但是运算放大器内部的电阻呈现得不太好,因为我无法很好地定位组件。
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\documentclass{article}
\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}
\draw (0,0) node[op amp,scale=2](oaq) {};
\draw (6,0) node[op amp,scale=2](oaq2) {};
\draw (oaq.-) to [short,-*] ++(-0.5,0);
\draw (oaq.+) to [short,-*] ++(-0.5,0);
\draw (oaq.-) to [short,-] ++(1.5,0) to [R=\SI{$Z_{ENT}_{d}$}{\ohm}] ++(0,-1.95) to [oaq.+,-*] ++(-1,0);
\draw (oaq2.-) to [short,*-] ++(1.5,0) to [R=\SI{$Z_{ENT}_{d}$}{\ohm}] ++(0,-0.85) to [short,-] ++(0,-0.05) [R=\SI{$Z_{ENT}_{cm}$}{\ohm}] ++(0,-1) to [oaq2.+,-*] ++(-1,0);
\end{circuitikz}
\end{document}
答案1
作为第一张图片的起点:
\documentclass[border=3.141592]{standalone}
\usepackage[siunitx]{circuitikz}
\begin{document}
\begin{circuitikz}
\ctikzset{amplifiers/plus={}}
\ctikzset{amplifiers/minus={}}
\draw (0,0) node[op amp,scale=2] (oaq1) {};
\node [font=\small, above left] at (oaq1.bin up) {$-$};
\node [font=\small, below left] at (oaq1.bin down) {$+$};
\draw (oaq1.-) to [short,-o] ++(-0.5,0)
(oaq1.+) to [short,-o] ++(-0.5,0)
(oaq1.out) to [short,-o] ++(0.5,0);
\draw (oaq1.bin up) -- ++(0.5,0) coordinate (aux)
to [R=$Z_{\mathit{ENT}(d)}$] (aux |- oaq1.+)
-- (oaq1.bin down);
\end{circuitikz}
\end{document}
附录:
作为第二幅带有差分输入放大器的图像的起点。放大器符号如下plain amp
(未标记输入符号):
\documentclass[border=3.141592]{standalone}
\usepackage[siunitx]{circuitikz}
\usetikzlibrary{calc}
\begin{document}
\begin{circuitikz}[font=\footnotesize]
\ctikzset{resistors/scale=0.5},
\draw (0,0) node[plain amp, scale=2] (oaq2) {};
\node [font=\small, above left] at (oaq2.bin up) {$-$};
\node [font=\small, below left] at (oaq2.bin down) {$+$};
\draw (oaq2.-) to [short,-o] ++(-0.5,0)
(oaq2.+) to [short,-o] ++(-0.5,0)
(oaq2.out) to [short,-o] ++(0.5,0);
\coordinate (aux1) at ($(oaq2.bin up)!0.5!(oaq2.bin down)$);
\draw (oaq2.bin up) -- ++(0.5,0) coordinate (aux2)
to [R=$Z_{\mathit{ENT}(d)}$] (aux2 |- aux1)
to [R=$Z_{\mathit{ENT}(d)}$] (aux2 |- oaq2.bin down)
-- (oaq2.bin down)
(aux2 |- aux1) to [short,*-] ++ (1.5,0)
node[ground] {};
\end{circuitikz}
\end{document}