# problem with a circuit with one JFET

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.

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

$I_{R1}=I_{R2}$

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

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

$V_G=3,10\,V$

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

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

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

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

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

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

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

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

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

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

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

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

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