In this document, the author says in the last page that his amplifier output is 3.8 W. But looking at the schematic, the input voltage of the transistor is at maximum 12 V. Taking into account that the impedance of the coaxial cables and connectors is most probably 50 Ω, I found by simple computation assuming a sine wave, or just using this volt-to-power convertor, that the output power is at most 360 mW. I don't think the guy is using SMA connectors and coaxial cables not matched to 50 Ω. As I saw the same problem in several other documents, I would like to know what I am missing.
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\$\begingroup\$ What is the provenance of that design or author that would cause eyebrows to be raised. There's plenty of stuff on the internet published by people who know very little or don't check their numbers or, who are trying to attract you into some deal or other. \$\endgroup\$– Andy akaCommented Nov 29, 2023 at 11:30
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\$\begingroup\$ @Andy. Basically you are right. But as these guys are able to build 10GHz stuff with their hands, I tend to believe I've missed something. \$\endgroup\$– MikeTeXCommented Nov 29, 2023 at 11:55
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\$\begingroup\$ @Andy, actually, look at the MGFK35V4045 Datasheet in the answer of Tim Williams below, and you'll see that this question could have been asked about it. \$\endgroup\$– MikeTeXCommented Nov 29, 2023 at 12:15
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\$\begingroup\$ Mike, that's my point. A data sheet from a recognized supplier naturally has provenance and is a better example than what you linked (in terms of getting proper responses on this site). \$\endgroup\$– Andy akaCommented Nov 29, 2023 at 12:19
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1 Answer
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There is an impedance-matching network, presumably from a lower impedance at the die itself (which indeed cannot pull drain voltage below 0V), to the characteristic-impedance ports/pins/traces.
"Internally impedance matched"
Probably the same question applies at the input as well, but it's a harder question to ask, or justify, without an RLC equivalent model of the gate.
Whether this means anything at 10GHz, when the part says 14, or if they're using a variant tuned to a different band, I don't know.
Supposing you had an oscilloscope with high enough bandwidth and light enough loading that you could measure the time-domain voltage at the pin, you would indeed see that the pin voltage is dipping below GND and peaking several times VDD. This requires additional current flow as well, which circulates within the device, and through the substrate/base ground plane connection.
answered Nov 29, 2023 at 11:44
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\$\begingroup\$ +1 for having pointed out the question is in fact about the MGFK35V4045 in the datasheet, as they obtain 3W with only 10V. But I don't understand very well your answer. You say that in fact, the signal goes well below ground, so we have in fact a wave with something like 20V pp in the datasheet experiment ? \$\endgroup\$– MikeTeXCommented Nov 29, 2023 at 12:18
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1\$\begingroup\$ Yes, something like that. A more complete answer would require 1. an actual example of the matching network internal to the device (manufacturers never provide such details; occasionally they can be observed when de-capped e.g. Fig.8 here mdpi.com/2072-666X/11/4/375 but extracting an RLC equivalent circuit is nontrivial), and 2. knowledge on your part to understand how the matching network works, or further 3. knowledge of network theory, in which it suffices that such a network exists (not necessarily how a particular one would be realized). \$\endgroup\$ Commented Nov 29, 2023 at 12:43
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1\$\begingroup\$ I would generally air on the side of simplicity, on this site, so this answer serves more along the lines of "a solution exists". Which seems adequate to explain the article, without even having to call into question the veracity of said article. \$\endgroup\$ Commented Nov 29, 2023 at 12:46
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1\$\begingroup\$ Which, note that it's a separate matter whether the article actually even does what it claims it does. Like I said, "a solution exists"; whether they are making use of that solution, I can't tell, but it is plausible that they did. \$\endgroup\$ Commented Nov 29, 2023 at 12:47
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3\$\begingroup\$ Figure 5 of the linked article shows the "copper snowflakes" that are added to the transmission Iines to make them reactive. This can allow the output voltage to be different from that at the output transistor. \$\endgroup\$ Commented Nov 29, 2023 at 17:50