I have this fairly simple NMOS and it is mentioned that I can determine the Source current \$I_3\$ with the formula \$ I_3 = \frac{K}{2}(V_{GS} - V_{th})^2 \$.

Here they equalize Gate Source Voltage \$ V_{GS}\$ with the Source potential \$ \varphi_S \$ and my question is why? Since the Gate Source Voltage is the potential difference between the Gate potential and the source potential \$ V_{GS} = \varphi_G - \varphi_S \$ this would mean that the Gate potential is zero. And since I don't see why this should be obvious or what am I missing, what circumstances would allow this assumption?



1 Answer 1


\$ V_{GS}\$ is what determines if the transistor is on or off. For an NMOS transistor \$ V_{GS}\$ is positive so the transistor turns on. If the transistor were fully off, \$\varphi_s\$ would be -5V and \$ V_{GS}\$ would be 5V which would cause the transistor to turn on.

  • \$\begingroup\$ And since the transistor doesn't have a powersource at the gate, it is turned off? \$\endgroup\$
    – Maxim
    Apr 19, 2019 at 18:39
  • 1
    \$\begingroup\$ No, the gate voltage is higher than the source voltage, it would turn on. Current would then flow through the resistor and lower the gate voltage so some calculation is require to find out what the final vaules of this circuit are, I will leave that to you. \$\endgroup\$
    – Voltage Spike
    Apr 19, 2019 at 18:41
  • \$\begingroup\$ Now I get it. Thanks! \$\endgroup\$
    – Maxim
    Apr 19, 2019 at 18:49

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