

Quarter Wave Tline  RF Cafe Forums

guest 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.
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IR 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 viceversa,
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:
Wavelength=C0/f
Where:
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
Where:
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
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another guest
Post subject: quarterwave 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!
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guest 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
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Rod 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.
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IR Post subject: Posted: Fri Jan 06, 2006 2:39
am
Site Admin
Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373 Location: Germany Hello,
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
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Guest 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) where 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) + [(Er1)/(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
Posted 11/12/2012



