much simpler. all anybody needs to understand is that when you take say RMS, you are taking the Mean of the Square, so that means the Square is what is important, and you have to work in Squares. the Root is just to convert back to the units you started with after your work in Squares is done. obviously the Squares here are because Watt=VV/Ohm so were by working in Squares were treating power as the most important quantity and doing our voltage calcuations so that they give the right power calculations, power being proportional to the Square of the voltage.

It is very simple. For +VCC, just add a DC voltage Between the ground and the diode cathode at the output. The DC voltage has the symbol like a battery with the positive terminal to the diode and the negative terminal to ground. For -VCC, do the same, but with the inverted battery.

can you please give a clear image of how that will be added in the opamp modelling and also how it will affect if not added?
Also will be perfect if you add a schematic to show the connection

Yes it is valid for any function

Let’s look at it this way. If Vin = 0V, R1 is connected to ground, in parallel with R2. There is no current going through them. So, in effect, the non-inverting input of the op amp is connected to ground and the circuit becomes an inverting amplifier. With V2 being a positive value of +2.5V, how can the op amp output be positive when it inverts?
When V1 = 0V, the output is V2 = -R4/R3 * 2.5V= – (4kohm / 10kohm) * 2.5V = -1V exactly as the calculator shows.

Hi Graham. If you are still grappling with this problem reply here. I have the solution… and it’s not a differential amp.

I am seeing the same contradiction to the following link as Mihai Beffa
https://www.daycounter.com/Calculators/Op-Amp/Op-Amp-Voltage-Calculator.phtml
I get the same results to that link in a simulator.

With your default values, an input of +2.5V the above link and simulator gives Vo = +1V but with an input of 0V, Vo = +2V not the -1V as your calculator requires.

What is the problem?

I am not sure what you are trying to say. The RMS value does not depend on time values, within reason. If you have a signal which is on for 3 days and then off for 3 days, then you can consider it as a DC signal depending on the corresponding processes in your system.

For example: You try to determine what connector you can choose, based on the RMS value of the current that has to go through one pin. If you have a PWM signal of say, 100 kHz, then yes, you can use the RMS value. But if you have a signal with a period of 10 seconds, then you have to look at the maximum value of the current over 10 seconds. That spec becomes now DC, not AC.