The signal conditioning of the input is a simple voltage divider, a non-inverting amplifier, and a level shifter. The OpAmp selected is a TL064, which has a GBW of 1MHz, a slew rate of 3.5V/?s, and is low cost. The simulation below shows the front end circuit:

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Right on the input there is a spark gap to help keep ESD events out of the amplifier. The high impedance of the input also helps to keep the circuit protected.

An AC/DC coupling switch is present; the AC path is done with a high voltage 0.1uF capacitor.

Let?s calculate the transfer function. At the V+ pin of the Opamp, we have:

$$V^{+}=V_{in}?{180k}/(820k+180k)= V_{in}?9/50$$

This voltage is amplified by the OpAmp, using the non-inverting amplifier circuit:

$$V_{Op}=V^{+}?(1+{20k}/{180k})= V^{+}?(1+1/9)= V_{in}?9/50?10/9=V_{in}/5$$

Then, the voltage needs to be shifted so it can be applied to the microcontroller, using the superposition theorem:


And finally:

$$\bo V_{\bo \out}=V_{in}/10+1.024V$$


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