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Light-Emitting Diodes

                                                    Light-Emitting Diodes  131

          Equation (6.30) shows the basic physics of the situation. In a good
          LED, we can neglect the nonradiative term compared to the radiative
          term. Thus, the transient response time,   step , is inversely proportion-
          al to the carrier concentration. This could be the carrier concentration
          due to doping or induced by the current pulse.
            The rate equation can now be expressed as

                               d       J 2 – J 1   N
                                  N =         –                      (6.31)
                              dt         qd       step
          This is a simple (in appearance!) differential equation in  N. Howev-
          er, since 1/  step also depends on  N, a closed-form solution will be pos-
          sible only under special circumstances. If  N   N 1 , we can treat   step
          as a constant. This condition corresponds to the limit of high doping
          or low injection. Then Eq. (6.31) can be solved analytically. Otherwise,
          only a numerical solution is possible.

          Case 1. Low-Injection Limit
          In the low-injection limit,   step is treated as a constant. Eq. (6.31) is a
          first-order differential equation with a driving term  N/  step . The so-
          lution is written as

                            N(t) = Ae –t/  step + Be –t/  step + C
          where A, B, and C are constants to be determined by the boundary
          conditions.


          Boundary conditions

          1. At time t = 0,  N = 0.
          2. At time t =  ,  N = [(J 2 – J 1 )/qd]·   step .


            Applying boundary condition (1), A + B + C = 0.
            Applying boundary condition (2), it follows that B = 0, and

                                  (J 2 – J 1 )
                              C =          ·   step = –A
                                     qd
          So the particular solution is expressed:
                            (J 2 – J 1 )          (J 2 – J 1 )
                    N(t) = –         ·   step e –t/  step +     ·   step
                               qd                    qd
                           (J 2 – J 1 )
                         =         ·   step (1 – e –t/  step )
                             qd



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