# how to reproduce a matching network of a vendor

I'm trying to create a board using the BGT24MTR chip from infinion. I noticed that they said that there is a Tx/Txx impedance.

To compensate for this impedance they created a matching network.

Since I'm trying to use a different board stack up, I have to redesign this matching system. I tried calculating the impedance of each of the lengths

Here's my breakdown of the major components.

Width Length  Ω     diff Ω   diff Ω single
.3    .55     68.5     82     61
1.15  1.6     31.19    22.2   16.5
.5    ∞       53


I thought that I was trying to use the smith chart to get the impedance to be purely real around 5/3 after the second rotation so that when it is renormalized it would be purely in the unit circle in the center, but every way I looked at using the smith chart led to results that are far away from being matched. Any suggestions on what they are doing to match this?

P.S. The new board stack up is FR408 1.6mm thick. 1oz copper on both sides (1.4mil,35um). Frequency is 24 GHz

• Can you show an example of the Smith chart you made? Commented Apr 25, 2016 at 21:14
• @ThePhoton added the smith chart Commented Apr 25, 2016 at 22:14

Your use of the smith chart is correct, but your numbers are a little off due to the wavelength you used (6.5mm).

In stripline, the fields are completely contained inside the substrate. The fields are partially in air around a microstrip line, resulting in an effective permittivity which is lower than the relative permittivity of the substrate. Wavelength can be calculated using the effective permittivity or the delay (Tpd). Both are available from your impedance calculator. Er effective is 3.0635. Tpd is 5.8e-9 s/m.

$Wavelength = \frac{c}{\sqrt{E_{r}} \cdot f} = \frac{1}{T_{pd} \cdot f}$
$Wavelength(w/ E_{r}) = \frac{c}{\sqrt{E_{r}} \cdot f} = 6.5 mm$
$Wavelength(w/ E_{eff}) = \frac{c}{\sqrt{E_{eff}} \cdot f} = 7.1 mm$
$Wavelength(w/ T_{pd}) = \frac{1}{T_{pd} \cdot f} = 7.1 mm$

The difference between 6.5mm and 7.1mm may seem small, but check out the difference in return loss.

$Z_{out}(\Lambda:6.5mm) = 33-j24 \Omega$ >>>> $ORL=9dB$
$Z_{out}(\Lambda:7.1mm) = 46-j14 \Omega$ >>>> $ORL=16dB$

For a sanity check, match from the load to the conjugate of the generator. In other words, start at a load of $50\Omega$ and go towards the chip. You'll end up around $20.8+j20.2\Omega$. This is typical for most output stages, since they're usually low resistance and capacitive.

Your method of matching will provide a decent ballpark, but I'd like to point out a few important sources of error.

1. Tx & Txx are intended to be 100ohm differential signals. The small spacing between them produces an additional shunt capacitance. You need to include the extra capacitance, or use a coupled line model.
2. The datasheet doesn't say how they measured the Tx/Txx load impedances. Did they measure at the PCB output port and try to de-embed everything up to the chip? I'm skeptical of the accuracy without knowing where they came from. Additionally, I think these are the impedances looking into the Tx/Txx ports. I would call the load impedance the conjugate of the printed values, or the impedance presented to the Tx/Txx ports. It's all unclear.
3. The package pad (footprint) will have a different impact on a different board stackup.
4. I don't see anything to cancel mismatch from the launchers. The launchers will perform differently on a different board stackup.

The matching network they used is simple and probably low loss. I guess you could call it a stepped-impedance match. I don't know. It's a plus to avoid shunt components or stubs, and the high Q resonances they might produce.

• A load impedance of 100 ohm is clearly stated in User's Guide to BGT24MTR11. " It is a differential output signal with a load impedance of 100 ohm, given that the off-chip compensation structures, shown in the data sheet, are in place. ". How could OP's use of smith chart be correct?
– ivan
Commented Jun 1, 2016 at 20:55
• Are you referring to the OP using single-ended 50ohm, instead of 100ohm differential? Commented Jun 1, 2016 at 22:14
• Disregard my previous comment, you are probably right.
– ivan
Commented Jun 1, 2016 at 23:24

Since you have a two-sided FR4 you won't be able to reproduce their matching network. What you can do is to use a microwave design software.

I suggest awr microwave office or agilent ads. There you should create two ports for TX and TXX and assign their respective impedances. You should also create a load that you want to use.

Then use microstrip libraries to create a matching network (there's plenty of examples in both programs).

Create S21 graph and tune microstrip' width and length manually or use a built-in optimizer (set constraints and goals first).

• Thanks for the answer. I'm curious whether you have any suggestions as to why those designers chose that matching structure? Was my intuition correct or were they pursuing a different goal? Commented Apr 26, 2016 at 20:53
• They chose that matching because it's the simplest one. What you was trying to calculate is called characteristic impedance of a transmission line (and you calculated it wrong - check conductor height). What you need to know is a load that you are going to use (e.g. antenna, PA etc.).
– ivan
Commented Apr 26, 2016 at 21:53
• Your Smith chart attempt is valid with lumped elements but these guys use microstrips. There is lumped-element matching and distributed-element matching (e.g. microstrip) which is a completely other thing.
– ivan
Commented Apr 26, 2016 at 21:59
• I think you are misunderstanding what I'm doing. First the main thrust of the question is to determine what they were doing because stub matching is very well documented. Second I wasn't adding lumped elements with the smith chart, but was moving away from the load and to the generator (in retrospect that is something I'm going to try, moving from the generator to the load), which is why I was rotating around a circle based upon the circle at the origin. Commented Apr 26, 2016 at 23:19
• @LegenDiary, please look at page 9 paragraph 4.1 of this document link
– ivan
Commented Apr 27, 2016 at 5:42
1. Compensation structure in datasheet RO4350 0.254mm Estimating in QUCS / tools->line calculation (http://qucs.sourceforge.net/):

w=0.3 l=0.55 -> Z0 = 68 Ohm; Angle = 25 Deg ≈ 0.7λ
w=1.1 l=1.65 -> Z0 = 31 Ohm; Angle = 83 Deg ≈ 0.23λ

1. Tuning smith chart software (http://fritz.dellsperger.net/smith.html): Effective lengths are longer. Wide section is probably quarterwave transformer, so i used 0.25λ:
Thin section is shortened. In this particular case it is better to use EM simulation software, I think Sonnet Lite is capable of doing that.

2. If you are going to use FR4 material at 24 GHz, 1.6mm is too thick. It will result in extremely wide lines, and they may act not as expected. I recommend to use 0.6mm, maximum 1.0mm.

3. Here is my compensation structure for BGT24MTR12:

It differs from original compensation structure. Thin section is longer and act as half-wavelength impedance repeater. It will have worse bandwidth than original RO4350 compensation structure.

Check new bgt24ltr11, although it may be more difficult to solder (UPDATE 2020: bgt24ltr is not 50 Ohm matched, TX/RX port impedance is around 100 Ohm and may have some little complex part. Datasheet/appnote provides matching structure to 50Ohm).