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I know optocoupler does the job of isolation very well. How to isolate supply ground?

If car battery is supplying the power to the MCU, then the ground will be common always and it defeats the purpose of having optocoupler for sensing voltage.

Is there any way to isolate supply ground, should be as cheap as possible or if not possible what to do?

Here is the circuit : enter image description here

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  • \$\begingroup\$ Google "isolated dc/dc converter". Won't be as cheap as a 7805. \$\endgroup\$ – The Photon Sep 25 '17 at 5:28
  • \$\begingroup\$ Highly expensive. Not recommended \$\endgroup\$ – Learner Learner Sep 25 '17 at 5:33
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    \$\begingroup\$ Isolation of power always comes at cost. But I sense some misconception in this question. Why do you need to have your processor floating relative to common chassis ground? Your original signal (IN1) seems to be referenced to the same ground. Why do you need to isolate anything? What am I missing? \$\endgroup\$ – Ale..chenski Sep 25 '17 at 6:12
  • \$\begingroup\$ That's the point, if Optocoupler LED's ground is connected to same ground as of MCU, this defeats the purpose of isolation. I think there is no way but to use this circuit as it is without isolation just that optocoupler will work as better option than voltage divider, nothing else. \$\endgroup\$ – Learner Learner Sep 25 '17 at 6:18
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I think your effort is seriously misguided. From your other question it looks obvious that you want to interface a logical signal in the range from 10 to 28 V to a 5-V tolerant MCU input. Optoisolator is not the device you need for this task, you need a simple signal limiter, something like a 1-k resistor with a 4.7-V Zener diode to ground. That's it, unless you have some other issues related to noise or something.

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If car battery is supplying the power to the MCU, then the ground will be common always and it defeats the purpose of having optocoupler for sensing voltage.

Yes and no.
For the purpose of isolation it obviously doesn't help due to the common ground.
For the purpose of protection it still provides some functionality.

If the ground of the optocoupler can carry enough current, your circuit can withstand a fault condition only damaging the optocoupler, but not the microcontroller or other circuitry.

However, I recommend this circuit.

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