[Blabber] Help with microstrip traces

Ian Harris imharris at gmail.com
Sun May 19 14:44:51 UTC 2013


Thanks Guan.  So wavelength of 2.4GHz = 0.125m, so traces longer than
1.25cm are going to be an issue.

Incidently, that looks like pretty sweet chip, the nRF51822.  You're sure
getting a lot of value for your money there.  How did you go about getting
the tool chain working?

cheers
Ian.


On 19 May 2013 10:35, Guan Yang <guan at yang.dk> wrote:

> 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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>>>>>
>>>>>
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