PCB track impedence

If you don't mind the resistive losses, then a resistor in series and a resistor in shunt should work. Easy enough to solve for the values.

Very much so. The PCB substrate dielectric constant is very important if you worry about trace impedance.

If you are asking if the software will do this automatically, then it depends on the software. Capture software alone won't, but simulation software can be set up to assist in matching. Matching resistors, or any component can just be included. The layout step is where you need to really worry about matching. If you are doing it, have fun. If someone else it, then I've seen notes left on the schematic about trace impedances and so on.

-Joel

Joel,

Thanks a lot.

I am currently designing the schematic around LPC313x and trying interface it with SDRAM. I read somewhere that we need termination resistor if trace propagation delay is more than the rise time or fall time of the driver but how do I know the expected trace impedance or trace propagation delay at the time of schematic design to include termination resistors or not. Also, LPC313x datasheet doesn't talk about rise and fall times of its SDRAM interface pins.

TIA

-m

e is higher than the impedance of the driver, a

Top-posting - as you have just done - makes extended threads even more confusing thannthey have to be.

Trace propagation delay is typically 2/3rds of the speed of light in most PCB dielectrics. The speed of light is 300mm per nanosecond, so typical propagation delays are around 200mm per nanosecond.

Trace impedances are around 75R on the outside of board, dropping to closer to 50R for buried tracks (with ground or power planes above and below them. Track widths and dielectic thickness make a difference, but it is is mostly the logarithm of a ratio, and you can't get far away from 50R.

As for the rise and fall times, the LPC313X datsheets still seem to be preliminary

df

but you might try looking at the SDRAM parts it is intended to drive, which will constrain the signals that the LPC313X are going to have to produce to drive them.

-- Bill Sloman, Nijmegen

No. A low-impedance driver is often used to drive a transmission line. ECL, for instance, has a few ohms output impedance and commonly drives

50 to 100 ohm traces or cables. The traces/cables are usually terminated at the far end to prevent reflections.

One *can* source terminate a transmission line, in which case the far end can be unterminated; there is a big reflection from the open end, which bounces back to the driver and is absorbed there.

How

Terminate the far end with the characteristic impedance of the line. The driver impedance and the line then form a voltage divider.

Yes. The geometry and the dielectric constant of the insulator affect the impedance.

Use this:

and select "passive circuits"

It depends on the situation. Sometimes you don't care about reflections, sometimes you do. If this were simple, anybody could do it.

John

higher than the impedance of the driver, a

If you keep your memory really close to the part it's interfacing to, you might not need terminations. Without the terminators, you save lots of power and trace length. You'll need to do simulations on what sort of lengths you can get away with. We've used DDR2 memory without terminators and ended up with very nice signal fidelity. Memory was located very close to the FPGA and pinout was optimized to keep traces short.

If one just "terminates" using a series resistor equal to the driver impedance, then (nominally) driving anything from an open line to a shorted line will make little difference.

er, line impedance?

then (nominally) driving anything from an open line to a

for loads at the end of the line. Intermediate loads may get a nasty step waveform.

John

^^^^^^line

Not true. A series termination (including the driver output impedance) equal to the line impedance will give a step which is resolved to the full signal at the far end. In the middle there will be a 50% step (doubled at the end). If the series termination isn't equal to the line impedance the step won't be 50% and the rest (doubled at the far end) will be bounce around forever. ;-)

...and you'll never drive that shorted line. ;-)

Yes; sorry about the slip.

You're lucky you're not me. Any slip I make, all sorts of rubes attack me for being imperfect.

John

How about the following illogic: the rubes would be perfectly correct..

Even rubes are right once in a while. With one glaring exception, of course.

John

Price you pay for trying to stay on top!