On page 400, example 6.26, we are asked to analyze a double-balanced circuit on its \(IP_2\).
I don't get where does (6.127) come from. Since there's no explanation on (6.127), I guess this equation is obvious. So I reduce the problem and try to derive it by myself.
The reduced problem is as follows.
\(M_1\) and \(M_2\) are two identical NMOS. \(I_{SS}\) is the biasing current source. \(V_{RF}^{+}+V_{OS1}\) and \(V_{RF}^-\) are differential input voltages. \(V_{OS1}\) is the offset. \(I_{D1}\) and \(I_{D2}\) are drain currents. The goal is to derive differential output current \(I_{D1}-I_{D2}\).
I don't think this is a small-signal model here because in (6.127), it seems that transconductance is not linearized and depends on bias.
We first write the relation between overdrive voltage and drain current for \(M_1\) and \(M_2\)
\[\begin{align} & I_{D1} = \frac{1}{2}\mu_{n}C_{ox}\frac{W}{L}(V_{GS0}+V_{RF}^++V_{OS1}-V_{TH})^2 \\ & I_{D2} = \frac{1}{2}\mu_{n}C_{ox}\frac{W}{L}(V_{GS0}+V_{RF}^--V_{TH})^2 \end{align}\]The differential input voltage is
\[\begin{align} \Delta V_{in} = V_{RF}^+ - V_{RF}^- \end{align}\]And drain current of \(M_{1}\) and \(M_{2}\) are offered by \(I_{SS}\), i.e
\[\begin{align} I_{D1}+I_{D2} = I_{SS} \end{align}\]The differential output current is
\[\begin{align} \Delta I_{out} = I_{D1} - I_{D2} \\ \end{align}\]Replace \(I_{D1}\) and \(I_{D2}\) in \((5)\) using \((1)\) and \((2)\), we have
\[\begin{align} \Delta I_{out} & = \frac{1}{2}\mu_{n}C_{ox}\frac{W}{L}[(V_{GS0}+V_{RF}^++V_{OS1}-V_{TH})^2 - (V_{GS0}+V_{RF}^--V_{TH})^2] \\ & = \frac{1}{2}\mu_{n}C_{ox}\frac{W}{L}(\Delta V_{in}+V_{OS1})(V_{GS0}+V_{RF}^++V_{OS1}-V_{TH}+V_{GS0}+V_{RF}^--V_{TH}) \end{align}\]\((7)\) is very similar to (6.127) in the book except for the last term.
To make the derivation trivial we rewrite \((7)\) in terms of over overdrive voltages
\[\begin{align} (\underbrace{V_{GS0}+V_{RF}^++V_{OS1}-V_{TH}}_{V_{od,1}}+\underbrace{V_{GS0}+V_{RF}^--V_{TH}}_{V_{od,2}}) = V_{od,1} + V_{od,2} \end{align}\]Noticing that in (6.127), last term is expressed in terms of \(I_{SS}\) and $ \Delta V_{in}+V_{OS1} = V_{od,1} - V_{od,2}$ (I don't think the sign before \(V_{OS1}\) matters because it is an offset), I guess we can prove this equality, which is
\[\begin{align} V_{od,1} + V_{od,2} = \sqrt{\frac{4I_{SS}}{\mu_{n}C_{ox}\frac{W}{L}}-(V_{od,1} - V_{od,2})^2} \end{align}\]We can replace \((1)\),\((2)\),\((4)\) in \((9)\)
\[\begin{align} \sqrt{\frac{4I_{SS}}{\mu_{n}C_{ox}\frac{W}{L}}-(V_{od,1} - V_{od,2})^2} & \\ & =\sqrt{\frac{4(I_{D1}+I_{D2})}{\mu_{n}C_{ox}\frac{W}{L}}-(V_{od,1} - V_{od,2})^2} \\ & = \sqrt{2(V_{od,1}^2 + V_{od,2}^2)-(V_{od,1} - V_{od,2})^2} \\ & = \sqrt{(V_{od,1} + V_{od,2})^2} \\ & = (V_{od,1} + V_{od,2}) \end{align}\]QED
标签：GS0,od,frac,OS1,Razavi,align,Differential,Microelectronics,RF From： https://www.cnblogs.com/lilulounote/p/18174644