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When \$V_\text{DS} = V_\text{P}\$, thus \$I_\text{D}\$ will achieve maximum value hence equal to \$I_\text{DSS}=10\text{mA}\$. By substituting \$I_D\$ into \$I_\text{D}=I_\text{DSS}\left(1-V_\text{GS}/V_\text{P}\right)^2\$, I got \$V_\text{GS} = 0V\$. However by doing this way I'm ignoring the effect of load resistance and it's a 10 marks question I don't think it is this simple.

I have also try to use \$I_D\$ into \$I_\text{D}=I_\text{DSS}\left(1-V_\text{GS}/V_\text{P}\right)^2\$ to approach the question but there are two unknowns \$I_D\$ and \$V_\text{GS}\$ and I'm stuck. There are others information provided such as load resistance, \$V_\text{DD}\$ and \$V_\text{GG}\$ which I have try to relate them using following formula (Correct me if I'm wrong) : $$ \begin{align} V_\text{DS} &= V_\text{DD} - I_D \cdot R_D\\ V_\text{GS} &= V_\text{GG} - I_G \cdot R_G \end{align} $$ However, I cannot find a way to relate them to \$I_\text{D}=I_\text{DSS}\left(1-V_\text{GS}/V_\text{P}\right)^2\$ to solve it.

  • \$\begingroup\$ Do you wan to have Vds = 4V? Yes? \$\endgroup\$
    – G36
    Jul 3, 2023 at 14:52
  • \$\begingroup\$ V_ds given in the question is equal to V_p so it should be -5.0V too \$\endgroup\$ Jul 3, 2023 at 15:01
  • \$\begingroup\$ I think that you are wrongly interpreting the question. In your circuit, you cannot have negative Vds. It is simply impossible. \$\endgroup\$
    – G36
    Jul 3, 2023 at 15:31
  • 1
    \$\begingroup\$ I'm actually new to this so I thought that would be the case. If only the magnitude is equal it makes much more sense. \$\endgroup\$ Jul 3, 2023 at 15:47

1 Answer 1


Disclaimer. I am assuming that \$V_\text{DS}=\lvert V_\text{P}\rvert\$ since only in this case the condition \$V_\text{GS} -V_\text{DS} \le V_\text{P}\$ under which the fundamental formula for \$ I_\text{D} \$ is applicable surely holds. Despite this, we'll see that there are some incongruences from the point of view of the schematics shown in the OP: precisely, the gate circuit cannot be the one shown.

Let's start by considering the anode mesh: by applying the Kelvin Voltage Law (KVL for short) we have $$ V_\text{DD}-R_\text{D}I_\text{D}-V_\text{DS}=0 $$ Since \$V_\text{DS}=5\text{V}\$ and \$V_\text{DD}=16\text{V}\$ are known you have that $$ R_\text{D}I_\text{D}=V_\text{DD}-V_\text{DS}\iff I_\text{D}=\frac{V_\text{DD}-V_\text{DS}}{R_\text{D}}=\frac{16\text{V}-5\text{V}}{10\text{k}\Omega}=1.1\text{mA} $$ Now you can use the fundamental \$I_\text{D} = f(V_\text{GS})\$ by plugging in it the found value of \$I_\text{D}\$ $$ \begin{split} I_\text{D} = I_\text{DSS}\left(1-\frac{V_\text{GS}}{V_\text{P}}\right)^2 &\iff \sqrt{\frac{I_\text{D}}{I_\text{DSS}}} = \left(1-\frac{V_\text{GS}}{V_\text{P}}\right)\\ &\iff {V_\text{GS}}= V_\text{P}\left(1 - \sqrt{\frac{I_\text{D}}{I_\text{DSS}}}\right)\\ & \iff {V_\text{GS}}= (-5\text{V})\left(1 - \sqrt{\frac{1.1\text{mA}}{10\text{mA}}}\right) \\ &\iff {V_\text{GS}}= (-5\text{V})\left(1 - \sqrt{{0.11}}\right)\\ &\iff V_\text{GS} \simeq -3,34\text{V} \end{split} $$ Notes

The result found is in contrast with the circuit in that in the schematics is shown a \$V_\text{GG}\$ voltage of \$-1\text{V}\$ (respect to the source): this clearly cannot be as by applying the KVL law to the input circuit the have that $$ V_\text{GG}-R_\text{G}I_\text{G}-V_\text{GS}=0 \iff V_\text{GG} \le V_\text{GS}\iff V_\text{GG} \le -3,34\text{V}. $$ Possibly, the text of the exercise was added to a picture draw for another one, and this caused a bit of confusion.


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