EE 3324 Electromagnetics Laboratory

Transcription

1 EE 3324 Electromagnetics Laboratory Experiment #2 Capacitors and Capacitance 1. Objective The objective of Experiment #2 is to investigate the concepts of capacitors and capacitance. The capacitance of coaxial and two-wire transmission lines will be measured and the basic properties of a parallel plate capacitor will be investigated. The physical properties of a parallel plate capacitor (plate separation, dielectric material) are varied to determine the effect on circuit behavior. 2. Introduction A capacitor is a device used to store energy in an electric field. The basic geometry of the ideal parallel-plate capacitor is shown in Figure 1. The plates are assumed to have an area A and the separation distance between the plates is defined as d. The material between the two conducting plates is a dielectric (insulator) characterized by permittivity, (F/m). A voltage V is applied to the capacitor which results in stored charge on the capacitor plates (total charges of +Q and!q). The capacitance C defines the relationship between the applied voltage and the stored charge. The vector electric field E (V/m) within the ideal capacitor is assumed to be uniform everywhere between the plates so that the potential difference between the plates may be written as (1) (2) Figure 1. Ideal parallel plate capacitor.

2 The assumption of a uniform electric field between the plates of the capacitor is accurate for large, closely-spaced plates. The uniform electric field within the ideal parallel plate capacitor requires that the charge density on the capacitor plates also be uniform. The boundary condition for the electric flux on the surface of the capacitor plate gives Inserting Equations (2) and (3) into the capacitance definition of Equation (1) yields the ideal parallelplate capacitance formula: The relationship between charge, voltage and capacitance holds true for both DC and time-varying voltages. Since the capacitance C is constant for a given conductor geometry, this implies that a time-varying voltage across a capacitor must yield a capacitor charge with the same time variation [Q(t) = CV(t)]. In an actual parallel plate capacitor, the charge density is larger at the plate edges than the interior region of the plate. This charge distribution causes fringing (or bending) of the electric field at the edges of the capacitor. For large, closely-spaced plates, the fringing effect is minimal and the ideal capacitor equation in (4) is an accurate approximation. As the spacing of the plates increases, the fringing effect becomes more significant and the accuracy of Equation (4) is reduced. For the ideal capacitor, the shape of the capacitor plates is immaterial. However, for actual capacitors, conductors with sharp corners produce larger charge densities and more pronounced fringing. When the atoms of the dielectric material insulating the conductors of the capacitor are exposed to the capacitor electric field, the equivalent charges in the atoms are displaced (polarized). The relative permittivity of the dielectric material is a measure of the degree of polarization within the dielectric. More polarization occurs in dielectrics with higher permittivities. The additional energy required to polarize the dielectric results in a higher stored energy within the capacitor. This effect is seen in the parallel plate capacitance formula of Equation (4). Capacitance exists whenever two conductors are in close proximity separated by an insulating material. Even devices which are not specifically designed to store energy have capacitance. For example, transmission lines designed to carry signals from one point to another via guided electromagnetic waves contain capacitance based on their physical structure. The coaxial transmission line shown in Figure 2 can be shown to contain the following capacitance per unit length: (5) (3) (4) Equation (5) represents the capacitance of an ideal coaxial capacitor in which the fringing effects at the cable ends are neglected. A two-wire transmission line formed by parallel conductors also has capacitance. Given equal wire radii of a and a center-tocenter spacing between the wires of d, the ideal two-wire transmission line capacitance per unit length (neglecting fringing at the conductor ends) is given by Figure 2. Coaxial transmission line.

3 (6) where, is the total permittivity of the insulating medium between the wires. 3. Equipment List LCR Meter Adjustable Parallel Plate Capacitors (283 mm 283 mm, 200 mm 200 mm) Dielectric plates (glass, polystyrene), Coaxial cable, two insulated copper wires Calipers, 120 ks resistor, 100 :F capacitor Breadboard Unit Digital multimeter 4. Procedure Note: Be sure to discharge any capacitor before making a capacitance measurement. 1. Use the LCR Meter to measure the capacitance of the segment of coaxial cable provided to you. Measure the length of your cable segment and compute its total capacitance using Equation (5) for the ideal coaxial capacitor (see Table 1 for the physical characteristics of your coaxial cable). Compare your measured capacitance and your computed capacitance with the manufacturer specification provided in Table 1. Cable a (mm) b (mm), r Nominal capacitance (pf/ft) RG 58C/U RG 59/U RG 213U RG 223U Table 1. Coaxial cable specifications. 2. Using calipers, measure the conductor dimensions (conductor diameter, insulation diameter) of the two straight insulated copper wires which are provided. Also measure the length of the two-wires. Form a two-wire transmission line by placing the conductors parallel to each other for the following spacings: [1] conductor insulation touching, [2] center-to-center wire spacing of 0.5 inch, and [3] center-to-center spacing of 1 inch. Measure the capacitance in each case with the LCR Meter. Discuss your results comparing them with the ideal two-wire line given in Equation (6). Note that your two-wire line has an inhomogeneous insulating

4 medium (air/dielectric). 3. Use the LCR Meter to measure the capacitance of the smaller (200 mm 200 mm) parallel plate capacitor for plate spacings of d = 1, 2, 3, 4 and 6 mm with air as the insulating medium between the plates. Use the insulating spacers on the corners of the capacitor plates to achieve the necessary plate spacing. Each spacer provides either 1 mm or 3 mm of plate separation, depending on its orientation. Align the leads from the LCR meter to the capacitor in such a way as to minimize their capacitive contribution to the measurement. Repeat these measurements for the larger (283 mm 283 mm) parallel plate capacitor. Note that the plate area of the larger capacitor is twice that of the smaller capacitor. Compare the measured capacitances of these two air-insulated capacitors with the values obtained using the ideal parallel plate capacitor equation. Discuss your results. How does fringing effect the percentage errors obtained for the two capacitors? 4. Connect the smaller and larger air-insulated parallel plate capacitors in series using plate spacings of 1 mm for both capacitors. Measure the capacitance of the series combination. Repeat the measurement for the capacitors in parallel. Compare your measured series and parallel capacitances with the appropriate circuit equation using the measured capacitances for the individual capacitors found in part Measure and record the thicknesses of the glass and polystyrene plates. Using the larger parallel plate capacitor, measure the capacitance of the glass-insulated capacitor and the polystyrene-insulated capacitor. Compare these values with the equivalent air-insulated capacitor and discuss the differences in capacitance with regard to polarization. Determine the approximate permittivities of glass and polystyrene based on your measurements. Compare these with the permittivity values of these materials found in your textbook. 6. Place both the glass and polystyrene panels between the capacitor plates in the form of a dielectric sandwich. Measure the capacitance of this inhomogeneous dielectric capacitor. Compare your measured results with what you would expect to measure based on the results for the individual plexiglass-insulated and fiberglass-insulated capacitors. Show the equivalent circuit of this inhomogeneous dielectric capacitor in your report. 7. Connect a parallel RC circuit with a capacitance of C = 100 :F and a resistance of R = 120 ks. Use the LCR meter to measure the actual values of R and C for your components. Note: You are using electrolytic capacitors so that the polarity of the voltage source across the capacitor is important. Be careful to connect the ground connection of your voltage source to the terminal labeled by a! on the capacitor. Apply a DC voltage of 12 V across the parallel circuit and measure the capacitor voltage with a multimeter. Disconnect the voltage source (open circuit the source connection so the capacitor can discharge only through the resistor) and measure the capacitor voltage at t = 0s to t = 45s in 5s intervals. (a.) (b.) From circuit analysis, we may write a simple expression for the timedependent voltage across the capacitor V(t) in terms of the initial voltage V o, the resistance R and the capacitance C. Using the same terms (V o, R and C) and the definition of capacitance, determine a similar expression for the capacitor charge with respect to time [Q(t)]. Plot the theoretical charge expression and the measured charge (computed from the measured voltage) verses time on the same plot. Based on each measurement of the capacitor charge, determine the following

5 5. Additional Questions quantities: (1) the time constant J for the parallel RC network, (2) the total amount on charge on each capacitor conductor before the source voltage was removed, (3) the total amount of charge on each capacitor conductor 30 seconds after the source was disconnected and (4) the time required for the charge on the capacitor to decay to 1% of its original value. 1. A parallel plate capacitor (d = 1mm, A = 20 cm 2 ) is filled with two dielectrics which are side by side. The relative permittivities of the dielectrics are 0 r1 = 2 and 0 r2 = 4. If each dielectric comprises one-half of the total capacitor volume, determine the overall capacitance? 2. Repeat question 1 given the same problem for a coaxial capacitor with a = 5 mm, b = 20 mm, and l = 10 m. The two dielectrics (0 r1 = 2 and 0 r2 = 4) are side by side and each comprises one-half of the total capacitor volume as shown below.