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I would like to make a match half wave length (λ) dipole antenna (or probably a short dipole antenna). So I would like to make it by measure using LCR meter.

What I am still confuse is about what is the correct formula of impedance (Z). Is in most explanations the formula is written that impedance (Z) = R + jX (Ω), which X is capacitance (and inductance if any). The resistance, the capacitance, and the inductance (if any). As there will be imaginary component there, then the unit will not be ohm (Ω). Then the question is, which one is the correct one for this two formulas (assuming no inductance)?

$$Z = R + jX_C = R + j\frac{1}{2\pi fC} \qquad(1)$$

$$Z = \sqrt{R^2 + (\frac{1}{2\pi fC})^2} \qquad(2)$$

LCR Meter

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  • $\begingroup$ you've got half a sentence in there twice. $\endgroup$ – Marcus Müller Aug 3 at 10:06
  • $\begingroup$ Nevertheless, your presumption that the unit of impedance can't be Ω is wrong: Z=(50-j10)Ω is a perfectly fine impedance. $\endgroup$ – Marcus Müller Aug 3 at 10:06
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    $\begingroup$ Your "digital LCR tester" will only give you the reactance (inductive or capacitive) at the meter's test frequency. This is probably very far from the frequency on which you want to use the antenna, so it will not be helpful. $\endgroup$ – Brian K1LI Aug 3 at 10:58
  • $\begingroup$ @MarcusMüller, are you saying that the 1st equation is correct? $\endgroup$ – Sitorus Aug 3 at 11:38
  • $\begingroup$ @BrianK1LI, frequency is a parameter we will assign. The C is the actual capacitance of the material. The rest we knew already. My point here is to know the actual parameter of the material, and how to use them to a formula. And what formula should I use when do calculation. $\endgroup$ – Sitorus Aug 3 at 11:40
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Impedance is the sum of resistance and reactance:

$$ Z = R + jX $$

$j$ is the imaginary unit, equal to $\sqrt{-1}$. Some equations use $i$ instead to mean the same thing.

Reactance is a concept that describes the effects of capacitors and inductors, as well as other components that introduce a phase shift between voltage and current, but don't dissipate energy.

For example, the reactance of a capacitor is given by:

$$ X_C = {-1 \over 2 \pi fC} \tag 1 $$

And for an inductor:

$$ X_L = 2 \pi fL \tag 2 $$

where $C$ is capacitance (farads), $L$ is inductance (henrys), and $f$ is frequency (hertz). From this you can see for capacitors and inductors (and in practice, most other things) the reactance depends on frequency, and capacitors have negative reactance whereas inductors have positive reactance.

The unit of impedance, resistance, and reactance is ohms in each case. Impedance is represented by a complex number, but the unit is still ohms.


That's the theory.

In practice, you won't be able to measure the impedance of your antenna with an LCR meter. As you can see in equations 1 and 2, reactance (and thus impedance) depends on frequency. The LCR meter works by measuring the impedance at some frequency, then working backwards though equations 1 or 2 to find the inductance or capacitance.

This works OK for inductors and capacitors (at least, when operating at frequencies where parasitic effects are negligible), but an antenna is something else so equations 1 and 2 don't apply.

Instead, people use an antenna analyzer to measure the impedance directly at the frequency where the antenna is intended to be used.

If you don't have an antenna analyzer, you can use an SWR bridge. This will tell you how close you are to 50 ohms, though not in what direction. There will be a dip in SWR around where the antenna is resonant. Making the antenna longer moves this dip lower in frequency, and making the antenna shorter moves it higher. By measuring SWR at several frequencies and iteratively adjusting the length of the antenna, it's possible to get the antenna to be the right length for the desired frequency.

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  • $\begingroup$ Say I would like to make a half wave dipole antenna with the element made of the inner part of RG6-75 ohm cable that will work for 2,100MHz (3G network). So, during planning and calculation of an antenna design, how do I estimate the capacitance? As you wrote the formula of Xc, the last thing we don't have is the capacitance part, which we have to adjust it for the 2,100MHz to match the impedance. From the calculation, then we may decide the required C. That my understanding. We do calculation before we cut the material. $\endgroup$ – Sitorus Aug 4 at 17:38
  • $\begingroup$ If I didn't get this this explanation wrongly, the C is capacitance between the two element. As I said in my comment above, I intended to use inner cable of RG6-75ohm which the diameter is almost 1mm. Hence we can calculate the capacitance, base on the explanation, to meet the requirement, match the impedance. $\endgroup$ – Sitorus Aug 4 at 17:43
  • $\begingroup$ Before this post, I have post question here about what is the minimum and the maximum of the gap to get the optimum performance. And one recommended me to read the link above. $\endgroup$ – Sitorus Aug 4 at 17:47
  • $\begingroup$ Phil, what would hamSE do without your technical expertise? :-) $\endgroup$ – Mike Waters Aug 5 at 1:40
  • $\begingroup$ @Sitorus Why are you trying to make a dipole with coax? $\endgroup$ – Phil Frost - W8II Aug 5 at 20:33
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  • $\begingroup$ I tried to understand those 3 videos. I can not say I understood all. But as far as I could understand, that measurement is after the antenna installed. Contrast, what I am asking here is during the process of planning the material to make antenna. I.e: I have 3mm diameter of copper. How should I cut it to make dipole antenna? Or I have aluminum pipe with 1 cm outer diameter, then how I treat it to make dipole antenna? Should I not measure them first before I cut? Even we know all resistivity, but there would be variation from one to another. $\endgroup$ – Sitorus Aug 3 at 15:33
  • $\begingroup$ Assumptions based on your comments: You have basic measurement equipment like the LCR meter (no VNA, ...). You plan using materials that are not specified by a manufacturer datasheet. Then probably it's best to start following basic design guides on dimensions. Measuring detailed material parameters is more in the realm of commercial companies having expensive equipment. Look for guides in the internet, like: <wikihow.com/Design-a-Simple-Antenna> or <nathan.chantrell.net/old-stuff/radio/radio-scanning/…> $\endgroup$ – MagnusO_O Aug 3 at 16:23
  • $\begingroup$ OK, let me read them. I will come back later. I am amateur, I didn't have any experience in antenna. So, I absolutely don't have any certified material to be used. I will just use what can I use near me. $\endgroup$ – Sitorus Aug 3 at 16:26
  • $\begingroup$ I have read both. For the wikihow.com, I got confused. Seems that it is wrong as one of the transmission line's conductor should connected to the ground (plate), not one of the monopole antenna is connected to the ground. Moreover, the transmitting antenna element of the monopole should be only one, not two. For the second tutorial, it is general. Moreover, it was not dealt with its balance. $\endgroup$ – Sitorus Aug 3 at 17:06
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In general, impedance consists of two parts: a resistive part that dissipates energy and a reactive part that stores energy in an electric or magnetic field. A complex number that includes a real part and an imaginary part is a mathematical tool for keeping track of two related properties of a system. On a Cartesian graph, the real part (resistance) is plotted along the $x$-axis, while the imaginary part (reactance) is plotted along the $y$-axis. By convention, inductive reactance has a positive sign and capacitive reactance has a negative sign. The distance from the origin of the graph to the point $(x,y)$ corresponding to the impedance of interest is the magnitude of the impedance, while the angle of a line segment from the origin to the point is the phase angle of the impedance, as shown in the diagram:

enter image description here

Your first equation should read: $Z=R-j\frac{1}{2\pi fC}$ (note the change of sign). If the reactive part of the impedance was inductive, the equation would be: $Z=R+j2\pi fL$. Your second equation is the magnitude of the impedance: $|Z|=\sqrt{R^2+X^2}$, where $X$ is the inductive or capacitive reactance calculated earlier. The companion of the magnitude of the impedance is its phase angle, which is $arctan\frac{X}{R}$.

Using the simple example of a half-wave dipole, the impedance at the feed point would: be purely resistive at the resonant frequency $(R+j0)$; show capacitive reactance below the resonant frequency $(R-jX_C)$; show inductive reactance above the resonant frequency $(R+jX_L)$. The graph below illustrates this behavior for a half-wave dipole designed for the 20m amateur band (14.1MHz):

enter image description here

Measuring the same dipole's feed point impedance with an $LCR$ meter operating at 100kHz would produce a measurement of $0-j88184 \Omega$, which is equivalent to an 18pF capacitor. Why? At this very low frequency, the 20m dipole is only 0.003 wavelengths long, so it will only radiate (dissipate) a very small amount of energy; thus, the resistive part of the impedance is very small. The capacitive coupling between the two dipole legs dominates the small inductance of the short wires, so the reactance is capacitive.

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  • $\begingroup$ I worry that I didn't understand your explanation fully. Just to make simpler. Can I assume that the impedance is Z=R+jXc? As I said, I would like to measure the actual R and capacitance. So, to input it to the formula, which formula should I use? Should I use your Z=R-j(1/2*pifC)? $\endgroup$ – Sitorus Aug 3 at 11:37
  • $\begingroup$ Since your original post says that you want to match a short dipole antenna, you may assume the impedance has a (negative!) capacitive reactance, but factors like nearby metal objects could invalidate your assumption. Under that assumption, yes, use $Z=R-j\frac{1}{2 \pi f C}$. To measure the actual resistance and capacitance, you must use the right instrument at the frequency where you will operate the antenna. For example, if your LCR meter measures at 1kHz, it will not tell you anything useful about the antenna impedance at 7MHz where you want to match it. $\endgroup$ – Brian K1LI Aug 3 at 11:50
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    $\begingroup$ Your measurement of a capacitor is correct, but an antenna is not a capacitor. An antenna is more like a transmission line. Measuring it as if it was a lumped element, like a capacitor or inductor, will lead to errors. $\endgroup$ – Brian K1LI Aug 3 at 15:31
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    $\begingroup$ Because the antenna has a complex impedance. $\endgroup$ – Marcus Müller Aug 3 at 16:20
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    $\begingroup$ Let me suggest Antenna Theory. It contains a wealth of useful information from basic to advanced topics. $\endgroup$ – Brian K1LI Aug 3 at 19:08

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