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
Jun 27, 2005 2:02 pm
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.
calculate the wavelength by using the following formula:
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.
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.
subject: quarter-wave linePosted: Thu Jan 05, 2006 4:36 pm
Just a minor addition, as the original poster
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.
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.
Posted: Fri Jan 06, 2006 2:39 am
Joined: Mon Jun 27, 2005 2:02 pm
As you are a student I gave you the theory behind the calculation of the
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.
Post subject: Posted: Fri Jan 06, 2006 3:54 pm
Electrical length is given by
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))
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
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
lambda = 0.1875 m
d = 0.046875m = 1845 mils
Use any software
I used AppCAD from Agilent
Using this method, we get d = 1838 mils