How can one determine the maximal cable length an Opamp can drive without getting unstable? The thing which leads me to this question is that the impedance of a cable changes over frequency. Means, a cable can be seen capacitive and/or inductive over frequency. What is the procedure and what parameters are important to look at in the datasheet of an Amplifier? Cable is NOT impedance matched. Cable end terminates in high input impedance.

See the following LTSpice Simulation of a 5m long Coaxial Cable: enter image description here

It can be seen that in the first plot that the phase alternates between -90 (capacitive) and +90 (Inductive)

  • 2
    \$\begingroup\$ Try the section on capacitive loading of the output. If the data sheet doesn't cover it, choose an op-amp that has a data sheet that does cover it. A normal cable won't ever appear to be net inductive. \$\endgroup\$
    – Andy aka
    Sep 13, 2020 at 13:57
  • \$\begingroup\$ You are right. But have a look on my edit in the first post please. At higher frequencies, it changes. That is why I was wondering. But more theoretical, apart of having a look if the datasheet specifies such informations, what are the parameters to look at? \$\endgroup\$
    – newstudent
    Sep 13, 2020 at 14:27
  • \$\begingroup\$ Try the section on capacitive loading of the output. Try putting a white background on the plots. \$\endgroup\$
    – Andy aka
    Sep 13, 2020 at 14:29
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    \$\begingroup\$ Unless you have a very very long cable, and a very very high frequency amplifier, like VHF/microwave, the cable will never be long enough to look inductive within the amplification bandwidth of the amplifier. Put some numbers into your question, so 1 metre or 1 km of cable, and a TL072 or an OPA855 amplifier, so we can answer with some context. A small series resistor will usually sort out any instability with capacitive loading, and is a good idea anyway with transmission line loading, 50 ohms is probably right for both. \$\endgroup\$
    – Neil_UK
    Sep 13, 2020 at 14:34
  • \$\begingroup\$ Neil: The simulations shows a 5m long RG188 Coaxial Cable. You can clearly see that starting from 20 MHz, the phase starts to change between -90 and +90 degrees. So you say that this one is not long enough to look inductive? \$\endgroup\$
    – newstudent
    Sep 13, 2020 at 14:52

2 Answers 2


For relatively short cables, capacitance just adds up with length.

If you deal with relatively long cables (longer than wavelength/10), you have to properly terminate the end (I.e. a 75ohms cable needs a 75 ohms termination impedance).

If you don‘t do that, then you will get different impedances at different cable lengths. The cable could e.g. be purely inductive at 3m and purely capacitive at 2m.

You would need to measure the impedance of a cable of a given length.

So there‘s no max. length of the cable in your context. Just some lengths that may have (too) high capacitive loading because you have a mismatch at the cable end.

BTW: The impedance of the cable does not change (much) over frequency if properly terminated. Just over length or frequency if not properly terminated.

  • \$\begingroup\$ Thanks Stefan! You said that the impedance is not changing much over frequency, but why do I see that in my simulations like above? There it changes much.. \$\endgroup\$
    – newstudent
    Sep 13, 2020 at 15:28
  • \$\begingroup\$ Because your cable is not properly terminated. If you connect 50 ohms at the end, it will disappear. \$\endgroup\$ Sep 13, 2020 at 15:29
  • \$\begingroup\$ I edited my answer accordingly. \$\endgroup\$ Sep 13, 2020 at 15:32
  • \$\begingroup\$ I know it is not terminated. I do NOT want to terminate it. That is why I am asking on how it is normally done to specify the max length of a cable for an Amplifier. Example: Lets say at 2m length at a frequency of 1 Mhz it is capacitive (max capacitance Opamp can handle). At 2.2m however it is inductive. Means I could go for 2.2m, as the Opamp does not see any large capacitance anymore ?! Do you know what I mean? \$\endgroup\$
    – newstudent
    Sep 13, 2020 at 15:38
  • \$\begingroup\$ Yes, this is the way to go. But there is no max. length, just some repeating impedance pattern with lambda/2 repeating period. \$\endgroup\$ Sep 13, 2020 at 16:32

The terminated coaxial cable is just an analog delay line. The delay is the same on all frequencies, for ideal cable. The signal speed is 1.966E8 meters / second (less than the speed of light, the multiplier is 0.666). This was used e.g. in oscilloscopes to see the beginning of the signal under study. delay

If the end of the cable is not terminated, it behaves as capacitance at lower frequencies.

The cable can basically be unterminated in two ways.

  1. Load >> 50 Ohm in this case the cable acts as a capacitor.
  2. Load << 50 Ohm (eg short circuit), then the cable behaves as inductance.

The delay of the 5m cable is about 25ns. (The period time of a 40MHz signal.) The actual capacitance of the unterminated cable and the specification of the particular op-amp determines how much capacitance and in what circuit it can handle. The characteristics of each type of op-amp determine the maximum allowed length of unterminated cable. The cable data contains this value: Capacitance: 100.7pF / m (RG58U).

The maximum capacitance is decided by the given op-amp. For example, quoting from the OPA347 datasheet, page 9, under "CAPACITIVE LOAD AND STABILITY":

The OPA347 in a unity-gain configuration can directly drive up to 250pF pure capacitive load. Increasing the gain enhances the amplifier’s ability to drive greater capacitive loads (see the characteristic curve Small-Signal Overshoot vs Capacitive Load). In unity-gain configurations, capacitive load drive can be improved by inserting a small (10Ω to 20Ω) resistor, RS, in series with the output [...]


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