antenna-theory.com

[NSmith]

Hello,

I am not well versed in antenna design but because I am the guy with the "most" experience it has become my job(I made one once and it worked).

I am trying to simulate a printed folded dipole in Genesys Momentum for use at 350MHz. From the tutorials on this site which make sense I want a total length of Lambda/2 connected at the ends. With each side folded that looks like a length of Lambda/4. In free space that works out to roughly 8.5"
According to my far field sims this thing is singing at > 600MHz and completely lossy at 350MHz.
But looking at the arrow current fields it seems like 350 should be right.

I was hoping the sims would help me feel better about the design but now I am second guessing everything I though I new. One thing in paticular is the adjustment of wavelength for a trace on FR4. Previously I would have only suspected that this impacted wavelength for a transmission line over, under or between ground planes. Is that wrong? Do I need to reduce my calculated 1/4 wavelength to under 5" for copper floating on FR4 with no ground plane nearby?

Thanks for reading
-Nick-

[NSmith]

My previous post was too wordy perhaps.

I use a free tool called Appcad( http://www.hp.woodshot.com ) that calculates wavelength for a printed transmission line based on the PCB dielectric. I assumed that if the trace was an antenna with all other metal pulled away that my wavelength would look like metal in free space.

This feels like a rookie question and I am a rookie. But I am not finding clear answers on how the dielectric must be accounted for specifically in a printed antenna.

Thanks for taking the time to read this.
-Nick

[NSmith]

...Thinking out loud...

On my Printed inverted F design at 900MHz my top length was 2.51". I thought 1/4 wave was supposed to include the shorting arm which would make the total length just over 3". In free space I calculated my 1/4 wave as 3.25" but I ended up trimming the open arm length down to tune the antenna.

Now having read the literature on this site about the mechanics of an inverted F, I see my logic was flawed. It should have been a 1/4wave on just the top arm. So I thought my answer may be in that design if I re-check with the addition of FR4 in my wavelength calc. C/(4*900e6*sqrt(4.6)) = 1.5".

That didn't work out either. So my air calc says my 1/4 wave is too short(25% short) and my FR4 calc says my 1/4 wave was too long(67% long)... frustrating.

Reasonably I suppose I just found a balance of inductance on one side and capacitance on the other, in spite of the wavelength. I sure wish I had a solid process for estimating what a 1/4 wave is on FR4 before spinning copper.