In pst-circ, how can you rotate the op-amp by 90 degrees?
For items that have two terminals like resistors and inductors, they rotate automatically. For the 3 terminal op-amp I'm having trouble.
I'm currently using PCTeX v5 and pst-circ v1.2a in WinXP.
I tried....
\OA[OAperfect = false]{90}(A1c)(A1a)(A1b)
but that did not work.
Also I gave this a shot...
\OA[OAperfect = false,TRot= 90](A1c)(A1a)(A1b)
but I think TRot is only for transistors.
This is what I have now
from this code
\documentclass[12pt]{article}
\usepackage{pst-circ}
\begin{document}
\begin{pspicture}(8,12)
\pnode(10,7.75){A1a} \pnode(13,8){A1b} \pnode(10,8.25){A1c}
\pnode(1,4){R7a} \pnode(3,4){R7b}
\pnode(4,4){R8a} \pnode(6,4){R8b}
\pnode(7,4){R9a} \pnode(9,4){R9b}
\pnode(10,4){Ca} \pnode(11,4){Cb}
\resistor[ dipolestyle= zigzag](R7a)(R7b){$R_7$}
\resistor[ dipolestyle= zigzag](R8a)(R8b){$R_8$}
\resistor[ dipolestyle= zigzag](R9a)(R9b){$R_9$}
\capacitor(Ca)(Cb){$C$}
\wire(R7b)(R8a)
\wire(R8b)(R9a)
\wire(R9b)(Ca)
\wire(Cb)(12,4)
\OA[OAperfect = false](A1c)(A1a)(A1b)
\wire(A1b)(13,10)
\wire(13,10)(3.5,10)
\wire(3.5,10)(3.5,4)
\wire(A1c)(6.5,8.25)
\wire(6.5,8.25)(6.5,4)
\wire(A1a)(9.5,7.75)
\wire(9.5,7.75)(9.5,4)
\end{pspicture}
\end{document}
What I would like is this...
which I made using circuitikz
pstricksmailing list, it seems like you have a very outdated version ofpst-circ- 1.2a. The most recent version is 2.02. Please include a minimal working example (MWE) that shows your code, as well as the output in the form of an image. – Werner Jan 25 '12 at 7:18