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TLDR: Why does my math not work. SITUATION: I have a 1.5KW 50 ohm RF dummy load for ham radio and I'm trying to design a sample port for monitoring with a spectrum analyzer. I have now realized that there are probably better methods for accomplishing this task. Simple capacitive coupling isn't really suitable as I would like a reasonably flat attenuation curve from 1.8Mhz to 100Mhz. I'm trying to be a good radio operator and get accurate numbers on higher harmonic spurs. I designed this about 6 months ago and had to walk away due to not understanding what is wrong, I've revisited it many times since then looking for what I don't understand.

PLAN: Embed the attenuator into the oil filled dummy load. Design with a series of 3 PI attenuators while lowering the impedance through each stage for 2 reasons. One is to minimize any change in impedance to the existing 50 ohms of the dummy load. Second is to minimize the power dissipation in the attenuator. Here is what I learned from Matching Pi Attenuators I have found other sites agreeing. I wrote a spreadsheet and verified it generates the same results per stage as the online calc. I then simulated the results in Kicad and results were way less at output than calculated. I then did a DC analysis by hand, I got the same answer as Kicad's sim.

I can't upload the spreadsheet here is a screen shot. I'm glad to share it if someone tells me how. Green boxes are inputs. Total power in, input and output impedance of each stage and the db attenuation of each stage. Resistor values are calculated with power dissipation. These numbers match for online calcs. Power figures have to be invalid as the Vpp calc at each stage is incorrect.

Spreadsheet I get about 8mVpp output, spreadsheet says 774mVpp. As a test, I modified the circuit an changed R100 to 1 ohm, disabled R3 and R6 and only get 140mVpp output. R9 would be the 50 ohms of the spectrum analyzer. Kicad sim circuit

Obviously I completely misunderstand something here. Please someone explain why this doesn't calculate so I can sleep again. lol
Thanks Cory

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    \$\begingroup\$ At 100 MHz, and well below it, those 100kohm values are simply unrealisable. There's a reason RF systems are designed around 50 ohms, and don't use resistor values that are more than an order of magnitude away from that. It may be a 1500 W load, but what power do you actually want to operate up to? \$\endgroup\$
    – Neil_UK
    Feb 27, 2023 at 8:20
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    \$\begingroup\$ Are your 8 mV and 140 mV measured or simulated? So, are your numbers wrong, or the physical implementation? At 100 MHz, you should still be able to make a reasonable wideband transmission line transformer for picking off lower powers. You might want to consider other attenuator topologies. Waveguide beyond cutoff is one, and reconsider capacitive input/output, if you use an output C low enough to drive 50 ohms. However, a series string of several 10s of K at several watts each, straight into the 50 ohm of a line to your spec ana is what I would do. I do that at GHz, though not kW! \$\endgroup\$
    – Neil_UK
    Feb 27, 2023 at 9:01
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    \$\begingroup\$ @CoryLytle A capacitve sampler can be flat, you make a C to C attenuator, see my comment, use an output C low enough to drive 50 ohms. Any 10:1 scope probe is a flat capacitive divider for most of the bandwidth, only using its resistor to control things in the DC-kHz range. \$\endgroup\$
    – Neil_UK
    Feb 27, 2023 at 9:59
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    \$\begingroup\$ @CoryLytle sorry to inform that every on-line calculator I've ever seen (including the one you linked) won't calculate the resistor values correctly when the input impedance and the output impedance are different values. They all seem to have been copied from some "ancestor" website (god knows which one) that has a basic error carried through to them all. For this reason I made my own website calculator but, it uses a T attenuator (aka a taper pad attenuator) rather than a pi section attenuator. I will get round to making a pi attenuator that does the job correctly but not this year it seems! \$\endgroup\$
    – Andy aka
    Feb 27, 2023 at 11:07
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    \$\begingroup\$ stades.co.uk/Impedance%20TX/Taper%20pad%20attenuator.html <-- if it plus a few words constitutes an acceptable answer for this site, please let me know @CoryLytle \$\endgroup\$
    – Andy aka
    Feb 28, 2023 at 10:23

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It appears that @Andyaka has discovered that many online attenuator calculators do not give proper results when the input and output impedance are not equal. I'm accepting this as the answer since they don't seem to work in real life circuits

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