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I know how the non-inverting configuration actually looks but I wonder why it can't be identical to the inverting mode (regarding connection of resistors). The feedback is the major issue I can see here but just for mathematical calculation of gain, this modified circuit would still produce the gain same as that of the inverting mode(with negative sign of course).

Why is the non-inverting configuration producing inverted output like in inverted configuration? What are the factors that differentiate the non-inverting and inverting configuration of the op amp?

schematic

simulate this circuit – Schematic created using CircuitLab

math

original image

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The op amp is a pretty basic amplifier, it measures the difference between its inputs, and provides a correction on its output based on its gain, be it internal, or limited by external components,

With the 2 inputs, if you give a slightly more positive voltage different things happen, for the + input, the output will increase slightly, for the - input, it will decrease slightly, the device is amplifying this difference by up to 1,000,000 times, so only a tiny difference is needed for it to respond.

For what you have drawn, it will act more like a comparator with hysteresis, when the + input is slightly positive, the output will increase, but this just sums with the + input making it more positive until the output slams against its supply rail, where as the normal non inverting approach uses its feedback to stop the run away,

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  • \$\begingroup\$ Thank you @Reroute. But I can't still figure out why I am getting gain like -ve feedback although it is connected in +ve feedback. \$\endgroup\$ – Giga-Byte Aug 27 '19 at 13:14
  • \$\begingroup\$ Your issue is the assumption of I1 + I2 = 0, that is only true for negative feedback, in this case any positive input pushes it furthur away from 0, and every change in output just moves it away furthur, same any negative input would push it negative until the output hits the negative rail, If the feedback was negative, then it would try and correct its input towards 0, this is not the case for your example. \$\endgroup\$ – Reroute Aug 27 '19 at 13:19
  • \$\begingroup\$ If you mean the op amp keeping both inputs at the same voltage, again, only possible with negative feedback, lets say you have a supply voltage of +-15V just to make things easy, and you feed in 1V to the + input, as its positive feedback and it will slam on a rail, the + input will be at (15-1)*(R/(Rf+R))+1, think of it like a voltage divider, the output is 15V and is stuck there, the input is 1V, so you end up with some voltage inbetween them, not 0V, \$\endgroup\$ – Reroute Aug 27 '19 at 13:27
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You've connected the feedback resistor to the noninverting input, which means that you now have positive feedback instead of negative feedback.

Remember, without feedback, the opamp has very high (theoretically infinite) gain. It is only the application of negative feedback that reduces the circuit gain to a definite value. That is why, even in the noninverting "mode", the feedback must still be negative — connected to the inverting input.

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  • \$\begingroup\$ I can understand the case of feedback here. But the confusing part is: +ve feedback still produces gain like -ve feedback. I have attached the mathematical calculation. \$\endgroup\$ – Giga-Byte Aug 27 '19 at 13:08
  • \$\begingroup\$ In your math, you assumed that the voltage at node A is ground, which is no longer true without negative feedback. Try again, leaving \$V_A\$ as a variable. \$\endgroup\$ – Dave Tweed Aug 27 '19 at 14:14

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