LPF EXAMPLE You can use the LPF program to calculate standard passive filters. ----------------------------------------------------------------------- As an example, design the 3-pole Tchebychev broadbanding filter used in the ClassE startup example: Press enter 3 [elements] Press

enter 0.4 [passband ripple] -The normalized coefficients are displayed Press Enter 50 [Source impedance] Enter 4.4e6 [Cutoff freq] -A Shunt-C and Series-L design is displayed: TCHEBYCHEV LOWPASS FILTER ORDER : 3 Passband Ripple = 0.400 dB Cutoff Frequency: 4.400000E+06 Hz SHUNT CAPACITOR INPUT SERIES INDUCTOR INPUT Ro = 5.000000E+01 Ro = 5.000000E+01 C1 = 1.078549E-09 L1 = 2.696373E-06 L2 = 2.021996E-06 C2 = 8.087983E-10 C3 = 1.078549E-09 L3 = 2.696373E-06 Rload = 5.000000E+01 Rload = 5.000000E+01 Although this filter is down 0.4 db at 4.4 Mhz, it was purposely selected to provide a non-reactive 50 +j0 load at the 3.8 Mhz design center for the CE amp. ----------------------------------------------------------------------- For a PDF modulator filter you might experiment with 4 or 6 element designs. at about 10KC cutoff (Band-limit incoming audio to about 6.5KC to minimize filter ringing transients). The higher the ripple, the more the attenuation at the switching frequency. Try 0.5 to 1 db. For the even order Tchebychev filters, the Source and Load Z are unequal, so choose Ro to be the RF amp dc impedance, but run the Shunt-C filter in REVERSE order. See the 5 ohm example below where the PDM modulator would connect to the (L4) 62.8 uhy inductor and the RF stage is shunted by the (C1) 6.68 uf cap.. TCHEBYCHEV LOWPASS FILTER ORDER : 4 Passband Ripple = 1.000 dB Cutoff Frequency: 1.000000E+04 Hz SHUNT CAPACITOR INPUT Ro = 5.000000E+00 Ro = 5.000000E+00 C1 = 6.681738E-06 L1 = 1.670435E-04 L2 = 8.470366E-05 C2 = 3.388147E-06 C3 = 9.012001E-06 L3 = 2.253000E-04 L4 = 6.280156E-05 C4 = 2.512062E-06 Rload = 1.879797E+00 Rload = 1.329931E+01 Regards and have fun, BobbyT NU2B 2 April 2006