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Unit 2
MOS Inverter Static Characterstics
Noise Margin Definition
1) High Noise Margin = $NM_H = V_{OH} - V_{IH}$
2) Low Noise Margin = $NM_L = V_{IL} - V_{OL}$
$V_{OH}$ = Max O/P voltage when O/P level is logic '1'
$V_{OL}$ = Min O/P voltage when O/P level is logic '0'
$V_{IL}$ = Max I/P voltage which can be interpreted as logic 0
$V_{IH}$ = Min I/P voltage which can be interpreted as logic 1
$V_{out}$ = f($V_{in}$)
$V_{out}^{'}$ = f($V_{in} + \Lambda V_{noise}$)
$V_{out}^{'}$ = f($V_{in}) + \frac{d}{dv_{in}}V_{out} V_{noise}$ + Higher order terms neglected
Resistive type nMOS Inverter :
Assumption - 1) Channel length modulation effect is neglected
2) $V_{SB} = 0$ i.e No barriers => NMOS $V_T$ is always $V_{TO}$
$I_{DS} = \frac{kn}{2} (V_{in} - V_{TO})^2$
$I_{DS} = \frac{kn}{2} [2(V_{in} - V_{TO})(V_{out})-V_{out}^2]$
Calculation of critical parameter :
1) Calculation of $V_{OH}$
When $V_{in} = 0, T_N$ is off,
we know that
$V_{out} = V_{DD} - I_RR_L$
$V_{out} = V_{DD} = V_{OH}$
=> $V_{OH} = V_{DD}$
2) Calculation of $V_{OL}$
When $V_{in} = V_{DD}$
and $V_{out} = = V_{OL}$
$I_R = \frac{V_{DD} - V_{out}}{R_L}$
But $I_R = I_D$
$\frac{V_{DD} - V_{out}}{R_L} = \frac{Kn}{2}[(V_{gs} - V_{TO})V_{DS} - V_{DS}^2]$
$\frac{V_{DD} - V_{OL}}{R_L} = \frac{Kn}{2}[(V_{in} - V_{TO})V_{OL} - V_{OL}^2]$
Quadratic equation in 1/OL correct sol is
$V_{OL} = V_{DD} - V_{TO} + \frac{1}{K_nR_L} - ((V_{DD} - V_{TO} + \frac{1}{K_nR_L})^2 - \frac{2V_{OD}}{K_nR_L})^{1/2}$
- Calculation of $V_{IL}$
When $V_{in} \gt V_T$ $I_N$ is in saturation
=> $V_{DS} \gt V_{gs} - V_{TO}$ or $V_{out} \gt V_{in} - V_{TO}$
$I_{DS} = \frac{Kn}{2}[V_{in} - V_{TO}]^2$
$I_R = \frac{V_{DD} - V_{out}}{R_L}$
=> $I_D = I_R$
$\frac{Kn}{2}[V_{in} - V_{TO}]^2 = \frac{V_{DD} - V_{out}}{R_L}$
At $V_{IC}$, $\frac{dV_{out}}{dV_{in} = 1}$
$V_{tc} = \frac{1}{K_nR_L} + V_{TO}$
- Claculation of $V_{IH}$
While $V_{in}\gtV_{out - V_{TO}}$ $T_N$ is in linear
$V_{out} \lt V_{in} - V_{TO}$
$I_R = I_D$
$\frac{V_{DD} - V_{out}}{2} = \frac{Kn}{2}[2(V_{in} - V_{TO})V_{out} - V_{out}^2]$
D.W.R.T. $V_{in}$
=> $V_{IH} = 2V_{out} + V_{TO} - \frac{1}{K_nR_L}$ ----------------(2)
Substitute $V_{IH}$ in (1) to get,
$V_{out} (at V_{IH}) = (\frac{2V_{DD}}{3K_nR_L})^{1/2}$ -------------------(3)
(3) in (2)
$V_{IH} = V_{TO} + (\frac{2 V_{DD}}{3K_nR_L})^{1/2} - \frac{1}{K_nR_L}$