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I am a rookie on both RF and PCB design and I am currently trying to do a 50 ohm 11 watt dummy load project for 100MHz-1GHz.

My calculations are done for the load but I have some questions about designing the PCB.

My new design is as below:

enter image description here

My instructor keeps saying that :

''There is a problem with lining up resistors on a row, because you will see a high impedance with reflections on first resistor, then traveling to the next, and the next ... Try designing so that the signal reaches all resistors at the same time. This is difficult with so many resistors. If the attenuator is then directly connected to the same point your series resistor can be of high impedance without affecting signal, because of the short length.''

I don't know how to do this and out of ideas, I tried to do it as a circle around the connector but still not accepted.

Attenuators written as AT1,AT2,AT3. The other resistors are for the load.

Project Specification:

  • The power I am trying to terminate is 5-10 watts.
  • SMD resistors are 560Nohms 1 watt resistors and the signal strength should not exceed -10dBm that is why I am using an 50dB attenuator, to achieve the desired signal strength.

PCB design Specification:

  • Trace Width = 0.5mm
  • Dielectric Thickness = 0.3mm
  • Trace Thickness = 0.036mm
  • Substrate Dielectric (Ɛr) = 4.6
  • Total board thickness = 0.39mm
  • Size of the resistors (x,y) = (6.4,3.2)mm
  • Dimensions of the board are (x,y) = (7.5,10)cm

EDIT

So I decided to do it as a star or circle shaped design to make the lines equal to each other. Here is what it looks like:

enter image description here Schematic of The Dummy Load and Attenuators

I just draw this schematic to show how my circuit looks like there might be errors.

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  • \$\begingroup\$ What are the dimensions of the board (to give us an idea of the trace lengths)? What size are those resistors? How much power are you trying to terminate? \$\endgroup\$ – The Photon May 12 '20 at 19:24
  • \$\begingroup\$ What power has this got to absorb? What's your specification for S11? There is no way that the layout you have shown has the slightest chance of achieving a reasonable spec. \$\endgroup\$ – Neil_UK May 12 '20 at 19:25
  • \$\begingroup\$ For RF the board layer stack makes big difference. What is your board thickness? How much power are you attempting to handle in the load? \$\endgroup\$ – Aaron May 12 '20 at 19:27
  • \$\begingroup\$ Show a schematic of what you're attempting to do with those attenuator components \$\endgroup\$ – Neil_UK May 12 '20 at 20:14
  • \$\begingroup\$ Why such a very thin board, can you use thicker? The vias are the least of your problems. What is the S11 (aka return loss, or VSWR) that you want? Have you got a box of 560 ohm resistors that you must use, or can you change values if required? Show a schematic of what you're attempting to do with those attenuator components What test equipment do you have? \$\endgroup\$ – Neil_UK May 13 '20 at 9:36
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The biggest problem you have is the parallel connection of all those 560 Ω resistors. You obviously need to space them apart so that they can all dissipate their heat. However to parallel them at RF, you need 560 Ω transmission lines, and you can't get impedance that high on microstrip, especially the very thin dielectric you've chosen. The eleven lines in parallel will cause a very large shunt capacitive defect, giving you an increasingly degraded return loss as the frequency rises.

This is how I'd approach the circuit

schematic

simulate this circuit – Schematic created using CircuitLab

Resistors are chosen from the E24 series, nearest to the value that would actually give 50 Ω match and equal power dissipation. As a result of that approximation, the final DC resistance comes out to about 49 Ω. Each horizontal line is a length of transmission line, with the approximate impedance (rounded to integer) above it. You'll note they are all 'reasonable' values, in the ballpark of 50 Ω, so straightforward to fabricate on the board. As transmission lines, they can be any length, allowing you to space the resistors at will around the board.

There are some practical details that will affect the high frequency match. High value resistors tend to be a bit capacitive, low value resistors a bit inductive, so this layout would only be a first cut, and measurement, or detailed modelling with data for those specific devices, would be needed to tweak up the match. There is plenty of scope to change the impedance of the series lines slightly to compensate for the parasitics of the components.

Note that the signal decreases along the chain, so R11 provides the ideal place to feed the final instrumentation attenuator, with a slight modification of value as required.

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  • \$\begingroup\$ Thank you for your answer. I edited my design and question. I want to do it with given resistors this is my first goal but if I can't manage I will do as you say with your resistors. Your answer explained a lot thank you so much. \$\endgroup\$ – Onur Can Saglam May 18 '20 at 8:32
  • \$\begingroup\$ FYI: The free space wave impedance is 377 Ω. It is impossible to construct a transmission line, which has a higher impedance. \$\endgroup\$ – Horror Vacui May 18 '20 at 8:53
  • \$\begingroup\$ Wouldn't too much power dissipated on the first 560Ohm resistor? Nice idea by the way to transfer the problem into another form. \$\endgroup\$ – Horror Vacui May 18 '20 at 8:54
  • \$\begingroup\$ Power in 560 ohms? Click on 'simulate this circuit', add a 23.5 V source (11 watts into 50 ohms), do a DC solve and read off the voltages. It's close to 1 watt per component, would be exactly 1 watt per but I've approximated to stock E24 values. Max impedance of line? Coax has an impedance of 60ln(a/b) where a and b are the conductor radii, so I kinda thought that a sufficiently extreme ratio would give any arbitrary impedance, but perhaps I'm wrong and need to think and read. We are both agreed that you can't get 560 ohm on microstrip. \$\endgroup\$ – Neil_UK May 18 '20 at 9:03
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    \$\begingroup\$ @HorrorVacui I've compressed the schematic a little so it's more readable. I was surprised to find there wasn't a better resolution linked under the image. I like the username BTW. \$\endgroup\$ – Neil_UK May 19 '20 at 16:28
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The free space wavelength at 1GHz is 30cm. Assuming an effective dielectric constant of 4, it will be 60cm. I believe you can add those resistances in a few centimeter easily. In that case you can neglect the wave effect in the layout. You should verify with simulation if it is really case. Some reflections will happen of course.

I see plenty of space around your resistors. I would definitively make it more compact. At RF design the trace lengths is your enemy due to their inductance if you do not use a transmission line. If you use a transmission line, then the junctions might be tricky to design.

If you worry about the different impedances seen between the resistors, you can connect them in a star fashion, so they will load the line at the same point. If the inductance of the connection is negligible this might be OK.

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  • \$\begingroup\$ I changed the design as a circle shape and editted the question. Wha do you think about it now? \$\endgroup\$ – Onur Can Saglam May 19 '20 at 12:29
  • \$\begingroup\$ @OnurCanSaglam It is better. You might consider a trade-off where the devices are closer to the center and each other, but have uneven trace length. As long as the trace length imbalance introduces negligible impedance difference, it could be a way to go. You might consider at which point do you want to connect AT2: middle of the star, like the others or as it is now. \$\endgroup\$ – Horror Vacui May 19 '20 at 14:48
  • \$\begingroup\$ What I can't figure out here is my 4 wide tracks should be 200 ohms and thin tracks going to the terminators have to be 600 ohms but these values are really high and I think nearly impossible to achieve. What can I do in this situation? @HorrorVacui \$\endgroup\$ – Onur Can Saglam May 24 '20 at 8:37
  • \$\begingroup\$ And one more question I'd like to connect AT2 to the middle of the star so it should be routed 50 ohms, should I thicken the the track goes right under the SMA or can I just connect the wide 50 ohm to the current bottom track? \$\endgroup\$ – Onur Can Saglam May 24 '20 at 8:39
  • \$\begingroup\$ @OnurCanSaglam as I mentioned in a comment to Neil's answer, the wave impedance in vacuum is 377Ohm. Since the wave propagation happens in a slower medium - your dielectric - the maximum possible characteristic impedance you can get will be smaller. 600 Ohm is theoretically impossible and probably there is no material where 200 ohm would be possible. In my answer I tried to highlight the fact that you can neglect the wave nature of the signal if your design is compact. I do not know the dimension of the resistor needed for 1W dissipation, but it seems to be possible w/o transmission lines. \$\endgroup\$ – Horror Vacui May 25 '20 at 14:38

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