PCB Impedance Control

Hi

If I am designing a pcb using impedance controlled layers can I treat the
power planes as a reference layer as well as the gnd layers?

Cheers

Jon

"maxascent" <maxascent@yahoo.co.uk> wrote in message 
news:ifidnQIYxf3YUUvbRVn_vw@giganews.com...
>
> Hi
>
> If I am designing a pcb using impedance controlled layers can I treat the
> power planes as a reference layer as well as the gnd layers?
>
> Cheers
>
> Jon

Hi Jon,
Yes. But with a caveat. When your signals switch reference layers, make sure 
there is a path for the reference current.

E.g. Take a 6 layer board, layer 2 ground, layer 5 power. If the signal goes 
from layer 1 to 6 through a via, you should have a bypass cap bewteen power 
and ground near this via. Think of your signal as differential, the 
complementary signal being the reference.

It's for this reason that I long ago ditched power planes and use multiple 
ground planes instead. I route and/or copper pour powers. Then the bypass 
cap in the example I gave can be replaced by a ground via, because layer 5 
is ground in my PCB.

HTH., Syms. 

Symon wrote:
> "maxascent" <maxascent@yahoo.co.uk> wrote in message 
> news:ifidnQIYxf3YUUvbRVn_vw@giganews.com...
>> Hi
>>
>> If I am designing a pcb using impedance controlled layers can I treat the
>> power planes as a reference layer as well as the gnd layers?
>>
>> Cheers
>>
>> Jon
> 
> Hi Jon,
> Yes. But with a caveat. When your signals switch reference layers, make sure 
> there is a path for the reference current.
> 
> E.g. Take a 6 layer board, layer 2 ground, layer 5 power. If the signal goes 
> from layer 1 to 6 through a via, you should have a bypass cap bewteen power 
> and ground near this via. Think of your signal as differential, the 
> complementary signal being the reference.
> 
> It's for this reason that I long ago ditched power planes and use multiple 
> ground planes instead. I route and/or copper pour powers. Then the bypass 
> cap in the example I gave can be replaced by a ground via, because layer 5 
> is ground in my PCB.
> 
> HTH., Syms. 
> 
> 

I have a number of pretty pics I made to illustrate the issue. I'll put 
'em on a.b.s.e tomorrow sometime.

As an aside, at high speeds (which I define as having a track longer 
than 1/4 rising/falling edge, YMMVG), there is no such thing as 
'ground'. The ground plane is a signal layer insulated by copper ;)
[Although it helps to keep large power currents away from the high speed 
return paths].

Cheers

PeteS

"PeteS" <axkz70@dsl.pipex.com> wrote in message 
news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
<snip>
>
> I have a number of pretty pics I made to illustrate the issue. I'll put 
> 'em on a.b.s.e tomorrow sometime.
>
> As an aside, at high speeds (which I define as having a track longer than 
> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
> ground plane is a signal layer insulated by copper ;)
> [Although it helps to keep large power currents away from the high speed 
> return paths].
>
> Cheers
>
> PeteS

Since without a ground one has a dipole antenna, could you qualify what you 
mean?  The impedance of the plane is measured in units of picoHenry per 
square so it's not a solid ground, certainly, but without a ground we'd be 
in a world of hurt at these high frequencies. 

"John_H" <newsgroup@johnhandwork.com> wrote in message 
news:13de6b5q88e3ke1@corp.supernews.com...
> "PeteS" <axkz70@dsl.pipex.com> wrote in message 
> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
> <snip>
>>
>> I have a number of pretty pics I made to illustrate the issue. I'll put 
>> 'em on a.b.s.e tomorrow sometime.
>>
>> As an aside, at high speeds (which I define as having a track longer than 
>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
>> ground plane is a signal layer insulated by copper ;)
>> [Although it helps to keep large power currents away from the high speed 
>> return paths].
>>
>> Cheers
>>
>> PeteS
>
> Since without a ground one has a dipole antenna, could you qualify what 
> you mean?  The impedance of the plane is measured in units of picoHenry 
> per square so it's not a solid ground, certainly, but without a ground 
> we'd be in a world of hurt at these high frequencies.
>
I think Pete means that at high frequencies the skin effect means that the 
current in the plane is all on one surface of the plane. Thus the bulk of 
the plane might as well be an insulator. Or something like that?
Cheers, Syms. 

On Fri, 31 Aug 2007 11:02:13 +0100, "Symon" <symon_brewer@hotmail.com>
wrote:

>"John_H" <newsgroup@johnhandwork.com> wrote in message 
>news:13de6b5q88e3ke1@corp.supernews.com...
>> "PeteS" <axkz70@dsl.pipex.com> wrote in message 
>> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
>> <snip>
>>>
>>> I have a number of pretty pics I made to illustrate the issue. I'll put 
>>> 'em on a.b.s.e tomorrow sometime.
>>>
>>> As an aside, at high speeds (which I define as having a track longer than 
>>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
>>> ground plane is a signal layer insulated by copper ;)
>>> [Although it helps to keep large power currents away from the high speed 
>>> return paths].
>>>
>>> Cheers
>>>
>>> PeteS
>>
>> Since without a ground one has a dipole antenna, could you qualify what 
>> you mean?  The impedance of the plane is measured in units of picoHenry 
>> per square so it's not a solid ground, certainly, but without a ground 
>> we'd be in a world of hurt at these high frequencies.
>>
>I think Pete means that at high frequencies the skin effect means that the 
>current in the plane is all on one surface of the plane. Thus the bulk of 
>the plane might as well be an insulator. Or something like that?

I think he means the return current is mostly localised directly under
the signal track; the copper under the "space" between tracks carries
relatively little current and could *almost* be replaced by insulator. 

- Brian

Hi - 

On Fri, 31 Aug 2007 13:38:11 +0100, Brian Drummond
<brian_drummond@btconnect.com> wrote:

>On Fri, 31 Aug 2007 11:02:13 +0100, "Symon" <symon_brewer@hotmail.com>
>wrote:
>
>>"John_H" <newsgroup@johnhandwork.com> wrote in message 
>>news:13de6b5q88e3ke1@corp.supernews.com...
>>> "PeteS" <axkz70@dsl.pipex.com> wrote in message 
>>> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
>>> <snip>
>>>>
>>>> I have a number of pretty pics I made to illustrate the issue. I'll put 
>>>> 'em on a.b.s.e tomorrow sometime.
>>>>
>>>> As an aside, at high speeds (which I define as having a track longer than 
>>>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
>>>> ground plane is a signal layer insulated by copper ;)
>>>> [Although it helps to keep large power currents away from the high speed 
>>>> return paths].
>>>>
>>>> Cheers
>>>>
>>>> PeteS
>>>
>>> Since without a ground one has a dipole antenna, could you qualify what 
>>> you mean?  The impedance of the plane is measured in units of picoHenry 
>>> per square so it's not a solid ground, certainly, but without a ground 
>>> we'd be in a world of hurt at these high frequencies.
>>>
>>I think Pete means that at high frequencies the skin effect means that the 
>>current in the plane is all on one surface of the plane. Thus the bulk of 
>>the plane might as well be an insulator. Or something like that?
>
>I think he means the return current is mostly localised directly under
>the signal track; the copper under the "space" between tracks carries
>relatively little current and could *almost* be replaced by insulator. 
>
>- Brian

Seems to me you're both right: the return current is bunched up under
trace, on the side of the ground plane adjacent to the trace.  This
"path" is surrounded by copper.  To paraphrase Doug Smith, a local EMC
consultant, the other side of the plane might as well be on the moon.

Bob Perlman
Cambrian Design Works
http://www.cambriandesign.com

"Bob Perlman" <bobsrefusebin@hotmail.com> wrote in message 
news:st8gd39018b9sr1aqur727crpgdqjc2nbf@4ax.com...
> Hi -
>
> On Fri, 31 Aug 2007 13:38:11 +0100, Brian Drummond
> <brian_drummond@btconnect.com> wrote:
>
>>On Fri, 31 Aug 2007 11:02:13 +0100, "Symon" <symon_brewer@hotmail.com>
>>wrote:
>>
>>>"John_H" <newsgroup@johnhandwork.com> wrote in message
>>>news:13de6b5q88e3ke1@corp.supernews.com...
>>>> "PeteS" <axkz70@dsl.pipex.com> wrote in message
>>>> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
>>>> <snip>
>>>>>
>>>>> I have a number of pretty pics I made to illustrate the issue. I'll 
>>>>> put
>>>>> 'em on a.b.s.e tomorrow sometime.
>>>>>
>>>>> As an aside, at high speeds (which I define as having a track longer 
>>>>> than
>>>>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. 
>>>>> The
>>>>> ground plane is a signal layer insulated by copper ;)
>>>>> [Although it helps to keep large power currents away from the high 
>>>>> speed
>>>>> return paths].
>>>>>
>>>>> Cheers
>>>>>
>>>>> PeteS
>>>>
>>>> Since without a ground one has a dipole antenna, could you qualify what
>>>> you mean?  The impedance of the plane is measured in units of picoHenry
>>>> per square so it's not a solid ground, certainly, but without a ground
>>>> we'd be in a world of hurt at these high frequencies.
>>>>
>>>I think Pete means that at high frequencies the skin effect means that 
>>>the
>>>current in the plane is all on one surface of the plane. Thus the bulk of
>>>the plane might as well be an insulator. Or something like that?
>>
>>I think he means the return current is mostly localised directly under
>>the signal track; the copper under the "space" between tracks carries
>>relatively little current and could *almost* be replaced by insulator.
>>
>>- Brian
>
> Seems to me you're both right: the return current is bunched up under
> trace, on the side of the ground plane adjacent to the trace.  This
> "path" is surrounded by copper.  To paraphrase Doug Smith, a local EMC
> consultant, the other side of the plane might as well be on the moon.
>
> Bob Perlman
> Cambrian Design Works
> http://www.cambriandesign.com

Thanks to everyone for devining what PeteS means.
I understand skin effect.
I understand the confinement of return current for high frequencies.
I don't understand the statement that "there is no such thing as 'ground'."

I was hoping Pete could conjecture what Pete meant.

- John_H 

John_H wrote:
> "PeteS" <axkz70@dsl.pipex.com> wrote in message 
> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
> <snip>
>> I have a number of pretty pics I made to illustrate the issue. I'll put 
>> 'em on a.b.s.e tomorrow sometime.
>>
>> As an aside, at high speeds (which I define as having a track longer than 
>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
>> ground plane is a signal layer insulated by copper ;)
>> [Although it helps to keep large power currents away from the high speed 
>> return paths].
>>
>> Cheers
>>
>> PeteS
> 
> Since without a ground one has a dipole antenna, could you qualify what you 
> mean?  The impedance of the plane is measured in units of picoHenry per 
> square so it's not a solid ground, certainly, but without a ground we'd be 
> in a world of hurt at these high frequencies. 
> 
> 

Wow. Didn't know I would set off such a discussion :)

At high speeds, return currents flow virtually exclusively on the 
reference layer at not much more than the width of the signal track; it 
also flows only in the skin area, the depth of which depends on a number 
of things.
It won't flow elsewhere [subject to normal current distribution laws] 
(unless it has to because there is no return path, which was alluded to 
in the first response) because of the high impedance involved. (At ~4nH 
/ inch there's a high energy barrier, although that value is highly 
topology dependent).

As such, if you have a signal track on a single layer point to point, 
then the return current will flow immediately adjacent on the nearest 
reference layer (so make sure it really *is* a reference layer), and for 
a 6 thou track, the return current will have ~95% in 6 thou on the 
reference layer.

As I said, at high frequencies, the notion of 'ground' as used in power, 
does not really exist; there are only signals and reference, whatever it 
happens to be.

Cheers

PeteS

Bob Perlman wrote:
> Hi - 
> 
> On Fri, 31 Aug 2007 13:38:11 +0100, Brian Drummond
> <brian_drummond@btconnect.com> wrote:
> 
>> On Fri, 31 Aug 2007 11:02:13 +0100, "Symon" <symon_brewer@hotmail.com>
>> wrote:
>>
>>> "John_H" <newsgroup@johnhandwork.com> wrote in message 
>>> news:13de6b5q88e3ke1@corp.supernews.com...
>>>> "PeteS" <axkz70@dsl.pipex.com> wrote in message 
>>>> news:H-ydnSZMEttzlkrbnZ2dnUVZ8vidnZ2d@pipex.net...
>>>> <snip>
>>>>> I have a number of pretty pics I made to illustrate the issue. I'll put 
>>>>> 'em on a.b.s.e tomorrow sometime.
>>>>>
>>>>> As an aside, at high speeds (which I define as having a track longer than 
>>>>> 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The 
>>>>> ground plane is a signal layer insulated by copper ;)
>>>>> [Although it helps to keep large power currents away from the high speed 
>>>>> return paths].
>>>>>
>>>>> Cheers
>>>>>
>>>>> PeteS
>>>> Since without a ground one has a dipole antenna, could you qualify what 
>>>> you mean?  The impedance of the plane is measured in units of picoHenry 
>>>> per square so it's not a solid ground, certainly, but without a ground 
>>>> we'd be in a world of hurt at these high frequencies.
>>>>
>>> I think Pete means that at high frequencies the skin effect means that the 
>>> current in the plane is all on one surface of the plane. Thus the bulk of 
>>> the plane might as well be an insulator. Or something like that?
>> I think he means the return current is mostly localised directly under
>> the signal track; the copper under the "space" between tracks carries
>> relatively little current and could *almost* be replaced by insulator. 
>>
>> - Brian
> 
> Seems to me you're both right: the return current is bunched up under
> trace, on the side of the ground plane adjacent to the trace.  This
> "path" is surrounded by copper.  To paraphrase Doug Smith, a local EMC
> consultant, the other side of the plane might as well be on the moon.
> 
> Bob Perlman
> Cambrian Design Works
> http://www.cambriandesign.com

Absolutely. The other side of the plane doesn't really exist to the 
return currents. There is some speculation that return currents take the 
  lowest energy state return path, which seems intuitively true, but I'm 
not sure we have proper evidence for.

There is a great deal of misunderstanding in the area of reference 
layers, planes and high speed signalling in general. When we take a 
signal through the board, we have to make sure we also take the return 
path through the board. I spent a long time doing calculations on diff 
pairs of varying topologies to figure out the optimal via sizes (and 
spacing, annular ring size etc) to take signals from one layer to 
another with minimal loss (and therefore minimal radiation). I got the 
loss down to < 0.3dB / via on a rather dense high speed board.

Cheers

PeteS