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Ozone Cell Alan Millner 01-18-2005 Review of the Field of a Pair of Oppositely Charged Plates in Vacuum By symmetry, no field outside If plates have charge Q, area A, then by Gauss’ Law D = 0E = Q/A between plates If separation is d, then V = dE = dQ/ 0A or Q= (0A/d) If the capacitance C = 0A/d then Q = C V Energy W = ½ QV = ½ 0E2 (dA) Or, W = ½ C V2 +Q -Q A d Polarization and Dielectric Constant Suppose we have a polarization between the plates P proportional to E Gauss’ Law says D = P+ 0E = Q/A Define dielectric constant in a material by P = ( -0) so D = E = Q/A So V = dE = dQ/ A , or Q = CV where capacitance C = A/d (substitute for 0) Note W = ½ C V2 = ½ E2 (dA) – Energy density times volume Design Constraints Need 4E7 volts/meter across O2 to make O3 Broad optimum around 30kHz Need insulating layer to avoid arc damage – Alumina from Kyocera is good Need 1:1000 round shape factor for support Over 4kV becomes difficult to insulate Fewer larger cells are less expensive NOW YOU DESIGN AN OZONE CELL Ozone Cell Construction Ozone Cell High Voltage Wire Steel shell Alumina 3mm ground HV Electrode metal 90mm dia Alumina 0.20mm gas space 0.11mm ground electrode Steel shell Ground wire Equivalent Circuit Ozone Cell Equivalent Circuit C2 R1 C1 C3 Ozone Cell Analysis Dimensions diameter Th1 back Th2 active Airgap mm Dielectric const K Mu0 Eps0 Area Cback Cdiel active Cairgap Cactive tot Ctotal m 90 3 0.2 0.11 0.09 0.003 0.0002 0.00011 9.5 1.25664E-06 8.85E-12 0.006361725 1.78E-10 2.68E-09 5.12E-10 4.30E-10 6.08E-10 Energy W= v*I dt W= v* (dQ/dt)* dt= vdQ For linear C, Q = C*V dQ= C dv W= C*vdv from v=0 to v=V W = ½ C V2 Or W = ½ V*Q (area under curve of V vs Q) Energy in a field V= E*d = Efield/gap Q=D*A= Dfield*area W= (dA) * ( E*dD) Volume V = dA W = volume * energy per unit volume = V *U If linear, D = E and U = E*dD = ½ E2 Displacement current I = dQ/dt Q = A*D = A * E So I = d(A* E )/dt, I/A = d(E )/dt We may identify d(E )/dt as displacement current density Like j=I/A = current density Note later: H*dr = {j+ d(E )/dt} dA so a loop around either current J or displacement current d(E )/dt produces the same H field Fringe fields If C = Co + Cfringe W = Wo + Wfringe Cfringe/Co = Wfringe/ Wo If Efringe <= Eo then Cfringe/Co <= Vfringe/ Vo Vo = dA Vfringe = d2 *P where P = perimeter Fringe fields example Our example of a circular capacitor, radius R Vo = d* *R2 Vfringe = d2 * 2R Vfringe/Vo = 2d/R In our example, better than 1%. Field Patterns Consider 2 dimes, arranged on an axis, 10 meters apart, oppositely charged to 1 coulomb What does the field pattern look like: 1. within 1 mm of the positive dime's surface? 2. one meter from the positive dime? 3. 5meters from the axis center? What is the field strength at each location above?