Resonant LC Circuits

[GK]

In practical terms though it's preferable to determine Q then L because size for size normally C is easier to deal with.

For 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.

[GK]

Having 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.

Who 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.

[The Electrician]

I don't know what else I can do.

You could just provide the formula the OP asked for.

[GK]

I don't know what else I can do.

You could just provide the formula the OP asked for.

Just dick waving now? The question was already answered before I got here and there was essentially two parts to the opening post, starting with:

"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, 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.

[IanB]

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.

Suppose 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?

[GK]

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.

Suppose 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?

Sure, 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.

[German_EE]

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 MHz
254 pF     1.997 MHz

2 pF        22.5 MHz
1 pF        31.8 MHz

The 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?

[GK]

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 MHz
254 pF     1.997 MHz

2 pF        22.5 MHz
1 pF        31.8 MHz

The 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?

Wow..... 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.
   

[dom0]

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.

[MrAl]

That looks too hard. The differentiation approach is easier...   

Hi,

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.

[MrAl]

Hi,

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?

I wasn't trying out your suggestion.  I didn't even read post #11, I just did the obvious thing.

...and i believe you.

But that doesnt answer the question about what software you used.

[dannyf]

i bet we can get a version that works more accurately too.

From 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.

[nctnico]

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.

[Dave Turner]

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.

[The Electrician]

i bet we can get a version that works more accurately too.

From 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.

You'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."

[dannyf]

I prefer differentiation. The results are the same.

The same only if the function is linear.

Otherwise, it is only 1st order approximation.

[The Electrician]

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.

Something is a little off.  Your expression returns a value of about 1 million picofarads for F=1 MHz, deltaF = 1000 Hz, deltaC = 1 pF:

[Dave Turner]

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.

[The Electrician]

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.

Yes, I noticed that. 

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.

I 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.

[dannyf]

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.

Not 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.

[T3sl4co1l]

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

[MrAl]

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 

Hello 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 :-)

[TSL]

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.

regards

Tim