Explanation:
Temperature range → 0 to 80'c
respective voltage output → 0.2v to 0.5v
required temperature range 20'c to 40'c
Where T = 20'c respective voltage
[tex]\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}[/tex]
[tex]\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}[/tex]
Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)
so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v
Therefore, the formula for the circuit will be
[tex]\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}[/tex]
The simplest circuit will be a op-amp
NOTE: Refer the figure attached
Vs is sensor output
Vr is the reference volt, Vr = 0.275v
[tex]\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}[/tex]
choose R2, R1 such that it will maintain required ratio
The output Vo can be connected to voltage buffer if you required better isolation.
