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5.1 Piezoresistivity                                                           87

                      If it is assumed that the block is made of a resistive material, then its resistance,
                  R, is given by

                                                      l ρ
                                                 R =                                     (5.3)
                                                     A

                  where ρ is the bulk resistivity of the material (Ωcm), l is the length, and A is the
                  cross-sectional area (i.e., the product of width w and thickness t).
                      Hence,

                                                      l ρ
                                                 R =                                     (5.4)
                                                     wt

                      Differentiating the equation for resistance gives
                                         l      ρ       l ρ       l ρ
                                   dR =    d +    dl −    dw −      dt                   (5.5)
                                            ρ
                                                        2
                                        wt     wt     wt        wt  2
                      and hence

                                          dR   dρ  dl   dw   dt
                                             =   +    −    −                             (5.6)
                                          R    ρ    l   w    t

                      By definition, ε = dl/l, so the following equations apply on the assumption
                                    l
                  that we are dealing with small changes, and hence dl = ∆l, dw = ∆w, and dt = ∆t:

                                   dw                    dt
                                       =ε  =−νε    and     =ε  =−νε                      (5.7)
                                    w     w      l       t    t      l
                  where ν is Poisson’s ratio. Note the minus signs, indicating that the width and thick-
                  ness both experience compression and hence shrink. It is worth noting that the
                  above example illustrates a positive Poisson’s ratio. 1
                      Therefore, from (5.6) and (5.7) we have

                                          dR   dρ
                                             =   +ε  +νε  +νε                            (5.8)
                                          R     ρ   l    l    l

                      From (5.1) the gauge factor is therefore

                                             dR R   dρρ
                                       GF =       =      + ( + ν12  )                    (5.9)
                                              ε      ε
                                               l      l
                      Equation (5.9) indicates clearly that there are two distinct effects that contribute
                  to the gauge factor. The first term is the piezoresistive effect ((dρ/ρ)/ε) and the sec-
                                                                                 l
                  ond is the geometric effect (1 + 2). As Poisson’s ratio is usually between 0.2 and 0.3,


            1.  Materials having a negative Poisson’s ratio do exist. That is to say, as you stretch them, the width and thick-
               ness actually increase. Examples of such materials include special foams and polymers such as polyte-
               trafluoroethylene (PTFE).
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