LC resonance mixed success - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: May 26, 2017, 08:53:36 pm
content: Read a bunch of web pages, watched w2aew #54 #55 #56 (all 3 re: L and C resonance), watched Dave's #396 (use scope to create bode plot), ran the theoretical numbers and built a circuit on breadboard. See attached results. I'm amazed how close to theoretical the result is.I wish there were a way to display the vertical scale log instead of linear. Maybe if I turn on the FFT channel I'll effectively get that. The Q looks pretty low to me but I don't have a really good intuition about that yet. Next I'll play with series and parallel resistance to see the effect on Q.I did it as a series circuit because prior to this, I was looking at the individual L and C as individual components in the LC circuit, to see their individual contribution to the overall circuit. That's not possible as a parallel circuit. Ultimately I will build this parallel but the resonance is the same so I just left the series configuration in place to do the "bode plot".To do it more thoroughly, I think I should configure it as a parallel circuit, that way the current is minimized and the voltage is maximized at resonance. That way it's easy to use the scope auto cursors to find the resonant point instead of all the zooming and manual manipulation I had to do. Also I should offset the display to see only the top half of the wave. That will increase visibility a lot. I also should use all 10 horizontal divisions instead of 8. As a first go at it, I felt I wanted the borders to verify correctness.I hope this is encouraging to anyone else getting started on LC resonant circuits!

author: Audioguru
author_rank: Super Contributor
timestamp: May 26, 2017, 09:00:10 pm
content: The Q appears to be very, very low.

author: electrolust
author_rank: Unknown
timestamp: May 27, 2017, 01:14:08 am
content: Didn't work out so well in parallel.In the first trace, I have the same L & C but in parallel. The resonance frequency shifted quite a lot. The effective capacitance is 53% of the measured capacitance.In the second trace, I added ~100nF of capacitance in parallel. By the scope auto measurement it shifts but not by a linear proportional amount. (could be some kind of x/x+y thing though.) But visually the resonance is around 5500, not 5000 (the scope is picking up a noisy peak). At 5500 the effective capacitance is the same 53% of the measured capacitance.I'm at a bit of a loss.The probe tip capacitance should be insignificant at ~10pF. And it would be in parallel anyway, so it should add to the net circuit capacitance. By the numbers, the parasitic capacitance of the inductor should be 1.8pF @ SRF (1400kHz). By the theoretical lumped circuit model (well under SRF), this parasitic capacitance, even if significant, would be in parallel.I thought maybe my capacitance measurement is actually ok, and that with the parallel circuit I'm adding parallel inductance, ie reducing the net overall inductance. I added 10" of loop to the capacitor (5" to each terminal), and I added 5" (total 10") to the probe ground clip. That didn't change the readings.My HH LCR meter reads just 6mH for the total circuit, vs 7.2mH for the inductor alone. 6mH doesn't match the resonance frequency, and I don't think the meter is reading this value correctly (too much capacitance for it to read the inductance properly), but it's much, much closer. My capacitor is a decade capacitance box. I can imagine it has a lot of parasitic inductance, however shouldn't that have affected the series LC circuit equally? I'll try again after the long weekend with a single axial lead capacitor.In general, at this order of magnitude of values, is it challenging to build breadboard prototypes? It's LF stuff, I wouldn't have imagined that to be the case. And again, in series it worked perfectly.

author: StillTrying
author_rank: Super Contributor
timestamp: May 28, 2017, 04:23:32 pm
content: How are you driving the parallel LC. To see the resonance effect you'll need to drive it (and read it) from a high-ish impedance, using a SG, a series resistor of 1k to 5k will do for starters.

author: Electro Fan
author_rank: Super Contributor
timestamp: May 28, 2017, 04:43:17 pm
content: Nice project. What model scope are you using?

author: JimRemington
author_rank: Regular Contributor
timestamp: May 28, 2017, 05:19:22 pm
content: +1 for reply #3. Don't forget that your signal source (probably low impedance) and probe are part of the LC circuit!

author: electrolust
author_rank: Unknown
timestamp: May 29, 2017, 08:28:02 am
content: Quote from: StillTrying on May 28, 2017, 04:23:32 pmI'm using a Siglent SDG2042X in 50 ohm mode.Quote from: Electro Fan on May 28, 2017, 04:43:17 pmKeysight 3000T.

author: StillTrying
author_rank: Super Contributor
timestamp: May 29, 2017, 10:48:55 am
content: Quote from: electrolust on May 29, 2017, 08:28:02 amYou need to add some impedance to the driving source so that the parallel LC canshort outthe signal, except where it nears resonance when the parallel LC becomes high impedance.

author: danadak
author_rank: Super Contributor
timestamp: May 29, 2017, 12:38:12 pm
content: To display vertical in log mode you would need a AC to RMS converter,something likehttp://www.analog.com/en/products/linear-products/rms-to-dc-converters.htmlYou would then use log f() for a math waveform on this DC output.If your scope supports multi f() math generation you could use absolute value followedby LPF and log10f() to generate math waveform for vertical (add the X 20into the expression to get db).Or if scope todays DSO acquire dataset and run a script on it to display plot.Regards, Dana.

author: electrolust
author_rank: Unknown
timestamp: May 29, 2017, 11:08:26 pm
content: Quote from: StillTrying on May 29, 2017, 10:48:55 amI added a 2.2k resistor, it works! So cool! The Q is much better looking as well, although I don't yet know if that's the effect of the added resistor or if I'm doing something else. I'll post more results tomorrow or Wednesday when I have time to get back to this.Is your graph a real graph or a simulation? Because where'd you happen to get a 7.2mH inductor?! Or do you have a variable tap / adjustable inductor on hand. Either way that graph was very very helpful. I do have to go back and understand why the source impedance matters. It's great that adding the 2.2k worked but I don't yet understand why.BTW it's also interesting that this works as an antenna. In the parallel mode it's very evident. With my workbench LED light on, it picks up all kinds of EMI (at LF 5KHz) and is unreadable. With the workbench LED off it still is noisy but not a disaster. Another experiment I'll try is a simply aluminum shield around the breadboard as well as tying the circuit ground to the breadboard chassis as a shield. I think it must have been just as receptive/noisy in series, it just wasn't noticeable because the shape of the wave was convex?

author: StillTrying
author_rank: Super Contributor
timestamp: May 30, 2017, 12:51:40 am
content: "Is your graph a real graph or a simulation?"Just a 2 minute sim, added to above post, the hardest part was swapping the plot's axis from log to lin so that it looks more like what you were seeing. I don't see the point in calculating these things by hand when LT can do it in 1ms."why the source impedance matters."Just think of it as a simple 2 resistor potential divider effect, the fixed source impedance, and the LC's impedance which varies depending on the frequency.

author: electrolust
author_rank: Unknown
timestamp: May 30, 2017, 08:06:21 am
content: Quote from: StillTrying on May 30, 2017, 12:51:40 amI figured it out independently ... it was clear after thinking about your circuit with the 2 branches (50r and 1k). Really great to have figured it out and then read that my "guess" was correct. Thanks again! I'll have to rewatch some of the videos and see if this was discussed and I just missed it.

author: electrolust
author_rank: Unknown
timestamp: June 01, 2017, 10:43:02 pm
content: Quote from: StillTrying on May 29, 2017, 10:48:55 amI've learned that the Rs (source impedance, R3 in your circuit) affects the resonant frequency. In a series circuit it does not, but in a parallel circuit it does. Please, mathematically what is the contribution of Rs to the resonant frequency?Answers I can find are extremely watered down. Most mathematical treatments are geared toward people like me (beginners) and don't even discuss the contribution of (DCR/L)^2 to the resonant frequency. (fres= (1/(LC - (DCR/L)^2))^0.5).I ask because I am unable to calculate proper C for a desired fres. After adding the DCR/L term to my spreadsheet I get new values, but they still do not agree with simulation (circuitlab, FWIW).Fromwikipedia, am I correct to think that the contribution of ESR is 1/(ESR*C)^2? And so with the combined parasitics a more real world resonance equation isAnd then if so, how to factor in Rs?

author: StillTrying
author_rank: Super Contributor
timestamp: June 01, 2017, 11:14:47 pm
content: Quote from: electrolust on June 01, 2017, 10:43:02 pmI don't really know, you know more about the maths of it than me. But I think it'll be the way the source impedance and the real world impedances of the LC interact. Have you checked at the SG side of the source resistance, to check that the SG is not doing something funny.In the sim the only real world values I had were 1R DC resistance for the inductor, and 0.5R ESR for the 100n cap.I've just tried giving them bigger and more parasitics and a 2X difference in source impedances, 1K and 2K2, the Q varies but not the frequency of the resonant peak.LTSpice is very easy to use for these kinds of simple things if you want to have a go of it yourself.

author: electrolust
author_rank: Unknown
timestamp: June 01, 2017, 11:41:52 pm
content: So in circuitlab the resonant frequency does change. Also lots of references that a parallel R does change the resonance. And then I foundVA3IULwho shows that Rs interacts with the circuit to create an effective parallel resistance. So that means that Rs should affect the resonant frequency and anyway circuitlab sim agrees.I tried to fire up ltspice on the mac but it appears incomplete. I don't see how to actually run a simulation. I will get it going on windows and see how it differs from circuitlab. I did notice circuitlab's inductor inputs are highly simplified compared to LTspice. I think in theory I could just add the extra parasitics as non-lumped circuit elements but I didn't get that far yet.

author: StillTrying
author_rank: Super Contributor
timestamp: June 02, 2017, 12:09:07 am
content: Here's the .asc. You can just right-click to change most of the values.I can't see any noticeable change in F on the plot with added parallel R, just the Q change.

author: electrolust
author_rank: Unknown
timestamp: June 02, 2017, 12:37:07 am
content: maybe i'm misreading my source material.https://www.allaboutcircuits.com/textbook/alternating-current/chpt-6/resonance-series-parallel-circuits/shows that DCR changes the frequency. That makes perfect sense since DCR affects the impedance of the inductor. Perhaps I confused DCR with parallel R. If so, the "root cause" of my confusion is that circuitlab shows a change in frequency with a change in Rs. I do now see (I think) that all the discussion I find about Rs effect is that it changes the Q, not the frequency.Let me pause here and make sure I am building the same circuit in both circuitlab and LTspice and see what's the difference. Here are my circuitlab results.1. the circuit2. Rs = 1k3. Rs = 2.2kThe x2 cursor in the 2nd graph shows the shifted frequency.

author: StillTrying
author_rank: Super Contributor
timestamp: June 02, 2017, 12:52:01 am
content: If you copy the other source resistance to the the other side of the generator surely you can get both plots on the one graph.If I have one L with DCR of 1R and the other with DCR of 100R, the RF of the 100R one is 1kHz higher.

author: electrolust
author_rank: Unknown
timestamp: June 02, 2017, 08:32:05 pm
content: In CL, I can't figure out how to put 2 plots on one graph.So I did verify/reproduce your results with LTSpice. ARGH!!!Yup, here's how it went wrong for me.1. I was using a simplified formula for resonant frequency that did not consider the inductor DCR, a significant factor at my frequency of interest.2. When I got this working (adding source resistance) on the bench, the resonant point differed from my calculation.3. Thanks to your (ST) help I played with simulation. I tried MacSpice, LTSpice. LTSpice on the Mac lacks the toolbar which is apparently present on Windows. Being new to it I didn't figure out how to add the dot directives. I then tried easyeda. For some inscrutable reason, they removed the simulation button, replaced with CTRL-R, which doesn't work on Mac. (and the normal Mac translation of CMD-R just reloads page.) Then I tried circuitlab. So easy to use!4. Circuitlab shows that a change in R_s changes the resonant frequency.5. This led me to lots of docs about DCR affecting res.freq. and about R_L affecting Q, and about R_s||R_L affecting Q. Tying that back to CL simulation, I drew the wrong conclusion that R_s affects res.freq.6. UGH. CL FTL (CircuitLab For theLose!)The LTSpice frequency peak agrees with my first calculation, that does NOT consider DCR. The difference in frequency is significant enough to be measurable so I will go back to the workbench and see what I actually get. Interestingly, ESR of the cap DOES affect the res. freq in LTSpice. DCR affects the shape of the curve below resonance but not the resonant point. That's not in agreement to what I read elsewhere so I'll have to review this further.

author: electrolust
author_rank: Unknown
timestamp: June 03, 2017, 09:12:29 am
content: OK huge update here.CL is correct, or at least it agrees with LTSpice AFAICT. R_sdoes notchange the res. freq. So phooey on me, not circuitlab!! I don't know what I was doing before in CL, but starting over I'm unable to reproduce the behavior I saw quite clearly.First image is the circuit.Second/Third is 1k vs 2k2 values of R_s in CL/LT.Fourth image is (CL) 500r R_s and stepping through the inductor DCR as 20r, 40r, 80r. 3 things to notice on this busy graph. 1) Resonance points move, 2) I resonance does not align with V resonance, 3) V resonance does not move in the same direction with DCR -- rather the V resonant point moves one way then the other, as DCR increases.Fifth image is the same as #4 but with just 80r DCR, to make clear the shift between I and V resonance.Sixth image is the same as #4 but in LTSpice. Note that the DCR here is "explicit", not lumped. I added a resistor representing DCR and stepped that, leaving DCR of the inductor itself at 0. I couldn't figure out how to step a "sub" parameter of a component in spice. Whereas in CL I'm stepping the DCR parameter of the lumped inductor.So, first of all, seeing that resonance didn't move with R_s really cleared up a lot for me, this is what I expected and I must have wasted 30 hours after some botched work in CL where I saw it was moving.Then, seeing that resonance does move as DCR changes reinforced more of my understanding. But the difference between I and V points took some more thought.How can it be that I and V points don't align, doesn't this violate V=IR? [spoiler alert] I think the answer is that when XC= XL(resonant point), if these were ideal components the net phase would be 0 (+90L and -90C). But now that I've added DCR, I've altered the phase relationship. I don't quite understand this relationship, but well enough to intuit that adding a similar series resistance to the capacitor, ie ESR, will shift the net phase back to 0 and the I and V points should align. And they do! Image 7.I know I've lost you all this far down, but now I have a vital question. Which is the actual resonance point, the minimum I or the maximum V?

author: MrAl
author_rank: Super Contributor
timestamp: June 03, 2017, 11:40:26 am
content: Hello there,The amplitude of your output with your most recent circuit is:Vout/Vin=sqrt(RL^2+w^2*L^2)/sqrt((Rs*w*C*RL+w*L)^2+(RL-Rs*w^2*C*L+Rs)^2)wherew=2*pi*f, f in Hertz,L is the inductor value in Henries,Rs the series resistance in Ohms,Vin your input voltage in peak or peak to peak or RMS or whatever you want units,RL is the inductor ESR value in Ohms.The peak amplitude should occur at angular frequency:w=sqrt((sqrt(2*Rs^2*C*L*RL^2+2*Rs*L^2*RL+Rs^2*L^2)-Rs*C*RL^2)/(Rs*C*L^2))where again w=2*pi*f.

author: electrolust
author_rank: Unknown
timestamp: June 04, 2017, 04:20:53 am
content: Quote from: MrAl on June 03, 2017, 11:40:26 amReformatting:I don't see how that's possible since it doesn't include terms for ESR and DCR. Also, I mean I guess it could be wrong but StillTrying showed me and I was able to verify, that the Rs term has no affect on resonant frequency. Without going through your formula, maybe the Rs terms cancel out, but the problem remains that it is only considering an ideal inductor and ideal capacitor.

author: MrAl
author_rank: Super Contributor
timestamp: June 04, 2017, 05:51:12 am
content: Quote from: electrolust on June 04, 2017, 04:20:53 amHello again,Not sure what you mean here about the ideal components, because RL is the ESR of the indcutor. You did not seem to want to include non idealities for teh capacitor so i did not include that.What we call the 'resonant' point can sometimes be different in electronic circuits because we often just want to calculate the 'peak' which is also sometimes called the resonant point. That is what i calculated, the peak that is the maximum output of the filter. The point where this happens is at w0 and that is calculated with this simpler formula:w0=sqrt(sqrt(2*Rs*C*RL^2+2*L*RL+Rs*L)/(sqrt(Rs)*C*L^(3/2))-RL^2/L^2)where again RL is the inductor series resistance (ESR) and Rs is the series resistance of the circuit, and we take the output from across the capacitor.The series resistance in a series RLC circuit always plays a part in the resonant point. If you dont think this is true, then graph it.

author: StillTrying
author_rank: Super Contributor
timestamp: June 05, 2017, 01:55:06 am
content: I don't know why the Mac version of LT wouldn't have a toolbar, clicking on the schem. should give you an option to add a .spice directive under the Edit Menu.Right clicking a component you should be able to change one of it's values to {x}, and you can then .step xIf I step the L's DCR from 1R to 2000R, I only see a very small change in the resonant freq.

author: MrAl
author_rank: Super Contributor
timestamp: June 05, 2017, 09:40:58 am
content: Quote from: StillTrying on June 05, 2017, 01:55:06 amHello there,Nice graph BTW :-)When we look at these problems we usually try to cover all or at least most of the bases. We like to know what kind of change we are dealing with if there is a change at all. So it's not really a matter of how much it changes as much as if it changes at all. If it changes, we want to know about it.However, for some values of Rs and RL we will see very little change. I think one combo i tried only gave something like 0.001 percent change in w0. Other combos gave more of a change.For values i tried like RL=1 and Rs=100 to 2000, i got f0 from about 5900Hz to 6200Hz. That seems significant to me, but what is more important is that we learn just how Rs and RL affect the circuit because if we left either one out of the calculation then we would never know it plays a part in it overall.Also note i did not look for the worst combination, which you might want to try just to see what you can come up with. We might actually see 2 or 3 times the w0 change from RL=0 for example. But even for small changes we like to know about it so that we can ourselves later make the informed decision when to ignore it and when we cant ignore it. Once we know, we know, and it's better to know. Sometimes small changes matter, sometimes not, but knowing they are there at all is better.The calculation for the values you used matches your simulation. I got 5933Hz to 6034Hz. I would want to know it changed that much especially if i was studying the theory of RLC circuits.

author: electrolust
author_rank: Unknown
timestamp: June 05, 2017, 10:11:00 pm
content: Quote from: MrAl on June 04, 2017, 05:51:12 amSorry, yes you did say that. I overlooked that and thought the RL was a parallel load in a generalized RLC parallel circuit.

author: The Electrician
author_rank: Frequent Contributor
timestamp: June 05, 2017, 10:41:14 pm
content: Quote from: electrolust on June 04, 2017, 04:20:53 amWhat definition are you using for resonance?For parallel circuits there can be more than one definition:http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html#c1You might also want to have a look at:https://en.wikipedia.org/wiki/RLC_circuit

author: electrolust
author_rank: Unknown
timestamp: June 05, 2017, 11:00:12 pm
content: Ah, thanks for that. Do you know if AoE covers that detail? At least in chapter 1, I don't think they do. They seem to only talk about resonance in terms of min/max impedance. I guess AoE is more of a 100-level text?Now that I understand the subject a bit more, I see that the wikipedia page is going to be a good jumping off point for me.Wikipedia uses Kaiser as a reference, but ouch that book is expensive!

author: electrolust
author_rank: Unknown
timestamp: June 08, 2017, 09:12:55 am
content: So. I read wikipedia and the hyper-physics site. Wikipedia in particular has a nice reference to a paper by Kenneth Cartwright, which is very approachable. In summary, there are three resonant points.f_0: series resonant point (X_C = X_L); works for parallel given low ESRf_p: (phase = 0) = (power factor = 1)f_m: max impedanceFrom the Cartwright paper I was easily able to construct a spreadsheet that allows me to calculate either f_m or C (given the other), for a given L+ESR. The result is in very good agreement with Circuitlab and LTSpice. And I can see both the f_p and f_m points on the plots. Now I have a good enough grasp of this to properly consider the effect of R_s.But I'm still at a loss as to why the minimum current does not occur at the f_m maximum impedance frequency. It also is not at the f_p frequency. Cartwright does not consider this at all; in his circuit he uses a 1A constant current source, which at least one good reason to do so is to remove the effect of R_s!Why doesn't V=IZ apply? Z is not plotted, V is plotted. It occurs at f_m, max impedance. Ergo, given V=IZ, V is at a maximum, Z is at a maximum, shouldn't I be at a minimum at this same frequency?EDIT: Possible answer: V is not the total circuit voltage, it's the voltage at the divider point. As Z (of L//C) changes, R_s remains constant. This means the division ratio changes with frequency. The shift in I_min away from f_m reflects that V is not changing at the same rate as Z!Re-attaching a couple of relevant plots from a previous post.

author: electrolust
author_rank: Unknown
timestamp: June 09, 2017, 08:25:33 pm
content: Quote from: MrAl on June 03, 2017, 11:40:26 amReformatting:This doesn't work for me. The numerator of the outer radical is negative and the denominator is positive. So I can't take the square root of that.It works. I wasn't including factors for the prefix (10^ -3,-6,-9) in my calculation.I'd love a better understanding of how this is derived. I'm sure it's too involved to make a forum post so a text reference would be great.

author: electrolust
author_rank: Unknown
timestamp: June 14, 2017, 08:31:55 pm
content: quick bump