CRLH Transmissionline
- use the via (conventional)
- use the Strip line
My question is how to create left-Inductor by using strip line, i cant understand what to do to connect the upper face to the bottom without a via.
I have read a paper about a virtual ground that is on the upper plane. Is that an idea for this problem? Someone know about this (create left-Inductor), please help me!
Thank you so much!
Hi gemmy94,
What do you mean by "strip line"? Usually the only metallization between layers of a substrate is in the form of a via.
So, the only way to create Left-Inductor is to use the via. I have read one paper about use the via-less
http://ieeexplore.ieee.org/xpl/freea...&userType=inst
I mean, is it possible to creat L-left without a via. It would be easier in implement.
Thank you for your care
I'm not sure what you mean by "Left-Inductor" -- that term doesn't mean anything to me. There are many ways to make inductors, such as thin strips, meanders, spirals, etc. See for example http://www.microwaves101.com/encyclopedias/inductors.
I presume "left inductor" is referring to "LL" (the shunt inductor) in the usual CRLH (composite right/left handed transmission line) scheme.
For a good reason, Edaboard rules say it's not allowed to use "Technical abbreviations unless they are universally understood, they will confuse members and make it hard for them to understand what you mean."
I believe the OP didn't consider this point.
I mean the Lleft in CRLH scheme, Mr Planar :) so do you get my question before?
Thank you for your document. It is very helpful.
See, the left-handed inductance is actually given by the meander line (fig. 2(a) of the paper). The idea here is that that stub after meander line is so large that it acts as a short to ground (the capacitance between stub and ground is so high that at operating frequency it act as a SC). The effect is the same as putting SC vias, but some applications don’t allow it and this design is an alternative.