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Ok I have the following circuit, with values of variables: circuit \$\displaystyle R_{1} = 1 M\Omega, R_{2}=10k\Omega,~~ k_{n}C_{ox}\frac{W}{L}=1mA/V^2\$ , and \$V_{T}=1V\$.

1) When \$v_{IN}=0\$, calculate minimal \$V_{R}\$ so that the transistor is turned on (\$v_{GS} \geq V_{T}\$)

2) If it's given that \$i_{D} >> i_{1}\$, evaluate the mode of operation of transistor \$M_1\$ and calculate the voltage gain \$\displaystyle a = \frac{\hbox{d}v_{OUT}}{\hbox{d}v_{IN}}\$


The first part is easy, \$\displaystyle v_{IN}=0 \implies v_{GS}=\frac{V_R}{2} \implies V_R \geq 2V_T=2V \implies \boxed{V_{Rmin}=2V}\$

The second part, however, is bothering me. We have that \$\displaystyle v_{DS} = v_{IN}\$ and \$\displaystyle v_{GS}=\frac{V_R+v_{IN}}{2}\$, because the two resistors are both \$R_1\$ so voltage is equally divided. Now, substituting them into the NMOS requirement for saturation: \$\displaystyle v_{DS} > v_{GS}-V_T \Longleftrightarrow \frac{v_{IN}}{2} > \frac{V_R}{2}- V_T > V_T - V_T = 0 \$ so in the end it follows that the transistor is in saturation if \$ v_{IN} > 0\$, and in ohm regime conversely.

My teacher, however, just said that the transistor is always in the triode (ohm) regime, and then proceeded to do the problem, using the current equation for ohm regime, etc.

Please tell me whether he was right and why. Thanks!

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For the transistor:

$$V_{DS}=V_{in}$$

$$V_{GS}= \frac{V_P-V_{in}}{2}$$

Saturation condition becomes:

$$\frac{3V_{in}}{2}\geq\frac{V_P}{2}-V_T$$

Given that \$\frac{V_P}{2}> V_T\$ (condition for transistor to be ON), and assuming that the DC voltage at the non-inverting input of the amplifier \$V_{in}=0\$, the above saturation condition can't be satisfied.

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