I have a TTP223 touch sensor, a BC347 Npn transistor and 10Kohm resistor. I want to make a touch button as substitution for my pwr physical button on the case of my pc.

On the motherboard I have 2 pin, PWR+ (5V) and PWR- (GND) and, if I short them, the Pc turn on: That’s the logic I have to replicate. Now, The problem is that the ttp223, when touched, apply a voltage (about 4.5V) to his output pin (I/O and GND). I don’t want this voltage because I only need to short the pin on the motherboard and not to apply any other voltage to them.

So, I used a transistor controlled by the touch sensor’s I/O pin on the base, the PWR+ connected to the collector and the PWR- to the emitter.. BUT IT DOESNT WORK .. whyyyy.. the logic is: the touch sensor apply a voltage to gate and source and this should close the interrupt between source and drain that is PWR- and PWR+ . So I don’t understand..

I measured with a tester and the voltage between source and drain is 5V (like it should be) when not touched and it drops to 4.8 when touched, BUT this is not enough to trigger the motherboard and turn on the PC. I think it should go really close to 0V.

So what I’m not considering? thanks for the help

Here the schematic


simulate this circuit – Schematic created using CircuitLab

So, what i need is: on touch --> Voltage between base and emitter --> short between PWR+ and PWR-

I also noticed that i dont need to short the PWR+ with PWR- only. It works with every GND PIN even one of another ATX power supply. So probably, in order to work, it just need to pull to 0V the PWR+ (from 5V) I also thought about relay but I prefer a transistor which makes no noise and its significantly smaller.. Someone also said to me that one of my problem is that, between base and emitter, can be 0,7V max.. and I apply 4,3-4,5V .. could that be the reason why it doesnt work? how can i reduce the voltage with a resistor? (how many ohms?) I want to link another question on this forum which works for him.. and it s almost my same situation -->

[Similar Question][1] [1]: https://electronics.stackexchange.com/questions/334362/capacitive-sensor-as-drop-in-replacement-for-mechanical-button

  • \$\begingroup\$ Put R1 between TTP's output and the BJT's base. And connect ATX PON (green) directly to BJT's collector. \$\endgroup\$ Feb 19, 2021 at 12:06
  • \$\begingroup\$ The R1 in the picture is wrong. If you think about it, the pushbutton is simply a 0 ohm short between your PWR+ and PWR- pins, so what good is a 10k there? Yes, the other issue might be that the sensor is directly connected to the transistor base, instead of a resistor. And yes, most of the time, power button shorts a pulled-up input to ground which you most likely have determined properly, but there really is no defined standard you can say it's always like this. \$\endgroup\$
    – Justme
    Feb 19, 2021 at 12:18
  • \$\begingroup\$ yes i knew it.. in fact the resistor there, was just one of the try .. i also tryed without it but still nothing.. @Justme do you think that 10K ohm in series with base its enough to pull 4,3V from sensor output to around 0,7 V? In case it isnt, how many ohms i need? \$\endgroup\$
    – Raffix
    Feb 19, 2021 at 12:40
  • \$\begingroup\$ @RohatKılıç i prefer using the pin on the motherboard intead of ATX wires but i think it s the same.. it's just easier to connect with the motherboard pins. Do u think 10K ohm is enough between output and base? or i need more? \$\endgroup\$
    – Raffix
    Feb 19, 2021 at 12:45
  • \$\begingroup\$ If you are lucky, TTP's output buffer is not broken. Because in your circuit, the output is loaded with a diode (B-E junction of the BJT). You might have burnt the TTP's output, BJT, or both. I'd replace all with newer ones and hook up the circuit as explained in my comment. \$\endgroup\$ Feb 19, 2021 at 13:42


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