Hi,
I’m using a T100V-020.0M 20 MHz oscillator (output’s got a 15pF load, fed into a binary counter). In the data sheet they say that I need a “50 Ohm trace <1’’ by design” for the output pulse (page 5 of the data sheet, I attached it to the post).
So I used this online calculator (http://referencedesigner.com/tutorials/si/si_06.php) to calculate parameters for trace of 50 Ohm impedance.

I have FR-4, thus H=1.5mm, 1oz copper - T=0.035mm and Er = 4.5. 2-side PCB, one side with signals, the other is GND. The calculator says that I need more than 100mil trace width. That’s a bit too much. I understand that I can use a 4 layer board instead, and the problem will be solved. However, it’s cost and time effective for me.

What would happen if instead of having a 50 Ohm trace I will have a 100-130 Ohm trace and if I keep the oscillator output very close to the pin that I’m reading the pulse from, so that the reflections are won't be too wide? How will it affect my output pulse?

Thanks a lot in advance.

 

The critical parameter is the length of the connection wire from source to load. Wave propagation travels at approx. 5ns per meter.
Keep you connections to under 0.2m and you shouldn't have to worry with signal degradation at 20MHz.

To minimise signal degradation you can try inserting 22-33Ω resistor in series with the source.

 

The critical parameter is the length of the connection wire from source to load. Wave propagation travels at approx. 5ns per meter.
Keep you connections to under 0.2m and you shouldn't have to worry with signal degradation at 20MHz.

To minimise signal degradation you can try inserting 22-33Ω resistor in series with the source.

thank you for the advice!
The pulse travels for less than 2cm. I assume that means I have nothing to worry about, right?

 

thank you for the advice!
The pulse travels for less than 2cm. I assume that means I have nothing to worry about, right?

Right.