I'm trying to transfer a small (1V) variable frequency signal over about a metre of coax. I'm not interested in anything above 100 kHz. One thing to note is that my signal already comes with a +24VDC offset. So I have this:-
based on this from a proper coax driver datasheet (Intersil HA-5002 driver):-
The HA-5002 is essentially a current amplifier, so I have my FET set up as a voltage follower. The Spice simulation seems to suggest that it works, with the expected 50% signal reduction at the end of the coax. The response is -5dBV over the 100kHz bandwidth. R1 (1.5k) runs a little warm as it's getting towards the top end of its power dissipation, but heat sinking a coax driver doesn't seem unusual.
This circuit seems very simple and the question is, therefore, is it too simple?
PS. I've looked at many driven coax arrangements here and elsewhere. The way it's done seems uncertain with 50Ω resistors, here or there. Sometimes there's just one 100Ω resistor at the source.
Best Answer
With 100 kHz at 1 meter, your transmission line length is 1/3000 the vacuum wavelength of the signal. Maybe 1/2000 of the wavelength in the transmission line. Unless you have some really extreme requirements about signal fidelity, you don't even need to worry about transmission line terminations in this system. If you want maximum signal transmission, you can certainly eliminate either the source (R3) or load (R4) termination from this system. Most likely you can elminate both with no ill effects.
Second, your schematic shows the signal source has 2 V peak-peak AC signal on top of the 24 V DC offset.
With this signal, I disagree with the two previous answers. You can arrange this so the FET will remain in saturation (the FET equivalent of "forward active mode" for BJTs) and your driver will have low output impedance throughout the signal's cycle. You won't see any meaningful asymmetry in the output. To do this you have to arrange things so the FET is always sourcing current into the rest of the circuit.
Currently you have maybe 13 mA bias current (conservatively assuming 4 V \$V_{gs}\$ for the 2n7002) through R1, and you're switching 10 mA signal into the transmission line, so your FET will remain saturated throughout the signal cycle.
If you eliminate the source termination (R3) the signal current driven by the FET will increase, and you'll have to decrease the value of R1 to have enough bias current through the FET to keep it saturated.