How to invert Logic Gate input in Circuitikz
I have drawn the following circuit, but how could the inputs of a Logic Gate be inverted? (As in the scanned picture, but it does not matter which gate exactly, just generally)
documentclass[a4paper,12pt]{article}
usepackage{tikz} %vectorgraphics
usetikzlibrary{shapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc}
%graph/flowchart
usepackage[siunitx]{circuitikz}
begin{document}
begin{circuitikz} draw
(0,0) node[nand port] (nand1) {}
(3,0) node[nand port] (nand2) {}
(5,0) node[nand port] (nand3) {}
(3,2) node[nand port] (nand4) {}
(5,2) node[nand port] (nand5) {}
(3,4) node[nand port] (nand6) {}
(5,4) node[nand port] (nand7) {}
(0,4) node[nand port] (nand8) {}
(nand1.out) |- (nand2.in 2)
(nand2.out) -| node[circ,midway]{} (nand3.in 1)
(nand2.out) -| (nand3.in 2)
(nand8.out) |- (nand6.in 1)
(nand6.out) -| node[circ,midway]{} (nand7.in 2)
(nand6.out) -| (nand7.in 1)
(nand4.out) -| node[circ,midway]{} (nand5.in 2)
(nand4.out) -| (nand5.in 1)
(-2,1) node(B1)[anchor=east] {B}
(-2,1) -| node[circ,midway]{} (nand1.in 1)
to[short,*-] (nand1.in 2)
(-2,1) -| (nand4.in 2)
(1,1) node[circ]{} |- (nand6.in 2)
(-2,3) node[anchor=east] {A}
(-2,3) -| node[circ,midway]{} (nand8.in 2)
to[short,*-] (nand8.in 1)
(-2,3) -| (nand4.in 1)
(0.5,3) node[circ]{} |- (nand2.in 1)
(6,1.97) node[anchor=east] {O2}
(6,0.0) node[anchor=east] {O3}
(6,4) node[anchor=east] {O1}
;end{circuitikz}
end{document}
circuitikz logic
add a comment |
I have drawn the following circuit, but how could the inputs of a Logic Gate be inverted? (As in the scanned picture, but it does not matter which gate exactly, just generally)
documentclass[a4paper,12pt]{article}
usepackage{tikz} %vectorgraphics
usetikzlibrary{shapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc}
%graph/flowchart
usepackage[siunitx]{circuitikz}
begin{document}
begin{circuitikz} draw
(0,0) node[nand port] (nand1) {}
(3,0) node[nand port] (nand2) {}
(5,0) node[nand port] (nand3) {}
(3,2) node[nand port] (nand4) {}
(5,2) node[nand port] (nand5) {}
(3,4) node[nand port] (nand6) {}
(5,4) node[nand port] (nand7) {}
(0,4) node[nand port] (nand8) {}
(nand1.out) |- (nand2.in 2)
(nand2.out) -| node[circ,midway]{} (nand3.in 1)
(nand2.out) -| (nand3.in 2)
(nand8.out) |- (nand6.in 1)
(nand6.out) -| node[circ,midway]{} (nand7.in 2)
(nand6.out) -| (nand7.in 1)
(nand4.out) -| node[circ,midway]{} (nand5.in 2)
(nand4.out) -| (nand5.in 1)
(-2,1) node(B1)[anchor=east] {B}
(-2,1) -| node[circ,midway]{} (nand1.in 1)
to[short,*-] (nand1.in 2)
(-2,1) -| (nand4.in 2)
(1,1) node[circ]{} |- (nand6.in 2)
(-2,3) node[anchor=east] {A}
(-2,3) -| node[circ,midway]{} (nand8.in 2)
to[short,*-] (nand8.in 1)
(-2,3) -| (nand4.in 1)
(0.5,3) node[circ]{} |- (nand2.in 1)
(6,1.97) node[anchor=east] {O2}
(6,0.0) node[anchor=east] {O3}
(6,4) node[anchor=east] {O1}
;end{circuitikz}
end{document}
circuitikz logic
2
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
1
With currentcircuitikz
you cannot achieve this. However, you can usetikz's
stockciruit
option to achieve it. Nevertheless, a very useful question though.
– Raaja
Mar 14 at 11:51
add a comment |
I have drawn the following circuit, but how could the inputs of a Logic Gate be inverted? (As in the scanned picture, but it does not matter which gate exactly, just generally)
documentclass[a4paper,12pt]{article}
usepackage{tikz} %vectorgraphics
usetikzlibrary{shapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc}
%graph/flowchart
usepackage[siunitx]{circuitikz}
begin{document}
begin{circuitikz} draw
(0,0) node[nand port] (nand1) {}
(3,0) node[nand port] (nand2) {}
(5,0) node[nand port] (nand3) {}
(3,2) node[nand port] (nand4) {}
(5,2) node[nand port] (nand5) {}
(3,4) node[nand port] (nand6) {}
(5,4) node[nand port] (nand7) {}
(0,4) node[nand port] (nand8) {}
(nand1.out) |- (nand2.in 2)
(nand2.out) -| node[circ,midway]{} (nand3.in 1)
(nand2.out) -| (nand3.in 2)
(nand8.out) |- (nand6.in 1)
(nand6.out) -| node[circ,midway]{} (nand7.in 2)
(nand6.out) -| (nand7.in 1)
(nand4.out) -| node[circ,midway]{} (nand5.in 2)
(nand4.out) -| (nand5.in 1)
(-2,1) node(B1)[anchor=east] {B}
(-2,1) -| node[circ,midway]{} (nand1.in 1)
to[short,*-] (nand1.in 2)
(-2,1) -| (nand4.in 2)
(1,1) node[circ]{} |- (nand6.in 2)
(-2,3) node[anchor=east] {A}
(-2,3) -| node[circ,midway]{} (nand8.in 2)
to[short,*-] (nand8.in 1)
(-2,3) -| (nand4.in 1)
(0.5,3) node[circ]{} |- (nand2.in 1)
(6,1.97) node[anchor=east] {O2}
(6,0.0) node[anchor=east] {O3}
(6,4) node[anchor=east] {O1}
;end{circuitikz}
end{document}
circuitikz logic
I have drawn the following circuit, but how could the inputs of a Logic Gate be inverted? (As in the scanned picture, but it does not matter which gate exactly, just generally)
documentclass[a4paper,12pt]{article}
usepackage{tikz} %vectorgraphics
usetikzlibrary{shapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc}
%graph/flowchart
usepackage[siunitx]{circuitikz}
begin{document}
begin{circuitikz} draw
(0,0) node[nand port] (nand1) {}
(3,0) node[nand port] (nand2) {}
(5,0) node[nand port] (nand3) {}
(3,2) node[nand port] (nand4) {}
(5,2) node[nand port] (nand5) {}
(3,4) node[nand port] (nand6) {}
(5,4) node[nand port] (nand7) {}
(0,4) node[nand port] (nand8) {}
(nand1.out) |- (nand2.in 2)
(nand2.out) -| node[circ,midway]{} (nand3.in 1)
(nand2.out) -| (nand3.in 2)
(nand8.out) |- (nand6.in 1)
(nand6.out) -| node[circ,midway]{} (nand7.in 2)
(nand6.out) -| (nand7.in 1)
(nand4.out) -| node[circ,midway]{} (nand5.in 2)
(nand4.out) -| (nand5.in 1)
(-2,1) node(B1)[anchor=east] {B}
(-2,1) -| node[circ,midway]{} (nand1.in 1)
to[short,*-] (nand1.in 2)
(-2,1) -| (nand4.in 2)
(1,1) node[circ]{} |- (nand6.in 2)
(-2,3) node[anchor=east] {A}
(-2,3) -| node[circ,midway]{} (nand8.in 2)
to[short,*-] (nand8.in 1)
(-2,3) -| (nand4.in 1)
(0.5,3) node[circ]{} |- (nand2.in 1)
(6,1.97) node[anchor=east] {O2}
(6,0.0) node[anchor=east] {O3}
(6,4) node[anchor=east] {O1}
;end{circuitikz}
end{document}
circuitikz logic
circuitikz logic
asked Mar 13 at 21:54
Roman StadlerRoman Stadler
575
575
2
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
1
With currentcircuitikz
you cannot achieve this. However, you can usetikz's
stockciruit
option to achieve it. Nevertheless, a very useful question though.
– Raaja
Mar 14 at 11:51
add a comment |
2
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
1
With currentcircuitikz
you cannot achieve this. However, you can usetikz's
stockciruit
option to achieve it. Nevertheless, a very useful question though.
– Raaja
Mar 14 at 11:51
2
2
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
1
1
With current
circuitikz
you cannot achieve this. However, you can use tikz's
stock ciruit
option to achieve it. Nevertheless, a very useful question though.– Raaja
Mar 14 at 11:51
With current
circuitikz
you cannot achieve this. However, you can use tikz's
stock ciruit
option to achieve it. Nevertheless, a very useful question though.– Raaja
Mar 14 at 11:51
add a comment |
1 Answer
1
active
oldest
votes
Using the fresh new circuitikzgit
uploaded minutes ago, you can now do this:
documentclass[border=10pt]{standalone}
usepackage[siunitx, RPvoltages]{circuitikzgit}
begin{document}
begin{circuitikz}
draw (0,3) node[american and port] (A) {P1};
node at (A.bin 1) [ocirc, left]{} ;
begin{scope}
ctikzset{tripoles/american or port/height=1.6}
draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
node at (B.bin 3) [ocirc, left]{} ;
end{scope}
draw (0,1.5) node[american or port] (C) {P3};
node at (C.bin 2) [ocirc, left]{} ;
draw (C.out) |- (B.in 2);
end{circuitikz}
end{document}
See: https://github.com/circuitikz/circuitikz/issues/166 and https://github.com/circuitikz/circuitikz/pull/168
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits withciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).
– Raaja
Mar 18 at 8:29
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using the fresh new circuitikzgit
uploaded minutes ago, you can now do this:
documentclass[border=10pt]{standalone}
usepackage[siunitx, RPvoltages]{circuitikzgit}
begin{document}
begin{circuitikz}
draw (0,3) node[american and port] (A) {P1};
node at (A.bin 1) [ocirc, left]{} ;
begin{scope}
ctikzset{tripoles/american or port/height=1.6}
draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
node at (B.bin 3) [ocirc, left]{} ;
end{scope}
draw (0,1.5) node[american or port] (C) {P3};
node at (C.bin 2) [ocirc, left]{} ;
draw (C.out) |- (B.in 2);
end{circuitikz}
end{document}
See: https://github.com/circuitikz/circuitikz/issues/166 and https://github.com/circuitikz/circuitikz/pull/168
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits withciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).
– Raaja
Mar 18 at 8:29
add a comment |
Using the fresh new circuitikzgit
uploaded minutes ago, you can now do this:
documentclass[border=10pt]{standalone}
usepackage[siunitx, RPvoltages]{circuitikzgit}
begin{document}
begin{circuitikz}
draw (0,3) node[american and port] (A) {P1};
node at (A.bin 1) [ocirc, left]{} ;
begin{scope}
ctikzset{tripoles/american or port/height=1.6}
draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
node at (B.bin 3) [ocirc, left]{} ;
end{scope}
draw (0,1.5) node[american or port] (C) {P3};
node at (C.bin 2) [ocirc, left]{} ;
draw (C.out) |- (B.in 2);
end{circuitikz}
end{document}
See: https://github.com/circuitikz/circuitikz/issues/166 and https://github.com/circuitikz/circuitikz/pull/168
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits withciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).
– Raaja
Mar 18 at 8:29
add a comment |
Using the fresh new circuitikzgit
uploaded minutes ago, you can now do this:
documentclass[border=10pt]{standalone}
usepackage[siunitx, RPvoltages]{circuitikzgit}
begin{document}
begin{circuitikz}
draw (0,3) node[american and port] (A) {P1};
node at (A.bin 1) [ocirc, left]{} ;
begin{scope}
ctikzset{tripoles/american or port/height=1.6}
draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
node at (B.bin 3) [ocirc, left]{} ;
end{scope}
draw (0,1.5) node[american or port] (C) {P3};
node at (C.bin 2) [ocirc, left]{} ;
draw (C.out) |- (B.in 2);
end{circuitikz}
end{document}
See: https://github.com/circuitikz/circuitikz/issues/166 and https://github.com/circuitikz/circuitikz/pull/168
Using the fresh new circuitikzgit
uploaded minutes ago, you can now do this:
documentclass[border=10pt]{standalone}
usepackage[siunitx, RPvoltages]{circuitikzgit}
begin{document}
begin{circuitikz}
draw (0,3) node[american and port] (A) {P1};
node at (A.bin 1) [ocirc, left]{} ;
begin{scope}
ctikzset{tripoles/american or port/height=1.6}
draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
node at (B.bin 3) [ocirc, left]{} ;
end{scope}
draw (0,1.5) node[american or port] (C) {P3};
node at (C.bin 2) [ocirc, left]{} ;
draw (C.out) |- (B.in 2);
end{circuitikz}
end{document}
See: https://github.com/circuitikz/circuitikz/issues/166 and https://github.com/circuitikz/circuitikz/pull/168
answered Mar 15 at 10:14
RmanoRmano
8,18121647
8,18121647
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits withciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).
– Raaja
Mar 18 at 8:29
add a comment |
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits withciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).
– Raaja
Mar 18 at 8:29
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
when will it be available in CTAN?
– Raaja
Mar 18 at 7:28
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
@Raaja I am completing a couple of things more (basically, manual overhaul and fillable shapes) and then I'll do a release. Three to five weeks, if RL doesn't create problems... ;-)
– Rmano
Mar 18 at 8:23
Good to know! Because, it seems many are interested in creating big-logic circuits with
ciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).– Raaja
Mar 18 at 8:29
Good to know! Because, it seems many are interested in creating big-logic circuits with
ciruittikz
(see: tex.stackexchange.com/questions/480034/…). And, I think this is a really nice contribution indeed :).– Raaja
Mar 18 at 8:29
add a comment |
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2
Nice question. I'll look over it tomorrow (it's nighttime here), but I don't think it could be easily done with the current implementation. But seems a valuable enhancement proposal, anyway.
– Rmano
Mar 13 at 22:41
I am considering it as a enhancement proposal; if you want you can open an issue at github.com/circuitikz/circuitikz/issues
– Rmano
Mar 14 at 7:24
1
With current
circuitikz
you cannot achieve this. However, you can usetikz's
stockciruit
option to achieve it. Nevertheless, a very useful question though.– Raaja
Mar 14 at 11:51