Microstrips operated in differential mode question

• posted

Could some electronics guru here please clarify the following:

For two tightly coupled parallel microstrips, with single, isolated mode characteristic impedance Z0, the characteristic impedance of the pair when operated in differential mode is 2.0*Z0. What would be the terminating impedances for each of this pair, Z0 OR 2.0*Z0.

Also, if a signal is propagating down ONLY one microstrip of a pair of parallel pair of microstrips, are they in odd(differential) or even mode ?

• posted

You can quote it either way. In any case, the termination has to be balanced--i.e. the resistor goes between the two lines, not from either to ground.

Driving one line puts half the amplitude in the differential mode and half in common mode.

The point of doing this, rather than using two uncoupled lines of half the impedance driven differentially, is that the wave vectors of the even and odd modes are different, which reduces the effect of perturbations on the mode structure. (Google 'coupled mode theory'.)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal Consultant ```
• posted

No. There are two impedances, odd mode and even mode, and neither is the same as the single-trace isolated Zo. The farther apart the traces, the closer odd and even are to Zo.

Play with the Saturn software, diff pair tab, to get a feel for this.

The usual terminations are

two resistors to ground, each equal to Zodd

or one line-to-line, 2 * Zodd

Neither. We'd need a new term for that. Both the odd mode and even mode impedances assume differential drive.

```--
John Larkin         Highland Technology, Inc

Science teaches us to doubt. ```
• posted

If you drive a differential pair in perfect odd mode, there is no point in terminating the even mode. Of course, imperfections, asymmetries, bends, taps and vias may cause some leakage between modes, so it may be good to also terminate the unintended mode.

You can terminate the odd mode with a pair of resistors of Z0/2 to GND or with a single single resistor between the strips. The centre of the terminating resistor is at AC GND anyway. The even mode, if it has significant power, must be terminated to GND. Note that if you do ground the centre of the odd mode terminator, it will also dissipate even mode power, but probably not all of it.

Z0 of the odd mode and Z0 of the even mode are different, the former being mostly dependent of the separation between the traces and the latter on their distance to other planes, GND or supplies.

If you drive just one strip of a pair, that's equivalent to half the even and half the odd mode. That's simple superposition.

Jeroen Belleman

• posted

I sometimes use the terminations as ECL pulldowns too, so they go to ground. One variation is a Y termination with three resistors, to optimize the AC and DC issues.

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John Larkin         Highland Technology, Inc

Science teaches us to doubt. ```
• posted

Thanks to all for providing me the insight into this very interesting problem. I am experimenting with SPICE simulations of crosstalk between parallel microstrips with a grounded guard line in-between two signal lines. Then these questions popped up in mind.

• posted

simulations of crosstalk between parallel microstrips

SPICE can't model the physics of coupled lines--it's an ODE solver with a f ew knobs on, but coupled lines are inherently a PDE problem. (You can do a Mickey Mouse version, similar to modelling coax with Ls and Cs, but to do t hat you have to know the answer already.)

Cheers

Phil Hobbs

• posted

few knobs on, but coupled lines are inherently a PDE problem. (You can do a Mickey Mouse version, similar to modelling coax with Ls and Cs, but to do that you have to know the answer already.)

I've been reading about coupled mode theory. Is there any difference between odd modes and differential signals?

And between even modes and common mode...

In each case there is one input signal. with even mode I drive both lines in phase. With odd I drive them 180 degrees out of phase.

Or is there something more I'm missing?

George H. and even modes

• posted

One is a property of the T-line and one is a property of what you're driving it with.

The even and odd modes propagate at different speeds, so on long lines they get out of phase.

The speeds are different for the same reason as metal waveguide modes. The Helmholtz wave equation is

(del**2 + k**2) E = 0.

(This is just the time Fourier transform of the scalar wave equation.)

The boundary conditions impose a certain minimum curvature in the XY plane, which reduces the amount of curvature available. The curvature in the two modes is different, which makes the values of k_Z different.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted

few knobs on, but coupled lines are inherently a PDE problem. (You can do a Mickey Mouse version, similar to modelling coax with Ls and Cs, but to do that you have to know the answer already.)

I totally agree. However, available ready-to-use field solvers are very non

-intuitive to use, and have steep learning curves. A lot of time and effort is spent trying to get a grip on what the results of simple test cases mean. For exa mple ASITIC.

• posted

Coupled mode theory is pretty easy--once you've measured the coefficient of coupling, it's a 2x2 matrix problem, which is trivial to do by hand. The mode amplitudes go as sines and cosines of the propagation distance.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted

a few knobs on, but coupled lines are inherently a PDE problem. (You can d o a Mickey Mouse version, similar to modelling coax with Ls and Cs, but to do that you have to know the answer already.)

I am happy that someone else is as confused as I was some years ago, while in grad school. There is no difference between differential and odd mode, from the physics point of view. The difference comes when classifying the impedance. Off mod e impedance Z0Odd is measured with one (of a parallel pair of microstrips) grounded The differential impedance is 2.0*Z0Odd. And so on.

• posted

So two terminations to ground could have weird waveforms I guess. But the differential signal should still be OK into a good differential receiver. Which is faster on a pcb differential microstrip?

Even with a diff termination, the different timing of the diff and common-mode signals will make the waveforms look weird.

Is that right? Gotta verify that next proto board.

```--
John Larkin         Highland Technology, Inc

Science teaches us to doubt. ```
• posted

th a few knobs on, but coupled lines are inherently a PDE problem. (You can do a Mickey Mouse version, similar to modelling coax with Ls and Cs, but t o do that you have to know the answer already.)

ode impedance Z0Odd is measured with one

This seemed to be OK.

George H.

• posted

Oh! So the idea of having a dual (coupled) transmission line and then only driving one line..(keeping one at ground) Is really not a good idea.

George H.

• posted

Depends what you want to do. If the mode coupling is very weak, the amplitu des are sines and cosines of almost unit strength, so that if you drive one side only, the power will shift almost 100% to the other side, then back a gain sinusoidally.

That's good for directional couplers and such like.

Cheers

Phil Hobbs

• posted

Here's my attempt to model a twisted pair in Spice. I guess the common-mode delay should be different, faster, than the diff mode.

This was to demonstrate the futility of bootstrapping a photodiode at the end of a longish cable, which it was successful in doing.

Improvements are welcome.

```--
John Larkin         Highland Technology, Inc

Science teaches us to doubt. ```
• posted

We successfully bootstrapped a 1400-pF MPPC array at the end of a 4-inch FFC cable, but the bandwidth wasn't great (about 12 MHz). Getting 100x reduction in effective capacitance requires that the round-trip phase shift be less than 0.01 radian.

Cheers

Phil Hobbs

• posted

In principle, yes. How noticeable it is depends on the accumulated phase shifts between the modes, i.e. how long the lines are and the difference in velocities. The velocity difference depends on the boundary conditions and dielectric properties. In a metal waveguide, the phase velocity is higher than c, because the magnitude of k is always omega/v_phase, but the transverse k uses up some of it, leaving less available for the axial k, so that v_phase goes up.

That's why there's a low-frequency cutoff, where the axial k becomes imaginary--the transverse k uses up all the available curvature.

In more complicated boundary conditions such as coupled microstrip, the modes are more complicated as well. I've never gone through the math on that in any detail. Ordinary microstrip has mode dispersion, because you get a bit of curvature in the vertical direction at higher frequency, so the field extends further into the air.

The mode splitting in coupled lines is qualitatively similar to what happens in quantum mechanical band structure, for similar reasons.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted

On a current board layout, I have a differential PECL driver, a microstrip pair, two termination resistors to ground, and five differential line receivers along the way, about an inch apart.

The termination resistors are also the pecl dc pulldowns.

Should work.

```--
John Larkin         Highland Technology, Inc

Science teaches us to doubt. ```

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