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I have the following problem:

Consider the circuit below

enter image description here

These component values are given: \$R_{G2}=1.5\text{M}\Omega\$, \$R_{G1}=1.2\text{M}\Omega\$, \$K=2.3\frac{\text{mA}}{\text{V}^2}\$,\$V_{to}=-1.8\text{V}\$, \$V_{cc}=5\text{V}\$.

What is the drain current, \$i_D\$?

Okay, so my thought of solving this would be the following.

First find the voltage-drop across \$R_{G2}\$ through voltage division.

\$V_{G1}=V_{cc}\times \frac{R_{G2}}{R_{G1}+R_{G2}}=5\text{V}\times\frac{1.5\text{M}\Omega}{2.7\text{M}\Omega} =2.78\text{V}\$

I interpret this as also being the voltage at the gate of the pmos. That results in \$V_{GS}=2.78-5=-2.22\text{V}\$

And I'm stuck here. My next step would be to find \$V_{DS}\$, but I am unsure of how to do so. Can anyone help me?

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  • \$\begingroup\$ Well, you know Vgs voltage so you can find Id current without any problem. Yes? \$\endgroup\$
    – G36
    Commented Jan 5, 2020 at 18:03
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    \$\begingroup\$ (BTW, the decimal separator is a dot in English, not a comma like in German or Dutch) \$\endgroup\$
    – Huisman
    Commented Jan 5, 2020 at 18:09
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    \$\begingroup\$ Since Rd was not given, you need to assume that the MOSFET is in the saturation region. What else you can do here? \$\endgroup\$
    – G36
    Commented Jan 5, 2020 at 18:13
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    \$\begingroup\$ Hmm, I think I messed up on my prefixes. I meant to write \$i_D=0,406mA = 0,41mA\$. \$\endgroup\$
    – Carl
    Commented Jan 5, 2020 at 18:29
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    \$\begingroup\$ Much better this time. \$\endgroup\$
    – G36
    Commented Jan 5, 2020 at 18:34

1 Answer 1

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$$\ I_{D} = \frac{\textrm{Kp}}{2}(V_{GS}-V_T)^2 $$ $$\ I_D =\frac{\textrm{2.3}}{2}((-2.22V)-(1.8V))^2 = 0.203 \textrm{m} A = 203 \mu A $$

Found this : https://vlsitips.blogspot.com/2012/06/vlsi-physical-design.html Operation Modes for PMOS

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  • \$\begingroup\$ But we can "define" K "differently" and we can get rid of this 2 in the equation and use Id = K(Vgs - Vt)^2 instead. Also, as the author points outs, his teacher is using this equation Id = K(Vgs - Vt)^2 for saturation. \$\endgroup\$
    – G36
    Commented Jan 5, 2020 at 19:39
  • \$\begingroup\$ OK. convention. \$\endgroup\$
    – Sisyphus
    Commented Jan 5, 2020 at 19:41
  • \$\begingroup\$ You right. I hope it's fine. You welcome \$\endgroup\$
    – Sisyphus
    Commented Jan 5, 2020 at 20:01

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