CMOS Inverter

Find out $V_{TH}$,

When $V_{in} = V_{TH}$

i.e. region c of operation of CMOS inverter

Both pmos and nmos transistors are in saturation

$I_L(sat) = I_D(sat)$

$\frac{K_P}{2}[V_{in} - V_{DD} - V_{TP}]^2 = \frac{K_n}{2}[V_{in} - V_{TD}]^2$

$V_{in}(1 - (\frac{K_P}{K_n})^{1/2}) = V_{Th} + (\frac{K_P}{K_n})^{1/2} - V_{DD} - V_{Tp}$

substitute $V_{in} = V_{Th}$

=> $V_{Th} = \frac{V_{TOD} + (\frac{1}{K_R})^{1/2}(V_{DD} + V_{TOP})}{(1 + (\frac{1}{K_R})^{1/2})}$

if $K_P = K_n$ and $V_{ToD} = |V_{Top}|$,

$V_{Th} = \frac{V_{ToD} + V_{DD} + V_{Top}}{2}$

$V_{Th} = \frac{V_{DD}}{2}$

Design of CMOS inverter:

$V_{Th} = \frac{V_{TOD} + (\frac{1}{K_R}(V_{DD} + V_{TOD}))^{1/2}}{1 + (\frac{1}{K_R})^{1/2}}$

Solve if for $K_R$ to get the desired values of V_{Th}

$K_R = \frac{K_n}{K_P}(\frac{V_{DD} + V_{TOD} - V_{Th}}{V_{Th} - V_{TOD}})^2$ -------------(1)

But for ideal inverter $K_R$ = 1 & $V_{Th} = \frac{V_{DD}}{2}$

$K_R = (\frac{K_n}{K_P})_{ideal} = (\frac{V_{DD} + V_{TOP} - V_{DD}/2}{\frac{V_{DD}}{2} - V_{TOD}})^2$

for a symmetric VTC,

Then it reduces to

($K_R = \frac{K_D}{K_P}$) = 1

i.e. $\frac{\mu_n C_{OX}}{\mu_p C_{OX}} = \frac{(W/L)_P}{(W/P)_n}$

if $C_{OX}$ of pmos = $C_{OX} of nmos$

$(W/L)_P = 25(W/L)_n$

For a symmetric CMOS inverter :

$V_{IL} = \frac{1}{8}[3V_{DD} + 2V_{TOD}]$

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

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

$NM_L = V_{IL} - V_{OL} = V_{IL} - 0 = V_{IL}$

$NM_H = V_{OH} - V_{IH} = V_{DD} - V_{IH}$

Asymmetric CMOS inverter :

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

2. $V_{OL} = 0$

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

4. $V_{IH} = \frac{K_R(2V_{out} + V_{TOP}) + V_{DD} + V_{TOP}}{1 + K_R}$

5. $V_{TH} = \frac{V_{TOD} + (1/K_R)^{1/2}(V_{DD} + V_{TOD})}{(1 + (1/K_R)^{1/2})}$

6. For Symm.

inv $V_{TH} = \frac{V_{OD}}{2}$

For Resistive load nmos Inverter :

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

2. $V_{IH} = V_{TO} + (\frac{8V_{DD}}{3K_nR_L})^{1/2} - \frac{1}{K_nR_L}$

3. $V_{OL} = V_{DD} - V_{TOD} + \frac{1}{K_nR_L} - ((V_{DD} - V_{TO} + \frac{1}{K_nR_L})^2 -\frac{2V_{DD}}{K_nR_L})^{1/2}$

4. $V_{OH} = V_{DD}$