8
\$\begingroup\$

I'm designing a circuit with RFICs using differential signaling. I'd like to keep noise to an absolute minimum. Let's use the mixer (LT5560) as an example. All of the reference schematics match to 50 ohms and use a balun to convert the differential inputs and outputs to single ended. I would like to use differential signaling to and from the mixer as my IC's on either side are differential. Converting to single ended and then back to differential seems wasteful. Why would none of the reference schematics take advantage of differential signaling when the datasheet strongly recommends it?

My main question is this: I know how to match the impedance of single ended signals, but how do I match the impedance of differential lines? My source impedance is not 50 Ohms.

LT5560 Reference Schematic

LT5560 Datasheet

edit: to clarify, I am not trying to match 50 Ohms, I need to match 2 different complex differential impedances. So, in this image, how do I calculate the correct component values to ensure maximum power transfer?

Unbalanced Network

\$\endgroup\$
3
  • \$\begingroup\$ Those schematics show inputs and outputs via coaxial connectors. Therefore baluns are needed. If your signals are going from/to some ICs with differential i/o's, you won't need the baluns. \$\endgroup\$ – The Photon Jul 27 '12 at 20:40
  • \$\begingroup\$ Okay that explains why all the references use baluns, but how do I ensure correct impedance matching of the differential lines? Do I use the same matching network twice? \$\endgroup\$ – RobbR Jul 27 '12 at 20:45
  • \$\begingroup\$ Unfortunately I'm not enough of an rf guy to give you a quick answer to that question. I took a quick look at the datasheet, and there's several pages explaining the matching requirements--too much for me to wade through in a few minutes, but maybe you can work it out with some careful reading. \$\endgroup\$ – The Photon Jul 27 '12 at 22:17
1
\$\begingroup\$

You need to use an impedance calculator tool to figure out the geometry for routing your differential pairs. If your schematics call for 50-ohm single-ended routing, you would want to use 100-ohm differential routing. Punch that into the tools along with your board and trace geometries and it will tell you how thick your traces should be, and how far apart to space your differential traces.

I believe the 100-ohm differential equates to 50-ohm single ended because you can think of the two 100-ohm impedances as being in parallel resulting in an effective 50-ohm single-ended equivalence. FWIW, the calculator I've used in the past is called Polar SI8000 which apparently has been superceded by Speedstack PCB.

Edit

I think you sould just remove the Balun, inductors, and capacitors if you are going chip to chip differential. It's not trivial to match differing characteristic impedances by putting in-line passive components on the transmission path. You would need to add an appropriate sized "stub resistance" at a carefully chosen location along the transmission path using engineering tools like a Smith Chart. Stuff like that is really only practically applicable over "large" distances with respect to the wavelength though. That's why I'm suggesting just routing pin to pin with differentially controlled impedance.

If you want to keep the capacitors and inductors for differential routing to optimize the power transfer, page 16 of the datasheet shows some formulas that "provide good starting values" for these components...

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The problem is not in the routing. I need to know the passive component values to match the different impedances in the schematic. I'm not matching to 50 Ohm, I'm trying to match 116 + j41 Ohms differential to 28.8 + j9.8 Ohms differential. \$\endgroup\$ – RobbR Jul 28 '12 at 6:19
  • \$\begingroup\$ Stubs are only one option, they are just a very easy way to add an inductor or capacitor if your wavelength is high enough. \$\endgroup\$ – Kortuk Jul 28 '12 at 15:49
1
\$\begingroup\$

After some research it turns out the answer was deceptively simple. The procedure is to simply design a standard single ended matching network, and divide the passive component values by 2 to match differentially. Here is a Genesys schematic showing the two configurations, with performance:

differential matching network

\$\endgroup\$
1
  • \$\begingroup\$ This is what I thought, but I really brain farted and doubted myself, good job. \$\endgroup\$ – Kortuk Jul 31 '12 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.