[SI-LIST] Re: Question regarding current loop
- From: Doug Brooks <doug@xxxxxxxxxx>
- To: Scott McMorrow <scott@xxxxxxxxxxxxx>
- Date: Wed, 26 Apr 2006 14:10:28 -0700
Scott,
That is the best and clearest description on this topic I have ever read.
I have no problem with it whatsoever if we are sending a pulse ( a 0-1-0
signal) down the trace whose pulse width is narrow compared to the
propagation time down the trace. And I think I buy it if we are sending a
step function down the trace.
Would you expand, though, on the specific case where we send a
repetitive 0-1-0-1..... (AC) signal down the (properly terminated) trace
whose half-period is significantly longer than the propagation time down
the trace. This (1) is an AC signal, but (2) there are periods of time when
there is no field still propagating (i.e. it has already propagated to the
end of the trace and been absorbed at the end.) It is not clear to me that
your explanation covers that case. (Or if it does, I didn't get it.)
Doug
At 06:02 PM 4/25/2006, Scott McMorrow wrote:
>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)
> * Lets attach that plane to ground.
> * 2) current flows into the transmission line
> * 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.
> * 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.
> * 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.
> * 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.
> * 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.
> * 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.
> * 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.
> * 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.
> * 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>http://www.teraspeed.com
>
>Teraspeed® 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
>>
>
>____________________________________________________________________________-
>Check out UltraCAD's differential impedance and skin effect calculators at
>http://www.ultracad.com
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