[SI-LIST] Re: Question regarding current loop
- From: <Christopher.Jakubiec@xxxxxxxxxxxx>
- To: <scott@xxxxxxxxxxxxx>, <doug@xxxxxxxxxx>
- Date: Wed, 26 Apr 2006 14:28:17 -0700
Well this clears a lot of things up for the amateur!
Thanks,
Chris
Infineon Technologies
=20
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
On Behalf Of Scott McMorrow
Sent: Tuesday, April 25, 2006 9:03 PM
To: doug@xxxxxxxxxx
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Question regarding current loop
Doug
You're diverging into the charge hose analogy again, but I'll entertain
your thought experiment.
* 1) driver switches from zero to one driving a trace over a
continuous plane (agreed)
o Lets attach that plane to ground.
* 2) current flows into the transmission line
o Well, some or most of the current does, depending upon
frequency. I know cases where all the power is transfered
down from the output pad of the die through to the substrate
ground, and never even passes out of the device. But in our
case, we'll agree that current travels through the output
transistors into the transmission line.
o There are actually several parts to this part of the
circuit.
+ There is a power and ground feed to the output
transistors
# The power and ground feed are connected to a
power distribution system, which acts as an
imperfect AC short.
+ The output transistor is "referenced" to a substrate
ground, or ground mesh.
+ The output transistor is attached at the pad to a
transmission line
+ The transmission line is referenced to a plane (lets
let it be ground for the sake of our discussion)
* 3) I prefer Maxwell's concept of electromagnetic field propagation
as the basis of AC current and charge flow.
o A field flows from the power/ground distribution system into
the output circuitry.
+ This field has a direction (therefore a direction for
AC current and power flow)
+ This field must be supported by some transmission
medium.
# In our case, it is the combination of the
power/ground distribution planes, grids and
capacitors, inside and outside the chip.
o The output circuitry redirects the EM field from the
power/ground system to the transmission line/ground system.
+ In so doing,
# some of the power reflects back into the
transistors and back into the PDS
# most of the power is transferred forward
# because the size of the output is very small, it
appears to act as a lumped element switch,
moving the current train from the big PDS to the
small transmission line.
* 4) If the transmission line it is driving is infinitely long,
energy will continue to be transferred from the device to the
transmission line, and the electromagnetic wavefront will continue
to travel in the forward direction.
o Because the signal is traveling down a "well-defined"
transmission line, with power being transferred in the
forward direction, the EM field induces instantaneous signal
and return currents in the trace and the underlying ground
plane.
+ All of the AC energy is contained in that wavefront
+ To support the AC wavefront, the trace only needs to
know where the ground plane is. The wave has been
disconnected from the driver.
* 5) If the transmission line is not infinite, but is terminated to
ground by it's complex impedance, then the AC wavefront (which has
a positive and negative component on the trace and plane,
respectively), will cause the traveling wave to be fully absorbed.
o This matched termination does not need any knowledge of the
driver, and no current of any kind flows back to the driver.
* 6) If instead of a unit step excitation, a band-limited repeating
1 0 pattern is driven, then this system can be decomposed using a
Fourier spectral analysis.
o The typical analysis shows a DC component, a fundamental AC
switching frequency, plus multiple harmonics above the
switching frequency.
+ For all energy carried by the fundamental and it's
harmonics (The AC components), there is no need for a
current to return from the terminator back to the
driver. All energy is absorbed in a complex impedance
matched termination.
+ This leaves us with DC.
# DC current does not induce a "return current"
field in the ground plane, since it is not time
varying. (It does, however, induce a magnetic
field)
# DC current does need to close the loop and find
it's way back to it's origin.
* In our case, the origin is the power
distribution system, and ultimately the
"battery" from which all current comes and
goes.
# So, when the DC current flowing through the
trace reaches the termination resistor, it does
not necessarily return back to the driver ground
to close the circuit, it takes the lowest
resistance path back to the battery (a.k.a.
power supply).
* 7) If we extend our thought experiment to two single-ended traces
well-separated on the board driven with a differential
excitation, then exactly the same steps can be followed until we
reach the terminator.
o When we reach the terminator, if both lines are terminated
to ground, the result is the same. Currents travel back to
the power supply through the termination resistors and the
ground plane.
o What if we detach the termination resistors from the ground
plane and tie them together, forming a differential
termination?
+ The positive current enters one terminal of the
resistor and negative current enters the other
terminal of the resistor, and a DC loop is formed.
+ NOTE: All AC current is still perfectly terminated by
a complex impedance matched terminator at the end of
the line, whether it is a terminator to ground or a
differential terminator.
+ As far as the AC wave is concerned the middle of the
resistor forms a virtual ground.
+ Implementation wise, this is a bit oversimplified.=20
For a differential terminator to work well at ultra
high frequencies, it has to be impedance matched to
what the wavefront sees. To do this the termination
must terminate all of the modes of propagation.
+ Normally these are simplified into even and odd mode.=20
But, when a 3rd conductor exists (the ground plane
itself) there are actually 3 eigenmodes, not 2, and
all three must be correctly terminated to provide a
reflection free match.
+ Practically, this is not an issue until we reach the
25 to 50 GHz region. The region depends strongly upon
the distance from the trace to the plane.
In all cases that I can think of, single-ended or differential, any
signal driven on a transmission line from a device can be decomposed
into AC components and a DC component. The AC components can always be
treated with EM wave theory and can be terminated at the end of the
line, without any subsequent currents flowing back to the originating
device. (Energy is lost from the system.) For single ended
termination, the real inductance of attachment vias and plane spreading
inductance will cause significant disruptions when termination occurs. =20
This causes significant bounce. But for differential termination, with
balanced currents, no such disruption occurs.
As far as I am aware, only AC components of a signal radiate, and return
to the universe. DC components do "flow" in the system too. But they
do not necessarily need to "flow" back to the driver. They flow back
through the lowest resistance path to the power supply. At DC the
driver is "just a switch" on a wire.
In summary, once an AC signal is "released" to a transmission line, it
carries it's own instantaneous return reference, which I like to call
the AC, or RF return path, or image plane. The path that the wave takes
is the past of lowest impedance at the frequency of interest, which
amounts to the path of least energy. This path can be terminated at the
end of the transmission line without any additional currents flowing
back to the driver or power distribution system. This is energy that is
"released" and then dissipated through termination and thermal
dissipation. In the limit, the universe forms a spherical transmission
line through which energy is radiated. Once it leaves the antenna, the
energy will not be returning to the driver, and travels off to infinity
and beyond.
DC current, however, does flow in a loop back to the power supply, along
the path of least resistance, which is the path of least energy. DC
current does not radiate, but does create a nice magnetic field.
So, if we go back to the initial contention that somehow in the
non-switching portion of differential pair signaling, the radiation loop
changes magically from between the trace and plane to between the traces
and back to the driver, this is simply not true. DC components of the
positive and negative differential signal find their way back to the
power supply along the path of least resistance. The loop could be
huge, but who cares? DC don't radiate. All AC components still travel
down the transmission line and are either terminated correctly or are
reflected back towards the driver.=20
Radiated energy is extremely low, because the traces are very close to
the image plane, which, as Lee Richey points out, pretty much fixes any
emissions problems anyway. (Formally, a trace over a plane can be
looked at like an antenna separated by it's mirror opposite 2X the plane
spacing away. The mirrored antenna effectively cancels out any
appreciable radiation, as long as the plane is unbroken.) Whatever
residual radiation is left over will have additional cancellation from
the other signal in the differential pair. Differential pairs will have
lower radiation than single-ended traces, but both will meet EMI
compliance if designed correctly. Because the image plane cancellation
of trace radiation is so good, there is negligible difference in
emissions between closely spaced and widely spaced differential traces.=20
As a result, EMI is generally never a good reason for choosing close vs.
wide spacing on differential traces. Density, routability, impedance
control, and manufacturability are good reasons.
Best regards,
Scott
Scott McMorrow
Teraspeed Consulting Group LLC
121 North River Drive
Narragansett, RI 02882
(401) 284-1827 Business
(401) 284-1840 Fax
http://www.teraspeed.com
Teraspeed(r) is the registered service mark of
Teraspeed Consulting Group LLC
Doug Brooks wrote:
> Apparently this is a pretty tricky concept.
> Let's break it down in very small increments and see what sort of
agreement=20
> we can get in stages.
> The first thing is to see if there is agreement at every step along
the way.
>
> To start:
> 1. Assume a driver switches from zero to one, driving a single ended=20
> transmission line (trace over a continuous plane).
> 2. At the first increment of time there will be a current into the
line=20
> from the driver that equals Vout/Zo
> 3. Since current is the flow of charge, there will be charge flowing
into=20
> the line.
> 4. If the driver doesn't change state, charge will continue to flow
into=20
> the line.
>
> Here's a punch line
>
> If charge leaves the driver (at any instant of time), the same amount
of=20
> charge must instantaneously be returning to the driver. (current flows
in a=20
> loop)
>
> Do we all agree with this statement? If not, why?
>
> Charge is something physical (it has mass). Do we all agree with that?
>
> If so, then as the signal travels down the line (forget about the end
of=20
> the line) how and where is this physical charge flowing?
>
> If we can agree on that, then the next question will be, when the
signal=20
> has propagated to the far end of the line, where (we assume) there is
a=20
> proper terminating resistor connected between the trace and the=20
> plane, charge is still flowing onto the line from the driver (which
is=20
> still driving a resistive load, Zo). The same amount of charge must=20
> instantaneously be returning to the driver. The same amount of charge
must=20
> be flowing through the terminating resistor (R =3D Zo). In the first=20
> increment of time after the signal reaches the terminating resistor,
where=20
> and how is this physical charge flowing?
>
> It's not clear to me that everyone sees this the same way.
>
> Doug
>
>
>
>
>
>
>
>
>
>
>
> At 10:05 AM 4/25/2006, Rick Brooks \(ricbrook\) wrote:
> =20
>> Only a current loop in a lumped (sized) circuit is the same at every
>> point (sums to zero).
>> For distributed circuits this is not true.
>> But a cutset at a single physical point in any circuit would have a
sum
>> of currents equal to zero.
>> EG, for a long Tline, the current at the load is not the same as at
the
>> driver.
>>
>> I don't know if this also misses the point.
>> Also, DC is an ideal concept just as infinite frequency is.
>>
>> just my 2 cents.
>>
>>
>>
>>
>>
>>
>> -----Original Message-----
>> From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]
>> On Behalf Of Doug Brooks
>> Sent: Tuesday, April 25, 2006 9:22 AM
>> To: si-list@xxxxxxxxxxxxx
>> Subject: [SI-LIST] Re: Question regarding current loop
>>
>> Ahhhhhh...............
>> Poor choice of words.
>> What I meant is that current is the *same* at every point in the loop
at
>> any point in time.
>> Doug
>>
>>
>>
>> At 09:09 AM 4/25/2006, Scott McMorrow wrote:
>> =20
>>> Doug
>>>
>>> Constant current flowing in a closed loop would be a DC circuit, by
>>> definition.
>>>
>>> Scott
>>>
>>>
>>>
>>>
>>> Scott McMorrow
>>> Teraspeed Consulting Group LLC
>>> 121 North River Drive
>>> Narragansett, RI 02882
>>> (401) 284-1827 Business
>>> (401) 284-1840 Fax
>>>
>>> <http://www.teraspeed.com>http://www.teraspeed.com
>>>
>>> Teraspeed(r) is the registered service mark of
>>> Teraspeed Consulting Group LLC
>>>
>>>
>>> Doug Brooks wrote:
>>> =20
>>>> Well,
>>>> I happen to believe these assumptions, but I'm not sure everyone
>>>> =20
>> does.
>> =20
>>>> Do you?
>>>> Doug
>>>>
>>>>
>>>>
>>>> At 07:05 AM 4/25/2006, Andrew Ingraham wrote:
>>>>
>>>> =20
>>>>> Doug,
>>>>>
>>>>> I'm having trouble with your assumption:
>>>>>
>>>>>
>>>>> =20
>>>>>> *If* current flows in a closed loop and *if* current is constant
>>>>>>
>>>>>> =20
>>>>> everywhere
>>>>>
>>>>> =20
>>>>>> in the loop
>>>>>>
>>>>>> =20
>>>>> Regards,
>>>>> Andy
>>>>>
>>>>>
>>>>>
>>>>> =20
>>>>
______________________________________________________________________
>>>> =20
>> ______-
>> =20
>>>> Check out UltraCAD's differential impedance and skin effect
>>>> =20
>> calculators at
>> =20
>>>> <http://www.ultracad.com>http://www.ultracad.com
>>>>
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_______________________________________________________________________
>>> =20
>> _____-
>> =20
>>> Check out UltraCAD's differential impedance and skin effect
calculators
>>> =20
>> at
>> =20
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>>> =20
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>
________________________________________________________________________
____-
> Check out UltraCAD's differential impedance and skin effect
calculators at=20
> http://www.ultracad.com=20
>
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