Homework assignment from , MEMS Capacitors lecture

Transcription

1 Homework assignment from , MEMS Capacitors lecture 1. Calculate the capacitance for a round plate of 100µm diameter with an air gap space of 2.0 µm. C = e r e 0 * A/d (1) e 0 = 8.85E-12 F/m e r = 1 (air) A = p*r 2 = 7.854E-9 m 2 d = 2 µm Into (1) yields: C = E-15 Farads or ff 2. If the capacitor in 1 above has the air gap replaced by water what will the capacitance be? Assuming the dielectric constant for water is 80 (from lecture on 05/02/06): Equation (1) yields 2.78E-12 F or 2.78 pf 3. Design an apparatus that can be used to illustrate the attractive force between two parallel plates when a voltage is applied. Using a portion of the structure from lecture on 05/02/06, page 10: Figure 1 - Pressure Sensor from Lecture (05/02/06) Page 1 of 7

2 Assume the following: A circular diaphragm made of aluminum is used with a diameter of 500 µm and a thickness of 2.5 µm The width and thickness of the insulator spacer is 2 µm The bottom plate under the diaphragm is highly doped poly The volume is not captured (there are reliefs in the metal and/or diaphragm) so the diaphragm is at 1 atm in air. When voltage is applied across the top and bottom plates, the electrostatic force will be (from the 05/02/06 lecture notes, page 15): F elec = (e r e 0 * A * V 2 ) / (2d 2 ) (2) A = The area of the diaphragm V = The applied voltage d = The distance between the plates Using the assumptions above and applying 50 Volts, yields the following result from (2): F elec = 543 µn The equation for the deflection at the center of the diaphragm (from the 05/02/06 lecture notes) is: Y = 3PR 4 * [(1/v) 2-1] / (16E(1/v) 2 δ 3 (3) E = Young s Modulus, 0.68E12 dynes/cm 2 for Al v = Poisson s Ratio, for Al δ = The diaphragm thickness, 2.5 µm P = The pressure exerted by the electrostatic force on the diaphragm (543 µn/a = 2.766E3 N/m 2 ) R = The diaphragm radius, 250 µm This yields a center deflection of: Ymax = 1.69 µm This deflection will be easily observed but not cause the diaphragm to contact the bottom surface. Page 2 of 7

3 4. Design and build an electronic circuit that can measure the capacitance of two metal plates (the size of a quarter or so) separated by a thin foam insulator for various applied forces. Using readily available components, the following capacitor was built: Figure 2 Home Made Capacitor The capacitor plates are 1 square and constructed from thin copper tape. A better material would be thick PCB with copper plating to create rigid plates. The dielectric material is 0.1 packing foam and easily deformed by external pressure. Using equation (1) to calculate C: C = e r e 0 * A/d e 0 = 8.85E-12 F/m e r = 2 (roughly assumed a value between air FR4) A = 1 x 1 = 1 in 2 d = 0.1 C = 4.5 pf Page 3 of 7

4 This capacitance value will be very difficult to measure with a high degree of accuracy using readily available equipment and components. However, since the foam compresses easily to about 0.01 inches under applied external pressure, the capacitance will increase to about 45 pf under this pressure. The circuit in Figure 3 below is a simple Schmitt trigger oscillator circuit using the 74LS14 (without the power supply and bypass capacitors shown). The theory of operation is as follows: Initially, the capacitor has no stored charge and hence the voltage at the input of the inverter is 0V. This causes a logic high (~4V) to appear at the inverter output. Resistor, R1 charges the capacitor, C2 until the input of the inverter is recognized as a logic high (~1.6V). The output of the inverter then changes state to a logic low causing C2 to discharge through R1 until the input of the inverter is recognized as a logic low (~0.8V). The cycle then repeats. R1 1k U1A V 0 C2 {Cvar} PARAMETERS: Cvar = 10e-12 Figure 3 - Simple Schmitt Trigger Oscillator The frequency of oscillation depends on the value of tau = R1 * C2 as follows: The charge phase: When the capacitor is charging towards 4V, the equation for the voltage across C2 is: t / τ VC2 = 4V (1 e ) (4) However, once this voltage reaches 1.6V, the inverter changes state and the capacitor starts to discharge. Setting V C2 = 1.6V and solving for t yields: t charge = * tau When the capacitor discharges from 1.6V to 0.8V, the equation for V C2 is: Page 4 of 7

5 V C Ve t /τ 2 = 1. 6 (5) Once V C2 reaches 0.8V, the charge cycle begins again. Setting V C2 = 0.8V and solving for t yields: t discharge = * tau Therefore, one period of this charge/discharge waveform is: T = t charge + t discharge = * tau This yields an oscillation frequency of 1/T = 0.83/tau Hz The circuit as shown in Figure 3 is limited to resistors of up to a couple of k-ohms at best due to the significant input current (~ 200 µa) required by the inverter. In order to keep the frequency of oscillation under 1MHz to minimize high-frequency effects (on the breadboard), a larger resistor is required. The circuit shown in Figure 4 below utilizes a simple op-amp buffer to supply the necessary inverter input current. Since the input bias current required for the opamp is in the na range, a much larger resistor can be used to charge the capacitor. VCC V1 5VDC V2 0 VEE VEE 5VDC R1 47k V+ V- OS1 OUT OS2 U2 ua U1A V 0 C2 {Cvar} PARAMETERS: Cvar = 10e-12 VCC Figure 4- Modified Schmitt Trigger Oscillator Page 5 of 7

6 The circuit shown in Figure 4 was simulated, built and finally tested. The results of this investigation are shown below in Table 1 and Figure 5: Table 1 - Calculated, Simulated and Measured Results R1 (meas) = 46.3E+3 ohms Ceramic Capacitors Decade R-C Box Capacitance Calculated Frequency Simulated Frequency Measured Frequency Measured Frequency % Difference % Difference (Farads) (Hz) (Hz) (Hz) (Hz) Meas to calc Meas to sim (Ceramic disc) (Ceramic disc) n/c n/a E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E+3 1.0E E E E E E-9 9.0E E E+3 3.0E-9 6.0E E E+3 3.3E-9 5.4E E E E-9 4.5E E E E-9 1.8E E E Oscillator Frequency as a Function of Capacitance Frequency (Hz) 1.E E E+3 Ceramic Disc Decade "C" Box Simulated (PSPICE) Calculated 1.E E E E E-9 Capacitance Value (F) Figure 5 - Comparison of Calculated, Simulated and Measured Results The calculations, simulation results and measured data (on 5% and 10% ceramic disc capacitors) agree fairly well for capacitance values of 100 pf and larger. Between 10pF and 100pF, the simulated results and the measured results agree to within + 22 and -21.7%w while the calculations fail to accurately predict the output frequency for such low values of capacitance. Page 6 of 7

7 For small values of capacitance under test, even in the 10 s of pf range, the circuit in Figure 4 does a reasonable job of estimating capacitance values using the simulation data. It is clear that the simulation model is more useful than the hand calculations as it is more accurate over the wide range of capacitor values tested. The home-made capacitor shown in Figure 2 caused the following output frequencies when placed in the circuit as the capacitor under test: No applied force => A great-deal of force => khz 1.8 khz This yields an estimated range on C of less than 10pF to about 10,000pF. Page 7 of 7