Hi,
I have designed a simple center fed dipole on Fr4 substrate at 2.4 GHz. Resonance at 2.45 GHz. But to achieve that I need to increase the dipole length to 42mm.
In Fr4 substrate, the wavelength is around 60mm and lambda/2 is around 30mm at 2.4 GHz. But for my dipole, the length is 42mm which is > lambda/2 (30mm).
Can someone please explain why my dipole length is larger?
The thing is that your dipole antenna is not submerged deep (several wavelengths) in the substrate. So that not all the near field energy is stored directly inside the FR-4 material, but also a great part is stored in the vacuum around. This is why here it is correct to introduce an 'effective' dielectric constant, which is exactly the ratio between electric lengths in vacuum and on your substrate; and this is the reason for what you have obtained.
The dipole length needs to be increased to achieve the desired resonance. A substrate like FR4, which is often 1.6 mm thick, alters the electromagnetic field distribution due to fringing effects. In addition, factors such as the conductor thickness, feed type, and transition structures introduce capacitive or inductive effects that electrically lengthen the dipole or modify its impedance. Therefore, it is necessary to adjust the dipole length to maintain proper matching at 2.4 GHz. To reduce these effects, it's better to use a thinner substrate (0.8 mm) with a lower relative permittivity, such as Taconic or Rogers materials (about 2.3).
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