Question 1: Can an op amp work with both of the inputs floating?

I assume that due to the input offset voltage if both the inputs are floating, the output would be +Vsat/-Vsat. Am I wrong?

Question 2: Can an op-amp work with any one of the inputs floating?

In both the cases there is no feedback to the op-amp.

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    \$\begingroup\$ Generally it's a bad idea to allow opamp inputs to float. Any noise picked up by the pins will be amplified through a large open loop gain, potentially leading to power comsumption and on-die crosstalk issues. See this Maxim app note on how to properly terminate an unused opamp: maximintegrated.com/en/app-notes/index.mvp/id/1957 \$\endgroup\$ – Peter Aug 22 '15 at 7:00
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    \$\begingroup\$ Can an Opamp be considered as "working" if the inputs are floating? The question should be more like; What conditions could be expected to occur if one or both inputs of an op amp are left floating? \$\endgroup\$ – vini_i Aug 22 '15 at 7:08
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    \$\begingroup\$ "working" means "amplifying". What could be amplified without an input? The question is a contradiction. \$\endgroup\$ – LvW Aug 22 '15 at 7:38
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    \$\begingroup\$ An input might float when an amplifier is switched between sources, so this question is useful. \$\endgroup\$ – CL. Aug 22 '15 at 7:53
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    \$\begingroup\$ The proper question is "is there an application for op-amps in which one or both inputs are left floating?" Obviously, applications whose circuit diagrams have a fully connected op-amp are not going to work if the circuit isn't implemented. \$\endgroup\$ – Kaz Aug 22 '15 at 14:28

I'll interpret your question a bit more broadly, since the results with one or both inputs floating are unlikely to be working usefully.

In general there are two categories of op-amps, bipolar ones which have significant bias current that is of guaranteed direction, and FET or MOSFET-input types which have smaller bias current of undetermined (and possibly changing) direction.

In the former case you can generally predict the output state with one input left open from the datasheet, provided the other input is kept within the common-mode range. For example, the LM324 has a bias current of 20nA typical. That means that if you leave the inverting input within the common mode range, then the non-inverting input will float up outside the common mode range in the positive direction and the output will be close to the negative rail. If the inputs are swapped, the output will be close to the positive rail.

If both inputs are left open the output state is harder to predict and may or may not be consistent for a given op-amp design. The schematic is published for the LM324, so one may see that the output ought to be low, but it isn't- it goes to the positive rail, so the schematic is not complete.

The second category of op-amps will have undetermined (and possibly changing) output depending on leakage, pickup, temperature, humidity etc.

In any case if the op-amp has proper supply voltages will be "working" but if you are feeding it garbage signals (floating around, picking up noise or outside the common mode range) it will provide garbage of some description at the output.

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    \$\begingroup\$ I would still consider it useful, as one could just use it as a RNG then ^_^ \$\endgroup\$ – user20088 Aug 22 '15 at 12:30
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    \$\begingroup\$ @vaxquis The cartoon is a good example of how well it would probably work. ;-) \$\endgroup\$ – Spehro Pefhany Aug 22 '15 at 12:46
  • \$\begingroup\$ @vaxquis: Noise diodes work a lot better for RNG :). \$\endgroup\$ – Peter Aug 23 '15 at 0:05
  • \$\begingroup\$ @PeterK nope. /turns the sarcasm off \$\endgroup\$ – user20088 Aug 23 '15 at 0:25

Can an op amp work with both the inputs floating?

If nothing is connected to the inputs, the op-amp doesn't care but the output will be nonsense and, by "working" this is meaningless.

can an op-amp work with any one of the inputs floating ?

Again this is a nonesense question because for an op-amp to "work" it has to do what it is intended to do and with an input disconnected all bets are off.

  • \$\begingroup\$ @vaxquis I was careful to avoid saying "not working". I made the point that to use the term "working" was meaningless. \$\endgroup\$ – Andy aka Aug 22 '15 at 16:46
  • \$\begingroup\$ true; still, my point was that it's not entirely meaningless - if somebody, for some reason, decides to e.g. detect noise/EM fluctuations/RF noise/stray values/whatever by leaving a floating pin (acting similarly to antenna and/or inductor and/or capacitor due to strays) - the op-amp with such pin(s) will not do meaningless work - it'll work just as usual, amplifying the difference between the terminals. Mind me, I've used such pins to do some "tricks" in my audio synth. \$\endgroup\$ – user20088 Aug 22 '15 at 17:31

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