Basic question on Balun impedance matching.

[Buriedcode]

Hi,

I'm current introducing myself to microwave design with a project - designing a test board for a 2.4Ghz transceiver using the ML2724. Thanks to the manufacturer, I have a few evaluation board schematics to look at, analyze/copy. These are great not just for copying subcircuits, but also for educational purposes, so I can see what each components does and why its value was chosen.

Although I have read up on baluns, one part bugs me. That is, differential matching. The balun they use is a 50/50 ohm hybrid ceramic balun (LDB212G4005C-001), so thats a balanced port impedance of 50 ohms. The TX output of the transceiver has two pins, one is a compliment of the other, both 12+j0 impedance.

Now, although I have a schematic for a matching network, I'm using smith charts to verify it, so I have a better understanding of whats going on. We have a differential output from the TX, two pins at 12+j0 - so thats a differential impedance of 24+j0 right? Now, the balun has a differential/balanced input impedance of 50 ohms. Does this mean that one could 'split' the matching network into two single ended matches? So instead of thinking 24+j0 -> 50+j0, I could think of it as 12+j0 -> 25+j0?

I have used a smith chart, and with a 'load' of 50ohms (input to balun) and using the components and values in the reference schematic, it brings me to 11.8+j0. - pretty damn close. However, the series capacitor must be double its value in order to get to 11.8+j0.

So..... I am confused, since I would have thought the above would be the case if the load impedance was 25 (half of our 50 ohm balanced input)... or is the differential impedance of the balun actually 100 ohms?

A picture is worth a thousand words, so I've included two. First is a snippet of the reference schematic. Two 12+j0 pins to a 50ohm balun, using 1.8nH and 1.5pF caps. The second was my own test on a smith chart, to convert just one pin, at 12+j0 (singled ended) to 50ohm. The inductance is the same, but why is the series capacitor double the value of the differential match? (3.0pf, instead of 1.5p). I read somewhere that this value, should in fact, be half of the differential match (0.75pf on the single ended). I'm guessing I'm not fully understanding it.

Any answers are welcome, thanks!

Buriedcode

 

[FvM]

The balun is clearly 50:50 ohm. If the transceiver is 24 ohm differential respectively 2x12 ohm single ended, you need two single ended matching network 12:25 ohm. You designed a 12:50 ohm matching network, which is incorrect. Apart from choosing the correct impedance, I think that a lowpass network should be preferred due to better harmonic suppression.

 

[Buriedcode]

Thanks for the reply! Ahh so it is a case of 'splitting' the differential balanced line into two single ended ones?

Yes, the second match on the smith chart is indeed a single ended one, using just one of the TX pins (12+j0) to 50 Ohms. This actually came about by accident, as I was just playing about with the chart to get from 25 to 12, but changed the load to '50' (wasn't sure if the balun was correct) and ended up with and inductor value identical to the reference design (left picture)... but with a cap double the value. This is handy, since it means I can 'do away' with the balun, just use single ended matching, sacrificing Tx power output from the transciever (halving its power) - but if I include a PA, I'm only losing 1-2dB for the convenience of NOT having to hand solder a 0603 size balun.

About the harmonic supression. This reference design was from 2003 And it has several low pass, and band pass 2.45Ghz ceramic filters after the TX chain, I guess that is how they suppressed the higher frequencies there. But I will look into a LP match, as it might prove better to lower band interference with my other transceivers I'm testing.

Can you explain the first picture? As I said, I tried matching 12 to 25 (25 ohm load) with the smith chart, but the values I get are (for a high pass match) 1.6nH shunt, with a series 5pF, and a shunt 22nH for the Tx bias. (actual input impedance is 12.3-j0.0). Now these are very different from the reference design, but the smith chart says its spot on. When converting these two single ended matches to differential, would I have to halve/double the values ? I'm asking because this document refers to it:

https://www.skyworksinc.com/uploads/documents/201156A.pdf

I have checked the PCB gerbers for the ref design, and I can't see any stubs or matching traces, its all very close together, so I can only assume that its the discrete LC parts which are doing all the matching. Also, thinking about it, given how close the parts are together, I'm not even sure a mismatch would be much of a problem, the balun is <5mm away from the chip via the caps - so traces between these parts are very short indeed. Output power is 3dB max for differential.

My main curiosity (remember, this is mainly academic, for learning) was the fact that my single ended match, 12->50 ohm, was the same as half the reference designs differential match, the only difference being, double the series capacitance.... that can't just be a co-incidence can it? Or have they made a mistake?

Thanks once again, sorry for asking lots of questions. I guess for my project is isn't 'critical', but I like to know whats going on

 

[FvM]

Perhaps the reference design uses a different (e.g. 100:50 ohm) balun? Lowpass matching network would be an obvious solution in my opinion, but of course it's not necessary. If harmonic filtering is done otherwise, it's O.K. For the matching calculation, there are two options:
- determine the impedances of the single ended network (12:25 in this case) and implement it as designed
- calculate for differential impedances (24:50) and scale the component values afterwards

 

[tony_lth]

Hi, FVM,

From microwave journal, there is an article with the following diff match.

From the same article, it demos the single-ended match.

For single-ended match, it matches from 12 ohms to 50 ohms.
I think it's not correct, because the balun is a 50 ohms (balanced) to 50 ohms (unbalanced).
So the match should be from 12 ohms to 25 ohms.

Pls confirm it.

 

[tony_lth]

And, from single-to-diff balanced, does the match need to split 3pF into two 1.5pF?
I think it should keep as 3pF on both arm.

 

[FvM]

It may be the case that the author confused bal 100: unbal 50 LDB212G4010C-001 with bal 50: unbal 50 LDB212G4005C-001.

And, from single-to-diff balanced, does the match need to split 3pF into two 1.5pF?
I think it should keep as 3pF on both arm.

As previously stated, the single ended impedance matching networks should be 12 to 25 ohm for LDB212G405C-001. I didn't calculate the circuit given next the Smith chart, simply assumed that 22nH, 3.0 pF and 1.8 nH are correct. In this case, 3.0 pF would be kept for both arms. A split operation is only involved for components shared by both arms.

So the publication is in fact double confused.

 

[tony_lth]

Hi, FvM,

Thank you very much.

BR,
Tony Liu