I have been working on a project and one of the ideas was to control the power supply by push button. In a normal case, the SW1 is open, the 5v_POWER supplies the 5V_IN; when the SW1 is closed, the 5V_IN is cut off. My solution is to use a PMOS SSM3J328R and a normal-off-momentary-on push buttonPMS9P1B10M2QE to implement the manual temporaray power-off feature. The rated current is 2A. I have googled for a high side switch circuit but so far I haven't found a proper solution for this scenario. Is this going to work or do I need something else to make this work? Thanks in advance. enter image description here

  • \$\begingroup\$ This will be a reset button. Releasing it returns power. \$\endgroup\$ – tobalt Jun 10 at 0:36
  • \$\begingroup\$ You need to fully explain how you expect everything to work. As is, closing SW1 will force Q2 to be off. Releasing SW1 will allow Q2 to turn on. That is all. So SW1 will be like a temporary power cut switch. Cuts power while you hold it down. I would suggest that to avoid blowing up Q2's gate you move C4 so it is from gate to source. Also add a zener protection diode from gate to source. Also make R8 bigger. Like 47k or 100k. \$\endgroup\$ – mkeith Jun 10 at 3:23
  • \$\begingroup\$ @mkeith Thanks for your comment. I will add a Zener diode and increase R8 as well. However, I don't quite understand why C4 may blow up Q2's gate in this circuit? \$\endgroup\$ – Ross Jun 10 at 4:03
  • \$\begingroup\$ I could be wrong but depending on how far the switch is from the cap and FET, the parasitic inductance could lead to voltage overshoot at the gate. That MOSFET has an absolute maximum gate voltage of 8V. Putting a cap from gate to source will help prevent high peak voltages. And the zener will help clamp them off if they somehow occur in spite of the cap. \$\endgroup\$ – mkeith Jun 10 at 6:17
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    \$\begingroup\$ @user_1818839 That means there will be a constant current running through the push-button all the time. Personally, I don't want to let the power of the whole system rely on the current over a pushbutton, especially when the system running in a harsh environment(dusty, hot..). Compared to the push button, I trust MOSFET more. \$\endgroup\$ – Ross Jun 11 at 1:49

R7-C4 have a time constant of only 18 us. This is not nearly long enough to act as a debounce network, so what is the purpose of these components?

I would place C4 in parallel with R8, and delete R7. That increases the debounce time constant to 1 ms. If this still is too short, increase C4 up to 1 uF, then start increasing R8 until you get the performance you want.

  • \$\begingroup\$ The R7 is simply used to make Vgs adjustable for more flexibility. During the debugging, we will increase the R8 and C4 values to match the switch debounce requirement as you suggested. \$\endgroup\$ – Ross Jun 10 at 4:22

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