Probing a resonating LC tank circuit, how much does the probe dampen it?  - Page 1
Forum section: electronics / beginners

author: VooDust
author_rank: Regular Contributor
timestamp: July 23, 2020, 05:47:25 am
content: The LC tank circuit consists of a 100nF ceramic capacitor and 10uH inductor. It resonates at about 162kHz on my breadboard (measured).The input jack of my oscilloscope says 11.5pF input capacitance. The 10X probe has 12pF in the datasheet. I assume then my input has a total capacitance of 23.5pF.At 162kHz, 23.5pF presents a resistance of 41.5kOhms to ground.Now, how would I determine how much this 41.5k resistance is responsible for dampening the oscillation? I don't know how I would calculate this. The series resistance of the capacitor is unknown, as well as of the inductor (but in which case it should be close to zero).I start the oscillation either by charging the 100nF cap to 1V first, then placing it into the circuit (result is *meh*), or running a DC current of ~180mA through the tank circuit first, then disconnecting the power (better, I think this equates to a lot more stored energy in the magnetic field of the inductor).I think I'd need calculus to derive this, but maybe someone knows the correct formula or some shorthand calculations.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 23, 2020, 06:41:18 am
content: The probe's capacitance is correct, actually. The equivalent circuit is roughly (R || C) + (R || C), an impedance divider. The first R and C are much higher impedance, while the second includes cable plus scope capacitance; a 9:1 impedance ratio gives the 10:1 voltage gain with flat frequency response.But actually it's more than that, it's (Ra + Rb || Ca) + (Rc + Rd || Cb), an impedance divider that has Ra and Rc at DC, ~Ca at MF, and Rb and Rd at HF.So it depends what frequency you test at, what the equivalent circuit is.162kHz is fairly low, so we can model it as a lossy capacitor. If Ra = 9M, Rb = 450 and Ceq = (Ca + Cb) = 12pF*, then we have two choices: the series equivalent of Ra || Ceq, or the parallel equivalent of Ra + Rb + Ceq. (Or, well, the parallel and series equivalents respectively, which are pretty obvious...)*Here I'm using "+" in the topological sense, i.e., capacitors in series. The numerical values act "in parallel", i.e., Ceq = Ca Cb / (Ca + Cb). Capacitance is a reciprocal impedance unit so I want to clarify this. The 12pF spec is the probe tip capacitance, which is the internal compensation capacitor in series with the cable and scope capacitances, hence the calculation. And to be clear, "||" means (topologically) in parallel, and takes higher precedence than "+".We can calculate the equivalents with these formulas:https://www.daycounter.com/Calculators/Parallel-Series-Impedance-Conversion-Calculator.phtmlThe first is equivalent to 12pF + 744 ohms, the second is 12pF + 500 ohms (or equivalent to 13M || 12pF).Pretty close actually, so we can guess neither is dominant, and we should account for both. We're going for series equivalent, so connect them in series: 744 + 500 = 1244 ohms. 12pF is still 12pF.Meanwhile, the 100nF || 10uH tank has a resonant impedance of ~10 ohms (sqrt(L/C)), and 12pF being 1/83,333th of 100nF, the Q would have to be comparable to 1/x of that ratio (say, >10k) to begin to have an effect -- and at that, only if the resistance is comparable (10 ohms), which it's not.Also, even if the 100nF is a precision 1% capacitor, that's a variation of +/-1nF, so the probe's reactive loading is another 1% of that already!So you're pretty safe to say it's having fuck all impact.Tim

author: VooDust
author_rank: Regular Contributor
timestamp: July 23, 2020, 06:59:35 am
content: Wow, that's a lot to process actually. I'll go through it step by step. Thanks a lot.

author: TimFox
author_rank: Super Contributor
timestamp: July 23, 2020, 02:23:15 pm
content: If your probe presents 12 pF in parallel with 10 megohm to the circuit, the only component that adds to the damping is the 10 megohm resistance. The capacitor has a reactance, which will affect the tuning of the circuit, but it does not dissipate energy (which is what damps the resonance). The post above demonstrates how to do the calculations.What type of ceramic capacitor is your 100 nF unit? The dissipation factor of an X7R capacitor is much worse than a C0G ceramic and the ESR must be included in the circuit damping.