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I'm trying to switch P-Mos given that:

  1. Vcc > Vgs max
  2. Enable_output is 0V or 3V
  3. I'm switching both GND and VCC on purpose.

sch 4. 8.4VR at 2uA ! enter image description here enter image description here The questions:

  • When Q2 is off, what is Vgs at Qsp1?
  • Is it going to be Vcc or Vcc-10v?
  • Will the Mosfet turn on and off with this schematic?
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The questions:

  • When Q2 is off, what is Vgs at Qsp1?
  • Is it going to be Vcc or Vcc-10v?
  • Will the Mosfet turn on and off with this schematic?

Qsp1 will not turn off reliably because it relies on leakage currents in the zener diode to try and reduce gate-source voltage to zero - use a 47 kohm resistor in parallel with the zener diode.

Qsp1 will turn on properly even if you added the needed 47 kohm I mention providing that Vcc is sufficient.

You need base resistors for both BJTs - you can't apply 3 volts to an unprotected base-emitter region. Given that you are using a p channel MOSFET for the upper switch, why not use an N channel MOSFET for the ground switch?

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  • \$\begingroup\$ hello, i've how do you make sure that its 47k is the right value. my question was exactly because i suspected that qsp1 will turn off because of zener leakage. im using NPN transistor to reduce components BOM as this is going to pcba. bjt resistor where not drawn here but it exist. what is the right way to switch high side pmos? (slow switching is acceptable) \$\endgroup\$ – Hasan alattar Dec 13 '18 at 9:59
  • \$\begingroup\$ 47 kohm is about right but if you wanted to be exact you would need to define how slowly it could be allowed to switch off and maybe go for something like 470 kohm but then you also need to consider that Q2 might be leaking a small current when off and so that small current through a 470 kohm might still switch on Qsp1 partially. \$\endgroup\$ – Andy aka Dec 13 '18 at 10:06
  • \$\begingroup\$ hello, i understand now what do you mean after researching and updated question about the zener diode. because the Vgs is acting like capacitor it needs time to fall and rise and zener leakage is very small current to turn gate off. so if possible can you tell me why 47k poped to your mind not 10k for example as 10k will provide faster switching right? \$\endgroup\$ – Hasan alattar Dec 13 '18 at 15:12
  • \$\begingroup\$ A resistor will form a potential divider with the 5 k resistor and, if the added resistor is too low in value there may not be enough VGS voltage to turn on the FET. I estimated 10x 5k = 50 k = ~47k just to prevent too much loss of VGS (10% loss). \$\endgroup\$ – Andy aka Dec 13 '18 at 15:16
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In this schematic the Qsp1 will be always on because the VGS = (VCC - 10) - VCC = -10 which is enough to turn on most Mosfets. You need to pull up the gate of the Qsp1 to ensure it's off when no signal then it will work properly.

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