I have built a grain moisture sensor using a 555 timer and an arduino,
as shown in the figure:
.
The sensor is composed of a two parallel aluminum plates (as in a capacitor):
The output of this circuit is connected to analog arduino pin A0. I have noticed that the higher the moisture of the grain, the lower is the voltage read by the analog pin. From basic phyisics:
\$ C =(ε_0 ε_R A)/d\$ – in which \$ ε_R\$ is the dielectric of the capacitor, so the higher the moisture the higher the capacitance of the sensor is.
I am having trouble understanding why voltage drops when moisture is higher. Why does the voltage drop when the moisture is higher?
Best Answer
From basic capacitor physics, we know that capacitance, charge, and voltage are related by
$$V = \frac{Q}{C}$$
and given that parallel plate capicatance is described by $$C = \frac{\epsilon A}{d},$$
If the total charge stays constant, then the voltage across the capacitor will decrease with increasing permittivity.
$$ V \varpropto \frac{\alpha}{\epsilon}$$
(\$ \alpha \$ represents the constants of charge, area, and distance.)