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  • 3 years ago

Parallel resonant "tank" circuits are common in radio equipment. But unfortunately there is always resistance that prevents them from being perfect: in every real inductor, the wire that makes up the inductance has some resistance, and there may be leakage in the capacitor that can be modeled as a resistance. Also, the Norton resistance of the system connected to the tank circut looks like a leakage through the capacitor. So a realistic model for a tank circuit is the following: In the space provided below write an algebraic expression in terms of the device parameters for the bandwidth Δω o

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  1. KenLJW
    • 3 years ago
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    s = jx : x = 2 pi f parallel RLC Tank 1) Z = R||sL||1/SC = (1/C) s/(s^2 + s/RC + 1/LC) |Z|^2 = (1/C^2) x^2/(x^2/(RC)^2 + (1/LC -x^2)^2) x^2 = 1/LC then Z = R for 3dB points |Z|^2 =R^2/2 |Z|^2 = x^2/(x^2 + (RC(1/LC - x^2)^2 =1/2 The two solutions above are the 3dB frequencies, there difference is the BW parallel LC tank with Rs in series with L 2) Z = (sL + Rs)||1/sC = (1/C)(s + Rs/L)/(s^2 + sRs/L1/LC) to make the denominator of eq 1 and 2 equal Rs/L = 1/RC R = L/CRs Q = xo L/Rs = 1/xo C Rs R = (xoL/Rs)(1/xoCRs)Rs = RsQ^2 place this R into equation 1 and find lower 3dB frequency and if it's greater than 10 Rs/L then eq 1 and 2 are a good approximation of each other Generally (sL + Rs)||1/sC = Rp||sL||1/sC Rp = RsQ^2 for 1/(LC)^1/2 > 10 Rs/L : L/C > 100 Rs^2

  2. KenLJW
    • 3 years ago
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    correction Lower 3dB point > 10 Rs/L

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