[Tex/LaTex] Orthogonal Edges Style

Question

I'm trying to draw a block diagram. I want to make the edge (red) orthogonal, like the black path made with -|. I know there are the commands \tikztostart, \tikztotarget and the ....!0.5!... opertator, but cant make them work inside the every edge/.style{}. Here is my code (dont know how to add the compiled version).

\begin{tikzpicture}[line width=1pt,-{Stealth[scale=3,length=3,width=2]},on grid,node distance=2cm,every node/.style={font=\huge},every edge/.style={draw,red,rounded corners}]
\node[draw,rectangle]   (B)                         {$B$};
\node[draw,circle]      (S)     [right=of B]        {$+$};
\node[draw,rectangle]   (I)     [right=of S]        {$\dfrac{1}{s}$};
\node[draw,rectangle]   (C)     [right=of I]        {$C$};
\node[draw,rectangle]   (A)     [below=of I]        {$A$};
\node[draw,rectangle]   (D)     [above=of I]        {$D$};

\draw           (B)     edge    [auto=left]     node    {}  (S)
                (S)     edge    [auto=right]    node                        {}          (I)
                (A)     edge    [auto]                          node                        {$A$}           (S);
\draw   (A) -| node {$A$} (S);
\end{tikzpicture}

Answer 1

Something like this?

Code:

    \documentclass[border=3mm,
               tikz,
               preview
               ]{standalone}
\usetikzlibrary{arrows.meta,chains,positioning}
\usepackage{amsmath}

    \begin{document}
%---------------------------------------------------------------%
    \begin{tikzpicture}[
      line width = 1pt,
         on grid,
      start chain = going right,
    node distance = 2cm,
       box/.style = {draw, rectangle, font=\huge, on chain},
         L/.style = {draw, red, -{Stealth[scale=3,length=3,width=2]}},
         T/.style = {draw, red, rounded corners,
                     to path={-| (\tikztotarget)},
                     -{Stealth[scale=3,length=3,width=2]}}
                        ]
\node[box]              (B) {$B$};
\node[draw,circle,
      on chain]         (S) {$+$};
\node[box]              (I) {$\frac{1}{s}$};
\node[box]              (C) {$C$};
\node[box,below=of I]   (A) {$A$};
\node[box,above=of I]   (D) {$D$};

\draw%[]
    (B) edge[L] (S)
    (S) edge[L] (I) 
    (I) edge[L] (C) 
    (D) edge[T] (S) 
    (A) edge[T] (S) 
        ;
\end{tikzpicture}
%---------------------------------------------------------------%
    \end{document}   

Idea for quadrature edges is stolen from here.

Edit: In above code is typing error: instead of positionings should be positioning. I correct this now.

Upgrade: In a case, that you like to have edges labeled, this can be simply done only for edges of type L, for example:

\draw   (B) edge[L] node {b} (S) 
        (S) edge[L] (I);

however, for edges of type T, the definition of edge should be changed as follows:

...
T/.style args = {#1/#2}{draw, red, rounded corners,
                 to path={-| node[pos=#1] {#2}
                 (\tikztotarget)},
                 -{Stealth[scale=3,length=3,width=2]}},
    T/.default = / ]

and edges width node are:

\draw   (D) edge[T=0.75/a] (S)
        (A) edge[T=0.25/a] (S);

or in case, when edge hasn't node:

\draw   (D) edge[T] (S)
        (A) edge[T=0.25/a] (S);

Another detail: since edges are defined to be in red color, than in case, that you like to have for example in black, you need add option text=black in edge node options. An complete example with edge labels:

        \documentclass[border=3mm,
               tikz,
               preview
               ]{standalone}
\usetikzlibrary{arrows.meta,chains,positioning}
\usepackage{amsmath}

    \begin{document}
%---------------------------------------------------------------%
    \begin{tikzpicture}[
      line width = 1pt,
         auto,
      start chain = going right,
    node distance = 2cm,
       box/.style = {draw, rectangle, font=\huge, on chain},
         L/.style = {draw, red, rounded corners,
                    -{Stealth[scale=3,length=3,width=2]}},
                    ]
\node[box]              (B) {$B$};
\node[draw,circle,
      on chain]         (S) {$+$};
\node[box]              (I) {$\frac{1}{s}$};
\node[box]              (C) {$C$};
\node[box,below=of I]   (A) {$A$};
\node[box,above=of I]   (D) {$D$};

\draw   (B) edge[L] node {b} (S)
        (S) edge[L] (I)
        (I) edge[L] (C)
        (D) edge[T=0.75/$d$] (S)
        (A) edge[T=0.25/a] (S)
        ;
\end{tikzpicture}
%---------------------------------------------------------------%
    \end{document} 

Beside above solution there exist more simple solution:

        \documentclass[border=3mm,
               tikz,
               preview
               ]{standalone}
\usetikzlibrary{arrows.meta,chains,positioning}
\usepackage{amsmath}

    \begin{document}
%---------------------------------------------------------------%
    \begin{tikzpicture}[
      line width = 1pt,
         auto,
      start chain = going right,
    node distance = 2cm,
       box/.style = {draw, rectangle, font=\huge, on chain},
         L/.style = {draw, red, rounded corners,
                     -{Stealth[scale=3,length=3,width=2]}},
    T/.style args = {#1/#2}{draw, red, rounded corners,
                     to path={-| node[pos=#1,text=black] {#2}
                     (\tikztotarget)},
                     -{Stealth[scale=3,length=3,width=2]}},
        T/.default = / ]
\node[box]              (B) {$B$};
\node[draw,circle,
      on chain]         (S) {$+$};
\node[box]              (I) {$\frac{1}{s}$};
\node[box]              (C) {$C$};
\node[box,below=of I]   (A) {$A$};
\node[box,above=of I]   (D) {$D$};

\draw[L]    (B) edge node {b} (S)
            (S) edge (I)
            (I) edge (C);
\draw[L]    (D) -| node[pos=0.75] {$d$} (S);
\draw[L]    (A) -| node[pos=0.25] {a} (S);
\end{tikzpicture}

Both gives the same result:

The second, simpler solution instead complicated edge with orthogonal path use separately draw of each orthogonal path between nodes, i.e. you need for each path write \draw[L] ... what in the first case is not needed. Both solution has pros and cons. Which is more suitable? This I left to user(s) :-).