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

What I get

from this code




\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){$R_9$}

\OA[OAperfect = false](A1c)(A1a)(A1b)






What I would like is this...

What I want which I made using circuitikz

closed as too localized by yo', Martin Schröder, Kurt, Qrrbrbirlbel, Stefan Kottwitz Dec 1 '12 at 23:02

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  • Can you show what you have tried, so that someone may try to replicate your problem and (possibly) solve it. – Frédéric Jan 25 '12 at 3:12
  • 3
    Based on the pstricks mailing list, it seems like you have a very outdated version of pst-circ - 1.2a.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
  • I edited the post based on your suggestions, still looking for a solution. Thank you. – ori333 Feb 6 '12 at 19:58
  • @ori333: I think the easiest way is to draw the tripole by hand. – Marco Daniel Mar 3 '12 at 23:00

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