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RC Circuit rise time

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    RC.jpg

    Calculate the rise time (the time require for the output to go from 10% to 90% of its final, or steady state, value) for the circuit shown above when a step input of 5V is applied for the following resistor values: 1k, 10k, 100k, 1000k with C equal to 0.01uF.


    2. Relevant equations
    I believe either

    Vt = Vo * exp(-t/RC)

    or

    Vt = Vo * (1 - exp(-t/RC))


    3. The attempt at a solution
    For the sake of simplicity, I'll just show you my 10k Resistor calculations. I took that second equation (not even sure if I should be using that?) and for Vt I multiplied my source voltage (which should be the final voltage of the capacitor before discharging, right?) by 0.1 for 10% of the final which is 0.5 so:

    0.5 = 5 * (1 - exp(-t/1E-4))

    I then distributed the 5 and took the natural log of both sides so:

    -0.693 = 1.61 - (-50000t)
    -2.303 = 50000t

    so t at 10% of the final = -46us

    Now the fact that I got a negative number indicates that I did something incorrectly. I honestly have no clue what to do as the question seems vague as to how to answer it. If anyone can help I'd appreciate it a lot.

    (also, as a note, for the first equation I got the number 14us at 10% of the final using .5 = 5(exp(-t/1E-4)). Don't think that's right either?)
     
  2. jcsd
  3. Feb 8, 2010 #2
    I may have figured it out, I simulated it in PSpice and worked around with a few new ideas and these numbers seemed to match up...

    Vt/vo = 1 - exp(-t/RC)

    The ratio of Vt/vo is the percentage of the final state so for 90% of the final it would be .9

    .9 = 1 - exp(-t/RC)
    -.1 = -exp(-t/RC)
    .1 = exp(-t/RC)
    ln(.1) = -t/RC
    RC * ln(.1) = -t
    1E-4 * ln(.1) = -t
    t = 230us

    and t for 10% is 10.5us

    Is that correct? Or am I way off? I seriously feel like I may be pulling all of this out of my backside :/
     
  4. Feb 9, 2010 #3
    You are right Fronzbot, you are using the correct equation and your algebra checks out. Be careful evaluating exponential functions as negative sign errors can do funny things like giving you negative time results that are incorrect. Below is an image of Vt/Vo with respect to time With a time constant of 0.1ms. The points where Vt/Vo is .1 and .9 are noted.

    http://img402.imageshack.us/img402/1924/rcplot.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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