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The circuit is a common source amplifier that uses the 2N3819. It is known that \$V_{DD} = 20\,V\$ and that \$R_g = 100 \,Ω\$ (the generator resistance \$V_g\$ applied in In). The entrance and exit are ports In and Out, respectively. For Q1 use the JFET parameters: \$I_{DSS}=8\,mA; V_P =-3\,V; R_0 = 30 \,k\Omega; Cgs = 5\, pF; Cgd = 2 \,pF\$. R4 is not the drawing value. \$R_4\$ is \$ 5,04\,k\Omega \$. Calculate the polarization of the JFET, indicating the DC values of the nodal, \$V_{GS}, V_{DS}\$ and \$I_D\$ voltages. enter image description here

I tried to solve it. I'd like you to confirm that it's right.

enter image description here \$I_{R1}=I_{R2}\$

\$\frac{V_{DD}-V_G}{R_1}=\frac{V_G}{R_2}\Leftrightarrow\$

\$\frac{20-V_G}{120\times 10^3}=\frac{V_G}{22 \times 10^3} \Leftrightarrow\$

\$ V_G=3,10\,V\$

\$I_D=\frac{V_S}{R_5}=\frac{V_G-V_{GS}}{R_5}\$

\$I_{DS}=I_{DSS}\bigg(1-\frac{V_{GS}}{V_P}\bigg)^2=\frac{V_G-V_{GS}}{R_5}\Leftrightarrow\$

\$I_{DSS}\bigg(1-\frac{2V_{GS}}{V_P}+\frac{V_{GS}^2}{V_P^2}\bigg)=\frac{V_G}{R_5}-\frac{V_{GS}}{R_5}\Leftrightarrow\$

\$\big(I_{DSS}-\frac{V_G}{R_5}\big)+\big(\frac{1}{R_5}-\frac{-2I_{DSS}}{V_P}\big)V_{GS}+\frac{I_{DSS}}{V_P^2}V_{GS}^2=0\Leftrightarrow\$

\$\big(8\times 10^{-3}-\frac{3,10}{4,7\times 10^3}\big)+\big(\frac{1}{4,7\times 10^3}-\frac{2\times 8 \times 10^{-3}}{-3}\big)V_{GS}+\frac{8\times 10^{-3}}{(-3)^2}V_{GS}^2=0\Leftrightarrow\$

\$7,34\times 10^{-3}+5,546\times 10^{-3}V_{GS}+8,889\times 10^{-4}V_{GS}^2=0\Leftrightarrow\$

\$V_{GS}=-4,334\,V\,or\,V_{GS}=-1,905\,V\$

\$V_{GS}>V_P\Rightarrow V_{GS}=-1,905\,V\$

\$I_D=\frac{V_G-V_{GS}}{R_5}=\frac{3,10-(-1,905)}{4,7\times 10^3}=1,065\times 10^{-3}\,A\$

\$ I_D=\frac{V_S}{R_5}\Leftrightarrow V_S=1,065\times 10^{-3}\times 4,7\times 10^3=5,01\,V\$

\$V_{DS}=V_D-V_S=14,63-5,01=9,62\,V\$

\$\frac{20-V_{D}}{R_4}=\frac{V_S}{R_S}\Leftrightarrow V_D=20-1,065\times 10^{-3}\times 5040=14,63\,V\$

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  • \$\begingroup\$ In Vgs > Vp, what is Vp? \$\endgroup\$ – analogsystemsrf May 10 '17 at 4:26
  • \$\begingroup\$ @analogsystemsrf It is a parameter that depends on the manufacturer and depends on Stefan-Boltzmann constant. \$\endgroup\$ – Carmen González May 10 '17 at 6:06
  • \$\begingroup\$ @analogsystemsrf Is my resolution correct? \$\endgroup\$ – Carmen González May 10 '17 at 6:22
  • \$\begingroup\$ @analogsystemsrf Can I ignore the current passing through Rg and R3 and then match the current of R1 to the current of R2? \$\endgroup\$ – Carmen González May 10 '17 at 6:51
  • \$\begingroup\$ Yes, Ig = 0A, so no voltage drop across R3 at DC. \$\endgroup\$ – G36 May 17 '17 at 11:44

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