Hi all,

I would like to ask some questions about the routing of my design that is composed of (i) two Flip-Flops (FF) and (ii) one splitter element. FFs are connected in serie that means the D1 of FF1 is connected to the generator via SMA connector, Q1 of FF1 is connected to the D2 of FF2 and Q2 of FF2 is connected to the oscilloscope also via SMA. The clock signal is shared and it is coming from clock generator (e.g. PLL) to the splitter that splits one clock signal to two equivalent clock signals that feed the FFs.

In this project I'm using the ECL logic in order to reach a high frequency data rate (e.g., ~2GHz). The ECL logic has differential inputs/outputs. In order to ensure correct behaviour of FFs, the impedance termination has to be done. I'm using the thevenin termination.

Now, my troubes are related to the layout issues, particularly to the routing.
In the attachement is the presentation of the project with added views (architectural, schematic, layout, ... ).
At the end of the presentation I have added my questions and open problems.

I would ask the PCB RF experts to help me to answer to these questions in order to finish the design and ensure that it will work properly.

Thanks in advance.
Kindly.

 

Use shortest connections possible.
Post your questions here.

 

Hi,
thanks for quick answer.
I put questions into the presentation in order to get an overview and the context of the project.

In your answer you do not precise whether the paths should be two single ended or one differential pair.

1) If single ended what about the problem of impedance ? Single ended traces (50 ohm) are too tick (0.65mm) and too close each other that will finally modify the impedance of given trace. In order to route correctly single ended traces the space between traces should be 4 or 5x bigger than its width. Allegro calculates it properly !

2) If differential pair how to route it ? Could paths be completly uncoupled ? What will happen in such case ? If not what values should be set in the constraint manager ?
In the presentation is shown the figure 7. What about the reflections problems ?

Thanks in advance.

 

Some basic comments about figure 7.

The two traces should enter Q and \Q the same way. You have Q entering the corner of the pad and \Q entering the side of the pad.
The two trace lengths should be the same. Between R6/R7 and Q \Q you have two 90 degree bends, but they are not the same radius. The overall length of the Q trace is longer than the \Q trace.
You question about reflections is correct. The traces should go straight through the center of the pads so there are no stubs. This makes an impedance bump in the trace, but this can not be avoided.
For traces this short, noise coupling probably is not a problem. So having the two traces very close to each other is not as important as them having the same impedance and length. I would route them as two independent 50 ohm lines, but others around here might have different experiences.
If you can, move FF2 downward so the traces go straight to it the way they leave FF1, without the two bends.

ak