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0.693??? RC Time constant

Discussion in 'General Electronics Chat' started by shaneshane1, Dec 15, 2007.

  1. shaneshane1

    shaneshane1 New Member

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    Hi

    I know that to find R in a RC Time constant is to do the following

    eg: 2(seconds) / 0.000470(farads) / 0.693 :confused: = 6140ohms

    what does 0.693 represent, i thought that it might have been 0.7V?
     
  2. kchriste

    kchriste New Member Forum Supporter

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    Last edited: Dec 15, 2007
  3. dknguyen

    dknguyen Well-Known Member

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    The way the resistor-capacitor charging equation works out, if you set the time duration to be equal to R*C (the RC time constant which also appears in the equation) then the capacitor will have charged to 63% of the full voltage, or discharge down to 100%-63% = 37% of the full voltage.

    The 0.693 number used for 555 timer calculations comes from how the voltage charges back and forth between 1/3V and 2/3V, while the RC time constant is for charging and discharging between 0V and +V.

    ...I think. I've never actually sat down and did the calculations. THat's just my reasoning. I'd like to know too actually.
     
    Last edited: Dec 15, 2007
  4. dave

    Dave New Member

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  5. Diver300

    Diver300 Well-Known Member

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    If you want to find the time constant, you don't need the 0.693

    The time constant is just RC

    The time constant is the total change divided by the initial rate of change.
    For example, 470 :mu:F and 10 k:eek:hm: resistor, time constant is 4.7 s.

    If the capacitor is charged to 5 V, the current is 5 / 10000 = 0.5 mA
    0.5 mA causes the capacitor to change at 0.0005 / 0.00047 = 1.06 V / second.
    Time constant is 5 / 1.06 = 4.7 s, and it doesn't matter what starting voltage you chose.

    (Time constants apply to all sorts of things, not just electricity)

    Because the voltage falls, the current falls, so the rate of change of voltage also falls. It's called exponential decay.

    The formula is V = Vstart * e^(-t/RC)
    where V is the voltage on the capacitor
    Vstart it what it starts at
    e is the mathematical constant = 2.717......
    t is the time in seconds
    R is the resitance
    C is the capacitance

    The way that time constants are often measure or used is with "half-lives", or the time to get to half the start value.

    If the time to get to half the start value is Th, then:-

    0.5 = e^(-Th/RC)

    This gives Th/RC = 0.693 and that is where 0.693 comes from.
     
  6. audioguru

    audioguru Well-Known Member Most Helpful Member

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    0.693 is used to calculate how long the capacitor charges to 67% of the supply voltage then discharges to 33% of the supply voltage in a 555 astable oscillator circuit.
     

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  7. GuiRitter

    GuiRitter New Member

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    The Secret

    hi y'all! i came here from google and i will reveal the secret! i also fought a lot to find out where it came from, i found it and i forgot it... but today i found it again! prepare yourself:

    it's the natural logarithm of two!

    don't believe me?
    Multivibrator - Wikipedia, the free encyclopedia

    from windows' calculator: ≈ 0,69314718055994530941723212145818

    i hope i made y'all happy!
     
    • Like Like x 1
  8. Abizzle

    Abizzle New Member

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  9. jpanhalt

    jpanhalt Well-Known Member Most Helpful Member

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    Welcome to ETO

    That was 6 years ago. Why are you still concerned about it in light of post #6?
     
  10. Abizzle

    Abizzle New Member

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    Sorry, I was recently looking into 555 timers and it's astable operation, and I saw the 0.693 number and I looked it up. This post came up first in the results, so I just wanted to share my results in terms of what I found was helpful. And also the fact that I didn't check Wikipedia :banghead:
     
  11. jpanhalt

    jpanhalt Well-Known Member Most Helpful Member

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    There are always first, second, and n^th order effects. Still, why are you concerned? Do you really think the published equations for the 555 apply to the 10^-4 decimal?
     
  12. Ratchit

    Ratchit Well-Known Member

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    The rise and fall of voltage in a RC circuit is determined by a differential equation (DE). That is because the current is dependent on the rate of voltage change across the capacitor. Attached is the derivation and solution of the DE. As you can see form the result, the constant 0.6321 comes from the voltage rise in one time constant and the constant 0.3679 comes from the voltage fall in one time constant. Ask if you have any questions.
    shaneshane1.JPG
    Ratch
     
  13. kinarfi

    kinarfi Well-Known Member

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    When I was much younger, learning electronics, ".693" was just one of those constant you learned, like the √2 and .707, the reciprocal if √2. Later I saw that .693 number being used in the calculation of the flow of water through a hole in a container, the illustration had a vertical pipe full of water with a hole in the side, at the bottom, and the statement that the pipe would be empty after 5 time constants, calculated by the size of the hole, the diameter, and the height of the water in the tube, just like the charge or discharge of a capacitor. I was impressed that an electronics number was used elsewhere. Back then, we didn't have computers, internet or Google to research such interesting discoveries, so I just took it to be one of those constants that applies to nature. Now , I have the computer, internet, E. T. O., and google, I just googled ".693" and learned a lot. Try it.
    Jeff
     
  14. Ratchit

    Ratchit Well-Known Member

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    Yes, 0.693 = ln(2), but that constant does not show up in the decay or rise of a RC circuit, which the OP was asking about. e^-1 = 0.37 or (1-0.37) - 0.63 does, however.

    Ratch
     
  15. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I think time constants (capacitor charging and discharging times) are easy to see on a graph:
     

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  16. Ratchit

    Ratchit Well-Known Member

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    True, and also easy to calculate once the formula is known. I derived it is post #11 of this thread.

    Ratch
     
  17. kinarfi

    kinarfi Well-Known Member

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    The OP did ask about .693 and Audioguru covered it with graphs, what I learned is that after 5 TC , the charge or discharge was considered done.
    If you start a charge at 0 volt with 100 volts applied, after 1 TC1 the voltage is 69.3 volts,
    during TC5, the capacitor charges 69.3% of remaining 0.888287 (0.615583) so now the voltage is 99.7273
    during TC6, the capacitor charges 69.3% of remaining 0.272704 (0.188984) so now the voltage is 99.91628
    The table carries the process out to TC11
    693 TC.JPG
    Jeff
     
  18. Ratchit

    Ratchit Well-Known Member

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    Your premise is false and your table is incorrect. Ln(2) =0.693147 is not involved in the energizing or de-energizing of a capacitor. The correct value for your example is 100(1-e^-1) = 63.2121 for the first time constant (TC) and 100(1-e^-5) = 99.3263 for 5 TC's.

    Ratch
     
    Last edited: Dec 2, 2017
  19. kinarfi

    kinarfi Well-Known Member

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    I just don't know, maybe electronics has change since I graduated, we only had slide rules back then and
    It was close enough
    693 TC#2.JPG
     
  20. Ratchit

    Ratchit Well-Known Member

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    I did 1/e on my old slide rule and got 0.368, accurate to 3 significant figures. Anyhow, '"good enough" depends on the application. The important point is that you do not think that the rise or decay of a RC circuit is logarithmic instead of exponential.

    Ratch
     
  21. AnalogKid

    AnalogKid Well-Known Member

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    Really? What course were you in where an 8th grade level math problem was allowed to have an error of almost 10%?

    ak
     

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