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LC matching network, are both components required ?

 
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Doa
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PostPosted: Mon Oct 20, 2014 6:26 am    Post subject: LC matching network, are both components required ? Reply with quote

Hello,
    and thank you for your great website.
    Through trial and error, I managed to obtain a better and more "stable to environment" VSWR on a 900mhz antenna by adding a 3.9pF capacitor between the feed line and ground. Does it need an inductor in serie to compensate for something ? Or is a parallel capacitor on its own fine ?
Thank you !
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admin
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Joined: 03 Jan 2007
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PostPosted: Mon Oct 20, 2014 5:08 pm    Post subject: Reply with quote

It's not a rule that you always need L and C to match an antenna. For some antennas, they need no impedance matching. Dual band antennas often need 4 elements to get impedance matching right.

The correct or best impedance match network just depends on what your antenna's impedance looks like. If you have an impedance of something like 50+j10 ohms (for instance), then some shunt capacitance may be all that's needed to get it close to 50 ohms.

A lot of this stuff is here:
http://www.antenna-theory.com/tutorial/smith/smithchart5.php

But if things work for you with one matching component, then that is not an issue.
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Doa
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PostPosted: Tue Oct 21, 2014 2:22 am    Post subject: Reply with quote

Thank you for your answer !

Is it possible the capacitor delivers (part of) the power rather than the antenna ? Asking because if I disconnect the antenna (but leave the capacitor between feed line and ground), the vector analyzer reports an S11 drop at the same center frequency as with the antenna albeit with less efficiency (-8dB rather than -22dB).

Thank you !
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tazy
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PostPosted: Thu Nov 06, 2014 5:49 am    Post subject: Reply with quote

Is it possible the capacitor delivers (part of) the power rather than the antenna ???







-------------------------------
tZy
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EA1DDO
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PostPosted: Thu Nov 06, 2014 10:25 am    Post subject: Reply with quote

Hi,

If you had to add capacitance is because your antenna is showing inductance, at reactance or positive imaginary part.

Z=R+jX

Applying capacitance (which is negative reactance) you kill the positive reactance and becomes zero.

+jX - jX = 0

Z=R

So, you donīt need to add anything else.

Regards, Maximo
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