# Large diameter of dipole antenna's rod....Can I use?

Discussion in 'Radio and Communications' started by Willen, Jul 27, 2012.

1. ### WillenWell-Known Member

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-Why you would not recommend coiled balun? Is it fake?
-It would be better if you show me a diagram in details based on my experience level plz.
- Are you saying about 1:1 balun? And if I have a 75 ohm, what will be the length of balun?

Last edited: Aug 2, 2012
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2. ### WillenWell-Known Member

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Nobody listened me !

Last edited: Aug 14, 2012
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3. ### PinocleNew Member

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Simple, the k for 100 MHz and d=1,8 cm is 0.941. (half wave dipole)
But. all theoric answers are good!
Regards

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5. ### WillenWell-Known Member

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How you find this 0.94 K factor for 1.8 cm diameter? Do you have any graph or equation? Please

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6. ### PinocleNew Member

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I'm using a MININEC soft based. (MININEC is a code for antenna design). But, for reference, here a graph. (the equation is very complex, with integral sin and cosine)
View attachment 66558

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7. ### WillenWell-Known Member

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But I found simple formula to find 'ratio' and then 'factor K'. Using this formula, I got almost 0.962 as a factor K for 1.8CM diameter.
Like:-
Ratio= (492/MHz) in inch/ Diameter in inch
*Eg: for 91 MHz, the Ratio is :- 64.56/0.708= 91.18
91.18 ratio represents almost 0.962
I AM IN CONFUSSION!!!
Check this link to see the simple equation www.braincambre500.freeservers.com/Open Half Wave Dipole Calculation.htm
I don't know what is happening! Just confusing...

Last edited: Aug 21, 2012
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8. ### PinocleNew Member

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Willen, the formula 492/f[MHz] is good, but is half wave in feet, no inch. (in meter 150/f[MHz]).

half wave for 91 MHz = 492 / 91= 5.4066 feet = 64.88 inchs = 1.65 m (only two digits is good for decimals in this case). (1 foot = 12 inch, 1 foot= 0.0254 m)

1/2Wave / diameter = 64.88 / (1.8 cm / 2.54) = 91.55 ==> k = 0.961 in the graph of the ....freeserver....

In my graph, Full Wave/d ~ 183.15 ==> k = 0.965, L = 62.35", for k=0.965, L = 62.60", for k= 0.941, L= 61.05". The diference between lenghts for the k is only 2.4% (of half wave)
View attachment 66577

In VHF antenna engineering an error of 3% is good!

The problems are:

1) 492 (in feet) and 150 (meters) is a aproximation for half speed of light.

2) The factor k is for resonance of the dipole. An exact THIN half wave dipole has Z=73.1 + j42.5 Ohms. For resonance, you need X=0, then Z= 73.1+j0 Ohms. For X= 0, you need lenght less than half wave, for that, you includes the factor k and this depends on the diameter. See wikipedia, dipole.

For reactance calculation are aproximations (with several methods). According to the method the value of k is different.

3) That formula is for dipole in free space (theoric value). When you put a dipole on earth, the impedance Z varies with the height, and then, you have Reactance and Resistance different than the free space impedance, and then, you need adjust the lenght for X=0.

4) But, with a tube of 1.8 cm in 91 MHz, you have more or less BW= 3.0 MHz for SWR = 1.5 ( 89.5 MHz < fzero < 92.5 MHz). Is a great BW ! Your dipole, more or less, will have a resistance of 75 Ohms, What cable will use? 50 Ohms (RG213). Your SWR ~ 1.4. If the cable is short not problem, but if your feeder is long, you need adjust the impedance with a gamma match (for example), for avoid excessives losses.

Best regards.
Nestor

9. ### WillenWell-Known Member

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Really complicated!
I think it is the best and easy way to cut accurate dipole by measuring SWR using SWR meter. So I am trying to make this SWR meter:- www.angelfire.com/electronic2/qrp/swrpwr.html

You said X=0 is different on ground and hight. I am testing my antenna above 10 feet from ground and after testing I will hang it almost in 35 feet height. So to get X=0 at 35 feet, is it better to cut little short? Or long? Approximately how much length?
Gamma match! Oh again complicated! ;(

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10. ### PinocleNew Member

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Yes, you need a wattmeter, a impedance bridge or a SWR meter. That SWR meter is good.

Well, your antenna will be vertical or horizontal? In the horizontal antenna, X will change more that a vertical dipole.

In the vertical, at 35 feet high, the antenna will have the lenght of the calculation for 91 MHz. A 10 feet high the resonance will be more or less in 91.125 MHz, but the difference will be more or less 1/16 inch, nothing (more long)

In the horizontal dipole, the difference will be mor or less 3/8 inch more long at 35 feet. But the way to adjust an antenna is in its working position.

In the next graph, do you have the reactance.
View attachment 66689

In that "fat dipole" , the bandwith is great, do you not need wil adjust the dipole lenght, but all depend of the transmission line. More SWR, mor feeder losses, but in this case the difference will be mor or less 0.1/0.2 dB, not important.

The gamma match? That is another history!

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11. ### WillenWell-Known Member

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It is almost impossibe to test Vertical Dipole at its real working position because i am going to hang it on 20 feet above from my house using Bambo mast. I think it is the last way:- I should have to accept some mismatch. You said- 1.8cm pipe has great BW, so i hope i won't face more mismatch. I hope, X factor of height won't affect more because of antenna's BW. Also I have only 5 watt FM Tx.

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Mismatch with a transmitter isn't as bad as with a receiver. You lose a little bit of power for xmit, but receiving power is drastically lower than transmit so the mismatch causes more problems. You can always slap an amplifier on a transmitter to account for losses but you can only amplifier a small signal a certain amount before the noise floor goes up with it sufficient to prevent reception. Just make sure you're covered thermally from the mismatch as mismatches tend to cause the final output transistor(s) to heat.

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13. ### unclejed613Well-Known Member

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you will NEVER get a perfect match in the real world. 1.1: is usually about the very best you can do, and 2:1 is ok. you lose about 0.5db of signal from a 2:1 mismatch, nothing to worry about. if it were 3:1 or higher, then you can worry (and even then, you only lose 1.5db of signal, but this is about where amplifier transistors begin to heat up).

Last edited: Aug 29, 2012
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14. ### davennActive Member

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cant see your pic ??

was there a question in there ?

Dave

15. ### Ian RogersSuper ModeratorMost Helpful Member

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He's a known spammer.... I deleted it...

16. ### davennActive Member

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thanks Ian

D