[Blabber] Help with microstrip traces

Robert Diamond rmd6502 at gmail.com
Tue May 14 13:54:14 UTC 2013


Aren't reactance and resistance at right angles to each other, so it becomes a vector addition?

On May 14, 2013, at 6:31 AM, Chris Stratton <cs07024 at gmail.com> wrote:

> It still combines the same way.
> 
> On May 14, 2013 9:28 AM, "Guan Yang" <guan at yang.dk> wrote:
> 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 is 0.017" (17mil).  Copper is, say 1oz (which the internet tells me is 1.37mil "high").  The dielectric constant is 4.6 (same source).
> 
> Now according to a calculator, 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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