Déformation photoinduite dans les films minces contenant des ...

Chapter 2. Experimental setup and sample preparation 30V =V ′ =rr + R V 0 = Y p0Y + Y 0 V 0 (2.4)ZZ + Z pV 0 =pY pY + Y pV 0 (2.5)and the differential output signal ∆V = V − V ′ is zero.When the system is excited at a resonance frequency the piezotube branch’s admittanceY 0pchanges:Y p = Y 0p + δY p (2.6)pastel-00527388, version 1 - 19 Oct 2010and the bridge is unbalanced:it results that the signal varies like −δV ′ :∆V = −δV ′ (2.7)dV ′= (Y + Y p) − Y p YV 0 =dY p (Y + Y p ) (Y + Y p ) V 0 (2.8)∆V = −δV ′ =Y Y p(Y + Y p ) 2 · δY pY p· V 0 (2.9)From this equation it easily seen that he quantity ∆V/V 0 is maximum for Y = Y p . Thuswe choose C ≃ C p and in the frequency range of interest we have:∆V ≃ − δC p· V exc ≃ 1 δC pV 0 (2.10)C p 2 C pThe signal ∆V depends on the tip-to-surface separation. At the resonance, if the tip isfar from the surface, its vibration amplitude is maximum, since no damping forces fromthe surface are present. We define ∆V R the voltage corresponding to this situation. Onthe other hand, if the tip is close to the surface the vibration amplitude is damped by