The DC analysis is regarding the amplifier calculations, but that is not relevant to the topic. The equivalent DC circuit of the amplifier is:
The known values are: \$Rg1=300k\Omega\$, \$Rg2=200k\Omega\$, \$Rs=100k\Omega\$, \$k_n=25\frac{µA}{V^2}\$, \$\lambda=0.02V^{-1}\$, \$Vtn=1V\$, \$Vdd=10V\$.
Now the problem is to find the bias point (drain current – \$I_D\$, voltage \$V_{GS}\$ and voltage \$V_{DS}\$.
First, I calculated the gate voltage as:
$$V_G=\frac{Rg2}{Rg1+Rg2}Vdd=4V$$
Then, I assumed that the transistor is operating in saturation mode, and set up these equations:
$$I_D=k_n(V_{GS}-Vtn)^2(1+\lambda V_{DS})$$
$$V_G=V_{GS}+Rs I_D$$
$$Vdd-V_{DS}-Rs I_D=0$$
The problem is, I cannot solve those equations, as there always seems to be one element missing. Any ideas on how to solve this?
Best Answer
For quick hand analysis, I would personally not include the impact channel length modulation.
Knowing, $$ V_{GS} = V_G - V_S\;\;\;\&\;\;\; V_{DS} = V_{dd} - V_S $$
and that the drain current equals,
$$ I_D=k_n(V_{GS}-Vtn)^2(1+\lambda V_{DS}) $$
Since gate current of Q1 is zero the drain current is also,
$$ I_D = \dfrac{V_S}{R_S} $$
Put it all together as,
$$ \dfrac{V_S}{R_S} = k_n(V_G - V_S-Vtn)^2(1+\lambda (V_{dd} - V_S)) $$ and solve for \$V_S\$.