这就是我目前所做的。
\begin{circuitikz}
\draw (0, 0) node[op amp] (opamp) {}
(opamp.-) to[short,-*] ++(-1, 0) coordinate(A)
(opamp.+) node[ground](B){}
(opamp.out) to[short,*-o] ++(1, 0) coordinate(C)
(A) to[R,l_=$4k\Omega$,-o] ++(-2, 0) -- ++(-1, 0) coordinate(D) node[ground]{}
(A) |- ++(1,1) coordinate[yshift=1ex] (L1) to[R=$40k\Omega$] ++(2,0) -| (opamp.out) to[short,-o] ++(1,0)
;
\end{circuitikz}
答案1
您已经非常接近所需的电路方案了...根据我之前的回答,可能的 MWE 是:
\documentclass[border=3.131592]{standalone}
\usepackage[siunitx]{circuitikz}
\begin{document}
\begin{circuitikz}
\draw (0,0) node[op amp] (oa) {}
(oa.+) to[short] ++(0,-1)
node[ground](B){}
(oa.out) to[short] ++(0, 2) coordinate(A)
to[R,l_=\qty{40}{\kilo\ohm}] (A -| oa.-)
to[short] (oa.-)
to[short,*-*] ++(-1,0) coordinate(B)
to[R,l_=\qty{4}{\kilo\ohm}, -o] ++(-2,0) node[left=1mm] {Terminal 2}
(B) to[short] ++(0, 1.5)
to[R,l_=\qty{15}{\kilo\ohm},-o] ++(-2,0) node[left=1mm] {\qty{-5}{\volt}}
(B) to[short] ++(0,-1.5)
to[R,l_=\qty{40}{\kilo\ohm},-o] ++(-2,0) node[left=1mm] {\qty{-2}{\volt}}
(oa.out) to[short,*-o] ++(1,0) node[right] {$V_o$}
;
\end{circuitikz}
\end{document}