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switching threshold of 2 input nand gate

Switching threshold of  2 input NAND gate

 

Question: Show that the Switching threshold of 2 input Nand gate is given by V_m=\frac{V_{DD}-\left|V_{tp}\right|+V_{tn}{\displaystyle\frac12}\sqrt{\displaystyle\frac{\beta_n}{\beta_p}}}{1+\frac12\sqrt{\frac{\beta_n}{\beta_p}}}

 

Solution:

switching threshold of 2 input nand gate

 

Voltage Transfer Characteristics vary with the transition. So VM also varies with the transition.

Say all tx have the same length L thus

VM=VA=VB=Vout

If the width of both PMOS and NMOS are the same WpA=WpB  and WnA=WnB

For a series NMOS we get  \beta_n=\frac12\beta_n

and For a parallel PMOS we get \beta_p=\frac12\beta_p

The Saturation current for NMOS is I_{Dn}=\frac{\mu_nc_{ox}W}{2L}(G_{gsn}-V_{tn})^2

and for PMOS is I_{Dp}=\frac{\mu_pc_{ox}W}{2L}(G_{gsp}-V_{tp})^2

Equating the two current 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\\\\\frac12\beta_n(G_{gsn}-V_{tn})^2=2\beta_p(G_{gsp}-\left|V_{tp}\right|)^2............(i)\\\end{array}

Again

\begin{array}{l}\\G_{gsn}=V_M\\\\G_{gsp}=V_{DD}-V_M\\\\\\\end{array}

 

Putting these values in equation (i) we get

\begin{array}{l}\frac12\beta_n(V_M-V_{tn})^2=2\beta_p(V_{DD}-V_M-\left|V_{tp}\right|)^2\\\\\frac14\frac{\beta_n}{\beta_p}(V_M-V_{tn})^2=(V_{DD}-V_M-\left|V_{tp}\right|)^2\\\\\frac12\sqrt{\frac{\beta_n}{\beta_p}}(V_M-V_{tn})=(V_{DD}-V_M-\left|V_{tp}\right|)\\\\\frac12\sqrt{\frac{\beta_n}{\beta_p}}V_M-\frac12\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}=V_{DD}-V_M-\left|V_{tp}\right|\\\\\frac12\sqrt{\frac{\beta_n}{\beta_p}}V_M+V_M=V_{DD}-\left|V_{tp}\right|+\frac12\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}\\\\V_M(1+\frac12\sqrt{\frac{\beta_n}{\beta_p}})=V_{DD}-\left|V_{tp}\right|+\frac12\sqrt{\frac{\beta_n}{\beta_p}}V_{tn}\\\\V_M=\frac{V_{DD}-\left|V_{tp}\right|+\frac12V_{tn}\sqrt{\frac{\beta_n}{\beta_p}}}{1+\frac12\sqrt{\frac{\beta_n}{\beta_p}}}(Showed)\\\\for\;N-Input\\\\V_M=\frac{V_{DD}-\left|V_{tp}\right|+\frac1NV_{tn}\sqrt{\frac{\beta_n}{\beta_p}}}{1+\frac1N\sqrt{\frac{\beta_n}{\beta_p}}}\end{array}

 

 

Read: Power consumption of a mos inverter

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