[Blabber] Help with microstrip traces

Guan Yang guan at yang.dk
Tue May 14 13:28:44 UTC 2013

Dave, characteristic impedance is not the same thing as DC resistance.

On Tue, May 14, 2013 at 8:26 AM, David Reeves <drreeves at gmail.com> wrote:

> Could you make 2 parallel traces of 100 ohm? From 1/((1/R1)+(1/R2)) that
> would be 50 ohm.
> On Tue, May 14, 2013 at 8:22 AM, Ian Harris <imharris at gmail.com> wrote:
>> Hi guys,
>> So I'm stuck.  I'm building a wifi module and I need to create a 50 ohm
>> trace from the module to the antenna.
>> Looking on the internet, I can find impedance calculators, and after
>> poking around it seems that I need to use a four layer board (since in two
>> layer boards the dielectric is just too thick).
>> So in a four layer board, I can find examples where the dielectric<http://www.expresspcb.com/expresspcbhtm/specs4layerstandard.htm>is 0.017" (17mil).  Copper is, say 1oz (which the internet
>> tells me <http://www.pcbuniverse.com/articles.php?a=4> is 1.37mil
>> "high").  The dielectric constant is 4.6 (same source).
>> Now according to a calculator <http://emclab.mst.edu/pcbtlc2/microstrip/>,
>> the required strip width (trial and error) for 50R is around 30mil.
>> However, on the demo board for this module, the antenna trace, while a
>> reasonably solid trace is *almost* the SMD pad width - but not quite -
>> and the SMD pad width is 27mil. If I had to guess, I'd say it's around
>> 20mil.
>> So how do they get the impedance down to 50R? The calculator says that a
>> 20mil trace / 1.37mil copper / dielectric (17mil) / 4.6 constant = 88R.
>>  I'm presuming that's not close enough.
>> Change copper from 1oz to 0.5oz makes almost no difference.
>> Any ideas?
>> kind regards
>> Ian.
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