opamp feedback calculation

I want to monitor the battery voltage( 120v DC) using microcontroller.For that i am using opamp circuit.
I want, when there is 120v DC at battery, the output of the opamp give 4v
for that purpose i use the formula

Vout= -(Rf/Rin) * Vin

4v=-(10k/Rin) * 120
Rin= 300k ohm

i simulate it on the proteus and it shows me the correct result which i desire.
I wanna ask is my calculations are correct or is this the right way to calculate the parameters?

its looking gud...

Yes, but the battery is 'floating' in that design, you can't ground one side of it.

If the battery has to be ground on one side, simply use a potential divider across it's terminals so you get 4V out with 120V in and if necessary, follow it with a unity hain buffer stage.

Brian.

Thanks
Kindly tell me What is the advantage of floating over ground?
If i use floating scheme, will it works well in the real world?

Can anyone tell me?

It depends matter how connected your battery in your case. Short R1 and R2, and the ground will not be floating. If you want to measure the voltage of the battery that is not connected to ground of your device, leave everything as it was.

Ya its in floating load and it must be avoided

I would look on the "floating battery" point from the other side.

A differential amplifier with finite common mode input impedance has no problems to measure a floating voltage. A floating voltage will be "pulled" to the amplifier ground by the input impedance. But the battery is unlikely to be floating in real life. You should ask for it's reference potential respectively the voltage difference between amplifier circuit ground and battery. The external battery connections and amplifier power supply must taken into regard.

The voltage difference must not exceed the common mode range of the amplifier circuit for correct measurement.

Another point is common mode rejection. Due to resistor mismatch and non-ideal OP behaviour, the superimposed common mode voltage will cause an error in voltage measurement.

FvM i am not getting your point.kindly tell me

Your calculation is correct. I was commenting the statements about "floating battery".

I said, there's no problem with "floating battery" in the shown circuit.

If the real circuit has external battery connections, draw a complete schematic.