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quarter wave Tline - RF Cafe Forums

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Post subject: quarter wave Tline Posted: Thu Jan 05, 2006 6:23 am

i'm a student and need to ask some questions...is quarter wavelength Tline always provide a AC choke (DC feed). also, what does quarter wavelength mean? does it mean that the microstrip's or CPW's electrical length be equal to 1/4 of the wavelength at a given frequency? and lastly, if i'm given an electrical length how do you calculate the actual length of the transmission line, meaning how long will it be in unit micrometers, meter, mils, etc. thank you very much.



Post subject: Posted: Thu Jan 05, 2006 8:12 am

Site Admin

Joined: Mon Jun 27, 2005 2:02 pm

Posts: 373

Location: Germany

Hello guest,

Quarter wavelength is simply as its name quarter of the wavelength of the signal

The quarter wavelength Tline for DC signal is simply a short, so DC signal is passing through it without resistance.

For AC (RF signal) the quarter wavelength uses to convert a short circuit to open circuit and vice-versa, or just simply to rotate a given point on the Smith Chart in 90 deg - since the entire Smith Chart has a cycle of 180 - half wave length.

You can calculate the wavelength by using the following formula:



C0=light velocity in free space (3*10^8 m/sec)

f= The frequency of the signal

The wavelength changes while passing through different media, e.g. different substrate. In order to find the actual wavelength through a given media, you have to know the velocity through the media Cr.

Cr=(1/sqrt Er)*C0


Er - The Dielectric Constant of the substrate.

Then you can use again the first formula and find the actual wavelength of the signal when passign through a substrate with given Dielectric Constant Er.

Hope this helps.


Best regards,

- IR


another guest

Post subject: quarter-wave linePosted: Thu Jan 05, 2006 4:36 pm

Just a minor addition, as the original poster mentioned microstrip:

The velocity of propagation for microstrip is an additional complication, because you can't just use the Er of the dielectric. The effective Er depends on the ratio of the line width to the substrate thickness, and the ratio is not linear. There are calculators available on the web.

Good Luck!



Post subject: Posted: Thu Jan 05, 2006 9:51 pm

thanks guys....but as IR said if the quarter wavelength is DC short and AC open then why do they but a bypass capacitor between the quarter wavelength to ground? also, no body mentioned about electrical length and how to calculate actual length of a tranmission line from a given electrical length. thanks again



Post subject: Posted: Fri Jan 06, 2006 1:39 am

The capacitor is used to block DC, prevent it from being shorted to ground. The cap value is probably large enough to have no effect on the quarter wave line.



Post subject: Posted: Fri Jan 06, 2006 2:39 am

Site Admin

Joined: Mon Jun 27, 2005 2:02 pm

Posts: 373

Location: Germany


As you are a student I gave you the theory behind the calculation of the wavelength.

I agree with Rod, that when microstrip is involved, then you must look at the actual physical dimesions of the substrate involved as: Thickness, width of the metal, roughness etc, and from. There are formulas to calculate the effective Er and line width from these values. The effective Er value can be later used to calculate the actual wavelength through the material.


Best regards,

- IR



Post subject: Posted: Fri Jan 06, 2006 3:54 pm

Electrical length is given by Beta*d

where Beta is the propogation constant,d is the physical distance.

Given an electrical length of 90 degrees( quarter wavelength) we will calculate actual length of transmission line

50 Ohm Microstrip,w = 12 mils,h = 6 mils,Dielectric constant = 3.48 RO4350 substrate,f = 1GHz

We want to calculate d (actual length of transmission line) Beta*d = 90 degrees Eqn (1)

Beta = (2 * pi )/lambda Eqn (2) d=?

Lambda = c / ( f *sqrt(Eeff)) Eqn (3)


c = 3*10^8 m/s,Eeff is the effective dielectric constant

There are 2 ways to find the value of Eeff

Option 1: (longer way)

For a microstrip transmission line with w/h <= 2

A = 2*pi*(Zo/Zf)(sqrt(Er+1)/2) + [(Er-1)/(Er+1)(0.23 + (0.11/Er))]

w/h = [8*(e^A)]/[e^(2A)-2]

Zf = wave impedance in free space = 376.8 Ohms

For our casewe can calculate a and w/h:

A = 1.3926

w/h = 2.26725

See Plot 2. at the link below

http://www.ansys.com/industries/mems/me ... ostrip.pdf

For w/h = 2.2675 we can read sqrt(Eeff) from the above plot for our dielectric constant of 3.48

sqrt(Eeff) is approximately = 1.6

Then using eqn (1),(2),(3)

lambda = 0.1875 m

Beta =1920

and finally

d = 0.046875m = 1845 mils

Option 2:

Use any software

I used AppCAD from Agilent

Using this method, we get d = 1838 mils

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