Reactance

[C2]

@420 MHz, if R = 50 and X = 50, I have a poor match (VSWR = 2.6) and I would need to add just 7.6pF to cancel that inductive reactance to get to resonance.

@420 MHz, if R = 50 and X = .1, I have a good match (VSWR=1.002), but I would need to add 3789pF to cancel that inductive reactance to get to resonance.

As Xc gets smaller* and smaller*, approaching zero (resonance), the effective capacitance gets larger and larger, approaching infinity. This seems puzzling to me :?

C = 1/(2*pi*F*Xc)

Where C is in Farads, F is in Hz, and Xc is the capacitive reactance.

:?



* I use the term smaller and smaller, even though the numbers are negative, what I mean is that Xc is approaching zero.

 

[freecell]

you didn't miss a decimal point there somewhere, did you?

 

[C2]

I don't know?

1
------------------------ = 7.6E-12 Farads = 7.6pF
2*pi*420,000,000.00*50


1
------------------------ = 3.789E-9 Farads = 3789pF
2*pi*420,000,000.00*0.1

Looks good to me, boss. :?

 

[C2]

BTW, freecell, I guess you weren't interested in the Excel spreadsheet analysis/charting package?

 

[Beetle]

But reactance doesn't "approach zero" as you near resonance! Capacitive reactance and inductive reactance are EQUAL, but with opposite signs, at resonance. They cancel each other, leaving only the RESISTANCE part. You could have a couple hundred ohms of purely inductive reactance, which would be cancelled by the same amount of purely capacitive reactance at ONE and ONLY one frequency. That's resonance; you have to select the values of L and C, probably making one, or both, variable for tuning purposes.

On an MFJ-259, for example, the "X" value that's displayed is the net reactance at whatever frequency is indicated. If you have "X = 50" displayed, you might have 300 ohms of inductive reactance and 250 ohms of capacitive reactance. The NET in this case is 50 ohms of inductive reactance.

 

[C2]

Beetle, let me just continue with what your saying...

I am seeing a "NET" reactance of -0.1 ohms, so all I need to do is remove a wee bit of inductive reactance to make it all better?

Everyone, at least from my perspective, seems to dance around this, for whatever reason, but I do appreciate the help.

The measurement is at one frequency as indicated.

If Xl is very small, I have to add a huge capacitor to get to resonance. I mean a really big fat one.

But if Xl is very large, I only need to add a wee tiny capacitor to get to resonance.

So inductors should go in series?

But aside from that, if Xc measured 0.00001 ohms, would that not emulate a huge, mongolian, supercalifragilisticexpialidosious capacitor? :shock: owie!

 

[thetnhillbilly]

Mathematically, it does have limits that approach zero and it has a negative counterpart. To graphically represent this we can use graphing software to graph the function. Let's do that.

Go to http://www.coolmath.com/graphit/index.html to use their free graphing software. They use the function format y=f(x), but we can substitute these variables into our function. Our new function is:
y=1/(2*pi*420000000*x)
so that y=C and x=Xc

Now we enter our function in the black field above the numbers, so enter:
1/(2*pi*420000000*x) then press enter

You are asking, where is the graph? Well it is there, just too small to see, so we need to zoom in. Take your mouse and drag a box around the origin; that zoomed it in (see your x and y min and max values changed). But you will probably have to do this over and over until you can see the function graphed. The min and max will be near 5E-5 to frame the graph.

You could use the function y=1/x and it would give you a good representation of the graph, just without the zooming.

Now you can see that for any value of Xc, C will never be zero. That is because mathematically you can not divide by zero, so Xc cannot be zero, and 1 divided by any number is will never be zero, so C cannot be zero.

For every unit change you make, you get less return on that unit. This is why in machining we have tolerances. We can never make anything perfect, just "perfect" with the tools and units we are measuring with. For example, if you use a ruler to measure and cut a perfect board, then measure with a micrometer, your board won't seem so perfect.

 

[Beetle]

If you're seeing a net reactance of -0.1 ohms (indicating a condition very slightly capacitive in the reactance department), adding an equal amount of inductive reactance will cancel it. Or, if the capacitance is variable, as in the AM tuner example I posted elsewhere, you can tweak the tuning capacitor to bring about a resonant condition on the frequency of interest.

Remember, even though you have some reactance at SOME frequency and therefore a non-resonant condition, your combination of L and C will be resonant at some OTHER frequency.

I think you're trying to reduce both capacitive reactance and inductive reactance to zero at the frequency of interest, and that simply won't happen.

 

[Lazybones1222]

I would post my picture of a bunny with a pancake on it's head - but you may think I am stupid. Which I believe I am after reading that. LOL

Ah hell here it is anyways:

 

[W5LZ]

There is only one frequency at which both Xl and Xc are zero. That frequency is very seldom even thought of as being useful when talking about radios. Edison tried it way back when and found out that the 'unamplified' transmission range was pitiful.
Think about it. If you figure out one, you'll have the other.
- 'Doc

(don't you hate it when people pull stuff like that?)

(I like the bunny.)

 

[C2]

Well, I used X = -0.1 to illustrate a point.

I am very well versed in the math and graphing (y = some constant/x), and it is the result as X approaches zero, which is defined as + or - infinity depending on which direction you are comming from. The proof of which was forgotten long ago with the rest of high school.

So, here I have my antenna and CW transmitter, and I measure R=50, X=0.1 (inductive). Naturally, I would like to adjust my tuner and add some capacitance to cancel out that reactance.

But, I don't have a cap that big. And, as I add capacitance, I find that X gets smaller, requiring me to add even more capacitance.

So, how is a tuner with a measly 300 or 500 pF variable cap able to cancel the inductive reactances found in an antenna circuit?

You see, the equations work when I'm trying to cancel out a capitive reactance, as you pointed out, Beetle. That is because the variable is in the numerator, so I do not get this inverse relationship.

Think about this, if you plot the reactive component values (picoFarads and microHenrys) through a very small frequency sweep that passes through resonance, the reactance goes from an infinite capacitance and immediately jumps to near zero inductance with the smallest change in frequency.

Back to my question with R=50, X=50, I can easily dial in 7.6 pF with my tuner, but then, if X=0.1, I can't cancel that out.

:?

 

[thetnhillbilly]

As Xc gets smaller* and smaller*, approaching zero (resonance), the effective capacitance gets larger and larger, approaching infinity. This seems puzzling to me :?

This was the puzzlement that I was trying to clear up. As Xc gets smaller (making you come closer to dividing by zero), you will get a larger (approaching infinity) C. I understand things better graphically and it seemed so obvious graphically, and that is why I thought it would help. I guess mabey I shouldn't be posting here, as I don't know jack about antenna theory, but I think I know math and your results are consistant with the formula. What is your question? Are you wanting someone to tell you the formula is wrong and you won't have to use that large of a cap? Are you wanting someone to tell you where to get a cap that large? Or do you want someone to tell you that perfection doesn't exist ouside of God?

 

[Beetle]

C2 -

It sounds like this "R=50, X=0.1" are values you picked to illustrate the point. Going on that assumption:

1. What you're seeing may be well within the error limits of the instrumentation and net X=0.000...0.

2. If you selected your inductor to have the value of inductive reactance that it has, select a different value.

Remember that the circuit has BOTH inductance and capacitance ALREADY; it must have if we're working a resonance problem. In your example, the capacitance is already some finite value, and it's typically adjustable by increasing or decreasing the total area of the meshed plates.

If your antenna reads "R=50, X=0.1", I suggest you connect it up and get on the air, rather than worrying about the difference. This represents an excellent match to a 50 ohm line. Of course, so does a dummy load...but that's another story.

 

[freecell]

"one frequency at which both Xl and Xc are zero." no, one frequency at which the values of Xl and Xc are equal and cancel each other. Xl - Xc = 0 or visa versa. in an electrical circuit there is no such thing as NO reactance but equal amounts of each effectively cancelling each other to establish the "resonance" of the circuit. reactance must be present in equal amounts for resonance to occur at a given frequency.

what is the tuner configuration that you're using? i believe what you're looking for is a series L/C circuit.

C2,
i don't have excel in any of my machines........

 

[C2]

C2 -

It sounds like this "R=50, X=0.1" are values you picked to illustrate the point.

Your right, these are made up values. But seeing as the idea is build resonant antenna systems, do these numbers seem impracticle?

The real numbers, and this is for my 102" whip on my vehicular, connected to my CB, at frequencies 27.395 and 27.405, respectively:
-- R ------- X---- VSWR
56.757 -1.015 1.137
59.170 +0.105 1.183

I don't want to get into the error tolerance for the device, but let's just assume it is +/- 0.003 for each reading.

Now, if I just wanted to find the conjugate match for channel 39, I calculate something like 0.006uH. For channel 40, however, I calculate something like 55,309pF.

It just seemed odd, and I'm not expecting the math to be shown wrong, and plotting the values for the capacitors and inductors is what brought me here anyway.