Split Plane

I agree, most of this topic is a non-issue. It only becomes one when layouts and/or designs get ugly, usually meaining lack of experience of the designer. Avoid what you can, do what you HAVE to, just do it smart. A bit of thinking before hand will help pass even the most ludicrous tests (and I have seen some doozies).

Please enlight me: how do you know the trace you are discussing about will have the "reference" plane either the ground plane or the ground signal plane ? The return path will be the smallest impedance path. Meaning it could be a part from the ground plane and a part of your solid signal ground plane. You haven't too many options in controlling precisely the return path as long you have plenty of vias between those two ground planes...

Vasile

So you design the stack-up so that the signal plane is close to the ground plane, and at least twice as far from the split power plane.

- Brian

The current will be inverse proportional to the impedance of the different paths.

Thinking about this again, as the frequency increases the dI/dt increases but the period decreases so the needed current stays constant and the charge per cycle decreases.

If that charge comes from the capacitance of the ground and power planes, the required area depends on the current and capacitance, and decreases with increasing frequency (to first order, anyway).

Considering the one plane case (no continuous ground plane like the recent discussion was considering.) For 50 ohms and 2 volts, the current is 40ma, the charge per cycle 40ma/f. This will be the same amount of charge per cycle as that on the signal line itself, and so can be connected to the capacitance of the signal line. That charge will be over an area of the trace width times the wavelength. The effect of the discontinuity is expected to start becoming significant as the wavelength approaches 10 times the gap width. Assuming gap width is close to trace width, the area needed is 10 times (trace width) squared, or about a radius of sqrt(10) times trace width around the point where the signal crosses the gap. (sqrt(10*trace width * gap width) is the trace width is not close to the gap width.) If one wants the voltage in the split plane to be about 1/10 of the signal voltage another factor of 10 to the area, or sqrt(10) to the width.

This would seem to indicate that if the traces as they cross the gap are spaced more than about 10 trace widths apart then the plane capacitance should be enough to cover the needed charge.

-- glen

Glen,

You don't appear to take into account what the plane capacitance is. Are we 3pF/sqin? 100pF/sqcm? The results should be significantly different for those extremes in plane capacitance. The factor of 10 for wavelength calculations is used widely for transmission line issues but I don't see where this applies to a signal crossing a split.

- John_H

(I wrote)

(snip)

I think the other terms cancel out. I was comparing the capacitance per unit area of the transmission line to the capacitance per unit area needed to support the current (charge) at the boundary. Lower capacitance per unit area will result in higher impedance (for the same trace width), and lower current.

I was considering it as a transmission line with a short length impedance discontinuity. The impedance should be the same on both sides of the gap, but different across the gap. The effect, then, depends on the wavelength relative to gap width.

-- glen

The gap width determines the coupling between the two parts of the plane across the gap. Board level decoupling is important in addition to plane-to-plane capacitance when looking at the effects across the split. If you have only one plane with one full slit and no decoupling from the one side to the other, bad things will happen. The situation doesn't "cancel out." If the planes are poor from a decoupling standpoint (including plane-to-plane capacitance) bad things will happen.

What's been suggested with respect to "Black Magic" and such is that the decoupling capacitance in well designed boards swamps out most of the problems the hand-wavers claim will kill a board, providing good signal fidelity and low EMI and crosstalk. If there's no capacitance to aid in the decoupling across that split, bad things will happen.

- John_H

(snip regarding crossing a split power or ground plane)

Yes. The question left is how bad those things might be. Someone had previously suggested that the capacitance of the plane itself would be enough to reduce the effect of those bad things.

What cancels out is the effect of capacitance per unit area. That is, the result is not dependent on it. That doesn't cancel out the bad things, only how you compare them.

There can't be no capacitance. Just like all wires have inductance, all ground planes have capacitance. The question is, then, how much is there and how much is needed.

Question: If you have a metal sphere of radius R in an otherwise empty universe, what is its capacitance? (Hint: it isn't zero.)

-- glen

Hi Glen, Do you not think that the trace inductance and magnetic field are important? You don't mention them at all in your post. Just thinking about the capacitance is not the whole story... HTH., Syms.

Inductance that is part of an impedance matched transmission line doesn't count. The situation at the end, where the trace crosses the gap is a little complicated. As the current spreads out in the split plane, it isn't quite a transmission line anymore. The half plane inductance should be pretty low, but it won't say it doesn't count.

At some point I made the assumption that there was something on the other side of the split plane to make a good capacitor. (Another split plane would do. It would work on either side.)

Consider a circular parallel plate capacitor fed at the center. Then consider it as concentric ring capacitors and the radial inductance of those rings. The inductance per (radial) length decreases with increasing radius, the capacitance per radial distance increases with increasing radius. I believe that makes capacitance win out over inductance for reasonable frequencies.

-- glen

And why is that?

What about the increased loop area because the current goes around the slot? Does this increase the generation of the magnetic field? Are you only considering the inductance of the plane, or the inductance of the trace also?

You really lost me there. Sorry.

How does the current get from the centre to these "concentric ring capacitors"? I think it travels out radially, and this current generates a magnetic field.

HTH, Symon.

For an impedance matched line, the inductance and capacitance balance out. This might be easier to see for a line that is not impedance matched.

It is common for telephone lines to be UTP cable of around

100 ohm impedance driven by a 600 ohm source. The frequency has to be low enough, but that isn't a problem for telephone lines. What does happen, though, is that the excess capacitance causes a high frequency drop off. A 600 ohm line would have more inductance (per unit length) or less capacitance (per unit length). This fix for this is to add series inductors periodically along the length, maybe every 5000 feet. This improves the frequency response in the voice band, with a sharp drop after that.

At low frequencies it goes around. As the frequency increases, the capacitance of the system is enough to support the current. At really high frequencies (wavelength much shorter than the dielectric thickness) the problem goes away.

The assumption is that the trace is properly impedance matched to its load and source.

Well, as an example, both split power and ground planes, with the signal line either between them or on one side.

And the voltage generates an electric field.

The result is a transmission line in the radial direction. The impedance, sqrt(L/C), decreases fast with radius, the velocity, depending on LC, stays constant with radius. The decreasing impedance makes it look capacitive to the load for radius greater than the trace width. (You can also see that through the increasing capacitance and decreasing inductance with increasing radius.)

The decreasing impedance with distance allows it to support the needed current, assuming the wires crossing the split aren't too close together.

-- glen