I need default close button, but many smd-buttons is default open. How create mosfet-button? It is for LCD display backlight switch off. When button is pressed the output must be 0 volts or disconnected. When button is not pressed the output must be 3v. I created an experimental circuit. It is working OK.

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Many old notebooks contain NDS9435. I also added a capacitor parallel to the switch to avoid bouncing. I don't know capacitor value- smd do not have a signature.

I think that switch-off current I = 3/3000 = 1mA is not optimal.

What values of resistor is optimal for this circut?


2 Answers 2


If I understand your requirements correctly, then make these changes and try it:

  1. Eliminate R1. Connect the gate directly to the bottom end of SW1. When the switch is pressed, the gate will be tied directly to the source. This will assure that the FET is off.

  2. Increase R2 to 10 K. This is optional; it reduces the current drawn when the switch is closed. In fact, R2 could be even larger, such as 47 K or even 100 K, but then the circuit is more susceptible to noise.

Note: That FET is not a logic-level type. Its ON resistance is not characterized in the tables for a gate voltage (Vgs) of only -3 V. You might need a different FET, depending on what kind of load is connected to the drain (output).

  • \$\begingroup\$ Thanks, I'd set R2 = 750K, and it's work success, and I'd decrease a capacitor. \$\endgroup\$
    – nick_n_a
    Dec 3, 2020 at 21:15

You should be able to connect a normally-open switch between the logic signal and ground, then connect a pullup resistor from the logic signal to +3V. Pressing the button causes the logic signal to be connected to ground and have a logic 0 value. Releasing the button allows the pullup resistor to pull the logic signal up to a logic 1 value.


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