Hi-Q inductors

I used a parabola due to laziness.

Take a parabola of equation

y = (L/2)*epsilon*(x**2-1),

where x goes from -1 to 1. the arc length s is

s = integral sqrt(1+(dy/dx)**2),

which is

s = (2640 feet)*integral[-1,1] (1+2 epsilon**2 x**2 + O(x**4)) = (5280 feet)(1+ 2/3 epsilon**2) + higher terms

Since by our choice of parabola,

height = (1/2 mile)*epsilon,

and setting the arc length to be 5281 feet, then up to terms of third order,

height = sqrt(3/8*5280) = 44.497 feet.

Cheers

Phil Hobbs

The exact solution isn't more complex. Assume y is the chord (5280') and h is the desired height and z is the unknown angle made from one side of the track to the center of it, taken to the center of the large circle, then:

h=y*tan(z/2)/2, where sin z/z=y/(y+1)=k=1-x^2/3!+x^4/5!-

But we already know z is small, so tan(z/2)=z/2, leaving us with: h=y*x/4, where y/(y+1)=1-x^2/3!+x^4/5!-...

This is really pretty simple. To find x, a quadratic on the approximation works. Truncate and multiply by 5! and collect terms to get:

set b=x^2, solve b^2-20*b+120*(1-k) quadratically to get b and then take the square root to about get x=.03370776.

So the whole thing boils down to a h=0.03370776*5280/4, with nothing more than a simple quadratic involved to create that funky number.

If you test everything by plugging in sin .03370776/.03370776 and get about .999810641, you will find that remarkably close to 5280/5281. Which is a good thing.

I don't consider the simplified quadratic to be more complex.

Jon

Surprisingly, it's looking like a powdered iron toroid, mix 6 (the yellow one) probably, can give a Q over 100 at 50 MHz with a tempco in the 35 PPM range. If I do that, I won't have to worry much about pcb proximity and microphonics and such.

John

If you can not find a capacitor with a suitable tempco, in addition to the main capacitor, add a small varactor in parallel (with a small capacitor in series to reduce varactor capacitance swing). The varactor control voltage is then controlled by a temperature sensor through a suitable adjustable network.

Of course, you could use a microcontroller to measure the temperature, read a tempco adjustment resistor, apply varactor capacitance curve and then feed the DC into the varactor :-) :-).

The traditional method of making a capacitor with an adjustable tempo was to use two fixed capacitors, one with a known positive and the other with a negative tempco. An adjustable differential capacitor was then just to select how much capacity was used from the positive tempco capacitor and how much from the negative tempco capacitor. This combination was then connected into the tank circuit.

The tempco could be adjusted by the differential capacitor, without adjusting the total capacitance and hence resonant frequency.

The problem is how to find a differential adjustable capacitors these days.

Yup, we do exactly that in our high-end delay generators. Life is slightly complicated by the fact that varicaps have horrible TCs of their own, and their TC varies with bias voltage. I was hoping for something simpler this time.

If the inductor TC is reliably positive and fairly repeatable, I might use one of my 18 pF N750 caps, in series with Cx, all across the LC resonator. I can trim Cx on a prototype to pull the N750 down as needed to net zero the oscillator TC, then go with that value in production. If the L is +35 PPM, I might expect to hit +-10 on production units, not too bad.

True. And production tuning of TC is a really expensive operation.

John

Not really but it's a design effort: You can use a dual varicap, they come in all sorts of flavors. Build an oscillator with the second varicap in there that runs all the time, same inductor to get its tempco ironed out as well. Then use a timer in the uC to measure the frequency of that extra oscillator.

I don't suppose that you have ever tried teaching math have you?

Not all.

How about the availability of N220 and N330 types?

Do you think this group requires a math teacher?

Jon

I think you can get them, under the same terms.

John

In case it helps you, imagine the 5280' forms a chord. You know that the arc above it is 5281'. But:

arc = r * angle (1/2)*chord = r * sin(angle/2)

therefore,

chord/arc = sin(angle/2)/(angle/2)

r is gone, chord/arc is given, and the equation fully decides angle exactly. Solving angle approximately by the quadratic is accurate and trivial.

The final lesson is the same length as above. Need it?

Jon

I might be interested in some tutoring in trig. But i was referring more traditional teaching (classroom instruction).

I think i may make a point of calling on you the few times i get stuck. Your explanation is straightforward, clear and memorable.

There may be a lot of people here who are very good at it.

I have tried teaching mathematics only when asked or as a component of formal computer science courses at the 200 and

Jon

What's really enjoyable about math like this is the number of different approaches there are. Obviously, Phil chose a different tact.

I completely followed his approach because I already know, from personal experience, that parabolizing a spherically ground mirror with such a large focal ratio (imagine that the arc is the reflecting surface and the chord is the original surface of the glass mirror blank _before_ grinding it and then the radius is the radius of curvature of the mirror) is not really necessary to do. The difference simply isn't. Telescope makers don't usually bother in such cases -- it's not all that easy to test if you got it right, anyway. So he's just as right to take that tact. But I know this because of early experiences making mirrors.

It's just that I don't consider the _exact_ approach, chord/arc=sin(angle/2)/(angle/2), all that complicated. As you point out, it's straightforward. Also, the eventual approximation is brought in nearer the tail end of the thinking process, rather than into the earliest step. And I've learned from _experience_ to prefer that, where possible.

Jon

I'm sometimes surprised by how little traditional closed-form math I do in daily life. CAD constructions can do geometry; Spice and Mathcad type stuff eliminate a lot of Laplace and root-locus sorts of things. There are tons of online and special-purpose calculators. Filter design and transmission line analysis is just a few clicks now. Things that used to take hours or days now take minutes, and most of the work I do is now numeric, not symbolic.

Understanding stuff like calculus and control theory is great, but I find I don't have to *do* it very often any more.

John

I don't recall arguing that folks need to frequently find closed solutions. And I wouldn't.

The infinite grid resistor problem that peeks out once in a while, for example, has me posting immediately about using a very powerful, very easy to use and remarkably powerful and broadly applicable finite difference approach to solving. (It solves charge distribution on very complex surfaces, just as easily.)

But to deal with your points, it does help "understanding stuff" even if you don't have to do it "often any more." In fact, to gain the benefit you probably need to have done that at some point. Lack of knowledge is ignorance; more informed is better than less informed.

(Though that says nothing about how much can be done while still ignorant of a lot. If folks had to be fully informed before doing anything, nothing would ever get done.)

Hacking works. But it's also nice to know a few things and to be able to find closed solutions where they are easily had.

Jon

Oh yeah. Back in high school we're not to round until we get to the final result, similar concept.