[SI-LIST] Re: Question regarding current loop
- From: Scott McMorrow <scott@xxxxxxxxxxxxx>
- To: Leonard Dieguez <ldieguez@xxxxxxxxxx>
- Date: Wed, 26 Apr 2006 18:45:42 -0400
Leonard,
Yes, I did mean path, not past.
We won't get to that until we talk about the multi-universe theory of
signal propagation ... kidding!
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® is the registered service mark of
Teraspeed Consulting Group LLC
Leonard Dieguez wrote:
> "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. "
>
> Scott, do you mean "Path" and not "past".
>
> Leonard.
>
>
>
>
>
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
> On Behalf Of Scott McMorrow
> Sent: Tuesday, April 25, 2006 6:03 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.
> 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.
> 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.
> 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.
>
> 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.
>
> 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
>
>> 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
>> transmission line (trace over a continuous plane).
>> 2. At the first increment of time there will be a current into the
>>
> line
>
>> from the driver that equals Vout/Zo
>> 3. Since current is the flow of charge, there will be charge flowing
>>
> into
>
>> the line.
>> 4. If the driver doesn't change state, charge will continue to flow
>>
> into
>
>> the line.
>>
>> Here's a punch line
>>
>> If charge leaves the driver (at any instant of time), the same amount
>>
> of
>
>> charge must instantaneously be returning to the driver. (current flows
>>
> in a
>
>> 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
>
>> 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
>
>> has propagated to the far end of the line, where (we assume) there is
>>
> a
>
>> proper terminating resistor connected between the trace and the
>> plane, charge is still flowing onto the line from the driver (which
>>
> is
>
>> still driving a resistive load, Zo). The same amount of charge must
>> instantaneously be returning to the driver. The same amount of charge
>>
> must
>
>> be flowing through the terminating resistor (R = Zo). In the first
>> increment of time after the signal reaches the terminating resistor,
>>
> where
>
>> 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:
>>
>>
>>> 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:
>>>
>>>
>>>> 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:
>>>>
>>>>
>>>>> Well,
>>>>> I happen to believe these assumptions, but I'm not sure everyone
>>>>>
>>>>>
>>> does.
>>>
>>>
>>>>> Do you?
>>>>> Doug
>>>>>
>>>>>
>>>>>
>>>>> At 07:05 AM 4/25/2006, Andrew Ingraham wrote:
>>>>>
>>>>>
>>>>>
>>>>>> Doug,
>>>>>>
>>>>>> I'm having trouble with your assumption:
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>> *If* current flows in a closed loop and *if* current is constant
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>> everywhere
>>>>>>
>>>>>>
>>>>>>
>>>>>>> in the loop
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>> Regards,
>>>>>> Andy
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
> ______________________________________________________________________
>
>>>>>
>>>>>
>>> ______-
>>>
>>>
>>>>> Check out UltraCAD's differential impedance and skin effect
>>>>>
>>>>>
>>> calculators at
>>>
>>>
>>>>> <http://www.ultracad.com>http://www.ultracad.com
>>>>>
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> _______________________________________________________________________
>
>>>>
>>>>
>>> _____-
>>>
>>>
>>>> Check out UltraCAD's differential impedance and skin effect
>>>>
> calculators
>
>>>>
>>>>
>>> at
>>>
>>>
>>>> http://www.ultracad.com
>>>>
>>>>
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>>
> ________________________________________________________________________
> ____-
>
>> Check out UltraCAD's differential impedance and skin effect
>>
> calculators at
>
>> http://www.ultracad.com
>>
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