signal S 1  applied thereto within a 1 Unit Interval (1UI) to output a delayed signal S 1   t-1 . Then, the signal comparator  710  compares the delayed signal S 1   t-1  with a second transmission signal S 1   t  on the transmission line through their exclusive OR operation to generate an output signal S 1   c . Thus, it is possible to check whether or not the signal on the transmission line is transited based on the output signal S 1   t  from the signal comparator  710 . That is, if the transmission signal S 1  is transited based on comparison between the delayed signal S 1   t-1  and the second transmission signal S 1   t , the signal comparator  710  outputs a first logic signal ‘1’, and otherwise, it outputs a second logic signal ‘0’. 
     Also, it is possible to check the transition shape of the transmission signal S 1  by using the output signal S 1   c  from the comparator  710  and the second transmission signal S 1   t . For example, if the output signal S 1   c  from the comparator  710  is ‘1’ and the second transmission signal S 1   t  is ‘1’, then the transmission signal S 1  is transited from ‘0’ to ‘1’, while if the output signal S 1   c  from the comparator  710  is ‘1’ and the second transmission signal S 1   t  is ‘0’, then the transmission signal S 1  is transited from ‘1’ to ‘0’. As described above, it is possible to check whether the transmission signal S 1  is transited and to detect its transition shape if so based on the output signal S 1   c  from the comparator  710  and the second transmission signal S 1   t . In other words, based on comparison between a signal, which is obtained by delaying a transmission signal by a predetermined time, and a signal on a current transmission line, a transition shape of that signal can be determined. 
     Differently from the above-stated embodiments, if a transmission signal has a preamble set, that is, if a circuit starts to operate after the transmission signal is set to a logic low level or a logic high level in advance, it is possible to check whether the transmission signal is transited and to detect its transition shape if so, only by sensing a level of that signal. In other words, in a system where a preamble is defined by a logic low level, if a signal being transmitted at is at a logic high level, this means that the transmission signal is transited from a logic low level to a logic high level at that time. 
       FIG. 7  is a circuit diagram illustrating another example of the signal mode determining unit in accordance with the invention. In particular,  FIG. 7  shows a circuit diagram of the signal mode determining unit  920  and also determination results in a table  930 , when the signal transition detector  910  checks whether transmission signals S 1  and S 2  on first and second transmission lines are transited and detects their transition shapes if so. 
     Referring to  FIG. 7 , the signal mode determining unit  920  determines signal modes between the first and the second transmission modes based on first to fourth output signals S 1   c , S 2   c , S 1   t , and S 2   t  provided from the signal transition detector  910 . To this end, the signal mode determining unit  920  is provided with a first AND gate AND 1  for performing an AND operation on the first and the second output signals S 1   c  and S 2   c , an exclusive OR gate XOR for performing an exclusive OR operation on the third and the fourth output signals S 1   t  and S 2   t , a second AND gate AND 2  for performing an AND operation on an output signal from the exclusive OR gate XOR and the first output signal S 1   c , and an OR gate OR for performing an OR operation on an output signal from the second AND gate AND 2  and the second output signal S 2   c . Therefore, signal modes between the two transmission lines can be determined based on a first mode signal Q 1  from the first AND gate AND 1  and a second mode signal Q 2  from the OR gate OR, wherein its results are shown in table  930 . 
       FIG. 8  is a circuit diagram illustrating an example of the delay element in accordance with the invention. Referring to  FIG. 8 , the delay element is provided with a control signal generator  1001  for receiving an output signal CONTROL from the signal mode determining unit, and first and second MOS capacitors  1002  coupled to a transmission line having a transmission signal loaded for responding to an output signal from the control signal generator  1001 . In this configuration, the delay element can be used only for configuring a circuit for delaying an ODD mode signal, and outputs a signal to activate the MOS capacitors  1002  from the control signal generator  1001  when the output signal CONTROL from the signal mode determining unit denotes an ODD mode, thereby delaying the transmission signal DATA. 
       FIG. 9  is a circuit diagram illustrating another example of the delay element in accordance with the invention. Referring to  FIG. 9 , the delay element is provided with a control signal generator  1011  for taking first and second output signals Q 1  and Q 2  from the signal mode determining unit, and first to fourth MOS capacitors  1012  coupled to a transmission line having a transmission signal loaded for responding to an output signal from the control signal generator  1011 . In this configuration, the delay element can be used for configuring a circuit for adjusting the degree of delay depending on each signal mode. How to determine operation modes based on the first and second output signals Q 1  and Q 2  from the signal mode determining unit will be described below. 
     First, if (Q 1 ,Q 2 ) is (0,0) and (Q 1 ,Q 2 ) is (1,0), a MOS capacitor is not activated, if (Q 1 ,Q 2 ) is (0,1), then a pair of MOS capacitors are activated, and if (Q 1 ,Q 2 ) is (1,1), two pair of MOS capacitors are activated. Therefore, the transmission delay of the transmission signal DATA can be adjusted depending on each signal mode. 
     In the above-described configuration, it should be noted that the number of transmission lines employed in the signal mode determining unit may be changed in various manner. For example, there may be only adjacent transmission lines, or transmission lines within any range set by considering the intensity of the influence of crosstalk occurring between adjacent transmission lines where necessary. In addition, while the delay element in the embodiments of the invention used a general RC delay with MOS capacitors, there may be a variety of methods to delay signals. For example, any of those methods may be properly selected depending on area or power consumption conditions. Since examples of circuit changes or modifications are obvious to those skilled in the art, details thereof will be omitted here. 
     As a result, the invention is more specialized and suitable for removing skew caused by crosstalk because it takes account of directly the reasons why skew is generated, differently from an existing skew compensating circuit such as a Delay Locked Loop (DLL). Thus, the invention is very advantageous in circuit area, power consumption, and so on. 
     In addition, the invention can improve the performance of a semiconductor device due to an increase in a timing margin by removal of skew, and also can integrate more transmission lines. 
     While the invention has been described with respect to the specific embodiments, it will be apparent to those skilled in the art that various changes and modifications may be made without departing from the spirit and scope of the invention as defined in the following claims.