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| Differential Bandpass Filter Design - RF Cafe Forums | ||||||
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Below are all of the forum threads, including all the responses to the original posts. aron_coop Post subject: Differential Bandpass Filter Design Unread postPosted: Sat Apr 10, 2004 12:02 pm Offline Lieutenant Joined: Sat Apr 10, 2004 11:35 am Posts: 1 I am tring to design a differential BPF to be used in a grad school project. It's suppose to have a bandwidth of 5.2Ghz to 5.9Ghz, less than 1dB of loss, Input/output match less than -10dB and have more than 30dB's of rejection for the bands 2Ghz-2.5Ghz and 10Ghz-12Ghz. I used the basic "T" equations to get my L and C values, athen built a Low-pass and a High-pass and cascaded them. To get the 30dB rejection I have to add additional stages on the Low-pass and High-pass filters. However, when I do this the bandwidth widens out to be greater than 5.2Ghz-5.9Ghz. Does anyone know why this is happening? And how can I correct this? Can some please help or advise. Also, here the equaltions that I used. They are for a "T" section, but I think they should work for a differential with some changes to the cap (take half the value) between the inductors for a LPF or to the inductor (double the value) between the caps on a HPF. EQUATIONS: LPF: Fh=5.8Ghz, Fc=11.6Ghz, Z0=50Ohms C=1/(pi*Z0*Fc) = 1/(3.14*50*11.6e9) = 0.55pF L=(Z0)^2 * C = (50^2)(0.55pF) = 1.37nH HPF: Fh=5.2Ghz, Fc=10.4Ghz, Z0=50Ohms C=1/(pi*Z0*Fc) = 1/(3.14*50*10.4e9) = 0.612pF L=(Z0)^2 * C = (50^2)(0.612pF) = 1.53nH Does anyone know if these equations are correct for a Differential BPF? And is the assumption of Fc = 2 Fh correct for a Differntial Circuit or just for the "T" model? Thanks, Aron Top Profile I.R Post subject: Unread postPosted: Sat Apr 10, 2004 2:53 pm Hi, You will of course have to realize your filter with distributed elements. There are equations for transforming the capacitance and inductance values to physical dimensions of transmission lines. You can easily find these equations in a text book :idea: The bandwidth can change because you change the characteristics of the filter when you add aditional sections. Do you take the Q element under consideration? You should use a design tool to synthesize your filter and by this you can save time and iterations. I suggest Eagleware: This is a great design tool for distributed and lumped filters with synthesis and simulation capabilities. The equations for the 'T' sections are: LPF: L=Zo/pi*fc C=1/pi*Zo*fc HPF: L=Zo/4*pi*fc C=1/4*pi*Zo*fc When you realize the filter with transmission lines (micro-strip or another), you will have to define the substrate and from that to derive few properties: Er (the dielectric coefficient), Loss tangent, resisitivity etc... you will have to define those and consider them in your final stage of design in order to match to the filter's requirements. I suggest you will use Rogers laminates 4350 or similar as the substrate due to its relatively low Er and stable characteristics. Top moe Post subject: check out this web site Unread postPosted: Tue Apr 20, 2004 4:38 pm http://www.maxim-ic.com/appnotes.cfm/ap ... /791/ln/en Top Pi Post subject: Unread postPosted: Wed May 05, 2004 3:40 pm there is an furmula. You can find out how many resonators you need for defined bandwith. What i can see these days, the engineers use software as crutches. Take a tea, switch off the computer, and start thinking. Purely theoretically. In my times we asked professor for consultation. If you add additional 50 resonators, it might be even wider. Top to moe Post subject: Unread postPosted: Wed May 05, 2004 3:50 pm some Maxim's appnotes are good. But only some. Maxim never was wireless company and therefore the datasheets are full of mistakes. e.g. The don't know location of minimum noise figure impedance and s-parameters. They measure only IP2 to come up with +56dBm IP2 for LNAs and similar stuff.. Posted 11/12/2012 | ||||||
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