LC tank circuits, real world complications - Page 1
Forum section: electronics / beginners

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 03:50:04 pm
content: In theory an LC tank circuit will oscillate strongly (with something pretty close to a clean sine wave) at resonance when driven by even a small signal (perhaps badly distorted or square) at the right frequency:f=1/(2*pi*sqrt(LC))But I've found in reality for a fixed L, when one tries increasing the capacitor value beyond certain limits, and driving at appropriately lower frequencies, resonance doesn't seem to result.I assume this is a matter of imperfections in real components as versus the ideal models the equations are based on, is this commonly encountered once one starts getting to and above ratios of C/L=F/H=0.01?For example I've a 3uH inductor which happily oscillates in parallel with a 680pF cap at around 3.5MHz, but substitute the cap for 25nF and try at 580KHz and it won't resonate. A C/L ratio of a little under 0.01 where it failed.I also did some tests varying L and C together whilst keeping a constant frequency (LC=1e-12,f=159kHz) and it was fine for L=1mH,C=100pF, or 100uH,10nF until 10uH,100nF, above this (1uH,1uF...) no resonance would occur though, again a ratio of around 0.01 at the last combination it would work for.Is this need for L and C to be within a certain ratio for resonance to occur a common phenomenon? What causes it?ThanksEDIT: C/L>0.01 corresponds to Q=(1/R)*sqrt(L/C) going below Q=10/R, is this going to be playing a major role?

author: Picuino
author_rank: Super Contributor
timestamp: June 05, 2023, 04:05:43 pm
content: Note that there will always be a resistor in series with the coil and the capacitor.Depending on the frequency, the w·L/R ratio will be higher or lower, causing a damping of the resonant circuit.Edit:For example, a 10uH coil at 500kHz:X_L = 2-pi-f-L = 31.4 Ohm.The same 10uH coil at 50kHz:X_L = 2-pi-f-L = 3.14 OhmIf the series resistance of the coil is 1 Ohm, in the first case it is much lower than the coil impedance.In the second case the impedance is almost equal to the resistance, so this will produce a very appreciable damping.

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 04:21:22 pm
content: Perhaps more important is the impedance of the source driving the LC circuit.A parallel-resonant circuit is best driven from a high impedance source to see a strong resonant peak.If the two reactances (L and C) are both, say 1000\$\Omega\$ at the resonant frequency (cancelling each other to get resonance), but you drive it from a 50\$\Omega\$ output-impedance generator, then theQof the total circuit (even with perfect L and C with no series resistance) is very low (Q= 1/20 in this example), and you will not see a peak.

author: Picuino
author_rank: Super Contributor
timestamp: June 05, 2023, 04:30:01 pm
content: Another experiment can be done by charging the capacitor separately and connecting it to the coil with a switch to see on an oscilloscope the more or less damped waveform.It can also be done with the collector of a transistor that forces for a very short period (100ns for example) a pulse on the resonant circuit already connected.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 04:33:27 pm
content: Thanks all, that clears up my understanding here.

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 04:46:17 pm
content: If you calculate the totalQfor the circuit, the resonant behavior is "underdamped" (shows a peak) forQ> 1/2, "critically damped" forQ= 1/2, and "overdamped" (low-pass filter) forQ< 1/2.For a parallel-resonant circuit driven from a finite impedance, re-draw the circuit with the parasitic resistances of the reactors as equivalent parallel resistors, combine those two resistors with the source resistance (in parallel), and compare that resistance valueRP totto the (equal magnitude) reactancesX0at resonance.Qtot=RP tot/X0.