1
$\begingroup$

I need a clarification about how to calculate the length of a dipole antenna.

For a 1/2 wave-dipole which of the formula is correct?

(140/λ) Or (300*10^6/λ)/2

Can a 1/2 wave-dipole dipole work when connected directly to a radio receiver using a coaxial cable or any matching transformer is required?

Improve this question
$\endgroup$
1
  • $\begingroup$ Repost this to the Amateur Radio Stack Exchange. They have experts there who will answer all your questions. $\endgroup$
    – niels nielsen
    Commented Aug 1 at 2:57

2 Answers 2

Reset to default
1
$\begingroup$

None of them is right. The length of a half wave dipole is λ/2. The lambda is the wavelength. It's used as the symbol of the wavelength so widely that sooner or later one makes errors if he uses the lambda for something else in radiowave calculations.

If you know only the frequency you must calculate the wavelength at first. Use the next formula: λ=300/f. It gives the wavelength in meters if you express the frequency f in megahertz. Of course, it's taken from the proper SI-unit formula λ=c/f.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So, it's the same thing as (300*10^6/λ) and (300*10^6/λ)/2 is to calculate the half-wave dipole. But in this case you need to put the value of frequency in hertz only. $\endgroup$
    – user10897549
    Commented Aug 1 at 10:06
  • $\begingroup$ Yes, but avoid using lambda for frequency because people easily see it as wavelength if they have watched and used common formulas long enough. In the original question you did NOT tell that lambda is frequency. $\endgroup$
    – Uno Vaincreek.
    Commented Aug 1 at 10:13
0
$\begingroup$

As Uno says, always use "λ" (lambda) for wavelength and "f" for frequency. This is universally understood.

The two formulas are actually very close to the same thing.

$$L = \frac{140}{f}$$

is assuming that you're expressing the frequency in MHz, and the length in meters. The constant 140 comes from dividing the speed of light (300e6 meters/second) by 1e6 (the number of cycles in one MHz), dividing by 2 to get a half wavelength, and finally multiplying by a correction factor of 14/15, or about 93%. This correction factor accounts for details like the "velocity factor" of the wire and "end effects" — it's a "rule of thumb" for getting better results for practical antennas.

The other formula is essentially identical to this, except that it ignores the correction factor, and it uses frequency expressed in Hz.

$$L = \frac{300e6/2}{f}$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think (300/MHz)/2 = dipole in meters, would be the best one. $\endgroup$
    – user10897549
    Commented Aug 3 at 13:59
  • $\begingroup$ That would depend on what's "best" for you. Like I said, the correction factor in the other formula is useful for building practical antennas. But if you start with that one, you can always trim the antenna shorter to tune it empirically. (Actually, you would just fold the ends over, so you can go back if you happen to go too far.) $\endgroup$
    – Dave Tweed
    Commented Aug 3 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.