40 MHz clock between 2 ICs

Question

I have an oscillator at 40 MHz which I would like to drive two ICs, 1.8V output. The ICs are in opposite directions so I have opted for star topology of the two clock traces. Trace length for the first IC is 13.75 mm while for the other is 6.25mm, so 20 mm in total, and the trace impedance is 50 ohm.

• I have used 50 Ohm impedance tracks as this is the recommendation by various sources I have read. Yet I have not seen any references for impedance in the datasheet. So is that the right track impedance?
• The documents I have read recommend a daisy chain approach in clock distribution. However in this situation is not possible (PCB layout, opposite directions, BGAs). Is a star topology a bad approach?
• In star topology, do I consider the traces as two different traces (13.75mm and 6.25 mm respectively, each terminated back at the clock pad) or as single 20 mm?
• The datasheet provides a typical rise time of 1.4ns for 15 pF load. Is this rise time considered for this specific load or up to this load?
• According to Max Conductor Length calculators, for rise time of 1.4ns max conductor length is around 50 mm before transmission line phenomena kick in. Is this applicable to this situation as well, since a) it is a star topology, b) trace lengths as individual and as a sum are below that value?

Any recommendations, sources and things to look for are always welcome!

Answer 1

Davide Andrea's answer is good, and I entirely agree with it, but let me address a couple of your specific questions:

In star topology, do I consider the traces as two different traces (13.75mm and 6.25 mm respectively, each terminated back at the clock pad) or as single 20 mm?

In terms of distance (delay) to each individual chip you consider them as two separate lines.

To calculate the capacitance loading the oscillator output you can just consider them one track. (Or two separate tracks but add up the capacitances)

The datasheet provides a typical rise time of 1.4ns for 15 pF load. Is this rise time considered for this specific load or up to this load?

It will scale roughly with the load capacitance, down to some minimum value when there's no external capacitive load.

If you are worried your load capacitance is too low, and it might cause too fast a rise and fall time, you could consider adding a location for a series resistor at the oscillator output. Initially load this location with 0 ohms. If you see ringing or other bad behavior, you can place 22 or 47 ohms instead.

According to Max Conductor Length calculators, for rise time of 1.4ns max conductor length is around 50 mm before transmission line phenomena kick in. Is this applicable to this situation as well, since a) it is a star topology, b) trace lengths as individual and as a sum are below that value?

You'd consider each line separately, but consider that reflections on one line will propagate into the other line, and the total length would be what matters for that phenomenon.

But in either case you're under the limit with your ~13 mm lengths, so you shouldn't have problems. If necessary you can also use the series resistor I recommended above to increase the rise and fall times and thus increase the line length before you encounter transmission line effects.

Answer 2

You may be overthinking this. The wavelength of 40 MHz is 8 meters. That's 400 times longer than your 20 mm. Therefore, the effects of the impedance of these short traces will be minimal.

Regarding 50 Ω trace impedance, you would need to also have a 50 Ω source impedance (it is not, it's much lower) and a 50 Ω load impedance (it is not, it's much higher). And, you would need that for each trace (a total of four 50 Ω resistors). But, if you did, the voltage at the receiving end would only be 50 % of the starting voltage, which is not good for digital signals.

So, no, don't try to make 50 Ω traces: it doesn't buy you anything. Instead, focus on cross-talk, emissions, and immunity. If you need the exact same timing to both recipients of this 40 MHz signal, then make both traces the same length, by routing one with "meandering" wiggles.