Transduction Based on Changes in the Energy Stored in an Electrical Field. Lecture 6-5. Department of Mechanical Engineering

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1 Transduction Based on Changes in the Energy Stored in an Electrical Field Lecture 6-5

2 Transducers with cylindrical Geometry For a cylinder of radius r centered inside a shell with with an inner radius of r, the capacitance is given πεl C l is the cylinder length In r r ( ) From QCV, we may write C ( r r ) QIn π εl Three fundamental parameters could be varied mechanically to induce a voltage, or which could be stimulated to change by application of a voltage: Permittivity ε, The cylinder length l and ln(r /r )

3 Transducers with cylindrical Geometry If an inner electrode radius is fixed The energy stored by the capacitor Q Q In( r r ) V dw dq current i W dw W Q W dr + dt Q t r dt C 4πεl Q r dw Q ln F πεl r dr 4πεlr Again these constitutive equations are nonlinear. As discussed before, we can certainly work with the nonlinear relations, but it is more common to use a linear approximation.

4 Transducers with cylindrical Geometry Suppose we are interested in a pressure, for example in which the change in r is used to produce a single voltage Then we consider Q r V ln πεl r Assume that the charge, voltage, and varying radius can each be expressed as a static term and a small perturbation, then we have lateral motion of the moving plate can be written as ~ r ~ 0 r ( Qo + Q) In + In ~ r + r 0 V V + 0 πεl We may rewrite the expression with Taylor series expansion ~ ~ Q ~ or V0 + V Qo In( r0 r ) + QIn( r0 r ) + πεl r0 for the first order only ~ r In + r 0

5 Transducers with cylindrical Geometry Balance the static term, we have QoIn( r0 r ) Vo πεl The terms which are first order small perturbation produce ~ V ~ QIn πεl Q or ( r r ) + 0 It is the second term that we wish to emphasize, so we may resort to electrical isolation of ac portion from dc bias, so that the charge on the capacitor is essentially constant. ~ V Q ~ r πεlr 0 r 0 ~ o o o Sensitivity ψ In( r0 r ) ~ V ~ r Q πεlr 0 V r 0

6 Gradient Transduction Using Two Dielectrics Now we consider the transduction mechanism based on changes in the energy stored in an electrical field uses two solid dielectric materials with different permittivities There are two cases: Normal gradient Tangential gradient This particular transduction mechanism is not often used, however, the use of gradient transduction could provide a means of noninvasively producing sound wave in solid materials----as actuator The effective capacitance C eff C + C where C i ε i wy i d

7 Gradient Transduction Using Two Dielectrics Q W CV C W * Since the voltage is specified in the system Instead of using the energy stored by the capacitor from We use the co-energy form The capacitance is independent of Q and V, the constitutive relation defining a capacitor is linear, the energy and co-energy are the same value. Using the co-energy, we have W W* Since y +y l W CV ( ε y + ε y ) [ ε y + ε ( l y )] wv wv

8 Gradient Transduction Using Two Dielectrics The net interface force F W y wv [ ε ε ] This equation directly relates the interface force with voltage. Once again the equation is nonlinear and it depends on the values of permittivity of the dielectrics. Greater interface force is generated for cases of widely varying permittivities Now let consider the second case: the tangential gradient capacitor The effective capacitance is + C eff C C

9 Gradient Transduction Using Two Dielectrics The energy stored in the capacitor W We determine the capacitance from each section C εa d and C ε A d Thus the energy stored by the capacitor W Q ( ε d + ε d ) ε ε A The net interface force will be F W d Q A ε ε ( C + C ) Q Q C C C This equation suggests that application of a voltage across the capacitor plates leads to a force at the interface between two dielectrics. The force will cause a movement of the interface. The exact movement is dependent upon the elastic compliances of the two dielectrics. Large force will be produced if the two dielectrics have widely different permittivities This mechanism could be used to impose a force at the interface which lunches compressive waves in the media eff

10 Nonlinear Dynamics of Parallel Capacitor Device If a parallel capacitor that is composed of a fixed plate a and a movable plate with a spring 3. Its motion is translational along the x-axis only and is characterized by lump parameters of mass m, friction coefficient b and spring factor k. An external force on system F is applied to plate. The electrical circuit of the system includes the capacitor and resistor R in series with the capacitor. The voltage at the terminal is V. R Our goal: to write electromechanical equations describing the behavior of the system

11 Nonlinear Dynamics of Parallel Capacitor Device For the electrostatic force: The mechanical force F M, opposing F, contains three components: force of inertia m(d x/dt ), force due to friction, b(dx/dt), and spring force kx:

12 Nonlinear Dynamics of Parallel Capacitor Device Therefore, we have F s mx + sbx q + kx + εa (A) s: differential operator On the other hand, the equating of voltages in the electrical circuit (KVL) gives: using The above equation can be rewritten as: Thus we have y srq + q V ε A (B) y is the gap at the new position: x+yδ

13 Nonlinear Dynamics of Parallel Capacitor Device The equation (A) and (B) describe the mechanical and electrical behavior of the system, whose transfer characteristic can be determined if the functional relation between the mechanical and electrical quantities is found. The linearization is possible since the plate displacement is small and the changes of electrical quantities are also small. In addition, a stable equilibrium point exits The differentiation of F gives: df q εa ( ) o s m + sb + k dx + dq The differentiation for V is written as: dv sr dq qo + dy εa yo + dq Rs εa q o charge for a stable equilibrium point δ x + εa o qo dq dx εa

14 Variable Area parallel Capacitor Replacing df, dx, dv, and dq with incremental changes F, x, V, and q, the system of two simultaneous, linear differential equations is obtained: ( ) V q A x Rs x A q F q A q x k sb m s o o o ε δ ε ε These equations in matrix form: ( ) V F q x A x Rs A q A q k sb m s o o o ε δ ε ε These equation can then be easily solved (C)

15 Temperature Instability In capacitive device, the temperature deformations quite often produce a change of capacitance comparable with that from the measurand. There are a number of techniques of compensation in the signal stages We will only discuss how a proper choice of materials can significantly improve temperature stability Let s consider a round, pressuresensitive device containing a metal casing with a conductive thin pressuresensitive diaphragm 4, which forms a variable capacitance with fixed electrode 3. The electrode is supported by posts, and is electrically isolated from casing.

16 Temperature Instability For the initial temperature δ o l - l For the changed temperature, a new gap δ t is δ o l (+α ) t -l (+α ) t The error γ in the reading of initial capacitance C o due to a uniform temperature change t. A deformation of electrode 3 ( ) t is neglected, the congruous area of electrodes remains the same.

17 Temperature Instability To have error γ0, we need to have α l or l α l. α α δ This condition can be satisfied for the given gap if α > α

18 Temperature Instability If temperature gradients exist during operation, the error due to temperature gradient could significant be which need to be considered Again, let consider a pressure sensitive diaphragm. If the temperature of the diaphragm changes due to, for example, the contact of hot air, while the casing temperature remain the same. The stress balance in the diaphragm is disturbed, which leads to the change of the transfer characteristic of the diaphragm If a plain diaphragm is buckled due to elongation of the materials under temperature change, the center displacement will be

19 Temperature Instability The length of the arc: Substitute into the first equation, we have Even if the diaphragm is made of low-α material, the magnitude of h is high h0. mm For example, for Ni-Span-C with α8.0 x 0-6 t5k and a50 mm This is a very high value for capacitive element such as microphone

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