about intergital capacitance

[hanway]

Hi all;
i am designing a low-pass filter, i found when estimating the capacitance of the planar intergital structue,capcaitance has great deal with the frequency.anyone can tell me how to reduce frequency effect and keep the capacitance a steady value in 1.9GHz¬2.2GHz ? Thanks in advance.

 

[radiohead]

This effect comes into play at frequencies where your physical features become a fraction of the wavelength. The only ways to get around this is using other layout. Why not use simple two-layer capacitor? This guarantees better response!

 

[hanway]

thanks, i was asked not to design the capicatance with multilayers. now i have no good ideas sbout the parallel capacitances. anyone can help me? fc=2GHz, and the wavelength is about 124mm, and the value of the parallel capacitance is required about 6pF(so big value ,isn't it ?).

 

[radiohead]

Well, you should remember the C=epsilon0*epsilonr*S/d formula. This is quite a good approximation for low frequency. You should simulate with planar EM simulator like momentum (ADS).

Is this cap used for decoupling? Why not use a radial stub instead then? Which stackup do you have?

 

[radiohead]

Look at this forum

 

[hanway]

thanks , now i am designing a lowpass filter with suspended material. I first designed the LC lumped lowpass filter. then i realized these L,C component in distributed structures . for example : shunt C is realized in patch , but the series C (the value is about 6pF) isn't easy to realized . it also can be realized in interdigital capactances ,but the value is not steady . by frequence increasing by 100MHz ,the value change 2pF . I have no good idea to make the value of capactance steady.

 

[radiohead]

You don't need series C for low pass filter.

Make a design with parameterized dimensions, use planar EM solver and do an overall optimization cycle. That will result in best performance.

 

[hanway]

I make the series L to parallel LC in order to produce the transmission zeros near the cutoff frequence because the filter require very high out-band attenuation.