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I am a newbie. I've searched many times for this topic and I found nothing! So here am I asking: How do the inputs of Op Amp become equal?

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As you can see in this internal Op Amp circuitry, inverting and non-inverting inputs are floating and there is no feedback to them. How does it work that way? \$V_{in+} = V_{in-}\$.

When I apply 3V to \$V_{in+}\$, \$V_{in-}\$ pulls up to 3V. I really cannot see it. If you guys can explain it to me in detail, I appreciate it. Thanks.

Alper91
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  • You can find a "very very detail" explanation of how 741 (winch is what's in your schematic) works in almost any textbook, but perhaps someone will indulge you. Also https://en.wikipedia.org/wiki/Operational_amplifier#Internal_circuitry_of_741-type_op-amp – Fizz Oct 14 '15 at 12:12
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    It doesn't "work that way" - you need negative feedback for the inputs to become very similar in value. If you have an op-amp where one input appears to drag the other around it might be because of protection diodes across the two inputs. – Andy aka Oct 14 '15 at 12:15
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    Your diagram shows an opamp schematic WITHOUT THE FEEDBACK so the inputs do not become equal ! The simplest for this to happen is to connect inverting input to output and non-inverting input to 0V (assuming Vs+ and Vs- form a symmetrical supply). – Bimpelrekkie Oct 14 '15 at 12:17

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To answer your question it helps to analyze what may happen after switch-on the power supplies (assumption) +/- Vs=+/-10V. More than that, you have to apply negative feedback (resistive netork between output node and inverting input). (The described timely sequence may be somewhat „formalistic“ - however, it helps to understand the feedback concept).

Example: Non-inverting gain stage with desired gain of "+2". That means: Feedback factor k=0.5 using a voltage divider with two equal resistors in the feedbacl loop. Open-loop gain (assumption): Aol=1E4.

1.) t=0: Apply at an input voltage Vin=1V. The opamp is not yet working in its linear range (feedback not yet active due to time constants within the circuit) and the output will immediately jump to Vs=+10V.

2.) t>0: The voltage at the inverting terminal will rise to 0.5Vs=5V>Vin=1V. Hence, the voltage at this inverting terminal dominates (is larger) and the output voltage will change in the direction to minus 10V.

3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inverting terminal a feedback voltage of +0.99980004V .

4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region because the diff. voltage is Vdiff=Vin-0.99980004=1-0.99980004=0.00019996V. As a result, the output voltage is Vout=Vdiff*Aol=0.00019996*1E4=1.9996001V.

5.) This is a stable equilibrium because: the classical feedback formula for a finite value of Aol also gives the output voltage Vout=Vin*[1E4/(1+0.5E4)]=1.9996001 V.

6.) That means: We have an equilibrium because the output voltage has a value which exactly meets the condition Vout=Vdiff*Aol. Any larger/smaller output voltage causes a small reduction/increase for Vdiff thereby correcting this deviation from the equilibrium.

7.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.2mV) that in can be neglected (assumed to be zero for calculations) in many cases.

LvW
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