Impedance of the Patch???

Thread Starter

lifeisworld

Joined Sep 26, 2009
5
Hi guys
This is my first tread, because i am new on this webside. I hope you can help me with this question:

This problem deals with the adaptation of a micro strip patch antenna with a 50 Ω micro-strip line. (I am using the print: Rogers Duroid 5870, 31 milli-inch).
patch given : W=25 mm, and l=18 mm.

a) For the patch i have too use the frequency f = 5.4 GHz.(see the attacht file) and the approximate formula for the impedance of a resonant patch is given by:

Zpatch = 90 (εr2 /(εr – 1)) (L/W)^2

- How do I find the impedance of the patch??

I have done it like this:

Zpatch = 90 (εr^2 /(εr – 1)) (L/W)^2
= 90( 1,258*10^10)^2 / ((1,258*10^10)-1)*(18*10^-3)/(25*10^-3)
= 8,152*10^11

I think iam doing something wrong when i am calculating εr. Can you help??
 

Attachments

kkazem

Joined Jul 23, 2009
160
Hi,

Using your formula, I get a result of 190.44 Ohms. Here is the calculation. But first, I can't figure out where you got the value of Er (epsilon r). It is listed clearly on the datasheet as 2.33, which is a unitless number. This is the dielectric constant. So, we have (and I made no attempt to verify your formula)
Z=90*((Er^2)/(Er-1))*(L/W)^2 =
Z=90*(2.33^2)/(2.33-1))*(.018/.025)^2 =
Z=90*(5.43)/(1.33)*(0.018/0.025)^2 = 190.4 Ohms.

If your formula is wrong, then the answer is wrong.

Good luck,
Kamran Kazem,
kkazem
 

Thread Starter

lifeisworld

Joined Sep 26, 2009
5
You are absolutely right. I just checked the datasheet and Er = 2.33. I was just a little confused since the datasheet says that ER = 2.33 for 1MHz. And we have to find the frequency at 5.4 GHz.

The Patch desired adjusted to 50 Ω using quarter-wave transformer. Imagining that the quarter-wave transformer (a micro-strip line) feeder the patch at l = 0 mm.

a) Find the required characteristic impedance Zq for the quarter-wave transformer.

b) Find the necessary width, Wq, the effective dielectric constant, εeq, and the necessary length lq for the quarter-wave transformer. Use Collins equations iteratively.

______________________________________________________
My suggestion:

a) Zq = sqrt(50*Zpatch found in the previous task) = sqrt(50*190,4)=97,6 Ω

b) Collins equation er given:
εeq = ((Er+1)/2)+((Er-1)/2))*(1+(12*d/Wq))^-0.5 - 0,217*(Er-1)*(T/sqrt(Wq*d))

But my problem is now that I dont have any equations to calculate Wq and lq. And as you can see I have you find Wq and lq before I can find εeq.
d have i found in the datasheet: d= 0,787 mm (31 milli inch as required in my task).

Do you know any equations to calcutale Wq and lq for a micro strip line???

Thanks
 

SgtWookie

Joined Jul 17, 2007
22,201
You might find HP's AppCad handy for things like this. It's available for download here:
http://www.hp.woodshot.com/

Specifically for RF and microwave designs. Microstrip, coplanar waveguide, stripline, plus a number of active RF circuit functions.
 

Papabravo

Joined Feb 24, 2006
12,405
You are absolutely right. I just checked the datasheet and Er = 2.33. I was just a little confused since the datasheet says that ER = 2.33 for 1MHz. And we have to find the frequency at 5.4 GHz.

....
What the datasheet actually shows is that they made the measurement at 1 MHZ and that the value changed very little all the the way to 10 GHz. That is the plus or minus .02. on the second line. What you did was multiply the 2.33 value by 5.4*10^9 to get the value you used, 1.258*10^10

When I read your original post I nearly fell off my chair thinking -- "Man those guys sure came up with some whupass dielectric material". My next thought was "Chaaa...and subtracting 1 one from that number will sure affect the result". Glad your problem was resolved so quickly.
 
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