**Switching threshold of CMOS inverter**

**Question: Show that the Switching threshold of cmos inverter is given by** V_M=\frac{V_{DD}-\left|V_{tp}\right|+V_{tn}\sqrt{\displaystyle\frac{\beta_n}{\beta_p}}}{1+\sqrt{\frac{\beta_n}{\beta_p}}}

Solution:

The switching threshold V_{M} also called the midpoint voltage is the point where the input voltage is equal to the output voltage(V_{in}=V_{out}) at V_{M} both nMOS and pMOS is in the** saturation region.**

The saturation current for a MOSFET is given by I_D=\frac{\mu c_{ox\;}W}{2L}(V_{GS}-V_t)^2

For nMOS

I_{Dn}=\frac{\mu_nc_{ox\;}W}{2L}(V_{GSn}-V_{tn})^2

For pMOS

I_{Dp}=\frac{\mu_pc_{ox\;}W}{2L}(V_{GSp}-V_{tp})^2

For an inverter the two currents are equals. so we get

\begin{array}{l}I_{Dn}=I_{Dp}\\\\\frac{\mu_nc_{ox}W}{2L}(G_{GSn}-V_{tn})^2=\frac{\mu_pc_{ox}W}{2L}(G_{GSp}-V_{tp})^2\\\\\beta_n(G_{GSn}-V_{tn})^2=\beta_p(G_{GSp}-V_{tp})^2............(i)\\\end{array}

Again

\begin{array}{l}\\G_{GSn}=V_{in}=V_M\\\\G_{GSp}=V_{DD}-V_{in}=V_{DD}-V_M\\\\\\\end{array}

Putting these values in equation (i) we get

\begin{array}{l}\beta_n(V_M-V_{tn})^2=\beta_p(V_{DD}-V_M-\left|V_{tp}\right|)^2\\\\\\\sqrt{\frac{\beta_n}{\beta_p}}(V_M-V_{tn})=(V_{DD}-V_M-\left|V_{tp}\right|)\\\\\\\sqrt{\frac{\beta_n}{\beta_p}}V_M-\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}=V_{DD}-V_M-\left|V_{tp}\right|\\\\\\\sqrt{\frac{\beta_n}{\beta_p}}V_M+V_M=V_{DD}-\left|V_{tp}\right|+\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}\\\\V_M(1+\sqrt{\frac{\beta_n}{\beta_p}})=V_{DD}-\left|V_{tp}\right|+\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}\\\\V_M=\frac{V_{DD}-\left|V_{tp}\right|+V_{tn}\sqrt{\frac{\beta_n}{\beta_p}}}{1+\sqrt{\frac{\beta_n}{\beta_p}}}(Showed)\\\\\end{array}

**Read: Switching threshold ofÂ 2 input NAND gate**