Resistive Touchscreen

Adapted from EECS16A Note 12, 13, 14

1D Model

We can simplify the touchscreen to a 1D structure in order to gain a better understanding of how it works.

We want to calculate the $L_{touch}$ in order to find the position of the touchpoint. As we know, it, $I = \frac{dQ}{dt}$ , where $Q$ and $I$ are charge and current respectively. We measure current in amps, which amounts to one coulomb per second.

The resistance of a conductive material is given by the equation $R = \rho \frac{L}{A}$ , where rho is the resistivity (a number unique to each material) L is the length and A is the cross-sectional area.

As you can see, you write the resistances of the two resistors in terms of the resistivity, which brings length into the equation, allowing you to solve for $L_{touch}$ using node voltage analysis. Simply solve for $u_{mid}$ in terms of $R_1$ and $R_2$ , and then you can plug the resistivities in.

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Last updated 3 years ago

1D Model

We can simplify the touchscreen to a 1D structure in order to gain a better understanding of how it works.

We want to calculate the

L_{touch}

in order to find the position of the touchpoint. As we know, it,

I = \frac{dQ}{dt}

, where

Q

and

I

are charge and current respectively. We measure current in amps, which amounts to one coulomb per second.

The resistance of a conductive material is given by the equation

R = \rho \frac{L}{A}

, where rho is the resistivity (a number unique to each material) L is the length and A is the cross-sectional area.

As you can see, you write the resistances of the two resistors in terms of the resistivity, which brings length into the equation, allowing you to solve for

L_{touch}

using node voltage analysis. Simply solve for

u_{mid}

in terms of

R_1

and

R_2

, and then you can plug the resistivities in.