Electrical – Converting “MPR121 capacitive keyboard” sensor filter data into capacitance values

Under a given voltage the ideal capacitor obeys the following relationship:

\$Q = CV\$ where Q is the charge, V is the voltage applied and C is the capacitance. If we sweep the voltage with time then we can rewrite this as,

$$\frac{dQ}{dt} = C \frac{dV}{dt} $$

where \$dQ/dt\$ is the current and \$dV/dt\$ is the scan rate. It turns out that an ideal capacitor would be a rectangle during a CV scan. Unfortunately resistances present in the system lead to non-ideal behavior and cause rounding of the corners for our perfect square. The equation above is really just what the paper uses but with different symbols; upon rearrangement we see that we can get $$C = \frac{ \frac{dQ}{dt} }{ \frac{dV}{dt} } = \frac{I}{s} $$

So how do we use this technique to get our capacitance? Easy, we'll take the current starting from 0 current to either a maximum cathodic or anodic current going in one direction (in either a positive direction or a negative direction) and divide that by the sweep rate to get the capacitance of the device. In an ideal case it shouldn't matter which direction you pick, either the anodic or cathodic, but non-idealities may make one current give a higher capacitance. You can tell by looking on figure 3 that it is not perfectly symmetrical about the 0 current horizontal line, therefore the integrated area will be larger in one direction rather than the other.

For the second part of your question, recall that an ideal capacitor should give completely flat horizontal lines once charged, but in this experiment there is a slope, but for the equation used we depend on the capacitance being constant no matter the applied voltage once charged: so the authors have simply forced their data to fit to an equation for the ideal capacitor. They are probably just taking the average of the current for the voltage span of interest and then plugging it into the equation.

It will be essential to know the size of the data read from each sensor, how many sensors and how fast of repetition rate that you need to get the data into the PC. Without that data it will be difficult to recommend an exact solution.

That said it may be reasonable to suggest that you configure your 8051 UART port's TXD and RXD lines to talk to a 1x6 header that can mate to one of these cables:

This can then plug into your PC's USB port. Using the appropriate virtual comm port driver from the FTDI web site the solution can easily transfer data in serial manner from your 8051 to the application in the PC. The PC app thinks it is simply talking to a COM port.

These serial adapter cables can easily transfer serial data at rates to 230.4 K baud. Using a standard 8N1 serial framing this baud rate translates to a just over 23000 bytes per second of raw transfer rate. If you had 12 sensors, each with two bytes of data and you wanted to update the data to the PC at 100 times per second this would equal an application data rate of 12*2*100 = 2400 bytes per second. As you can see there would be plenty of remaining bandwidth on the virtual comm port channel to wrap a protocol around your application data and send it to the PC.