Answer:
The minimum permitted output voltage is 0.5V or 2V_{OV}
Explanation:
The transistor is as indicated in the attached figure.
From the data
[tex]\mu_nC_{ox}=350 \mu A/V\\V_{tn}=0.5V\\g_{m1}=2mA/V\\R_{out}=200k \Omega \\V_{An}'=7.5\\L_{min}=0.18 \mu m\\L_{min}<L<3.5L_{min}\\V_{OV}=0.25 V[/tex]
Now as both the transistors are identical as NMOS and thus
[tex]g_{m1}=g_{m2}=g_m\\r_{o1}=r_{o2}=r_o[/tex]
Due to these properties
[tex]R_{out}=g_{m}r_o^2\\r_o=\sqrt{\frac{R_{out}}{g_{m}}}\\r_o=\sqrt{\frac{200*1000}{2/1000}}\\r_o=10\times 10^3 \Omega\\r_o=10k\Omega\\[/tex]
[tex]I_D=\frac{V_{OV}\times g_m}{2}\\I_D=\frac{0.25\times 0.002}{2}\\I_D=250 \mu A[/tex]
Also
[tex]L=\frac{I_D\times V_{OV}}{r_o}\\L=\frac{250 \mu A \times 10 k\Omega}{5V/\mu}\\L=0.5 \mu m[/tex]
Now W/L is given as
[tex]W/L=\frac{g_m^2}{2\mu_nC_{ox}I_D}\\W/L=\frac{(2/1000)^2}{2*350*10^{-6}*250*10^{-6}}\\W/L=160/7[/tex]
Now in order to obtain the maximum negative swing at the output, V_G is selected such that the voltage at the drain of Q_1 is the maximum permitted which is given as [tex]V_{OV}=0.25V[/tex]
[tex]V_G=0.25+V_{OV}+V_t\\V_G=0.25+0.25+0.5\\V_G=1 V\\[/tex]
The minimum permitted output is
[tex]V_O_{min}=V_G-V_t\\V_O_{min}=1-0.5\\V_O_{min}=0.5V\\V_O_{min}=2V_{OV}\\[/tex]
So the minimum permitted output voltage is 0.5V or 2V_{OV}
