Let's solve the problem **step-by-step** as requested: --- ### **Given Parameters** - \( V_{DD} = 1.1\,V \) - \( V_{TN} = 0.25\,V \) - \( \mu_n C_{ox} = 0.20\,mA/V^2 \) - \( W/L = 25 \) - \( \beta_n = \mu_n C_{ox} (W/L) = 5\,mA/V^2 \) - \( \lambda = 0 \) (ignore channel length modulation) --- ### **a) Determine the transistor region of operation as a function of \( V_x \) (0 to \( V_{DD} \)):** #### **Node Voltages:** - **Source of \( M_0 \):** \( V_x \) - **Drain of \( M_0 \):** \( V_{DD} \) - **Gate of \( M_0 \):** Set by the resistor divider: \[ V_G = V_{DD} \frac{R_2}{R_1 + R_2} \] Let's denote \( V_G = V_{G} \). #### **Region Conditions for nMOS**: - **OFF (Cutoff):** \( V_{GS} < V_{TN} \) - **Triode (Linear):** \( V_{GS} > V_{TN} \) and \( V_{DS} < V_{GS} - V_{TN} \) - **Saturation:** \( V_{GS} > V_{TN} \) and \( V_{DS} \geq V_{GS} - V_{TN} \) #### **Key Voltages:** - \( V_{GS} = V_G - V_x \) - \( V_{DS} = V_{DD} - V_x \) Now analyze as \( V_x \) increases from 0 to \( V_{DD} \): --- #### **1. OFF Region (\( V_{GS} < V_{TN} \)):** \[ V_G - V_x < V_{TN} \implies V_x > V_G - V_{TN} \] --- #### **2. Triode/Saturation Boundary:** - **Triode:** \( V_{DS} < V_{GS} - V_{TN} \) \[ V_{DD} - V_x < (V_G - V_x) - V_{TN} \implies V_{DD} < V_G - V_{TN} \] But since \( V_{DD} \) is a constant (1.1V), typically \( V_G - V_{TN} < V_{DD} \), so **normally, the triode region does not occur unless gate voltage is very high** (which is not the case with resistive divider). - **Saturation:** \( V_{DS} \geq V_{GS} - V_{TN} \) \[ V_{DD} - V_x \geq (V_G - V_x) - V_{TN} \implies V_{DD} \geq V_G - V_{TN} \] This is typically always true here. --- **Summary of Regions as \( V_x \) varies:** - **For** \( V_x < V_G - V_{TN} \): **Saturation** - **For** \( V_x \geq V_G - V_{TN} \): **OFF** --- ### **b) Derive \( I_x(V_x) \) (piecewise for each region):** #### **Saturation Region (\( V_x < V_G - V_{TN} \)):** \[ I_x = \frac{\beta_n}{2} (V_{GS} - V_{TN})^2 = \frac{5}{2} (V_G - V_x - 0.25)^2 \,\,\, [mA] \] #### **OFF Region (\( V_x \geq V_G - V_{TN} \)):** \[ I_x = 0 \] --- ### **c) Sketch \( I_x(V_x) \) and Mark Transition Point:** - **Transition occurs at:** \( V_x = V_G - V_{TN} \) - **For \( V_x < V_G - V_{TN} \):** Parabolic decrease in \( I_x \) - **At \( V_x = V_G - V_{TN} \):** \( I_x = 0 \) - **For \( V_x > V_G - V_{TN} \):** \( I_x = 0 \) **The graph would look like:** - A downward-opening parabola (starting at \( V_x = 0 \)), reaching zero at \( V_x = V_G - V_{TN} \), and staying at zero for higher \( V_x \). --- ## **Summary Table** | Region | Condition | \( I_x(V_x) \) equation | |----------------|---------------------------|---------------------------------| | Saturation | \( V_x < V_G - V_{TN} \) | \( \frac{5}{2} (V_G - V_x - 0.25)^2 \) | | OFF | \( V_x \geq V_G - V_{TN} \) | \( 0 \) | --- ## **Final Answers (in terms of \( V_G \)):** ### **a) Region of operation as a function of \( V_x \):** - For \( V_x < V_G - V_{TN} \): Saturation - For \( V_x \geq V_G - V_{TN} \): OFF ### **b) Piecewise Current Equation:** \[ I_x(V_x) = \begin{cases} \frac{5}{2} (V_G - V_x - 0.25)^2, & V_x < V_G - 0.25 \\ 0, & V_x \geq V_G - 0.25 \end{cases} \] ### **c) Sketch:** - See the description above: Parabola from \( V_x = 0 \) to \( V_x = V_G - 0.25 \), then flat at zero. **If you have values for \( R_1 \) and \( R_2 \), substitute to get \( V_G \) for a numerical answer.** Let me know if you want to plug in resistor values for a full numerical solution!

a) The transistor operates in saturation when \( V_x < V_G - V_{TN} \) because \( V_{GS} > V_{TN} \) and \( V_{DS} \geq V_{GS} - V_{TN} \). It turns off (cutoff) when \( V_x \geq V_G - V_{TN} \), as \( V_{GS} < V_{TN} \). b) In saturation, \( I_x = \frac{\beta_n}{2}(V_{GS} - V_{TN})^2 \); off, \( I_x=0 \). c) The \( I_x \) vs. \( V_x \) graph is a parabola decreasing to zero at \( V_x= V_G - V_{TN} \), then flat at zero, marking the transition from saturation to cutoff.

a) The transistor is in saturation when \( V_{GS} > V_{TN} \) and \( V_{DS} \geq V_{GS} - V_{TN} \), i.e., for \( V_x < V_G - V_{TN} \). It turns off (cutoff) when \( V_x \geq V_G - V_{TN} \), as \( V_{GS} < V_{TN} \). b) In saturation, \( I_x = \frac{\beta_n}{2} (V_{GS} - V_{TN})^2 \); off state, \( I_x=0 \). c) The \( I_x(V_x) \) curve is a parabola decreasing to zero at \( V_x = V_G - V_{TN} \), then remains zero, indicating the transition from saturation to cutoff.