WEBVTT NOTE Resonance and Q 1 00:00:00.000 --> 00:00:14.675 Resonant frequency, computed. f 0 equals (1) divided by (2 pi the square root of (LC)) Note what is absent: R. 2 00:00:14.675 --> 00:00:42.425 Resistance does not affect resonant frequency at all, it affects Q, and therefore bandwidth, but not where resonance falls. The exam supplies an R value in these questions precisely to see whether you use it. 3 00:00:42.425 --> 00:01:57.002 Two worked answers the pool asks: R = 22 Ω, L = 50 µH, C = 40 pF: √(50 times 10 to the power of minus 6 times 40 times 10 to the power of minus 12) = √(2 times 10 to the power of minus 15) = 4.472 times 10 to the power of minus 8 f 0 = 1 / (2π times 4.472 times 10 to the power of minus 8) = 3.56 MHz R = 33 Ω, L = 50 µH, C = 10 pF: √(50 times 10 to the power of minus 6 times 10 times 10 to the power of minus 12) = 2.236 times 10 to the power of minus 8 f 0 = 1 / (2π times 2.236 times 10 to the power of minus 8) = 7.12 MHz The R was irrelevant both times. 4 00:01:57.002 --> 00:02:06.658 What happens at resonance. There is a diagram here. Impedance magnitude plotted against frequency for a series RLC circuit. 5 00:02:06.658 --> 00:02:20.866 The curve falls steeply to a minimum equal to the resistance at the resonant frequency, then rises again. Below resonance the circuit is capacitive; above resonance it is inductive. 6 00:02:20.866 --> 00:02:56.506 Series and parallel behave oppositely, and Extra tests both properties of both., Impedance magnitude; Series resonance, approximately the circuit resistance; Parallel resonance, approximately the circuit resistance., Input current; Series resonance, maximum; Parallel resonance, minimum., Circulating current in L and C; Series resonance,; Parallel resonance, maximum., Voltage and current phase; Series resonance, in phase; Parallel resonance, in phase. 7 00:02:56.506 --> 00:03:11.185 Both read "approximately equal to circuit resistance", the pool asks each separately and they have the same answer, because at resonance the reactances cancel and only resistance remains. 8 00:03:11.185 --> 00:03:25.708 The value of that resistance differs enormously between the two cases (small in series, large in parallel), but the statement is the same. The parallel pair is the counterintuitive one: 9 00:03:25.708 --> 00:03:34.708 Question. What is the magnitude of the current at the input of a parallel RLC circuit at resonance? 10 00:03:35.608 --> 00:03:38.288 Answer. Minimum. 11 00:03:38.888 --> 00:03:48.888 Question. What is the magnitude of the circulating current within the components of a parallel LC circuit at resonance? 12 00:03:49.788 --> 00:03:53.108 Answer. It is at a maximum. 13 00:03:53.708 --> 00:04:09.818 Large current sloshes back and forth between L and C, while almost none flows in or out. The tank circuit is storing energy and exchanging it internally; the source only has to make up the losses. 14 00:04:09.818 --> 00:04:27.901 And at resonance, voltage and current are in phase, the definition of a purely resistive impedance. Q. Q is the ratio of energy stored to energy lost per cycle, and it has two formulas that are reciprocals of each other. 15 00:04:27.901 --> 00:04:40.148 Getting them the wrong way round is the most common error in this subelement. Q series equals (X) divided by (R) Q parallel equals (R) divided by (X) 16 00:04:40.148 --> 00:04:47.135 Question. How is the Q of an RLC parallel resonant circuit calculated? 17 00:04:48.035 --> 00:04:55.515 Answer. Resistance divided by the reactance of either the inductance or capacitance. 18 00:04:56.115 --> 00:05:13.768 Either, because at resonance X_L = X_C, so it does not matter which you use. A memory hook: in a series circuit the resistance is in the way, so more R means lower Q, R in the denominator. 19 00:05:13.768 --> 00:05:24.755 In a parallel circuit the resistance is a leakage path across the tank, so more R means higher Q, R in the numerator. 20 00:05:24.755 --> 00:05:27.498 Half-power bandwidth. 21 00:05:27.498 --> 00:06:00.287 BW equals (f 0 ) divided by (Q) 7.1 MHz, Q of 150: 7.1 times 10 to the power of 6 / 150 = 47.3 kHz 3.7 MHz, Q of 118: 3.7 times 10 to the power of 6 / 118 = 31.4 kHz Half-power means the −3 dB points, the same convention as every filter specification. 22 00:06:00.287 --> 00:06:14.395 What raising Q costs you. Two consequences, both asked, and both worth understanding rather than memorising. 23 00:06:14.395 --> 00:06:21.235 Question. What is the result of increasing the Q of an impedance-matching circuit? 24 00:06:22.135 --> 00:06:26.251 Answer. Matching bandwidth is decreased. 25 00:06:26.851 --> 00:06:29.090 Directly from BW = f 0/Q. 26 00:06:29.090 --> 00:06:47.091 A high-Q matching network matches beautifully at one frequency and poorly a little either side, which is exactly the mobile-antenna bandwidth problem from the General track, seen from the circuit side. 27 00:06:47.091 --> 00:06:53.632 Question. What is an effect of increasing Q in a series resonant circuit? 28 00:06:54.532 --> 00:06:58.698 Answer. Internal voltages increase. 29 00:06:59.298 --> 00:07:16.350 The voltage across the inductor and across the capacitor is Q times the applied voltage. They are equal and opposite, so they cancel in the total, but each is individually real. 30 00:07:16.350 --> 00:07:21.938 A Q of 100 with 100 V applied means 10 kV across the coil. 31 00:07:21.938 --> 00:07:32.716 Question. What can cause the voltage across reactances in a series RLC circuit to be higher than the voltage applied to the entire circuit? 32 00:07:33.616 --> 00:07:36.336 Answer. Resonance. 33 00:07:36.936 --> 00:07:57.828 This is why antenna tuner capacitors arc, why mobile loading coils flash over, and why a high-Q circuit is a high-voltage circuit whatever the supply says. Check yourself. L = 20 µH, C = 100 pF, R = 15 Ω. 34 00:07:57.828 --> 00:08:17.696 Resonant frequency? A parallel resonant circuit has R = 5 kΩ and X = 50 Ω. What is Q, and the bandwidth at 14 MHz? Why does a high-Q antenna tuner need bigger capacitor spacing than a low-Q one? 35 00:08:23.696 --> 00:08:47.963 √(20 times 10 to the power of minus 6 times 100 times 10 to the power of minus 12) = 4.472 times 10 to the power of minus 8; f 0 = 3.56 MHz. R is irrelevant. Parallel, so Q = R/X = 5000/50 = 100. 36 00:08:47.963 --> 00:09:15.216 BW = 14 times 10 to the power of 6 / 100 = 140 kHz. Internal voltages are Q times the applied voltage, so a high-Q network develops much higher voltages across its reactances, and arcs at spacings a low-Q one tolerates.