Q. Consider a resistive load inv ckt with $V_{DD} = 5V$, $K_N^{'} = 20 \mu A/V^2$ , $V_{TO} = 0.8V$, $R_L = 200 K$, W/L = 2. Calculate the critical volatge, $V_{IL}, V_{IH}, V_{OL}, V_{OH}$ on $V_{TC}$ & Find NM.

Sol. 1) $V_{OH} = V_{DD} = 5V$

2) $V_{IL} = V_{TOD} + \frac{1}{K_nR_L}$

$K_n = K_N^{'}(W/L) = 40 \mu A/V^2$

$V_{IL} = 0.8 + \frac{1}{40*10^{-6}*200*10^3}$ = 0.925 V

3) $V_{OL} = 0.147 V$

4) $V_{IH} = 1.97 V$

5) $NM_H = V_{OH} - V_{IH}$ = 5 - 1.97 = 3.03 V

6) $NM_L = V_{IL} - V_{OL} = 0.925 - 0.147 = 0.778 V$

At least 25% of total power supply

Q. Calculate the NM of CMOS inv with $V_{DD} = 5V, V_{TOD} = 1V, V_{TOP} = -1V, K_n = 200 \mu A/V^2, K_P = 200 \mu A/V^2$.

Sol. 1) $V_{OH} = V_{DD}$

Since $V_{in} = 0$ nmos cutoff

pmos conduction

$V_{out} = V_{OH} = V_{DD} = 5V$

2) $V_{OL} = 0$

Since when $V_{in} = V_{DD} = 5V$

pmos is cutoff & nmos conducting

=> $V_{out} = 0V = V_{OL}$

3) $V_{IL} = \frac{2V_{out} + V_{TOP} - V_{DD} + K_RV_{TOD}}{1 + K_R}$

$V_{IL} = 2.125 V$

4) $V_{IH} = \frac{1}{8}[5V_{DD} - 2V_{TOD}]$

$V_{IH} = 2.875 V$

5) Since it is symmetry,

$V_{IH} + V_{IL} = V_{DD}$

i.e. 2.125 + 2.8 + 5 =5V

6) $NM_H = V_{OH} - V_{IH} = 5 - 2.875 = 2.125 V$

$NM_L = V_{IL} - V_{OL} = 5 - 2.875 = 2.125 V$

Q. Consider CMOS inverter with following parameters NMOS $V_{TOD} = 0.6V, U_nC_{OX} = 60 \mu A/V^2$ & $(W/L)_n = 8$. PMOS $V_{TOP} = -0.7V, K_P = 25 \mu A/V^2$ & $(W/L)_n = 12$. Calculate the noise margin and switching threshold ($V_{TH}$) of this ckt, $V_{DD} = 3V$

Sol 1) $V_{OH} = V_{DD} = 3V$

2) $V_{OL} = 0V$