How to invert Logic Gate input in Circuitikz
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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
usepackagetikz %vectorgraphics
usetikzlibraryshapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc
%graph/flowchart
usepackage[siunitx]circuitikz
begindocument
begincircuitikz 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
;endcircuitikz
enddocument
circuitikz logic
asked 56 secs ago
Roman StadlerRoman Stadler
133
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Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
usepackagetikz %vectorgraphics
usetikzlibraryshapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc
%graph/flowchart
usepackage[siunitx]circuitikz
begindocument
begincircuitikz 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
;endcircuitikz
enddocument
circuitikz logic
asked 56 secs ago
Roman StadlerRoman Stadler
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
usepackagetikz %vectorgraphics
usetikzlibraryshapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc
%graph/flowchart
usepackage[siunitx]circuitikz
begindocument
begincircuitikz 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
;endcircuitikz
enddocument
circuitikz logic
asked 56 secs ago
Roman StadlerRoman Stadler
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
usepackagetikz %vectorgraphics
usetikzlibraryshapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc
%graph/flowchart
usepackage[siunitx]circuitikz
begindocument
begincircuitikz 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
;endcircuitikz
enddocument
circuitikz logic
circuitikz logic
asked 56 secs ago
Roman StadlerRoman Stadler
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 56 secs ago
Roman StadlerRoman Stadler
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 56 secs ago
Roman StadlerRoman Stadler
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 56 secs ago
Roman StadlerRoman Stadler
133
asked 56 secs ago
Roman StadlerRoman Stadler
133
133
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Roman Stadler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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