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Please give me the below content same it is without any changes Got it — I’ll use your solution and present the same answers clearly and compactly. --- ## Given * (V_{DD}=1\ \text{V}) * (V_{tn}=0.35\ \text{V},; V_{tp}=-0.35\ \text{V}) * (\mu_n C_{ox}=540\ \mu\text{A}/\text{V}^2,; \mu_p C_{ox}=100\ \mu\text{A}/\text{V}^2) * ((W/L)_n=1.5,; L_n=L_p=65\ \text{nm}) --- ## (a) (W_p) for (V_M=V_{DD}/2) and silicon area 1. ((W/L)*p = \dfrac{\mu_n C*{ox}}{\mu_p C_{ox}}(W/L)_n = \dfrac{540}{100}\times 1.5 = 8.1). 2. (W_p = 8.1 \times 65\ \text{nm} = 526.5\ \text{nm}). 3. (W_n = 1.5 \times 65\ \text{nm} = 97.5\ \text{nm}). 4. Areas: * Area(_n) = (97.5\ \text{nm}\times 65\ \text{nm}=6337.5\ \text{nm}^2) * Area(_p) = (526.5\ \text{nm}\times 65\ \text{nm}=34222.5\ \text{nm}^2) * **Total silicon area** = (6337.5+34222.5=40560\ \text{nm}^2). --- ## (b) (V_{OH},V_{OL},V_{IH},V_{IL},NM_L,NM_H) * (V_{OH}=V_{DD}=1.00\ \text{V}) * (V_{OL}=0.00\ \text{V}) For the matched symmetric inverter ((V_M=0.5) V), solving the (dV_O/dV_I=-1) condition (formulas given) yields the pair (correctly assigned): * (V_{IL}=0.382\ \text{V}) * (V_{IH}=0.618\ \text{V}) Noise margins: * (NM_L = V_{IL}-V_{OL} = 0.382\ \text{V}) * (NM_H = V_{OH}-V_{IH} = 0.382\ \text{V}) --- ## (c) Output resistances (R_{OH}) and (R_{OL}) Use (R_{on}\approx \big[k'(W/L)(V_{GS}-V_t)\big]^{-1}) with the triode small-(V_{DS}) approximation: * (k'_n(W/L)_n = 540\times 1.5 = 810\ \mu\text{A}/\text{V}^2) * (k'_p(W/L)_p = 100\times 8.1 = 810\ \mu\text{A}/\text{V}^2) * (V_{GS}-V_t = 1-0.35 = 0.65\ \text{V}) So [ R_{OH} = R_{OL} = \frac{1}{810\times10^{-6}\times0.65}\approx 1900.09\ \Omega \approx \mathbf{1.90\ \text{k}\Omega}. ] --- ## Final summary (same as your solution) | Quantity | Value | | -----------------: | :--------------------- | | (W_p) | (526.5\ \text{nm}) | | Total silicon area | (40560\ \text{nm}^2) | | (V_{OH}) | (1.00\ \text{V}) | | (V_{OL}) | (0.00\ \text{V}) | | (V_{IL}) | (0.382\ \text{V}) | | (V_{IH}) | (0.618\ \text{V}) | | (NM_L) | (0.382\ \text{V}) | | (NM_H) | (0.382\ \text{V}) | | (R_{OH}) | (1.90\ \text{k}\Omega) | | (R_{OL}) | (1.90\ \text{k}\Omega) | ---

Question:

Please give me the below content same it is without any changes Got it — I’ll use your solution and present the same answers clearly and compactly. --- ## Given * (V_{DD}=1\ \text{V}) * (V_{tn}=0.35\ \text{V},; V_{tp}=-0.35\ \text{V}) * (\mu_n C_{ox}=540\ \mu\text{A}/\text{V}^2,; \mu_p C_{ox}=100\ \mu\text{A}/\text{V}^2) * ((W/L)_n=1.5,; L_n=L_p=65\ \text{nm}) --- ## (a) (W_p) for (V_M=V_{DD}/2) and silicon area 1. ((W/L)*p = \dfrac{\mu_n C*{ox}}{\mu_p C_{ox}}(W/L)_n = \dfrac{540}{100}\times 1.5 = 8.1). 2. (W_p = 8.1 \times 65\ \text{nm} = 526.5\ \text{nm}). 3. (W_n = 1.5 \times 65\ \text{nm} = 97.5\ \text{nm}). 4. Areas: * Area(_n) = (97.5\ \text{nm}\times 65\ \text{nm}=6337.5\ \text{nm}^2) * Area(_p) = (526.5\ \text{nm}\times 65\ \text{nm}=34222.5\ \text{nm}^2) * **Total silicon area** = (6337.5+34222.5=40560\ \text{nm}^2). --- ## (b) (V_{OH},V_{OL},V_{IH},V_{IL},NM_L,NM_H) * (V_{OH}=V_{DD}=1.00\ \text{V}) * (V_{OL}=0.00\ \text{V}) For the matched symmetric inverter ((V_M=0.5) V), solving the (dV_O/dV_I=-1) condition (formulas given) yields the pair (correctly assigned): * (V_{IL}=0.382\ \text{V}) * (V_{IH}=0.618\ \text{V}) Noise margins: * (NM_L = V_{IL}-V_{OL} = 0.382\ \text{V}) * (NM_H = V_{OH}-V_{IH} = 0.382\ \text{V}) --- ## (c) Output resistances (R_{OH}) and (R_{OL}) Use (R_{on}\approx \big[k'(W/L)(V_{GS}-V_t)\big]^{-1}) with the triode small-(V_{DS}) approximation: * (k'_n(W/L)_n = 540\times 1.5 = 810\ \mu\text{A}/\text{V}^2) * (k'_p(W/L)_p = 100\times 8.1 = 810\ \mu\text{A}/\text{V}^2) * (V_{GS}-V_t = 1-0.35 = 0.65\ \text{V}) So [ R_{OH} = R_{OL} = \frac{1}{810\times10^{-6}\times0.65}\approx 1900.09\ \Omega \approx \mathbf{1.90\ \text{k}\Omega}. ] --- ## Final summary (same as your solution) | Quantity | Value | | -----------------: | :--------------------- | | (W_p) | (526.5\ \text{nm}) | | Total silicon area | (40560\ \text{nm}^2) | | (V_{OH}) | (1.00\ \text{V}) | | (V_{OL}) | (0.00\ \text{V}) | | (V_{IL}) | (0.382\ \text{V}) | | (V_{IH}) | (0.618\ \text{V}) | | (NM_L) | (0.382\ \text{V}) | | (NM_H) | (0.382\ \text{V}) | | (R_{OH}) | (1.90\ \text{k}\Omega) | | (R_{OL}) | (1.90\ \text{k}\Omega) | ---

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Asked by: Jack
Created at: 2025-10-21 12:09:30

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# Given * \( V_{DD} = 1\ \text{V} \) * \( V_{tn} = .35\ \text{V} \), \( V_{tp} = -.35\ \text{V} \) * \( \mu_n C_{ox} = 540\ \mu\text{A}/\text{V}^2 \), \( \mu_p C_{ox} = 100\ \mu\text{A}/\text{V}^2 \) * \( (W/L)_n = 1.5 \), \( L_n = L_p = 65\ \text{nm} \) --- ## (a) \(W_p\) for \(V_M = V_{DD}/2\) and silicon area 1. Calculate \( (W/L)_p \): \[ (W/L)_p = \frac{\mu_n C_{ox}}{\mu_p C_{ox}} (W/L)_n = \frac{540}{100} \times 1.5 = 8.1 \] 2. Determine \( W_p \): \[ W_p = 8.1 \times 65\ \text{nm} = 526.5\ \text{nm} \] 3. Determine \( W_n \): \[ W_n = (W/L)_n \times 65\ \text{nm} = 1.5 \times 65\ \text{nm} = 97.5\ \text{nm} \] 4. Areas: * \( \text{Area}_n = 97.5\ \text{nm} \times 65\ \text{nm} = 6337.5\ \text{nm}^2 \) * \( \text{Area}_p = 526.5\ \text{nm} \times 65\ \text{nm} = 34222.5\ \text{nm}^2 \) * **Total silicon area**: \[ 6337.5 + 34222.5 = 40560\ \text{nm}^2 \] --- ## (b) \( V_{OL}, V_{OH}, V_{IL}, V_{IH}, NM_L, NM_H \) - \( V_{OH} = V_{DD} = 1.00\ \text{V} \) - \( V_{OL} = .00\ \text{V} \) For the symmetric inverter with \( V_M = .5\ \text{V} \), solving the \( \frac{dV_O}{dV_I} = -1 \) condition yields: * \( V_{IL} = .382\ \text{V} \) * \( V_{IH} = .618\ \text{V} \) Noise margins: * \( NM_L = V_{IL} - V_{OL} = .382\ \text{V} \) * \( NM_H = V_{OH} - V_{IH} = .382\ \text{V} \) --- ## (c) Output resistances \( R_{OH} \) and \( R_{OL} \) Using the approximation: \[ R_{on} \approx \left[ k' (W/L) (V_{GS} - V_t) \right]^{-1} \] with small \( V_{DS} \) in triode region. Calculate: \[ k'_n(W/L)_n = 540 \times 1.5 = 810\ \mu\text{A}/\text{V}^2 \] \[ k'_p(W/L)_p = 100 \times 8.1 = 810\ \mu\text{A}/\text{V}^2 \] \[ V_{GS} - V_t = 1 - .35 = .65\ \text{V} \] Thus: \[ R_{OH} = R_{OL} = \frac{1}{810 \times 10^{-6} \times .65} \approx 190.09\ \Omega \approx \mathbf{1.90\ \text{k}\Omega} \] --- ## Final summary | Quantity | Value | |---------------------:|:------------------------------| | \( W_p \) | \( 526.5\ \text{nm} \) | | Total silicon area| \( 40560\ \text{nm}^2 \) | | \( V_{OH} \) | \( 1.00\ \text{V} \) | | \( V_{OL} \) | \( .00\ \text{V} \) | | \( V_{IL} \) | \( .382\ \text{V} \) | | \( V_{IH} \) | \( .618\ \text{V} \) | | \( NM_L \) | \( .382\ \text{V} \) | | \( NM_H \) | \( .382\ \text{V} \) | | \( R_{OH} \) | \( 1.90\ \text{k}\Omega \) | | \( R_{OL} \) | \( 1.90\ \text{k}\Omega \) |

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