How can a dominant state be the one with bigger voltage difference?
$begingroup$
I am studying about CAN buses, and there is one thing I just can't find an explanation for. I understand that the idea of a wired-and connection is, that if any node is driving the bus to the dominant state, the bus will get to the dominant state regardless of the number of nodes transmitting a recessive state.
However, I find that in CAN the dominant and recessive states are as shown in the image below.
I could easily image an implementation like this, if it was the other way around, and the dominant state was the one where the wires are on the same voltage level:
But this implementation would result in the states being swapped. So how is it possible for the dominant state to be the state with the voltage difference?
can logic-level
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
$endgroup$
add a comment |
$begingroup$
I am studying about CAN buses, and there is one thing I just can't find an explanation for. I understand that the idea of a wired-and connection is, that if any node is driving the bus to the dominant state, the bus will get to the dominant state regardless of the number of nodes transmitting a recessive state.
However, I find that in CAN the dominant and recessive states are as shown in the image below.
I could easily image an implementation like this, if it was the other way around, and the dominant state was the one where the wires are on the same voltage level:
But this implementation would result in the states being swapped. So how is it possible for the dominant state to be the state with the voltage difference?
can logic-level
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
$endgroup$
add a comment |
$begingroup$
I am studying about CAN buses, and there is one thing I just can't find an explanation for. I understand that the idea of a wired-and connection is, that if any node is driving the bus to the dominant state, the bus will get to the dominant state regardless of the number of nodes transmitting a recessive state.
However, I find that in CAN the dominant and recessive states are as shown in the image below.
I could easily image an implementation like this, if it was the other way around, and the dominant state was the one where the wires are on the same voltage level:
But this implementation would result in the states being swapped. So how is it possible for the dominant state to be the state with the voltage difference?
can logic-level
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
$endgroup$
I am studying about CAN buses, and there is one thing I just can't find an explanation for. I understand that the idea of a wired-and connection is, that if any node is driving the bus to the dominant state, the bus will get to the dominant state regardless of the number of nodes transmitting a recessive state.
However, I find that in CAN the dominant and recessive states are as shown in the image below.
I could easily image an implementation like this, if it was the other way around, and the dominant state was the one where the wires are on the same voltage level:
But this implementation would result in the states being swapped. So how is it possible for the dominant state to be the state with the voltage difference?
can logic-level
can logic-level
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
asked Nov 13 '18 at 11:56
TamasKotanTamasKotan
211
211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because CAN is not driven in the way you're imagining.
Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.
This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
$begingroup$
CAN drivers are like this
simulate this circuit – Schematic created using CircuitLab
For example.
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f406528%2fhow-can-a-dominant-state-be-the-one-with-bigger-voltage-difference%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because CAN is not driven in the way you're imagining.
Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.
This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
$begingroup$
Because CAN is not driven in the way you're imagining.
Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.
This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
$begingroup$
Because CAN is not driven in the way you're imagining.
Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.
This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
Because CAN is not driven in the way you're imagining.
Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.
This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
answered Nov 13 '18 at 13:04
Dave Tweed♦Dave Tweed
118k9145256
118k9145256
add a comment |
add a comment |
$begingroup$
CAN drivers are like this
simulate this circuit – Schematic created using CircuitLab
For example.
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
$endgroup$
add a comment |
$begingroup$
CAN drivers are like this
simulate this circuit – Schematic created using CircuitLab
For example.
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
$endgroup$
add a comment |
$begingroup$
CAN drivers are like this
simulate this circuit – Schematic created using CircuitLab
For example.
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
$endgroup$
CAN drivers are like this
simulate this circuit – Schematic created using CircuitLab
For example.
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
answered Nov 13 '18 at 14:57
analogsystemsrfanalogsystemsrf
14k2717
14k2717
add a comment |
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f406528%2fhow-can-a-dominant-state-be-the-one-with-bigger-voltage-difference%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
