What role does the 30k resistor play in the LF398?

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How would that resistor eliminate any offsets introduced by the op-amps in a circuit like this one:

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On this explanation, I don't get this line:

These diodes then require the 30k resistor to avoid overloading the output amplifier.


When the sample switch is open, there is no feedback around the first buffer, which means that any voltage difference between its inputs will cause its output to saturate. The diodes prevent this from happening, which would slow it down. They keep the output voltage of that buffer within ± one diode drop of the signal input.

However, when the sample switch is closed, you want feedback around both buffers in order to cancel out internal offsets caused by the switch and/or the second buffer. In this mode, the feedback insures that both inputs of the the first buffer are equal, and the outputs of the two buffers are (nearly) equal. The diodes do not conduct in this mode.

The resistor is required in order to isolate the outputs of the two buffers from each other when the switch is open. Each buffer has a low-impedance output, and if they were connected directly together (even through the diodes), large amounts of current would flow. The resistor limits this current to a value that does not affect the accuracy of the second buffer's output.

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  • \$\begingroup\$ what do you mean by the "the two buffers"? \$\endgroup\$ – ielyamani Feb 9 '16 at 13:49
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    \$\begingroup\$ The two triangular symbols at the top of the diagram. They are essentially opamps wired in a voltage-follower configuration, and the shorthand term for this is "buffer". The first one provides the current to charge/discharge the sample capacitor in the "sample" state, while the second one prevents it from being charged/discharged in the "hold" state. \$\endgroup\$ – Dave Tweed Feb 9 '16 at 13:52
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    \$\begingroup\$ Converting useful part of faulty answer to a comment: I found a description here where we learn: The feedback loop is extended from input to output. This requires the opposed diodes that "catch" the output of the first op-amp when the feedback loop is broken in the "hold" state. These diodes then require the 30k resistor to avoid overloading the output amplifier. \$\endgroup\$ – BartmanEH Feb 9 '16 at 14:24
  • \$\begingroup\$ @BartmanEH the link doesn't work, could you please update it, and you're welcome to answer the question, the whole thing is still blurry to me \$\endgroup\$ – ielyamani Feb 9 '16 at 14:28
  • \$\begingroup\$ The link works fine -- is your browser blocking it for some reason? \$\endgroup\$ – Dave Tweed Feb 9 '16 at 14:32

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