Series and Parallel Resonance Series Resonant circuit L 4.7mH C 0.001µF R 47Ω Vin 1V (p-p) fr = ? VC = ? XC = ? VL = ? XL = ? Q=? IT = ? βf = ? Θ=? Find fr • fr = 6.28 1 2π πΏπΆ 1 = 1 6.28 4.7×10−12 = 4.7×10−3 1 1.362×10−5 0.001×10−6 = = 73.412kHz Find reactances • XL = 2πfL = (6.28)(73.142X103)(4.7X10-6) = 2.168kΩ • 1 1 XC = = 2πππΆ 6.28 73.142×103 0.001×10−6 1 = 2.168kΩ −4 4.613×10 = Since XC and XL are equal, along with being 180° out of phase, the net reactance is zero which makes the total impedance equal to the resistor ∴ ZT = R Find total current and voltages • IT = πππ ππ = πππ π = 0.3535 = 47 7.521mA Since this is a series circuit, the current found for the total will also be the current flowing through the reactive components. • VC = (XC)(IT) = (2.168kΩ)(7.521mA) = 16.306V • VL = (XL)(IT) = (2.168kΩ)(7.521mA) = 16.306V As you can see, the resonant circuit appears to amplify the voltages. Find the Q of the circuit • Q= ππΏ ππ ο¨ Since there is no value given for a resistance of the coil, we have to use the only resistance in the circuit to find this value ππΏ 2.168ππΊ ∴Q= = = 46.128 π 47πΊ Solve for Bandwidth and Cutoff frequencies ππ π 73.142ππ»π§ 46.128 • βπ = π2 − π1 = = = 1.592kHz ο¨This means the frequency will vary ±796Hz βπ . The entire range is also known as 2 Bandwidth. • f2 = βπ fr + = 73.412kHz + 796Hz = 74.208kHz 2 βπ fr - = 73.412kHz - 796Hz = 72.616kHz 2 • f1 = • θ = 0° since XL and XC are canceling, which means at resonance the circuit is purely resistive. Parallel Resonant circuit C 162.11pF L 100µH VA 10v(p-p) 50% rs 7.85Ω fr = ? IL = ? XL = ? Q=? XC = ? Zeq = ? IC = ? IT = ? βf = ? Solve for fr • fr = 6.28 1 2π πΏπΆ 1 = 1 6.28 1.621×10−14 100×10−6 = 1 799×10−9 162.11×10−12 = 1.25MHz = Find the reactances • XL = 2πfL = (6.28)(1.25X106)(100X10-6) = 785.394Ω • 1 1 XC = = 2πππΆ 6.28 1.25×106 162.11×10−12 1 = 785.417Ω −3 1.273×10 = Since this is a parallel circuit, we presume the applied voltage will be across each reactive component. Find branch currents and the equivalent impedance • • • ππ΄ 10 IC = = = 12.732mA ππ 785.417 ππ΄ 10 IL = = = 12.732mA ππΏ 785.398 ππΏ 785.398 Q= = = 100.051 ≅ ππ 7.85 100 • Zeq = QXL = (100.051)(785.398) = 78.58kΩ ο¨ Since this is the only way we are going to get the total impedance, we now need to use it to find the total current. Find total current • IT = ππ΄ πππ = 10 78.58ππΊ = 127.259µA Again, we can see the magnification of the current due to resonance. Solve for Bandwidth and Cutoff frequencies • βπ = π2 − π1 = ππ π = 1.25ππ»π§ 100.051 = 12.494kHz ο¨This means the frequency will vary ±6.247kHz • f2 = • f1 = βπ 2 βπ fr + = 1.25MHz + 6.247kHz = 1.256MHz 2 βπ fr - = 1.25MKz – 6.247kHz = 1.243MHz 2