How to reduce the Q value of the antenna?

[051021031]

Hello, Everyone,


I want to ask if I can reduce the Q value of the antenna. Because usually for the narrowband antenna, it has quite high Q value (I think it should be larger than 20, some maybe 40 or 60).
Now I want to reduce the Q value of the antenna, and further increase the antenna bandwidth. For example, if I can add a resistor in parallel with the antenna to reduce the Q value of antenna? Now I have a narrowband antenna, but I want to measurement some wideband signal, I want to improve the bandwidth of the antenna in PCB.

If it is possible to do it. Thanks in advance.

Jack

 

[karesz]

Hi 051021031,
yes, so is it.
Usually you have ca. 20Kohm (with the IC parallel), you can apply some other smaller value, discrete, resistor_maybe up to ca. 1KOhm, but dont forget your signal/reader distance will be smaller/shorter too....
In all cases it is not a verysimple problem, than you have usually a impedance matching circuit too.
007_ alias Karesz

 

[biff44]

Adding a resistor might make it "look" like there is more bandwidth, but you are just decreasing the overall efficiency of the antenna to make it look that way. Another way would be to modify the structure to look more "lumped element" in nature. For example, if the antenna is a quarter wave long, if you add a lumped capacitor to the end and instead used an 1/8th wave element, you would have more bandwidth. That is why you see strange "top hat" structures on the top of some antennas.

 

[051021031]

Thank you very much for your reply.

Actually at present, I don't care how much the efficiency of the antenna is dropped. My target is improving the bandwidth of the antenna.

I am quite new in the field of antenna. Would you please give me some reference about the relationship between the bandwidth and the efficiency, and the methods to do it?

I the bandwidth is increased for the narrow-band antenna, the power efficiency should be dropped, I think there should be some equations to explain the principle.

Thank you again for your comments.