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I'm a little bit confused on the Vgs(voltage-gate-source) of a mosfet...from what i understand so far the Vgs is simply the potential difference of the voltage applied from gate to source...

FOR EXAMPLE: using a p-channel mosfet assume that Vg=10 volts and my Vs=18 volts so i get a VGS= -8

using a n-channel mosfet assume that Vg=10 volts and my Vs=0 (vs is always 0 since the source is at the ground level) so i get a VGS= 10

i hope someone will clear this confusion of mine...TIA

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  • \$\begingroup\$ What is your confusion? \$\endgroup\$ – nidhin Apr 14 '18 at 10:39
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    \$\begingroup\$ im confused if all the things i said was right i dont know if those are right \$\endgroup\$ – electronic noob Apr 14 '18 at 10:40
  • \$\begingroup\$ im sorry im just new to this electronics stuff \$\endgroup\$ – electronic noob Apr 14 '18 at 10:41
  • \$\begingroup\$ Yes you are correct. Vgs = Vg-Vs. For both n and p channel MOSFETs. \$\endgroup\$ – nidhin Apr 14 '18 at 10:45
  • \$\begingroup\$ Yes you are correct. N channel FETs are usually easier to understand, I still have to stop and think about the polarity of P channel FETs too. Sometimes the negative sign is omitted in P channels, and simply written as the absolute value, |Vgs| = 8 V. However, in this case you do have to double check that the gate is indeed negative when compared to the source. \$\endgroup\$ – A.S. Apr 14 '18 at 10:47
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You didn't really ask a question but I can state that what you said all sounds correct except that N-channel mosfets do not have to have the source attached to ground. Often times they will be used as a high side switch with some kind of driving/bootstrap circuit to elevate the gate over whatever drain voltage is being controlled.

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  • \$\begingroup\$ i didn't know that we can use n channel mosfet for high side switching \$\endgroup\$ – electronic noob Apr 14 '18 at 10:42
  • \$\begingroup\$ Yes they can, a bootstrap is a fancy way of saying a capacitor that is charged to a high voltage. If you are trying to switch a high side FET on a 20 V rail with a 3.3 V controller, a capacitor is charged to 20 V with a charge pump to turn on the gate. \$\endgroup\$ – A.S. Apr 14 '18 at 10:49
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    \$\begingroup\$ thanks for confirming the things i said...and i know now why i wreck my p channel mosfet because i exceeded the maximum vgs rated for that mosfet which is +/- 20 volts.....because i applied a VG=10 and VS=35 and their difference is high that i reach a VGS=-25 \$\endgroup\$ – electronic noob Apr 14 '18 at 10:52
  • \$\begingroup\$ Place a zener diode that clamps below your max Vgs across the gate to source connection and limit the current through the zener with a resistor. Some FETs have this zener built into the package. Here is an example of controlling a high side load with GPIO. acroname.com/sites/default/files/u9/electronicloads_02.jpg \$\endgroup\$ – lucky bot Apr 14 '18 at 19:14
  • \$\begingroup\$ @electronicnoob - " i know now why i wreck my p channel mosfet because i exceeded the maximum vgs rated for that mosfet which is +/- 20 volts". Yup. And you won't forget it easily, and you'll have a better idea that reading the data sheet carefully is a good idea. Welcome to real-world engineering. While paying attention to theory is a really good idea, a certain amount of hands-on experience (and embarrassing failure) is part of the game. "Good judgement comes from experience. Experience comes from bad judgment." \$\endgroup\$ – WhatRoughBeast Jan 27 at 19:30

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