Resonant LC Circuits - Page 1
Forum section: electronics / projects

author: German_EE
author_rank: Super Contributor
timestamp: February 12, 2015, 08:05:22 pm
content: Wire an inductor and a capacitor in parallel and you get a resonant circuit, use 1/2PI*SQRT(LC) to calculate the resonant frequency, we all know this as it's basic electronics. The problem is that there are an infinite number of LC combinations, 25,33uH and 1000pF will resonate at 1 MHz but so will 2,533mH and 10pF. The first example has a large capacitor and the second has a large inductor,So, here's the problem. Given that only frequency is known give me a formula that works out an LC value where a 1pF change in C will give a 1 KHz change in F. Warning, this is NOT an easy problem and after a day in front of a spreadsheet I've given up for the night

author: PA0PBZ
author_rank: Super Contributor
timestamp: February 12, 2015, 08:22:04 pm
content: SQRT(LC) is non-linear so your desired step will only work for one specific C.

author: mark03
author_rank: Frequent Contributor
timestamp: February 12, 2015, 08:33:54 pm
content: Ummm... calculus?Hint: take the partial of C w.r.t. f and set this equal to 10^-15 (1 pF per 1 kHz).

author: nfmax
author_rank: Super Contributor
timestamp: February 12, 2015, 08:40:16 pm
content: This is most easily solved using some calculus.We have f = 1/(2*pi*sqrt(L*C)), or equivalently f = C^-1/2 * 1/(2*pi*sqrt(L))Differentiate with respect to C, givesdf/dC = -1/2 * C^-3/2 * 1/(2*pi*sqrt(L))Substituting the original formula for f givesdf/dC = -1/2 * f * 1/CWe want df/dC to be 1kHz/1pF = 10^-9, so solving for C (since f is 1MHz) gives 500pF, with the corresponding L being 50.6606 microH. Increasing C by 1pF gives a frequency shift of -998.5Hz. Not quite exact, as it is a nonlinear relationship, but close enoughMax

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 13, 2015, 12:44:52 am
content: Could you tell us about your application? It seems like something's going to bite you in nonlinearity if you do what it sounds like you're trying for...Since F(C) = 1 / (2*pi*sqrt(L*C)),dF/dC = -1 / (4*pi*sqrt(L*C^3)) = -F / (2*C)So, for -1kHz/1pF, at 1MHz, you need 500pF tank capacitance. Which will be with 50.66uH, and a resonant impedance of 318.3 ohms.So I get the same result as nfmax. Hey, we used the same example frequency too..Tim

author: GK
author_rank: Super Contributor
timestamp: February 13, 2015, 01:58:16 am
content: QuoteThe ratio of L to C for a given fc is simply determined by the desired tank circuit Q (and thus bandwidth). For a given fc, >C & <L = >Q and visa versa. It is really very simple. Suppose we want a 10MHz (fc) parallel tank circuit with a bandwidth (BW) of 1MHz.Q= fc/BW= 10MHz/1MHz=10If this tank circuit is driven by a source resistance (Rs) of 1k, we need to design for a reactance (Xp) of:Xp=Rs/Q= 1k/10=100 ohms.For the inductance:L = Xp/(2*pi*fc)= 100/(6.28*10MHz)= 1.6uHFor the capacitance:C=1/(2*pi*fc*Xp)=1/(6.28*10MHz*100)=159pFQuoteI am not sure how/why you would ever start any practical design by specifying the actual value of C ahead of system Q.

author: c4757p
author_rank: Super Contributor
timestamp: February 13, 2015, 02:27:35 am
content: Quote from: German_EE on February 12, 2015, 08:05:22 pmThere's your problem, dude...

author: branadic
author_rank: Super Contributor
timestamp: February 13, 2015, 11:02:41 am
content: QuoteThis is true in a perfect world. In real world application there is the requirement for a given Q and/or impedance at resonant frequency. Therefore you have to take stray / distributed capacitance of the inductor, stray inductance of your capacitor as well as the dc resistance of your inductor and parasitc parallel resistor of the capacitor into account. This decreases the number of combinations dramatically, that fullfil your requirements.

author: German_EE
author_rank: Super Contributor
timestamp: February 13, 2015, 01:10:09 pm
content: Thank you for all the replies. I knew that this would not be easy but the last time I did any calculus was in the 1970's and since then I've avoided phrases like "Differentiate with respect to C" and focused on the more practical side. However, I do have a solution, throw lots of expensive hardware at the problem.A friend of mine has access to the computing facilities at the Technical University here in Darmstadt. A 1024 core machine will consider the problem for a brief moment sometime this afternoon and then supply the correct LC combination for every frequency between DC and 100 MHz in 1 KHz steps. The cost is a pack of beer as he does this sort of thing for fun.

author: KedasProbe
author_rank: Frequent Contributor
timestamp: February 13, 2015, 01:37:47 pm
content: uuhhhm brute force calculations, really??

author: c4757p
author_rank: Super Contributor
timestamp: February 13, 2015, 02:08:17 pm
content: Yes, really. Calculus is, apparently, impractical, that is until it's actually required for something...

author: MrAl
author_rank: Super Contributor
timestamp: February 13, 2015, 03:30:16 pm
content: Hi,The solution using straight calculus results in a linearization near the value of C. So when the actual C is used it may turn up the wrong result, although it may still be close.Another way to solve this, and it is at least a little more interesting (ha ha) is as follows...One way to interpret the problem is what value of C is required to lower the resonant frequency by a given amount, 1000Hz, for a given value of L. This may require numerical precision for the constants that exceeds the normal computer 16 digit precision floating point. If solved numerically, more precision will most likely be required.The formula for the resonant frequency as others have given is:f1=1/(2*pi*sqrt(L*C))Now we want a frequency that is 1000Hz lower than this 'f' (other ways of looking at this too though) with a value of C that is dC greater than C, where dC is the small increment change in capacitance (given as 1e-12), so we have:f2=1/(2*pi*sqrt(L*(C+dC)))This f2 will result in a frequency that is 1000Hz less than f1, so we can form an equation:f1-1000=f2which of course expanded is:1/(2*pi*sqrt(L*C))-1000=1/(2*pi*sqrt(L*(C+dC)))The values of C that satisfy this equation are the solutions to this EQUATION, but only one of them may be the real solution so each real C solution has to be tested to make sure it satisfies the initial problem.The solution to this equation isnt that easy because it results in a fourth order algebraic equation in C, which means there will be four solutions, probably two real and two complex, so the two real will have to be tested unless one of the two always turns out to be the right result all the time.One way to start to solve this is to multiply by 2*pi, then square, then separate terms with no square roots, then square again. This probably results in a regular algebraic equation with no radicals and can then be solved. Alternately try an online solver like that found at Wolfram.Another way of looking at this is what value of C is required to raise the frequency by 1000Hz.If this is done right the new value of the frequency will be exactly 1000Hz above or below the original frequency before any change in C.If you try this, please post results, thanks :-)

author: ivan747
author_rank: Super Contributor
timestamp: February 13, 2015, 04:19:48 pm
content: Quote from: c4757p on February 13, 2015, 02:08:17 pmAnd it always will be like this, for me.

author: dannyf
author_rank: Super Contributor
timestamp: February 13, 2015, 04:38:43 pm
content: Quotegiven that F=1/2PI/sqrt(LC),dF/dC = 1/2PI / sqrt(L)*dC^-0.5/dC = 1/2PI / sqrt(L) * (-0.5)*C^(-0.5-1) = -0.5 * 1/2PI / sqrt(L) *C^-1.5 = -0.5* 1/2PI/sqrt(LC) / C = -0.5*F/C.

author: nctnico
author_rank: Super Contributor
timestamp: February 13, 2015, 05:44:38 pm
content: At resonance the impedances of the capacitor and inductor must be equal. A change of 1kHz versus 1MHz is 0.001 so the capacitance change must be the same magnitude (1000x) but the capacitor only drives half the total impedance so 1000/2=500 times. C= 500*1pf=500pf.Zc=318,310 Ohm (500pf at 1MHz)Zl=Zc => L= 318,310 / (1M * 2pi) = 50.661uHNote that f0=1/2*pi sqrt(L*C) is Zl=Zc solved for f0. Using the impedances directly is much easier in this case.Edit: several additions

author: dannyf
author_rank: Super Contributor
timestamp: February 13, 2015, 06:05:25 pm
content: QuoteA 8086 + lotus 123 or anything better will do just fine.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 13, 2015, 06:32:37 pm
content: No calculus is needed:

author: MrAl
author_rank: Super Contributor
timestamp: February 13, 2015, 09:28:13 pm
content: Quote from: The Electrician on February 13, 2015, 06:32:37 pmHi,I didnt think anyone would get around to trying out my suggestion is post #11 so fast. It looks interesting, but what kind of software did you use?

author: Zero999
author_rank: Super Contributor
timestamp: February 13, 2015, 09:57:16 pm
content: This reminds me of a spreadsheet I made for calculating the change in capacitance required to give a certain frequency change in an FM transmitter circuit.

author: skipjackrc4
author_rank: Regular Contributor
timestamp: February 13, 2015, 10:09:00 pm
content: Quote from: The Electrician on February 13, 2015, 06:32:37 pmThis was the first thing that popped into my head when I read the question, but I couldn't be bothered to type it out. No need for 1024 core machines running an exhaustive search.

author: IanB
author_rank: Super Contributor
timestamp: February 13, 2015, 10:12:53 pm
content: That looks too hard. The differentiation approach is easier...

author: dannyf
author_rank: Super Contributor
timestamp: February 13, 2015, 10:17:45 pm
content: Solver in excel -> a 30 second job.

author: GK
author_rank: Super Contributor
timestamp: February 13, 2015, 10:54:37 pm
content: Quote from: German_EE on February 13, 2015, 01:10:09 pmIs this just a theoretical exercise or does it actually pertain to a real life design problem? In a practical situation "the correct LC combination" has little meaning if both the circuit impedance and resonant circuit Q isn't specified in addition to fc. I gave you the relationships in my previous post. I cannot think of any situation where I would start by specifying my value of capacitance and not care about whatever the value of Q (and thus the reactances of both L and C at fc) turned out to be.

author: Dave Turner
author_rank: Frequent Contributor
timestamp: February 13, 2015, 11:53:03 pm
content: It doesn't really matter. As an exercise it is worth it in it's own right.In practical terms though it's preferable to determine Q then L because size for size normally C is easier to deal with.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 14, 2015, 12:37:34 am
content: Quote from: MrAl on February 13, 2015, 09:28:13 pmI wasn't trying out your suggestion. I didn't even read post #11, I just did the obvious thing.Quote from: GK on February 13, 2015, 10:54:37 pmHaving fc and the df/dC requirement of a 1000 Hz change in frequency for a 1 pF change in C, completely determines L and C. Adding additional constraints of a specified Q and impedance level will require that the OP decide where the circuit losses are. He will need to decide if a single resistor in series with L will be the determining loss element, or if both L and C will be lossy, or perhaps a resistor in parallel with both.He may have wanted to solve the lossless circuit first, then deal with losses later, but in the first post he only mentioned a parallel combination of L and C, the lossless case. That case is easily solved without the use of calculus.

author: GK
author_rank: Super Contributor
timestamp: February 14, 2015, 12:39:13 am
content: Quote from: Dave Turner on February 13, 2015, 11:53:03 pmFor a given fc L is determined by Q and the circuit impedance just as C is. You cannot juggle the value of L for a more desirable C (or the other way around) without changing the other variables. I gave the equations in their most basic form an even posted a worked example. I don't know what else I can do.

author: GK
author_rank: Super Contributor
timestamp: February 14, 2015, 12:44:47 am
content: Quote from: The Electrician on February 14, 2015, 12:37:34 amWho said otherwise? That doesn't change the fact that it is a possibly strange and impractical specification/requirement to start with and more so now that the OP specifies a frequency range from "DC to 100MHz"! I guess this is just a theoretical exercise for the OP.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 14, 2015, 01:04:23 am
content: Quote from: GK on February 14, 2015, 12:39:13 amYou could just provide the formula the OP asked for.

author: GK
author_rank: Super Contributor
timestamp: February 14, 2015, 01:14:03 am
content: Quote from: The Electrician on February 14, 2015, 01:04:23 amJust dick waving now? The question was already answered before I got here and there was essentially two parts to the opening post, starting with:QuoteSo, in addition to what was already covered, I simply gave a elementary perspective on selecting the ratio of L to C. I hope that hasn't caused any offense and perhaps some may have actually found the information useful.And despite the protestation that it doesn't matter, I still think it is a valid question to ask the OP whether he/she is working on a real world problem here or if this is just a theoretical exercise. If the former, I'd be curious to know what it is.

author: IanB
author_rank: Super Contributor
timestamp: February 14, 2015, 02:57:26 am
content: Quote from: GK on February 14, 2015, 01:14:03 amSuppose you wanted to be able to trim the resonant frequency, and your available trim capacitors were only of a certain size and adjustment range? Then maybe you would want to set up the circuit so that you could get the desired adjustment range with the available trim capacitors?

author: GK
author_rank: Super Contributor
timestamp: February 14, 2015, 03:10:09 am
content: Quote from: IanB on February 14, 2015, 02:57:26 amSure, but you still have to establish a practical and ball park value for the capacitance to start with, not just pluck it out of thin air or a components catalog! The proposed exercise of computing L and C for a 1 kHz change in fc for a 1pf change in C for fc ranging "DC to 100MHz" in 1kHz steps has no obviously transferable practical merit.

author: German_EE
author_rank: Super Contributor
timestamp: February 14, 2015, 08:59:34 am
content: The reason for needing this information is a ham radio project, a device called a preselector that filters incoming RF before it hits the receiver. When I worked out the LC circuit I found that the tuning was not linear and depended on the coil value, it could be 25 KHz/pF at one end and 100 Hz/pF at the other. So, there must be a point where 1pF = 1 KHz (a nice nominal figure) and I'd like to know how many values of inductance I would need.An example, using a 25 uH inductor:255 pF 1.993 MHz254 pF 1.997 MHz2 pF 22.5 MHz1 pF 31.8 MHzThe old way of tackling this problem was to alter the shape of the variable capacitor blades but something new needs to be found when capacitors are digitally switched. I wonder if it is possible to create a switched capacitor that has a non-linear response?

author: GK
author_rank: Super Contributor
timestamp: February 14, 2015, 09:58:02 am
content: Quote from: German_EE on February 14, 2015, 08:59:34 amWow..... big can of worms opened here, but (hopefully) needless to say, a pre-selector circuit is exactly the kind of application where you can't just hope to start plucking practical values for your tank circuit reactances without first giving thought at least to tank circuit Q, as, after all, this is what defines your pre-selectors raison d'être - its selectivity.

author: dom0
author_rank: Super Contributor
timestamp: February 14, 2015, 11:52:39 am
content: A simple way to this might be to switch capacitors. E.g. you have 5, 10, 20, (<-- likely mostly parasitic) 40, 80, 160, 320, 640 pF switchable and just switch those in which you need (i.e. a linear combination with binary coefficients) to achieve the given frequency.

author: MrAl
author_rank: Super Contributor
timestamp: February 14, 2015, 12:09:51 pm
content: Quote from: IanB on February 13, 2015, 10:12:53 pmHi,Ha ha, and i had fun solving it by hand using a fourth order equation :-)See post #11.I agree though, the calculus method is faster and i bet we can get a version that works more accurately too.

author: MrAl
author_rank: Super Contributor
timestamp: February 14, 2015, 12:12:24 pm
content: Quote from: The Electrician on February 14, 2015, 12:37:34 am...and i believe you.But that doesnt answer the question about what software you used.

author: dannyf
author_rank: Super Contributor
timestamp: February 14, 2015, 12:33:49 pm
content: QuoteFrom F0 and C, you can solve for L:1/2pisqrt(L) = F0*sqrt(C);Once you have C+1pf, and L, you can solve for F1 (corresponding to frequency associated with L and C+1pf):F1=(1/(2pisqrt(L))/sqrt(C+1pf) = F0*sqrt(C) / sqrt(C+1pf);deltaF = F1 - F0 = F0(sqrt(C) - sqrt(C+1pf))/sqrt(C+1pf) = F0*(1-sqrt(1+1/C))/sqrt(1+1/C)Using tylor expansion to sqrt(1+1/C).Anyway, that's a solution from a highschool dropout.

author: nctnico
author_rank: Super Contributor
timestamp: February 14, 2015, 08:01:41 pm
content: Going from f0=1/(2*pi sqrt(LC)) is the wrong (complicated) approach. It is much easier to solve these kind of puzzles using the impedances and/or conductances.

author: Dave Turner
author_rank: Frequent Contributor
timestamp: February 14, 2015, 09:41:20 pm
content: Try this -C= -delta C/ (f/((f+ delta f)^2 -1)Where C is the capacitor in Farads, f is the frequency in Hertz, delta C is the change in Capacitance (eg 10^-12 or 1 picofarad) and delta f is the change in frequency in Hertz.I prefer differentiation. The results are the same.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 14, 2015, 09:45:15 pm
content: Quote from: dannyf on February 14, 2015, 12:33:49 pmYou're solving a different problem than the one the OP posed. You are starting from Fo and C. The OP is starting from Fo, deltaFo, and deltaC (his deltaC=1pF); he doesn't know C yet.As he said: "So, here's the problem. Given that only frequency is known give me a formula that works out an LC value where a 1pF change in C will give a 1 KHz change in F."

author: dannyf
author_rank: Super Contributor
timestamp: February 14, 2015, 10:03:00 pm
content: QuoteThe same only if the function is linear.Otherwise, it is only 1st order approximation.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 14, 2015, 10:05:37 pm
content: Quote from: Dave Turner on February 14, 2015, 09:41:20 pmSomething is a little off. Your expression returns a value of about 1 million picofarads for F=1 MHz, deltaF = 1000 Hz, deltaC = 1 pF:

author: Dave Turner
author_rank: Frequent Contributor
timestamp: February 14, 2015, 11:51:33 pm
content: Let me recheck and attempt to show my derivation.An interesting point (which should be obvious) is that if you try to solve for a positive delta of both frequency and capacitance (given fixed L) then C solves as a negative value. The simple way to look at this is if the result for a frequency of 1 MHz + 1kHz at a 1pF change shows -500pF (to 3 significant figures) then plug 499pF into the equation father than 501pF.Please note note that I'm looking at this as a math problem to answer the original question as opposed to a practical solution. Whichever way one solves the equations the required precision (7-8 significant figures) is beyond anything I'm aware of without using trimmers.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 15, 2015, 12:42:02 am
content: Quote from: Dave Turner on February 14, 2015, 11:51:33 pmYes, I noticed that.Quote from: Dave Turner on February 14, 2015, 11:51:33 pmI left my results with so many significant figures because when I did the calculation for the change in frequency of 1 kHz, I wanted to see that the deviation was indeed 1.0000000 kHz.Even with trimmers (ordinary ones, that is), one can't get a capacitance to such precision.It's a math problem alright.

author: dannyf
author_rank: Super Contributor
timestamp: February 15, 2015, 12:57:13 am
content: QuoteNot sure why that's interesting: given that F is proportional to 1/sqrt(C), you know that any increase in C will yield a decrease in F. So a positive delta for both frequency and capacitance is not possible.Let alone interesting.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 15, 2015, 01:41:53 am
content: Pick up a copy of RDH4 (Radiotron Designer's Handbook, 4th ed.), either in print or for free (it's out of copyright and PDFs are available online). Both subjects are covered: how to choose the tuning range, and how to compensate and track over that range.Tim

author: MrAl
author_rank: Super Contributor
timestamp: February 15, 2015, 02:36:14 am
content: Quote from: German_EE on February 12, 2015, 08:05:22 pmHello again,Providing the frequency first is too easy :-)If you specify the inductance L, the change in capacitance 'dc' (1e-12 here), and the change in angular frequency 'dw' then we have again (similar to post #11):1/sqrt(LC)-dw=1/(sqrt(L*(C+dc)))or:1/sqrt(C)-sqrt(L)*dw=1/sqrt(C+dc)Solving this for C gives the capacitance needed when L is given, and since this equation is independent of frequency you dont have to specify that. This allows finding C for any L, period, that matches the required dw and dc.This may be more useful because very often we cant set the value of L exactly, and specifying the frequency instead of the inductance means we might need a very unusual value of L for the circuit.This is more interesting anyway :-)

author: TSL
author_rank: Regular Contributor
timestamp: February 15, 2015, 06:39:56 am
content: It seems that the solution to this problem would be to use varactor diodes to do the tuning.This is a common way to tune preselectors on many bands.Google "varactor tuned preselector " for many examples.By reducing your "capacitance required" to a tuning voltage will simplify your circuit.regardsTim