2ndOLPFilt

2nd Order LC Low-Pass Filter. Fo = 1/[2π√(L·C)] Fo = 1/[2π√(10 µH · 0.1 µF)] Fo = 159.155 kHz Q = 2πfRC = 2π·Fo·R·C Q = 2π{1/[2π√(L·C)]}·R·C Q = {R/[√(L·C)]}·C Q = R√(C/L) For the saved values Q = R√(0.1 µF / 10 µH) Q = |R|/10 Q = [0.7071 · 100 (Ω)] / 10 Q = 7.071 Peaking intensifies as Q increases. Maximally flat magnitude response results when R = 10/√2 = 7.071 Ω. In this case Q = |R|/10 = (10/√2)/10 = 1/√2 = 0.707. In the circuit a quick way to do this is to adjust the total value of R to 10 Ω while keeping the position at 70.71%. At high frequencies roll-off rate is -40 dB/decade asymptotic slope.