|
 |
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.
Top
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 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:
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
Top
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!
Top
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
Top
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.
Top
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
Top
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) + [(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
Posted 11/12/2012
|
|
|
|
|
