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A Variational Inequality Theory Chapter | 4 125



















                           FIGURE 4.7 Double-diode clipper: Ideal diode, E 1 = 0.1, E 2 = 0.6.

                                                  ⎧
                                                  ⎪ E 1   if u(t) < E 1
                                                  ⎨
                                           V o (t) =  u(t)  if E 1 ≤ u(t) ≤ E 2      (4.112)
                                                  ⎪
                                                  ⎩       if u(t) > E 2 .
                                                    E 2
                           This shows that the circuit can be used to transmit the part of a given input-signal
                           u that lies above some level E 1 and below some level E 2 (see Fig. 4.7).
                           Example 61 (A double-diode clipper/Ideal diodes and nonlinear resistor). Let
                           us again consider the circuit in Fig. 4.6 involving a load resistance R> 0, two
                           ideal diodes, an input-signal source u, and two supply voltages E 1 and E 2 .We
                           assume that E 1 <E 2 . We denote by i the current through the resistor R, and
                           we set i = i 1 + i 2 , where i 1 denotes the current through diode D 1 , and i 2 is the
                           current through diode D 2 . We suppose here that the resistor is nonlinear with
                           ampere–volt characteristics described by the relation
                                                              5
                                                       V R = Ri ,
                           where V R is the difference of potential across the resistor. The Kirchoff voltage
                           laws yield the system
                                             ⎧
                                                             5
                                                E 1 + R(i 1 + i 2 ) − u =+V 1
                                             ⎨
                                                                                     (4.113)
                                                E 2 + R(i 1 + i 2 ) − u =−V 2 ,
                                             ⎩               5
                                                        (i 1 ) is the difference of potential across
                           where V 1 ∈ ∂  R +  (−i 1 ) =−∂  R −
                                                 (i 2 ) is the difference of potential across diode D 2 .
                           diode D 1 , and V 2 ∈ ∂  R +
                           Let us set
                                                     5
                                             R(i 1 + i 2 )      E 1 − u         i 1
                                 F(i 1 ,i 2 ) =         ,q =            ,I =        ,
                                             R(i 1 + i 2 ) 5    E 2 − u         i 2
                                            .                                        (4.114)
                                   ≡   R − ×R +
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