Why time constant is 63.2% not a 50 or 70%?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;







up vote
3
down vote

favorite
1












I am studying about RC and RL circuits . Where the time constant is equal to the 63.2% of the output voltage? Why it is defined as 63% not other values?



I know it's a dumb question but could anyone explain elaborately. Does circuit starts working at 63% of output voltage?why not at 50%?






circuit-analysis circuit-design math time-constant







share|improve this question

















  • 4




    1-e^-1 = 0.6321...
    – Andrew Morton
    28 mins ago










  • It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
    – Chu
    4 secs ago
















up vote
3
down vote

favorite
1












I am studying about RC and RL circuits . Where the time constant is equal to the 63.2% of the output voltage? Why it is defined as 63% not other values?



I know it's a dumb question but could anyone explain elaborately. Does circuit starts working at 63% of output voltage?why not at 50%?






circuit-analysis circuit-design math time-constant







share|improve this question

















  • 4




    1-e^-1 = 0.6321...
    – Andrew Morton
    28 mins ago










  • It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
    – Chu
    4 secs ago












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I am studying about RC and RL circuits . Where the time constant is equal to the 63.2% of the output voltage? Why it is defined as 63% not other values?



I know it's a dumb question but could anyone explain elaborately. Does circuit starts working at 63% of output voltage?why not at 50%?






circuit-analysis circuit-design math time-constant







share|improve this question













I am studying about RC and RL circuits . Where the time constant is equal to the 63.2% of the output voltage? Why it is defined as 63% not other values?



I know it's a dumb question but could anyone explain elaborately. Does circuit starts working at 63% of output voltage?why not at 50%?






circuit-analysis circuit-design math time-constant




circuit-analysis circuit-design math time-constant






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 33 mins ago









Bala Subramanian

23114




23114







  • 4




    1-e^-1 = 0.6321...
    – Andrew Morton
    28 mins ago










  • It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
    – Chu
    4 secs ago












  • 4




    1-e^-1 = 0.6321...
    – Andrew Morton
    28 mins ago










  • It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
    – Chu
    4 secs ago







4




4




1-e^-1 = 0.6321...
– Andrew Morton
28 mins ago




1-e^-1 = 0.6321...
– Andrew Morton
28 mins ago












It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
– Chu
4 secs ago




It coincides with 1/bandwidth and it's the time value in the first order lag $small frac11+omegatau or frac11+tau s$ . In radioactive decay they use 50% ('half life').
– Chu
4 secs ago










3 Answers
3






active

oldest

votes

















up vote
10
down vote













It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is $e^-1 = 0.36788$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.



The "unit of time" is referred to as the "time constant" of the system, and is usually denoted Ï„ (tau). The full expression for the system response over time (t) is



$$V(t) = V_0 e^-fracttau$$



So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.



In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or R/L in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.






share|improve this answer





























    up vote
    3
    down vote













    The decay of an RC parallel circuit with capacitor charged to Vo



    v(t) = $Vo(1-e^-t/tau)$ , where $tau$ is the time constant R$cdot$C.



    So v($tau$)/Vo is approximately 0.63212055882855767840447622983854



    In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.



    Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).



    Suppose you want to know the time when the voltage is 0.5. It is (from the above)



    t = -$ln(0.5)tau$ or about 0.693RC



    Either way you do it, some irrational numbers pop up and dealing with RC=$tau$ is the "natural" way.






    share|improve this answer


















    • 1




      That is a very rough approximation.
      – Arsenal
      17 mins ago






    • 1




      @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
      – Spehro Pefhany
      16 mins ago










    • @Arsenal Good point.
      – Spehro Pefhany
      11 mins ago

















    up vote
    0
    down vote













    this comes from the e constant value 1-e^(-1) = 0.63






    share|improve this answer




















      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',
      convertImagesToLinks: false,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f396653%2fwhy-time-constant-is-63-2-not-a-50-or-70%23new-answer', 'question_page');

      );

      Post as a guest

















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      10
      down vote













      It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is $e^-1 = 0.36788$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.



      The "unit of time" is referred to as the "time constant" of the system, and is usually denoted Ï„ (tau). The full expression for the system response over time (t) is



      $$V(t) = V_0 e^-fracttau$$



      So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.



      In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or R/L in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.






      share|improve this answer


























        up vote
        10
        down vote













        It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is $e^-1 = 0.36788$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.



        The "unit of time" is referred to as the "time constant" of the system, and is usually denoted Ï„ (tau). The full expression for the system response over time (t) is



        $$V(t) = V_0 e^-fracttau$$



        So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.



        In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or R/L in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.






        share|improve this answer
























          up vote
          10
          down vote










          up vote
          10
          down vote









          It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is $e^-1 = 0.36788$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.



          The "unit of time" is referred to as the "time constant" of the system, and is usually denoted Ï„ (tau). The full expression for the system response over time (t) is



          $$V(t) = V_0 e^-fracttau$$



          So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.



          In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or R/L in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.






          share|improve this answer














          It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is $e^-1 = 0.36788$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.



          The "unit of time" is referred to as the "time constant" of the system, and is usually denoted Ï„ (tau). The full expression for the system response over time (t) is



          $$V(t) = V_0 e^-fracttau$$



          So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.



          In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or R/L in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 14 mins ago

























          answered 20 mins ago









          Dave Tweed♦

          108k9130233




          108k9130233






















              up vote
              3
              down vote













              The decay of an RC parallel circuit with capacitor charged to Vo



              v(t) = $Vo(1-e^-t/tau)$ , where $tau$ is the time constant R$cdot$C.



              So v($tau$)/Vo is approximately 0.63212055882855767840447622983854



              In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.



              Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).



              Suppose you want to know the time when the voltage is 0.5. It is (from the above)



              t = -$ln(0.5)tau$ or about 0.693RC



              Either way you do it, some irrational numbers pop up and dealing with RC=$tau$ is the "natural" way.






              share|improve this answer


















              • 1




                That is a very rough approximation.
                – Arsenal
                17 mins ago






              • 1




                @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
                – Spehro Pefhany
                16 mins ago










              • @Arsenal Good point.
                – Spehro Pefhany
                11 mins ago














              up vote
              3
              down vote













              The decay of an RC parallel circuit with capacitor charged to Vo



              v(t) = $Vo(1-e^-t/tau)$ , where $tau$ is the time constant R$cdot$C.



              So v($tau$)/Vo is approximately 0.63212055882855767840447622983854



              In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.



              Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).



              Suppose you want to know the time when the voltage is 0.5. It is (from the above)



              t = -$ln(0.5)tau$ or about 0.693RC



              Either way you do it, some irrational numbers pop up and dealing with RC=$tau$ is the "natural" way.






              share|improve this answer


















              • 1




                That is a very rough approximation.
                – Arsenal
                17 mins ago






              • 1




                @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
                – Spehro Pefhany
                16 mins ago










              • @Arsenal Good point.
                – Spehro Pefhany
                11 mins ago












              up vote
              3
              down vote










              up vote
              3
              down vote









              The decay of an RC parallel circuit with capacitor charged to Vo



              v(t) = $Vo(1-e^-t/tau)$ , where $tau$ is the time constant R$cdot$C.



              So v($tau$)/Vo is approximately 0.63212055882855767840447622983854



              In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.



              Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).



              Suppose you want to know the time when the voltage is 0.5. It is (from the above)



              t = -$ln(0.5)tau$ or about 0.693RC



              Either way you do it, some irrational numbers pop up and dealing with RC=$tau$ is the "natural" way.






              share|improve this answer














              The decay of an RC parallel circuit with capacitor charged to Vo



              v(t) = $Vo(1-e^-t/tau)$ , where $tau$ is the time constant R$cdot$C.



              So v($tau$)/Vo is approximately 0.63212055882855767840447622983854



              In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.



              Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).



              Suppose you want to know the time when the voltage is 0.5. It is (from the above)



              t = -$ln(0.5)tau$ or about 0.693RC



              Either way you do it, some irrational numbers pop up and dealing with RC=$tau$ is the "natural" way.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 6 mins ago

























              answered 20 mins ago









              Spehro Pefhany

              195k4139385




              195k4139385







              • 1




                That is a very rough approximation.
                – Arsenal
                17 mins ago






              • 1




                @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
                – Spehro Pefhany
                16 mins ago










              • @Arsenal Good point.
                – Spehro Pefhany
                11 mins ago












              • 1




                That is a very rough approximation.
                – Arsenal
                17 mins ago






              • 1




                @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
                – Spehro Pefhany
                16 mins ago










              • @Arsenal Good point.
                – Spehro Pefhany
                11 mins ago







              1




              1




              That is a very rough approximation.
              – Arsenal
              17 mins ago




              That is a very rough approximation.
              – Arsenal
              17 mins ago




              1




              1




              @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
              – Spehro Pefhany
              16 mins ago




              @Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
              – Spehro Pefhany
              16 mins ago












              @Arsenal Good point.
              – Spehro Pefhany
              11 mins ago




              @Arsenal Good point.
              – Spehro Pefhany
              11 mins ago










              up vote
              0
              down vote













              this comes from the e constant value 1-e^(-1) = 0.63






              share|improve this answer
























                up vote
                0
                down vote













                this comes from the e constant value 1-e^(-1) = 0.63






                share|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  this comes from the e constant value 1-e^(-1) = 0.63






                  share|improve this answer












                  this comes from the e constant value 1-e^(-1) = 0.63







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 20 mins ago









                  Matthijs Huisman

                  163




                  163



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f396653%2fwhy-time-constant-is-63-2-not-a-50-or-70%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What does second last employer means? [closed]

                      Image
                      Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address only
                      Read more

                      Installing NextGIS Connect into QGIS 3?

                      Image
                      Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS 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  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve thi
                      Read more

                      One-line joke

                      A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala
                      Read more