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I stumbled across this claim in the datasheet of the TLV3541 opamp:

The TLV354x output stage is capable of driving standard back-terminated 75-Ω video cables, as shown in Figure 23. By back-terminating a transmission line, the device does not exhibit a capacitive load to its driver. A properly back-terminated 75-Ω cable does not appear as capacitance; the device presents a 150-Ω resistive load to the TLV354x output.

Non-inverting amplifier driving coax cable through 75 Ohm series resistor

This goes against my understanding very distinctly.

My thinking goes like this:

If the cable is electrically short (say, 1m) and open at the far end, then it will appear as a capacitance (~100 pF). The near-end series resistor will contribute favorably to the phase margin by isolating the opamp from the cable capacitance somewhat, but it does not reduce that capacitance to zero. Notably bandwidth at the far end of the cable is compromised (because R+C form a low-pass filter). (Plus, 75 Ohm might not even be enough for stable operation into 100pF, according to Figure 8 of the same datasheet, which recommends 120 Ohm instead.)

By comparison, a proper far-end termination indeed makes the cable appear purely resistive (regardless of length). Forward termination eats reflections at the near end, and thus, if the far-end termination was perfect, would not be strictly necessary.

Is TI wrong here, or am I?

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    \$\begingroup\$ Whoever voted to close: please advise what is unclear, or what detail is missing from the question. \$\endgroup\$
    – polwel
    Commented Apr 17, 2022 at 8:14
  • \$\begingroup\$ Having a capacitor directly on the output of an amplifier is not the same as having a resistor followed by a capacitor. The effect on stability is not the same. Putting a series resistor on the output of the amp will help mask any capacitive loading that the cable or load may present (due to damage, miscofiguration, sabotage or whatever). There are lots of amplifiers which are unstable with capacitive loading but which WILL be stable if a small series resistor is added between the output and the capacitor. This is even true for some older linear regulators. \$\endgroup\$
    – user57037
    Commented Apr 17, 2022 at 8:21
  • \$\begingroup\$ The beauty of the back terminated cable is that even if the receive side has incorrect impedance, the reflection will be totally absorbed by the driver side 75 Ohm resistor and no standing wave will develop. \$\endgroup\$
    – user57037
    Commented Apr 17, 2022 at 8:23
  • \$\begingroup\$ Agreeing with your 1st point (isolation resistor improves stability by introducing a zero in the loop gain). Even with back termination (but no receiver-side termination), won't standing waves still be possible? They are usually inoffensive though, agreed. \$\endgroup\$
    – polwel
    Commented Apr 17, 2022 at 8:34
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    \$\begingroup\$ I guess you are right. The reflections will be created at the far end of the load, but they will be completely absorbed when the get back to the series resistor at the driver. So in some sense they are possible, but they can't keep bouncing back and forth multiple times to build up. \$\endgroup\$
    – user57037
    Commented Apr 17, 2022 at 17:37

1 Answer 1

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In video distribution practice a cable is always terminated at the destination end with 75 ohms.

Your statement that a short cable looks like a capacitor is not true if it is terminated, it will look like a 75 ohm resistor. The amplifier will therefore see a resistive load of 150 ohm (75 + 75) and be stable.

The gain of 2 designed into the amplifier is to take into account the effect of the source and destination terminations so the overall gain is unity.

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  • \$\begingroup\$ Thanks for the answer. I stand by my statement that a short open cable looks capacitive. I agree that a far end termination hides the capacitance (also mentioned this in the question). Worrying about the capacitive load of the cable is only ever relevant if the far end is not terminated with 75 Ohm. Your answer does not explain how forward termination would do the same (IMHO it doesn't). \$\endgroup\$
    – polwel
    Commented Apr 17, 2022 at 8:07
  • \$\begingroup\$ About the gain: yup, you are correct, in my confusion I did not realize that Fig. 8 refers to unity gain. For G=2, 75 Ohm should be sufficient, I agree. \$\endgroup\$
    – polwel
    Commented Apr 17, 2022 at 8:08

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