Bandpass Filter Terminology f r Rejection A r Upper and Lower Rejection Frequencies Shape A r : SF = B r (A r )/B p Center Frequency Upper and Lower Cutoff Frequencies (3 dB) Passband Insertion Loss Ripple Bandwidth (3 dB Passband) Note: An attenuation must be specified in order to determine shape factor: “Shape Factor at 30 dB attenuation.”

Pole Placement and Pass Band

Bandpass Filter Math Horizontal axis is Logarithmic: f p+ /f 0 = f 0 /f p- and f r+ /f 0 = f 0 /f r- Center Frequency is Geometric Meanf 0 2 = f p+ f p- = f r+ f r-

Standard Design Curves Attenuation vs Shape Factor Separate Curves for different Ripple Inside Passband Outside passband Separate sets of curves for different Number of Poles (3 poles shown) DO NOT USE q MIN ON CHARTS

Use Loss Curve to Determine Q u (MIN) Figure 7-14 Q L /Q u vs Insertion Loss Per Pole

Filter Specs Center Frequency Passband Allowable Passband Insertion Loss/Ripple Required Out of Band Attenuation Rejection Bandwidth Number of Poles Design Ripple Component Unloaded Q Use Curves to Determine

Use Tables (p 230) to determine k and q values...

C 1A C 1,2 C 1B C 4A C 4B C3C3 C2C2 C 2,3 C 3,4 LL LL Four Pole, Parallel, Capacitive, Top Coupled Filter RSRS RLRL Coupling Capacitors Tank Capacitors End Loading This is what the author refers to as the “design impedance level” Equation 7-20 for  = 1, (used by author to compute X T = X L )

Design Example FM IF Filter – 200 Khz Channel Spacing Requirements: 1.Center Frequency – 10.7 Mhz 2.Acceptance ( 3 dB) Bandwidth – 160 Khz 3.Rejection – 30 dB at 240 Khz BW; SF(30 dB) = Max Insertion Loss – 4 dB 5.Ripple – 0.1 dB max 6.R S = 75  R L = 50  200 Khz 160 Khz 240 Khz 4 dB 3 dB 30 dB 10.7 Mhz 200 Khz 0 dB f

Determine Poles, Ripple, Q u (min), k, q 160 Khz 240 Khz 4 dB 3 dB 30 dB 10.7 Mhz 0 dB f Appendix B: Want > 30 dB attenuation at SF = 1.5 Not possible for 2,3,4 poles 5 poles, curve 6 – 1 dB ripple... Too much 6 poles, curve 4 – 0.01 dB ripple OK We could choose curve 5, 0.1 dB ripple and 32 SF = 1.5 Fig 7-14 Loss Per pole = 4 dB/6 poles = 0.66 dB/pole Q L /Q u (min) = Q L = 10.7/0.16 = Q u (min) = 891 (Difficult!!) Table dB Chebychev, n = 6 q 1,n = k 1,2 = k 5,6 = k 2,3 = k 4,5 = k 3,4 = 0.518

Determine Component Values Choose X T in the range of 50 – 500  - - Lets pick 135  (just for fun)

Determine Taps for End Loading C 6A = 1460 pFC 6B = 117 pF C 1A = 1195 pF C 1B = 119 pF These could also be implemented as autotransformer tap points on the input/output inductors.

Critique of Author’s Methodology “design impedance level” is really not a design choice, but a threshold for limiting tank reactance based on an arbitrary lower limit for inductors (50 uH). The author’s method requires equal source and load resistances, which is not always possible or desirable. Using the author’s method, the tank reactance is determined by the source and load impedances. The use of reactive voltage dividers on the input and output tanks allows the tank reactance to be chosen independently.