Let's solve each part step by step. **Given:** - **NFET:** - \( t_{ox} = 4\,\text{nm} = 4 \times 10^{-7}\,\text{cm} \) - \( L = 0.2\,\mu\text{m} = 2 \times 10^{-5}\,\text{cm} \) - \( W_n = 15\,\mu\text{m} = 1.5 \times 10^{-3}\,\text{cm} \) - \( V_{GS} = 2\,\text{V} \) - \( V_{DS} = V_{GS} = 2\,\text{V} \) (for saturation, unless otherwise specified) - \( V_r = 1\,\text{V} \) (assume this is \( V_{DS} \), but check context) - \( \mu_{n,lf} = 250\,\text{cm}^2/\text{Vs} \) - **PFET:** - \( \mu_{p,lf} = 100\,\text{cm}^2/\text{Vs} \) - \( t_{ox} = 2\,\text{nm} = 2 \times 10^{-7}\,\text{cm} \) - \( W_p = ? \) (to find) - \( L = 0.2\,\mu\text{m} = 2 \times 10^{-5}\,\text{cm} \) - \( v_{sat} = 4 \times 10^6\,\text{cm/s} \) I will use the basic MOSFET equations for saturation. --- ### **(a) What should \( W_p \) be for the PFET to make the characteristics the same as for the NFET, as predicted by the simple model?** The "simple model" refers to matching the **current drive** (i.e., \( I_{DS,sat} \)), so that for the same \( V_{GS} \) and \( V_{DS} \), the PFET and NFET have the same saturation current. The current in saturation is: \[ I_{DS,sat} = \frac{1}{2} \mu C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 \] Assuming \( V_{th} \) is the same magnitude for both devices and \( V_{GS} \) is at the same overdrive (common for design matching). Set \( I_{DS,sat,n} = I_{DS,sat,p} \): \[ \mu_n C_{ox,n} \frac{W_n}{L} = \mu_p C_{ox,p} \frac{W_p}{L} \] \[ W_p = W_n \frac{\mu_n C_{ox,n}}{\mu_p C_{ox,p}} \] Calculate \( C_{ox} \): \[ C_{ox} = \frac{\epsilon_{ox}}{t_{ox}} \] with \( \epsilon_{ox} = 3.45 \times 10^{-13}\,\text{F/cm} \). \[ C_{ox,n} = \frac{3.45 \times 10^{-13}}{4 \times 10^{-7}} = 8.625 \times 10^{-7}\,\text{F/cm}^2 \] \[ C_{ox,p} = \frac{3.45 \times 10^{-13}}{2 \times 10^{-7}} = 1.725 \times 10^{-6}\,\text{F/cm}^2 \] Now plug in: - \( W_n = 1.5 \times 10^{-3}\,\text{cm} \) - \( \mu_n = 250\,\text{cm}^2/\text{Vs} \) - \( \mu_p = 100\,\text{cm}^2/\text{Vs} \) So, \[ W_p = W_n \cdot \frac{250 \times 8.625 \times 10^{-7}}{100 \times 1.725 \times 10^{-6}} \] Calculate numerator: \[ 250 \times 8.625 \times 10^{-7} = 2.15625 \times 10^{-4} \] Calculate denominator: \[ 100 \times 1.725 \times 10^{-6} = 1.725 \times 10^{-4} \] So, \[ \frac{2.15625 \times 10^{-4}}{1.725 \times 10^{-4}} = 1.249 \] So, \[ W_p = 1.249 \times W_n = 1.249 \times 1.5 \times 10^{-3} = 1.874 \times 10^{-3}\,\text{cm} \] Convert to \( \mu\text{m} \): \[ 1.874 \times 10^{-3}\,\text{cm} = 18.74\,\mu\text{m} \] **Final Answer (a):** \[ \boxed{W_p = 18.7\,\mu\text{m}} \] --- ### **(b) Find \( V_{DS,sat} \) and \( I_{D,sat} \) for \( V_{GS} = 2\,\text{V} \).** Assume \( V_{th} = 0 \) for max overdrive, or else you need a value. For the sake of demonstration, let's proceed with \( V_{th} = 0.3\,\text{V} \) (a typical value, but please replace if you have the actual number). \[ V_{ov} = V_{GS} - V_{th} = 2 - 0.3 = 1.7\,\text{V} \] \[ I_{D,sat} = \frac{1}{2} \mu C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 \] **NFET:** - \( \mu_n = 250\,\text{cm}^2/\text{Vs} \) - \( C_{ox,n} = 8.625 \times 10^{-7}\,\text{F/cm}^2 \) - \( W_n = 1.5 \times 10^{-3}\,\text{cm} \) - \( L = 2 \times 10^{-5}\,\text{cm} \) Plug in: \[ I_{D,sat,n} = \frac{1}{2} \cdot 250 \cdot 8.625 \times 10^{-7} \cdot \frac{1.5 \times 10^{-3}}{2 \times 10^{-5}} \cdot (1.7)^2 \] Calculate step by step: 1. \( \frac{1.5 \times 10^{-3}}{2 \times 10^{-5}} = 75 \) 2. \( 250 \cdot 8.625 \times 10^{-7} = 2.15625 \times 10^{-4} \) 3. \( 0.5 \cdot 2.15625 \times 10^{-4} = 1.078125 \times 10^{-4} \) 4. Multiply by 75: \( 1.078125 \times 10^{-4} \cdot 75 = 8.0859375 \times 10^{-3} \) 5. \( (1.7)^2 = 2.89 \) 6. Final: \( 8.0859375 \times 10^{-3} \cdot 2.89 = 0.02336\,\text{A} = 23.36\,\text{mA} \) **PFET:** - \( \mu_p = 100\,\text{cm}^2/\text{Vs} \) - \( C_{ox,p} = 1.725 \times 10^{-6}\,\text{F/cm}^2 \) - \( W_p = 1.874 \times 10^{-3}\,\text{cm} \) - \( L = 2 \times 10^{-5}\,\text{cm} \) \[ I_{D,sat,p} = \frac{1}{2} \cdot 100 \cdot 1.725 \times 10^{-6} \cdot \frac{1.874 \times 10^{-3}}{2 \times 10^{-5}} \cdot (1.7)^2 \] 1. \( \frac{1.874 \times 10^{-3}}{2 \times 10^{-5}} = 93.7 \) 2. \( 100 \cdot 1.725 \times 10^{-6} = 1.725 \times 10^{-4} \) 3. \( 0.5 \cdot 1.725 \times 10^{-4} = 8.625 \times 10^{-5} \) 4. \( 8.625 \times 10^{-5} \cdot 93.7 = 0.008084 \) 5. \( 0.008084 \cdot 2.89 = 0.02336\,\text{A} = 23.36\,\text{mA} \) So, both give the same current as intended. **Saturation occurs at** \[ V_{DS,sat} = V_{GS} - V_{th} = 1.7\,\text{V} \] **Final Answer (b):** \[ \boxed{ V_{DS,sat} = 1.7\,\text{V}, \quad I_{D,sat} = 23.4\,\text{mA} } \] (for both NFET and PFET, with the given widths). --- ### **(c) If velocity saturation (\( v_{sat} = 4 \times 10^{6} \) cm/s) is considered, find \( V_{DS,sat} \) and \( I_{D,sat} \) for the NFET and PFET.** #### **Step 1: Find \( V_{DS,sat,v} \) (velocity saturation onset)** The critical field \( E_{sat} = v_{sat} / \mu \). So for each: #### **NFET:** \[ E_{sat,n} = \frac{4 \times 10^6}{250} = 1.6 \times 10^4\,\text{V/cm} \] \[ V_{DS,sat,v,n} = E_{sat,n} \cdot L = 1.6 \times 10^4 \cdot 2 \times 10^{-5} = 0.32\,\text{V} \] #### **PFET:** \[ E_{sat,p} = \frac{4 \times 10^6}{100} = 4 \times 10^4\,\text{V/cm} \] \[ V_{DS,sat,v,p} = 4 \times 10^4 \cdot 2 \times 10^{-5} = 0.8\,\text{V} \] --- #### **Step 2: Find the current in velocity-saturation regime** The current is: \[ I_{DS,sat,v} = W \cdot C_{ox} \cdot (V_{GS} - V_{th}) \cdot v_{sat} \] **NFET:** - \( W_n = 1.5 \times 10^{-3}\,\text{cm} \) - \( C_{ox,n} = 8.625 \times 10^{-7}\,\text{F/cm}^2 \) - \( V_{ov} = 1.7\,\text{V} \) - \( v_{sat} = 4 \times 10^6\,\text{cm/s} \) \[ I_{DS,sat,v,n} = 1.5 \times 10^{-3} \cdot 8.625 \times 10^{-7} \cdot 1.7 \cdot 4 \times 10^6 \] Stepwise: 1. \( 1.5 \times 10^{-3} \cdot 8.625 \times 10^{-7} = 1.29375 \times 10^{-9} \) 2. \( 1.29375 \times 10^{-9} \cdot 1.7 = 2.199375 \times 10^{-9} \) 3. \( 2.199375 \times 10^{-9} \cdot 4 \times 10^6 = 8.7975 \times 10^{-3}\,\text{A} = 8.80\,\text{mA} \) **PFET:** - \( W_p = 1.874 \times 10^{-3}\,\text{cm} \) - \( C_{ox,p} = 1.725 \times 10^{-6}\,\text{F/cm}^2 \) - \( V_{ov} = 1.7\,\text{V} \) - \( v_{sat} = 4 \times 10^6\,\text{cm/s} \) \[ I_{DS,sat,v,p} = 1.874 \times 10^{-3} \cdot 1.725 \times 10^{-6} \cdot 1.7 \cdot 4 \times 10^

Let's carefully solve each part step-by-step with detailed explanations. --- ## **Given Data:** ### NFET: - Oxide thickness: \( t_{ox,n} = 4\, \text{nm} = 4 \times 10^{-7}\, \text{cm} \) - Channel length: \( L = 0.2\, \mu m = 2 \times 10^{-5}\, \text{cm} \) - Width: \( W_n = 15\, \mu m = 1.5 \times 10^{-3}\, \text{cm} \) - Mobility: \( \mu_{n,lf} = 250\, \text{cm}^2/\text{V·s} \) - \( V_{GS} = 2\, \text{V} \) - \( V_{th} \) (assumed): to proceed, assume \( V_{th} \approx 0.3\, \text{V} \) ### PFET: - Mobility: \( \mu_{p,lf} = 100\, \text{cm}^2/\text{V·s} \) - Oxide thickness: \( t_{ox,p} = 2\, \text{nm} = 2 \times 10^{-7}\, \text{cm} \) - Width: \( W_p = ? \) - Channel length: same as NFET: \( 2 \times 10^{-5}\, \text{cm} \) - \( V_{GS} = 2\, \text{V} \) - \( V_{th} \) (assumed same as NFET): 0.3 V ### Additional: - Velocity saturation limit: \( v_{sat} = 4 \times 10^{6}\, \text{cm/s} \) --- ## **(a) Find \( W_p \) for the PFET to make the characteristics the same as the NFET** ### **Approach:** In the simple quadratic model, the saturation drain current: \[ I_{D,sat} = \frac{1}{2} \mu C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 \] Matching the currents (assuming same \( V_{GS} \) and \( V_{th} \)): \[ \Rightarrow W_p = W_n \times \frac{\mu_{n} C_{ox,n}}{\mu_{p} C_{ox,p}} \] ### **Calculate \( C_{ox} \) for NFET and PFET:** \[ C_{ox} = \frac{\epsilon_{ox}}{t_{ox}} \] where \( \epsilon_{ox} \approx 3.45 \times 10^{-13}\, \text{F/cm} \). - For NFET: \[ C_{ox,n} = \frac{3.45 \times 10^{-13}}{4 \times 10^{-7}} = 8.625 \times 10^{-7}\, \text{F/cm}^2 \] - For PFET: \[ C_{ox,p} = \frac{3.45 \times 10^{-13}}{2 \times 10^{-7}} = 1.725 \times 10^{-6}\, \text{F/cm}^2 \] ### **Calculate \( W_p \):** \[ W_p = W_n \times \frac{\mu_{n} C_{ox,n}}{\mu_{p} C_{ox,p}} = 1.5 \times 10^{-3} \times \frac{250 \times 8.625 \times 10^{-7}}{100 \times 1.725 \times 10^{-6}} \] Numerator: \[ 250 \times 8.625 \times 10^{-7} = 2.15625 \times 10^{-4} \] Denominator: \[ 100 \times 1.725 \times 10^{-6} = 1.725 \times 10^{-4} \] Ratio: \[ \frac{2.15625 \times 10^{-4}}{1.725 \times 10^{-4}} \approx 1.249 \] Thus: \[ W_p \approx 1.5 \times 10^{-3} \times 1.249 \approx 1.874 \times 10^{-3}\, \text{cm} \] Converting to \( \mu m \): \[ W_p \approx 1.874 \times 10^{-3} \text{cm} = 18.74\, \mu m \] --- ## **Final answer for (a):** \[ \boxed{ \text{W}_p \approx \mathbf{18.7\, \mu m} } \] --- ## **(b) Find \( V_{DS,sat} \) and \( I_{D,sat} \) for \( V_{GS} = 2\, \text{V} \)** ### **Assuming \( V_{th} = 0.3\, \text{V} \):** - Overdrive voltage: \( V_{ov} = V_{GS} - V_{th} = 2 - 0.3 = 1.7\, \text{V} \) ### **Calculate \( C_{ox} \) (already done):** \[ C_{ox,n} = 8.625 \times 10^{-7}\, \text{F/cm}^2 \] ### **Drain current in saturation:** \[ I_{D,sat} = \frac{1}{2} \mu C_{ox} \frac{W}{L} V_{ov}^2 \] Calculate the ratio \( W / L \): \[ \frac{W_n}{L} = \frac{1.5 \times 10^{-3}}{2 \times 10^{-5}} = 75 \] ### **NFET:** \[ I_{D,sat} = 0.5 \times 250 \times 8.625 \times 10^{-7} \times 75 \times (1.7)^2 \] Step-by-step: 1. \( 0.5 \times 250 = 125 \) 2. \( 125 \times 8.625 \times 10^{-7} = 1.078125 \times 10^{-4} \) 3. \( 1.078125 \times 10^{-4} \times 75 = 8.08594 \times 10^{-3} \) 4. \( (1.7)^2 = 2.89 \) Final: \[ I_{D,sat} = 8.08594 \times 10^{-3} \times 2.89 \approx 0.0234\, \text{A} = 23.4\, \text{mA} \] ### **PFET:** - \( W_p \approx 18.7\, \mu m = 1.87 \times 10^{-3}\, \text{cm} \) \[ \frac{W_p}{L} = \frac{1.87 \times 10^{-3}}{2 \times 10^{-5}} \approx 93.5 \] Calculate: \[ I_{D,sat} = 0.5 \times 100 \times 1.725 \times 10^{-6} \times 93.5 \times (1.7)^2 \] Step-by-step: 1. \( 0.5 \times 100 = 50 \) 2. \( 50 \times 1.725 \times 10^{-6} = 8.625 \times 10^{-5} \) 3. \( 8.625 \times 10^{-5} \times 93.5 \approx 8.07 \times 10^{-3} \) 4. Multiply by 2.89: \[ 8.07 \times 10^{-3} \times 2.89 \approx 0.0233\, \text{A} = 23.3\, \text{mA} \] **Result:** \[ \boxed{ V_{DS,sat} \approx 1.7\, \text{V} \quad \text{and} \quad I_{D,sat} \approx 23.4\, \text{mA} } \] --- ## **(c) Considering velocity saturation (\( v_{sat} = 4 \times 10^{6}\, \text{cm/s} \))** ### **Step 1: Find the voltage \( V_{DS} \) at velocity saturation** \[ V_{DS,sat,v} = E_{sat} \times L \] where \[ E_{sat} = \frac{v_{sat}}{\mu} \] - For NFET: \[ E_{sat,n} = \frac{4 \times 10^{6}}{250} = 1.6 \times 10^{4}\, \text{V/cm} \] \[ V_{DS,sat,v} = 1.6 \times 10^{4} \times 2 \times 10^{-5} = 0.32\, \text{V} \] - For PFET: \[ E_{sat,p} = \frac{4 \times 10^{6}}{100} = 4 \times 10^{4}\, \text{V/cm} \] \[ V_{DS,sat,v} = 4 \times 10^{4} \times 2 \times 10^{-5} = 0.8\, \text{V} \] ### **Step 2: Calculate the saturation current with velocity saturation** \[ I_{D} = W \times C_{ox} \times (V_{GS} - V_{th}) \times v_{sat} \] - **NFET:** \[ W_n = 1.5 \times 10^{-3}\, \text{cm} \] \[ C_{ox,n} = 8.625 \times 10^{-7}\, \text{F/cm}^2 \] \[ V_{ov} = 1.7\, \text{V} \] \[ I_{D,n} = 1.5 \times 10^{-3} \times 8.625 \times 10^{-7} \times 1.7 \times 4 \times 10^{6} \] Compute: 1. \( 1.5 \times 10^{-3} \times 8.625 \times 10^{-7} = 1.29375 \times 10^{-9} \) 2. \( 1.29375 \times 10^{-9} \times 1.7 = 2.199 \times 10^{-9} \) 3. \( 2.199 \times 10^{-9} \times 4 \times 10^{6} = 8.796 \times 10^{-3}\, \text{A} \) \[ \boxed{ I_{D,n} \approx 8.80\, \text{mA} } \] - **PFET:** \[ W_p \approx 18.7\, \mu m = 1.87 \times 10^{-3}\, \text{cm} \] \[ C_{ox,p} = 1.725 \times 10^{-6}\, \text{F/cm}^2 \] \[ V_{ov} = 1.7\, \text{V} \] \[ I_{D,p} = 1.87 \times 10^{-3} \times 1.725 \times 10^{-6} \times 1.7 \times 4 \times 10^{6} \] Compute: 1. \( 1.87 \times 10^{-3} \times 1.725 \times 10^{-6} = 3.226 \times 10^{-9} \) 2. \( 3.226 \times 10^{-9} \times 1.7 = 5.486 \times 10^{-9} \) 3. \( 5.486 \times 10^{-9} \times 4 \times 10^{6} \approx 21.944 \times 10^{-3} = 21.944\, \text{mA} \) --- ## **Final summarized answers:** | Part | NFET \( V_{DS} \) | NFET \( I_D \) | PFET \( V_{DS} \) | PFET \( I_D \) | |---------|---------------------|-------------------|---------------------|-------------------| | (b) | ~1.7 V | ~23.4 mA | ~1.7 V | ~23.4 mA | | (c) (velocity saturation) | 0.32 V | ~8.8 mA | 0.8 V | ~22 mA | --- **Summary:** - **(a)** \( W_p \approx 18.7\, \mu m \) - **(b)** \( V_{DS,sat} \approx 1.7\, \text{V} \), \( I_{D,sat} \approx 23.4\, \text{mA} \) - **(c)** With velocity saturation, \( V_{DS} \) drops, and \( I_D \) reduces accordingly. --- If you have any specific clarifications or data (like exact \( V_{th} \)), please include them!