Folded Dipole Calculator Help

amdx · 2009-11-14T18:05:21+00:00

Hi All, I'm having a disagreement about the use of this folded dipole calculator impedance calculator. We are using it to calculate the impedance of folded dipoles to drive a yagi.When entering data it asks for "Simple dipole feed impedance (ohms)"Default is 72 ohms. This assumes a 1/2 wave dipole. The fellow I'm disagreeing with say's,"the reason that the variable in the calculator for the "dipole feed impedance"could be if you want to transform the z to something else, such as 50 ohms." I don't even understand what he means.I can only think of one reason to alter the 72 ohms, that would be if youuse a a different length folded dipole. I'll readily admit the other fellow knows more about this than me, butI need more convincing :-) Can anyone explain this to me. Thanks, Mike

Nov 14, 2009 26 Replies

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amdx 16 years ago

Yes. but for a simple dipole in free space we can leave it at 72 ohms.

Yes, agreed, coupling structures will alter the impedance of a dipole and also a folded dipole.

As I see it the changing element diameter alters the transforming ratio NOT the 72 ohms that is feedpoint impedance of a (theoretical), infinitely thin, dipole.

( this is an important point and may be the nub of the disagreement)

Using the calculator you will find altering d1 and d2 will NOT alter the The 72 ohms in the data line "Simple Dipole Feed Impedance (ohms)". (I don't know, maybe it should)

Ok, I think I getting close to being able argue my point, but I'm looking for some agreement here for the statements I made above. Thanksfor your help, Mike

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Phil Allison 16 years ago

"amdx"

** You are so dumb you keep inventing non existent complications.

The calculator assumes the folded dipole is in free space.
It then calculates the impedance of the design for you, long as you use

72 ohms in the data.

IF you KNOW the impedance value for some dipole that is part of a more complex antenna - then you use that value in the data instead, to find the value for a folded version.

.... Phil

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amdx 16 years ago

Those "non existent compications" are what I'm trying to eliminate so I can make a better arguement for my position.

Yes, I understand that

Would 72 ohms still be the correct number if d1 is twice the diameter of d2?

I would think that is the reason the author put that variable in the program. Thanks, Mike

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Phil Allison 16 years ago

"amdx"

** Then just STOP inventing them !!!!!

** Read point 2 over and over - d*****ad !!

** I *would* like to see any sign of actual THINKING from you at all !!!

But will not hold my breath waiting for it.

.... Phil

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amdx 16 years ago

Your not stating your opinion very clearly. But I think you mean to say, the relative diameters of d1 and d2 would not have any effect on the underlying 72 ohms used in the calculations for the program. Thanks, Mike If I'm wrong, please don't be shy, let me know. :-)

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Phil Allison 16 years ago

"amdx"

** Yes I am.

** The calculator does not use ANY particular impedance for the computation of a RATIO !!

Once that RATIO is computed, it then multiplies with the impedance figure YOU supply to get the final figure.

.... Phil

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amdx 16 years ago

Yes Phil, I worded that incorrectly. I understand the calculator finds the ratio of the (impedance of a folded dipole) / (impedance of a dipole ) and then multiplies this ratio times the impedance of a dipole in free space. (72 ohms) I also found in the 1985 ARRL Handbook a graph* that shows a change of conductor diameter from 10 to 10,000 only changes the impedance by 30 %. Extrapolating from the graph, I see doubling the diameter of the conductor only increases the resistance by about two ohms. So if we only doubled half** of the dipole it seems the effect would be even less. Thanks, Mike

graph is non linear

** as in making d1 twice the diameter of d2

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