I am connecting a signal source of 10kHz sine wave with a driver which is a voltage follower made by an OpAmp. The follower has one input and two identical outputs. One output is connected with a device(LOAD) to drive the device(LOAD); the other is connected with an oscilloscope for measuring.

The OpAmp used is ADA4610


simulate this circuit – Schematic created using CircuitLab

The first picture: Voltage follower is not connected with the device(LOAD) and its output signal is pure. No reflection

The second picture: Voltage follower is connected with the device(LOAD). It can be observed that there are harmonic noises showing up. With reflection

Updated observation:

  • When I change the load to 1kOhm resistor, the output of OpAmp does not show any distortion.
  • When actual load is used (around 50Ohm), distortion shows up.
  • My suspect is that large output current could lead to distortion. However, I did not find related information in the datasheet.

My question is:

  • Are the harmonics caused by reflection?
  • Would very low frequency signals have reflection as well?
  • How to export pure sine waves without distortion in this case?
  • \$\begingroup\$ You mean 1 voltage followers with 2 traces going out, or 2 voltage followers with 2 outputs with the same input? \$\endgroup\$ – MdxBhmt Aug 21 '15 at 13:50
  • \$\begingroup\$ One follower with 2 traces out. \$\endgroup\$ – richieqianle Aug 21 '15 at 13:51
  • \$\begingroup\$ Is your signal getting distorted maybe because you are leaving the specified operating areas of the op amp which causes the signal to be distorted (like the peak is getting a bit cut off making it more like a rectangle (well the harmonics don't fit for that exactly)) And the second picture is not after the op amp but still the same point you measure, right? \$\endgroup\$ – Arsenal Aug 21 '15 at 14:16
  • \$\begingroup\$ @Arsenal The second one is also after the OpAmp. Comparing with the second one, the only difference is that the first one is connected with a device. \$\endgroup\$ – richieqianle Aug 21 '15 at 14:51

No, a reflection is a linear phenomenon and cannot produce harmonics.

The harmonics are generated by nonlinearities in your buffer amplifiers (voltage followers). -72 dB relative to the fundamental is actually pretty good. What kind of opamps are you using? You may need to upgrade to a precision high-bandwidth unit in order to do any better.

  • \$\begingroup\$ It is funny that using the same OpAmp, if the OpAmp's output is not connected with device, the output does not have harmonics. Do you mean if the OpAmp consumes current, its output may get distorted? \$\endgroup\$ – richieqianle Aug 21 '15 at 14:53
  • \$\begingroup\$ Ah, I misunderstood. I thought the second picture was at the output of the opamp. I take it that this measurement is still at the output of the device (the input of the opamp). In that case, yes, you are seeing the distortion caused by the input bias current of the opamp, which is very often nonlinear, interacting with the output impedance of the source device. Show us your exact circuit. Are you using a FET-input opamp? \$\endgroup\$ – Dave Tweed Aug 21 '15 at 14:56
  • \$\begingroup\$ I have conducted some small experiments and updated the results with exact circuit in the thread. Could you have a look? I think the problem could be caused by OpAmp overload. Also, do you think 15kHz signal can have reflection? \$\endgroup\$ – richieqianle Aug 22 '15 at 3:13
  • \$\begingroup\$ OK, so I still didn't understand your setup correctly. See, that's why diagrams are so much better than words, especially when English isn't your first language. Now I understand that both measurements were taken at the output of the opamp, with and without the 50-ohm load. \$\endgroup\$ – Dave Tweed Aug 22 '15 at 13:20
  • 2
    \$\begingroup\$ No. Forget about reflections. As I said up front, reflections (and other transmission line effects) are linear and cannot product harmonics. And in any case, transmission line effects only become significant when the length of the line is a large fraction of the wavelength of the signal, which would be thousands of meters in this case. \$\endgroup\$ – Dave Tweed Aug 22 '15 at 14:17

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