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Answer 1

Consider a CMOS Inveter circuit i,e Combination of both PMOS and NMOS

Assume NMOS is On(Saturated),PMOS is linear

Then Drain current expressed as follows

$I_{d}=\frac{\mu C_{0x}}{2}\frac{W}{L}\left [ (V_{GS}-V_{t})V_{DS}-\frac{V_{DS}^{2}}{2} \right ]$

$I_{d}=K\left [ (V_{GS}-V_{t})V_{DS}-\frac{V_{DS}^{2}}{2} \right ]$

which can be modified as follows for above assumed conditions for both N and P MOS as follows

$I_{d}=K_{p}\left [ \left ( V_{dd}-V_{i} -V_{tp} \right )\left ( V_{dd}-V_{o} \right )-\frac{1}{2}\left ( V_{dd}-V_{o} \right )^{2}\right ]=\frac{K_{n}}{2}\left ( V_{i}-V_{tn} \right )^{2}$

let

$V_{dp}=\left ( V_{dd}-V_{o} \right )$

$\beta =\frac{K_{n}}{K_{p}}$

$\Rightarrow \left [ \left ( V_{dd}-V_{i} -V_{tp} \right )V_{dp} -\frac{1}{2}V_{dp}^{2}\right ]=\frac{K_{n}}{2K_{p}}\left ( V_{i}-V_{tn} \right )^{2}$

$\Rightarrow \left [ \left ( V_{dd}-V_{i} -V_{tp} \right )V_{dp} -\frac{1}{2}V_{dp}^{2}\right ]=\frac{\beta }{2}\left ( V_{i}-V_{tn} \right )^{2}$

$\Rightarrow \frac{1}{2}V_{dp}^{2}-\left ( V_{dd}-V_{i} -V_{tp} \right )V_{dp} +\frac{\beta }{2}\left ( V_{i}-V_{tn} \right )^{2}=0$

Obtaining solution for above equation

$\alpha _{1},\alpha _{2}=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\alpha _{1},\alpha _{2}=\frac{\left ( V_{dd}-V_{i} -V_{tp} \right )\pm \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-4*\frac{1}{2}*\frac{\beta }{2}\left ( V_{i}-V_{tn} \right )^{2}}}{2*\frac{1}{2}}$

$\alpha _{1},\alpha _{2}=\left ( V_{dd}-V_{i} -V_{tp} \right )\pm \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-\beta \left ( V_{i}-V_{tn} \right )^{2}}$

$V_{dp}=\left ( V_{dd}-V_{i} -V_{tp} \right )- \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-\beta \left ( V_{i}-V_{tn} \right )^{2}}$

$V_{dd}-V_{0}=\left ( V_{dd}-V_{i} -V_{tp} \right )- \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-\beta \left ( V_{i}-V_{tn} \right )^{2}}$

$\Rightarrow V_{0}=V_{dd}-\left ( V_{dd}-V_{i} -V_{tp} \right )+\sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-\beta \left ( V_{i}-V_{tn} \right )^{2}}$

$\Rightarrow V_{0}=V_{i} +V_{tp} + \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}-\beta \left ( V_{i}-V_{tn} \right )^{2}}$

If $K_{n}=K_{p}=K$$\Rightarrow \beta =1$

$\Rightarrow V_{0}=V_{i} +V_{tp} + \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}- \left ( V_{i}-V_{tn} \right )^{2}}$

Finding V_OH and V_iL

$\Rightarrow V_{0}=V_{i} +V_{tp} + \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}- \left ( V_{i}-V_{tn} \right )^{2}}$

$\Rightarrow V_{0}=V_{i} +V_{tp} + \sqrt{\left ( V_{dd}-V_{i} -V_{tp}+V_{i}-V_{tn}\right )\left ( V_{dd}-V_{i} -V_{tp}- \left ( V_{i}-V_{tn} \right )\right )}$

$\Rightarrow V_{0}=V_{i} +V_{tp} + \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )}\rightarrow 1$

we Know that

$\frac{\partial V_{0}}{\partial V_{in}}=-1$

$\frac{\partial\left (V_{i} +V_{tp} + \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )} \right )}{\partial V_{in}}=-1$

$\Rightarrow 1+0+\frac{\partial\left ( \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )} \right )}{\partial V_{in}}=-1$

$\Rightarrow \frac{ 1 }{2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )}}*\left ( V_{dd} -V_{tp}-V_{tn}\right )(-2)=-2$

$\Rightarrow \frac{ 1 }{2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )}}*\left ( V_{dd} -V_{tp}-V_{tn}\right )=1$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )=2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )}$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )^{2}=4\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )=4\left ( V_{dd}-2V_{i} -V_{tp}+V_{tn} \right )$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )=\left (4 V_{dd}-8V_{i} -4V_{tp}+4V_{tn} \right )$

$\Rightarrow 8V_{i}=\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )$

$\Rightarrow V_{iL}=\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}\rightarrow 2$

Substitute eqn 2 in 1 we will get

$\Rightarrow V_{0H}=\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}+V_{tp} + \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-2(\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}) -V_{tp}+V_{tn} \right )}$$\Rightarrow V_{0H}=\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}+V_{tp} + \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( V_{dd}-\frac{3 V_{dd} -3V_{tp}+5V_{tn} }{4} -V_{tp}+V_{tn} \right )}$$\Rightarrow V_{0H}=\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}+V_{tp} + \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( \frac{4V_{dd}-3 V_{dd} +3V_{tp}-5V_{tn}-4V_{tp}+4V_{tn} }{4} \right )}$

$\Rightarrow V_{0H}=\frac{\left (3 V_{dd} -3V_{tp}+5V_{tn} \right )}{8}+V_{tp} + V_{dd} -V_{tp}-V_{tn}$

Finding V_OL and V_iH

$\Rightarrow V_{0}=V_{i} -V_{tn}- \sqrt{\left ( V_{dd}-V_{i} -V_{tp} \right )^{2}- \left ( V_{i}-V_{tn} \right )^{2}}$

$\Rightarrow V_{0}=V_{i} -V_{tn}- \sqrt{\left ( V_{dd}-V_{i} -V_{tp}+V_{i}-V_{tn}\right )\left ( V_{dd}-V_{i} -V_{tp}- \left ( V_{i}-V_{tn} \right )\right )}$

$\Rightarrow V_{0}=V_{i} -V_{tn} -\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )}\rightarrow 3$

we Know that

$\frac{\partial V_{0}}{\partial V_{in}}=-1$

$\frac{\partial\left (V_{i} -V_{tn} -\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )} \right )}{\partial V_{in}}=-1$

$1-0-\frac{\partial\left ( \sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )} \right )}{\partial V_{in}}=-1$

$\Rightarrow \frac{ -1 }{2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )}}*\left ( V_{dd} -V_{tp}-V_{tn}\right )(2)=-2$

$\Rightarrow \frac{ -1 }{2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )}}*\left ( V_{dd} -V_{tp}-V_{tn}\right )=-1$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )= 2\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )}$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )^{2}= 4\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )=4\left ( -V_{dd}+2V_{i} +V_{tp}-V_{tn} \right )$

$\Rightarrow \left ( V_{dd} -V_{tp}-V_{tn}\right )=\left ( -4V_{dd}+8V_{i} +4V_{tp}-4V_{tn} \right )$

$\Rightarrow \left ( 5V_{dd} -5V_{tp}+3V_{tn}\right )=8V_{iH}$

$\Rightarrow V_{iH}=\frac{\left ( 5V_{dd} -5V_{tp}+3V_{tn}\right )}{8} \rightarrow 4$

Substitute equation4 in equation3

$\Rightarrow V_{0L}=\frac{\left ( 5V_{dd} -5V_{tp}+3V_{tn}\right )}{8} -V_{tn} -\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right )\left ( -V_{dd}+2\left ( \frac{\left ( 5V_{dd} -5V_{tp}+3V_{tn}\right )}{8} \right )+V_{tp}-V_{tn} \right )}$

$\Rightarrow V_{0L}=\frac{\left ( 5V_{dd} -5V_{tp}+3V_{tn}\right )}{8} -V_{tn} -\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right ) \left ( \frac{ -4V_{dd}+5V_{dd} -5V_{tp}+3V_{tn}+4V_{tp}-4V_{tn}}{4} \right ) }$

$\Rightarrow V_{0L}=\frac{\left ( 5V_{dd} -5V_{tp}-5V_{tn}\right )}{8} -\sqrt{\left ( V_{dd} -V_{tp}-V_{tn}\right ) \left ( \frac{ -4V_{dd}+5V_{dd} -5V_{tp}+3V_{tn}+4V_{tp}-4V_{tn}}{4} \right ) }$$\Rightarrow V_{0L}=\frac{\left ( 5V_{dd} -5V_{tp}-5V_{tn}\right )}{8} -\left ( V_{dd} -V_{tp}-V_{tn}\right )$

$\Rightarrow V_{0L}=\frac{\left ( 5V_{dd} -5V_{tp}-5V_{tn}\right )-8\left ( V_{dd} -V_{tp}-V_{tn}\right )}{8}$

$\Rightarrow V_{0L}=\frac{\left ( -3V_{dd} +3V_{tp}+3V_{tn}\right )}{8}$