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I am a little confused about the Vgs(th) for example if I use a value a little greater than the specification about the Vgs(th) not the max rating enter image description here enter image description here

for example I am using a P-Channel MOSFET if I apply 3.3V on the gate it will open the circuit? or does nothing because it value is greater than the VGS(th) -0.45 to -1.5

what is the difference between SQJ411EP and this SUB75P03-07 talking in Vgs(th) in case I want to use them as switch

My specs are:

Vcc 5V

Vgs 3.7v

I am driving a 4A load

Thanks

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  • \$\begingroup\$ You're applying 3.3 V to the gate but what is the source connected to? \$\endgroup\$ – The Photon Sep 5 '17 at 21:51
  • \$\begingroup\$ my source is connected to 5V \$\endgroup\$ – Ale Sep 5 '17 at 21:52
  • \$\begingroup\$ 3.3 V - 5 V is -1.7 V, so you are applying more than the minimum Vgs(th) to this FET. \$\endgroup\$ – The Photon Sep 5 '17 at 22:32
  • \$\begingroup\$ applying more than minimun VGS(th) is it ok? it wont damage the Gate? \$\endgroup\$ – Ale Sep 5 '17 at 22:34
  • \$\begingroup\$ Not if you don't exceed the absolute maximum rating. \$\endgroup\$ – The Photon Sep 5 '17 at 22:35
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If VGS = +3.7V your pMOSFET will be OFF. If it's -3.7V, then your pMOSFET will be ON (see last row of your table).

To be clear: VGS = +3.7V means that, with your data (the source is at 5V), the gate is at 8.7V with respect to ground. With VGS = -3.7V, it means that the gate is 1.3V with respect to ground.

If you apply 3.3 V at the gate (I assume with respect to ground), then VGS = -1.7V which is larger (in absolute value) than the absolute value of the threshold voltage, i.e. the MOSFET is ON. However, it might dissipate more because of the large Rds. There is no zoom of the output characteristics (see vishay datasheet) for such low current values (4A), but it seems to me that with VGS = -1.5, you'll have a VDS of about 100mV or so (you'll dissipate 400mW).

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  • \$\begingroup\$ so if the value is larger than threshold voltage it will switch? now If I changed the gate voltage to 5v. gate - source -> 5-5 = 0V. the p-channel is still able to switch? \$\endgroup\$ – Ale Sep 5 '17 at 22:13
  • \$\begingroup\$ If |VGS|>|VTH|, and sign (Vgs) = sign(Vth) (for enhancement-mode MOSFETs, but I don't remember of having ever seen a depletion-mode discrete MOSFETs) the MOSFET is on. However, if the difference with the trashold voltage is very small, the MOSFET might not be "fully on" (i.e. as you want as a switch). It could be in its saturation region, i.e. where a) it works as amplifier b) it dissipates a lot of power. \$\endgroup\$ – next-hack Sep 5 '17 at 22:18
  • \$\begingroup\$ If you put the gate voltage to 5V, then VGS = 0V, then the pMOSFET is OFF. \$\endgroup\$ – next-hack Sep 5 '17 at 22:18
  • \$\begingroup\$ and if the gate voltage is 0V, and the VGS = -5, then the pMOSFET is on? because the VGS is larger than the threshold voltage? right? :D \$\endgroup\$ – Ale Sep 5 '17 at 22:23
  • \$\begingroup\$ Yes. (It's larger in absolute value, and it has the "correct" sign). \$\endgroup\$ – next-hack Sep 5 '17 at 22:26

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