In OP27's datasheet, p. 14, it shows a graph when use OP27 as an 'unity-gain buffer', and give it a step input (I think it's a positive step). But it doesn't explain more about the waveform on the right.

enter image description here

I have two questions:

  1. It uses the word 'drawn', does it means under a positive step input, at that time, the current will flow into the signal generator?
  2. What happens at the points circled in red? Particularly, the point 2 and 3.

1 Answer 1


The waveform on the right is for when the feedback resistor is very low. Under these circumstances the output injects current thru the input protection diodes.

This current transfers from inverting input to non-inverting input (via the protection diodes) and, if the applied voltage source (the actual input) on the non-inverting input is low-impedance, the output current limits.

  • \$\begingroup\$ Please see this link, e2e.ti.com/blogs_/archives/b/thesignal/archive/2012/03/21/…, it seems he mean the current flows from non-inverting to inverting terminal. And can you explain more about the point 2 and 3? \$\endgroup\$
    – diverger
    Apr 10, 2015 at 8:26
  • \$\begingroup\$ Once a protection diode is forward biased strange things are going to happen and there isn't enough information in the data sheet for the device to put a full story together about points 2 and 3 on the diagram but, I suspect that the source impedance of the input signal has something to do with it. \$\endgroup\$
    – Andy aka
    Apr 10, 2015 at 9:26
  • \$\begingroup\$ Can you explain the waveform based on your current flow direction, particularly the 'glitch'? That is at which point the current will be 'drawn' by the signal generator. \$\endgroup\$
    – diverger
    Apr 10, 2015 at 11:40
  • \$\begingroup\$ IMHO, The second rising slope apparently is limited by the slew rate. So the first one, which is more faster should be the result of the 'feed through'. \$\endgroup\$
    – diverger
    Apr 10, 2015 at 11:56

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