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I wanted to make sure I had the correct view on this circuit.

From my understanding D1 and D2 are input protection diodes that clip the input to prevent any large voltage differences to damage the input stage transistors.

In this configuration where J4, J2 are closed and J1, J3 are open, is this circuit an inverting amplifier, or is it a differential amplifier?

The reason I'm confused here is because during say a negative cycle only D1 is forward biased, but doesn't the input get split between non-inverting and inverting input (same for positive cycle where only D2 is biased?) Or does the signal just go to ground since that is what the non-inverted input is connected to?

Sorry if my question seems silly just haven't been able to find a similar example to this (some of my dad's old work but he's not around.)

enter image description here

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Under normal operation neither diode will conduct appreciably. This is because under normal operation, the difference between the inverting and the non-inverting inputs is millivolts (or less). The diodes only conduct under abnormal operation, such as when the op amp is saturated.

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  • \$\begingroup\$ "not conduct" is a rather absolute statement. Even when equilibrated, the opamp will force its input offset voltage across the diodes. This can lead to many pA's or even nA's of current (for Schottkys). This could be unacceptable for high impedance measurements. \$\endgroup\$
    – tobalt
    Jun 16, 2021 at 17:14
  • \$\begingroup\$ I wrote "not conduct appreciably". If pAs or nAs are too much, then that circuit shouldn't be used. But the question seems to why are they used. Given that they are used in that circuit, the designer felt they did not introduce too much error for the purpose the circuit was designed for. \$\endgroup\$
    – Math Keeps Me Busy
    Jun 16, 2021 at 17:23
  • \$\begingroup\$ True, but instead of mentioning only the benefits of these components, the drawbacks should be mentioned too. These drawbacks are also part of what they do and what the designer has to consider. \$\endgroup\$
    – tobalt
    Jun 16, 2021 at 17:25
  • \$\begingroup\$ So during normal operation the diodes don't conduct, meaning this is just an inverting amplifier correct? But if the op amp were to saturate the diodes clamp the input difference i.e. both negative and positive cycles get clipped? Just want to make sure I get it \$\endgroup\$
    – Fernando
    Jun 16, 2021 at 17:28
  • \$\begingroup\$ Yes. Minor quibble, both positive and negative might get clipped. An op-amp might saturate in only one direction, although it often saturates on both. \$\endgroup\$
    – Math Keeps Me Busy
    Jun 16, 2021 at 17:34

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