# Why is 50ohm sometimes used as a differential impedance?

I'm creating a design where I convert an RF single-ended signal to a differential signal with a 5400BL15B050E RF balun that is then fed to a ADL5802 RF mixer. The balun datasheet specifies the single ended characteristic impedance should be matched to $$\50\,\Omega\$$ and the differential balanced impedance should also be matched to $$\50\,\Omega\$$. Why would the balun choose a balanced differential impedance of $$\50\,\Omega\$$ and not $$\100\,\Omega\$$? Practically, this means that my differential microstrip traces need to be much wider than the single-ended traces (in my case almost 3x as wide). This makes routing trickier since (at least for my PCB stackup) the trace width is larger than the mixer pin spacing, and so I need to add tapering to make all the connections. Would it not have been more sensible to have the differential impedance be $$\100\,\Omega\$$ so that the balanced trace widths could be roughly the same as the single-ended ones?

Is there a reason why $$\50\,\Omega\$$ makes sense here?

Edit

Prompted by pericynthion's comment below, I'd also be interested to know why the mixer would use a $$\50\,\Omega\$$ differential impedance.

• Baluns are available with a variety of impedance ratios. Instead of asking "why would the balun choose a balanced differential impedance of 50 Ω and not 100 Ω" I think it would be better to ask why the active mixer IC designers chose to match its input ports to 50 Ω diff. Mar 29, 2020 at 4:00
• @pericynthion that's a fair point. Any thoughts on that, or that seems strange to you as well? Mar 29, 2020 at 4:05
• No, I'm curious though.. Mar 29, 2020 at 5:05
• @pericynthion I've added that into the main question. Mar 29, 2020 at 5:50