I want to replace a capacitive touch sensor on a circuit board (I think it’s a TTP223) with a tactile switch to overcome over/under sensitivity issues. I have wired a switch to ground and the CMOS output pin on the touch IC. This works, but I was informed that there may be potential to damage the unit by doing this. Can anyone confirm why this might be, or even if this is true? And if it is true, what would be a better approach?
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2 Answers
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The potential for damage is probably due to overloading the driver IC pin, depending on how it works. It might not be expecting a short. Instead of connecting to ground, I think you can make the switch connect a 100pF capacitance to GND in parallel with the touch sensor. I think that's how it works: a finger increases the capacitance and capacitances in parallel add up.
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It shouldn't, but if you want, you can put a 1k resistor between the sensor IC output and the rest of the circuit, and the switch to ground on the side of the resistor connected to the rest of the circuit.
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\$\begingroup\$ Can you be a little more specific, new to electronics. When you say “between the sensor IC output and the rest of the circuit”, you don’t mean cut the trace on the PCB and insert a resistor, do you? Where exactly would it go? \$\endgroup\$ Commented Feb 19, 2022 at 5:10