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I have this doubt which is freaking me out. First of all I am all confused between open and closed and ideal characteristics of the OPAMP. My main doubt is this: Considering the OPAMP as Inverting Amplifier.

  1. Which mode are we using the OPAMP in? Single mode or differential mode?
  2. What is getting amplified? The input source or the difference between the two voltages at the the two terminals ?
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  • \$\begingroup\$ I fixed the layout of your question, but I'm not sure I understand what you are asking. Can you include a circuit diagram with your question? \$\endgroup\$ – jippie Oct 26 '14 at 9:02
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Each opamp is a high-gain differential amplifier. Therefore, the unit can do nothing else than to amplify the voltage difference between both input nodes - independent if one of the input nodes is grounded or not. This voltage difference is in the micro-volt range and does not allow real applications.

However, because of the large open-loop gain (Aol) of the opamp unit we always use negative signal feedback. For simple amplifier purposes the feedback network consists of resistors only. Referred to the applied signal voltage, the so-called closed-loop gain Acl is reduced correspondingly and the applied input voltage now can assume realistic values in the millivolt or even volt range.

However, the voltage difference directly BETWEEN the opamps input nodes is still in the micro-volt range. During calculation of opamp amplifiers (with feedback) this tiny voltage can be neglected in comparison with the two other voltages within the circuit (applied signal input voltage, opamp output voltage). This approach leads to the concept of "virtual ground" (for inverting opamp applications).

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You are mixing up the model of an ideal opamp with a real one.

If you use an ideal opamp, the voltagegain is infinite. Hence, in a circuit where the opamp's operating point is on the linear section of its i/o-Diagram, nothing is amplified.

With a real opamp, the differential voltage between its inputs is amplified by a factor of 100000 or more.

Considering an inverting configuration, there's a feedback from the output to the input, limiting the overall amplification of the circuit. But while on an ideal op amp the differential input voltage will drop to zero, the differential input voltage of a real opamp will just be very small, as it can be read and calculated from the opamp's datasheet.

So, in a inverting configuration the whole circuit amplifies the input voltage. The real opamp amplifies its differential input.

And I don't know, what you consider single mode. The ideal opamp gives a s*** on which input is grounded or connected to the ballon in the clouds. It just gives an infinite positive voltage on its output when it finds any positive differential voltage on its inputs, and vice versa.

For real opamps there are of course nifty things like CMMR, but I don't know if this relates to your question. Can you give a source for your "mode thing"?

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  • \$\begingroup\$ thanks a lot. i probably got mixed up with open and closed loop operations. \$\endgroup\$ – hamza Oct 31 '14 at 19:30

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