[Blabber] Help with microstrip traces

Guan Yang guan at yang.dk
Sun May 19 14:35:24 UTC 2013


A rule of thumb is to worry about this for traces longer than 1/10 of a wavelength.


For traces longer than that you would want the correct width, or to do your best. You can still have a 120 mil wide trace on a board except at the ends, where you need some kind of fanout.


Also, read the datasheet for your components carefully, especially if you don't really know what you're doing, like me. For example, I use an integrated balun that specified a minimum trace length from transceiver to balun, which goes against the general advice of making the traces as short as possible.

On Sun, May 19, 2013 at 10:28 AM, Ian Harris <imharris at gmail.com> wrote:

> So it turns out you can indeed get 50R trace by keeping the dielectric to
> around 12mils.  With a thin 2-layer board, the dielectric is going to be
> around 0.6mm -  resulting in a trace of 75R.
> I guess my question is: How much difference is this going to make? If the
> trace to the antenna is only a few mm long, how important is this impedance
> matching? Guan I have seen your work on the Nordic chip which presumably
> operates at around 2.4GHz as well, was this an issue for you?
> cheers
> Ian.
> On 14 May 2013 09:54, Robert Diamond <rmd6502 at gmail.com> wrote:
>> 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<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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