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I have the following diagram of an ideal OP and $$ y=-Ae \tag 1 $$

schematic

simulate this circuit – Schematic created using CircuitLab

I'm not used of using arrows for voltages, so I don't grasp the meaning of the arrow \$e\$. I tried to redraw the diagram below. Is \$e\$ the voltage difference between the inputs \$+\$ and \$- \$? I call them \$v_2\$ and \$v_1\$ below, so is \$e=v_2-v_1\$?.

Is the following diagram equal to the one above?

schematic

simulate this circuit

And also, why the negative sign in equation \$(1)\$? Using my diagram we have \$e=v_2-v_1\$ and \$v_2=0\$ so \$ y=Ae=A(v_2-v_1)=-Av_1. \$ But this is not equal to equation \$(1)\$...

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  • \$\begingroup\$ A voltage is the difference between two point. An arrow is not unusual to make clear which two points. The convention is that the point of the arrow has a higher potential then the tail. Some times the voltage polarity is unknown. Then you just choose an arbitrary arrow direction. If your subsequent calculations produce a negative voltage this means that the original guessed arrow direction was wrong. \$\endgroup\$ – Oldfart Oct 5 '19 at 12:09
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The arrow simply define the voltage direction notation. It's really super easy:

-------------->

|            |
attach     attach
voltmeter  voltmeter
"-" here   "+" here

As you can see, that's necessary – it's not inherently clear which way around you'd measure voltage, and if you have an equation like \$y=-Ae\$, it'd be ambiguous.

Having that arrow is also a necessity, because what you do, introducing a ground potential at the positive input of the opamp, can't always be done – absolute voltages don't exist – everything is potential relative to an arbitrarily chosen point. And the arrow does nothing but define "the voltage I measure here is between the potential at the start of my arrow, and its tip – so subtract the start potential from the tip potential to get the voltage".

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  • \$\begingroup\$ Note that this can be different across different languages/cultures; I know I've read that the arrow points towards the negative end in some European engineering traditions. (which causes some Annoyances when trying to use CircuiTikZ...) \$\endgroup\$ – Hearth Oct 5 '19 at 12:14
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I think you're overthinking the difference between using arrows (upper schematic) and the more common + / - signs and ground symbols (lower schematic). Basically they're different ways of indicating the same thing, a voltage and it's polarity.

However in the upper schematic with the arrows, it is undefined what connects to "ground level potential. All voltages are referenced to the line connecting to the + input of the opamp but nothing says that the voltage on that line is the ground reference. In the bottom schematic that is indicated clearly with the ground symbols.

The "e" in the bottom schematic is confusing and it is not following conventions, it has no + and - so I cannot see what "e" means.

Yes the circuits are the same.

To "easily" understand that there must be a minus sign in the transfer, you must realize that the opamp will steer it's output such that the voltage difference between its (+ and -) inputs is zero. That means that the opamp forces the voltage at the - input to be zero (because the + input connects to ground so it is zero volt by definition).

Knowing that, now work put what the output will be when the input has a positive voltage. Z1 and Z2 work as a voltage divider so the voltage at the output of the opamp must become negative to satisfy the zero volt at the - input.

Learn more by reading Opamps for Everyone.

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