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physical/intuitive understanding of RL/RC time constants?

Discussion in 'Electronic Basics' started by Alan Horowitz, Oct 12, 2004.

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  1. when a current just starts flowing into a RL or RC circuit, how does
    the voltage "know" that it should be increasing exactly 63% during
    each time-constant period?

    And whence the number 63%?
     
  2. Uncle Al

    Uncle Al Guest

    1-(1/e). Crack your textbook.

    <http://www.iop.org/EJ/S/UNREG/aS0aO...e/-featured=jnl/0143-0807/23/1/304/ej2104.pdf>
    "Demonstration of the exponential decay law using beer froth"

    Google
    "exponential decay" 63 20,500 hits

    Uncle Al gotta think of everything.
     
  3. John T Lowry

    John T Lowry Guest

    Definition of time-constant period.

    John Lowry
    Flight Physics
     
  4. This all goes back to the solution of the differential equation for
    the RC or RL system. e is a natural constant that has some very sweet
    properties in many applications of mathematics, and simplifying
    differential equations is one of them. Read through this tutorial and
    see how the rate constant k in this tutorial is an example of a time
    constant.
    http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/intro.html
     
  5. I haven't had a chance to read other responses, but here's mine:

    Take the case of an RC:

    ,---,
    | |
    V| \
    --- / R
    - \
    --- |
    - +----->
    | |
    o --- C
    / ---
    o |
    '---+----->

    Assume C is discharged and V has just been applied by closing the switch...

    The current through R is based on V, less the voltage on C (which counters V),
    so:

    I(R) = ( V - V(C) ) / R

    The above is a function of time, because V(C) is a function of time. So, what's
    V(C)? Well, that needs to be arrived at more slowly.

    First, we know that this is true:

    Q = C*V

    Well, actually, that's an average statement. More exactly, it's:

    dQ = C * dV

    In other words, the instantaneous change in Coulombs is equal to the capacitance
    times the instantaneous change in voltage. Both sides can now be divided by an
    instant of time to give:

    dQ / dt = C * dV / dt

    Since dQ/dt is just current (I), for the above capacitor this becomes:

    I(C) = C * dV(C) / dt

    So how does this help? Well, we know that the current from R must accumulate on
    C. So, we know that:

    I(C) = I(R) = ( V - V(C) ) / R

    so, combining, we get:

    C * dV(C) / dt = ( V - V(C) ) / R

    Rearrangement of this gives:

    dV(C) / dt + V(C) / (R*C) = V / ( R*C )

    Which is the standard form for ordinary differential equations of this type.

    The standard form with general terms looks like: dy/dx + P(x)*y = Q(x). In our
    case, though, y = V(C), x = t, P(x) = 1 / (R*C), and Q(x) = V / (R*C).

    The solution to this includes multiplying by what is called "the integrating
    factor", which is:

    u(x) = e^(integral (P(x)*dx)) = e^(integral (dt/RC)) = e^(t/(R*C))

    (This is a VERY POWERFUL method to learn, by the way, and it is probably covered
    in the first few chapters of any ordinary differential equations book.) So,
    going back to look at the general form and multiplying both sides:

    u(x)*dy/dx + u(x)*P(x)*y = u(x)*Q(x)

    But the left hand side is just d(u(x)*y)/dx, so:

    d(u(x)*y)/dx = u(x)*Q(x)

    or,

    d(u(x)*y) = u(x)*Q(x) * dx

    In our case, this means:

    d( e^(t/(R*C)) * V(C) ) = V / ( R*C ) * e^(t/(R*C)) * dt

    Taking the integral of both sides, we are left with:

    e^(t/(R*C)) * V(C) = integral [ V / ( R*C ) * e^(t/(R*C)) * dt ]
    = V / ( R*C ) * integral [ e^(t/(R*C)) * dt ]

    setting z = t/(R*C), we have dz = dt/(R*C) or dt = R*C*dz, thus:

    e^(t/(R*C)) * V(C) = V / ( R*C ) * R*C * [e^(t/(R*C)) + k1]
    = V * [e^(t/(R*C)) + k1]
    = V * e^(t/(R*C)) + V * k1
    V(C) = V + V * k1 / e^(t/(R*C))
    = V + V * k1 * e^(-t/(R*C))
    = V * [ 1 + k1*e^(-t/(R*C)) ]

    From initial conditions, where V(C) = 0V at t=0, we know that k1=-1, so:

    V(C) = V * [ 1 - e^(-t/(R*C)) ]

    Time constants are usually taken to be:

    e^(-t/k)

    with (k) being the constant. In our V(C) case, this means that k=R*C. So
    that's the basic constant and it's in units of seconds.

    So, what's the voltage after one such constant of time? Well:

    V(C) = V * [ 1 - e^(R*C/(R*C)) ] = V * [ 1 - 1/e ] = V * .63212

    Ah! There's that 63% figure. Actually, more like 63.212%.

    Two time constants would be:

    1 - 1/e^2 = .864665

    and so on....

    Oh... and there are other methods you can use to solve the simple RC formula,
    but the method I chose is a very general and powerful one worth learning well.

    Jon
     
  6. Suppose you are trying to fill up a box with balls. However, for some
    strange reason, you've decided that each time you throw in balls, you'll
    throw in 1/2 of the balls that will fit in the remaining space.

    At the first second, you have 1/2 the balls. Next second, you'll have
    that plus 1/2 of the remaining space, which is 1/2 + 1/4 = 3/4. The
    third second, you'll have that plus 1/2 the remaining space, ie, 1/2 +
    1/4 + 1/8 = 7/8...

    So, the number of balls at any time t will be:

    B(t) = 1 - (1/2)^t

    Thus, after 3 seconds, there will be B(3) = 1 - (1/2)^3 = 1 - 1/8 = 7/8,
    just like above.

    Now, apply that same reasoning, only instead of using the ratio 1/2, use
    the ratio 1/e (since we are applying arbitrary rules)

    Then

    B(t) = 1 - (1/e)^t

    After the first second, you'll have

    B(1) = 1 - (1/e)^1 = 1 - 1/e = 0.632 (that is, 63%)

    Strange coincidence, isn't it? It happens because when you are charging
    a capacitor through a resistor, you are throwing balls, in the form of
    charges, into a box (the capacitor), and the number of charges you throw
    at any given time (the current) depends on how many charges are already
    on the capacitor (the voltage).

    Each step of the formula above is one time constant, RC. By dividing out
    the RC, you can get the answer given seconds, ie

    B(t) = 1 - (1/e)^(t/RC) = 1 - e^(-t/RC)

    Where B is the percentage 'filled' the capacitor is (ie, what percentage
    it is of the input voltage).

    Why is 1/e used instead of 1/2? That has to do with the fact that we
    must have a continuous solution, not a solution based on ratios of
    existing values; the rate of change of the current (ie, how many balls
    we throw in per unit time) is proportional to the voltage remaining,
    which is continuously changing. Using 1/e instead of 1/2 allows us to
    generalize to this, in the same way as the compound interest formula
    allows us to compute 'continuously compounding' interest.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  7. Ban

    Ban Guest

    Let's try another way. You can actually experiment yourself.

    Take a 1000uF electrolytic cap and charge it up to +5V, then disconnect the
    power supply. Put your voltmeter on the cap terminals and read +5V. Now take
    a 10k resistor and put it across the terminals as well. The cap discharges
    slowly. In the first moment the discharge current was 5V/10k = 0.5mA. With
    this current the cap would be discharged in 10s, this is the time constant
    "tau" = RC

    But since the voltage is dropping also the discharge current drops. Now you
    can use a stopwatch and read the voltage after 10s and you find it to be
    1.84V, which is (1-0.624)5V. So this is where your 63% come from. Since we
    are discharging, the value is 37% of the initial voltage. You can also note
    down the values for 20s, 30s etc. until your meter has no more resolution
    and find the corresponding values for multiple time constants.
    BTW you do not need to do this experiment yourself but use a simulator or
    solve the equations others have already written in their answers.
     
  8. john jardine

    john jardine Guest

    The voltage knows nothing about how it's "supposed" to behave. It just does
    its thing without a care in the world.
    The thing it does though, will always result in exactly the same voltage
    shape, because with a fixed R and C and supply voltage it can do no other.
    As the C voltage grows, the voltage across the R must drop. If the R voltage
    drops then the charging current must drop. If the charging current drops,
    then the C voltage must rise at a slower rate, ... and so on and so on ...
    Everything slows down more and more as time goes on. A bit of thought and
    you'll notice that the C can never actually charge exactly to the supply
    voltage.
    As this RL RC (dis)charging process must always result in this particular
    shape or curve and this quite 'natural' curve turns up across all branches
    of science, engineering and finance, it wasn't long before the
    mathematicians found they could usefully model, or describe the curve
    accurately, using an equation based on the 2.718 "e" value used for working
    out 'natural' logarithms.
    Hence the maths numbers and formulae that are taught are a good descriptive
    model or analogue of what's happening in the circuit but have nothing to do
    with the actual circuit workings.
    Be wary when relying purely on maths models. They confer 'expertise' into
    how something works, without offering 'understanding' of how something
    works. The difference can be crucial.
    regards
    john
     
  9. Steve Nosko

    Steve Nosko Guest

    **** Y I K E S !!!



    --
    Steve N, K,9;d, c. i My email has no u's.


     
  10. Steve Nosko

    Steve Nosko Guest

    Alan,

    John Popelish got a good start with "e is a natural constant that has
    some very sweet
    properties in many applications of mathematics, and simplifying..."



    Then, it looked as thought John Jardine was going to steal my thunder with
    "the voltage knows nothing about how it's "supposed" to behave. "



    This could resolve to a mater of faith Alan.



    Indeed, the voltage/current "knows" nothing.



    After observing what happens in such circuits, "we" (those who must
    understand all things) very carefully examined what was going on and
    "discovered" that there were mathematical expressions or equations which
    would model what happens in nature. "We" came up with theories about what
    was going on and what was causing it to happen. "We" then found ways to
    make the math fit reality. In the case of time constants, we have a natural
    phenomena which is very nicely described by the equations stated elsewhere
    in this thread (the 1/e thingy). It is just like the F=MA equation. "We"
    discovered that the force applied to a mass is equal to the mass times the
    acceleration. The Mass knows nothing about force, acceleration or
    mathematics. We found that this math describes nature.



    It is exactly like a model airplane (or whatever). We make the model to
    look like the real thing. The real thing knows not of the model that we
    built, but if we did a good job, I or you can now look at the model and
    "know" just how the real thing looks.



    The math behind all of our sciences is just like this. *WE* found math
    which models reality and because we did such a good job, we can now "do the
    math" and "know" how the real thing should behave.



    To be a little more specific, in the case of the time constant. we have
    theories about current flow, charge, capacitance, inductance magnetism and
    resistance which are borne out by countless experiments and then by
    subsequent usage. These theories have all had mathematics fitted to them,
    and by golly everything fits. We can now plug-in values to equations till
    the cows come home and holy-cripes! The real thing does just what the math
    predicted. Based upon the properties we have observed for each type of
    component, this math works out such that this 1/e thingy fits just right.



    In other words, the answer is: "It just does!"

    73,
     
  11. ;)

    Jon
     
  12. Guest

    Nice job Robert, I really liked it
    Art
     
  13. Reminds me of Galileo writing in "The Assayer," saying:

    "Philosophy is written in this grand book-I mean the universe-which stands
    continually open to our gaze, but it cannot be understood unless one first
    learns to comprehend the language and interpret the characters in which it is
    written. It is written in the language of mathematics, and its characters are
    triangles, circles, and other geometric figures, without which it is humanly
    impossible to understand a single word of it; without these, one is wandering
    about in a dark labyrinth."

    (By the way, to anyone who has NOT actually read The Assayer from beginning to
    end, I highly recommend it!)

    Mathematics is a wonderful world all of its own, independent of nature, yet
    where it often turns out that insights in that world happen to happily suggest
    relationships found in this world and where proper deductions there imply proper
    deductions here. The language is sufficiently rigorous that someone two
    millennia before me can describe a circle using it and I can read it today,
    knowing absolutely nothing about their lives, their fads or interests, their
    politics or style of dress, and come away with exactly the same image in mind
    with exactly the same deductive power. In short, mathematics is a quantitative
    language that speaks across culture, time, and place. And there is nothing we
    have to compare with that.

    The processes of science work to achieve a relatively objective process that
    works well. It requires the use of objective language sufficient for rigorous
    quantitative deductions (by anyone adequately trained in the language) to
    specific circumstances, insists that such language both explain past results
    well and (more importantly) also make accurate and repeatable predictions,
    requires quantitative prediction for discernment, and requires time and patience
    for the resulting critical opinion of others skilled in the field to arrive at a
    consensus. But mathematics *is* a key part of this objective language used in
    science because of its demonstrated congruencies with nature.
    One thing to keep in mind is that ideas like "density," a useful relationship
    between volume and mass, are truly discovered through hard work and through
    trying to find some kind of useful discernment regarding sinking and floating.
    One doesn't just naturally _know_ about density, as our direct senses tell us
    nothing of the kind. It's discovered and then taught and learned. And such
    relationships are about parsimonious tools for prediction.

    And yes, we have been fortunate that some math describes some nature.
    It can also be that the model ignores some of the unimportant details of the
    "real thing" and still be quite useful. Or that it ignores some important
    details, but that so long as we keep those boundaries and limitations in mind
    the model is still quite useful for many other things.

    I like your example.
    We can also disappear into the mathematical universe and discover brand new
    relationships there and have some expectation that where such new territory is
    true there, it will probably be found true in the real world as well. One can
    make important discoveries using mathematics and use them to suggest what can be
    searched out and found here. Surprising, at times.
    Yup. In the capacitor case, for example, I idealized it as a simple
    differential equation. Real capacitors are more complex, but the ideal is often
    close enough in practice to be useful.

    Enjoyed seeing your thunder!

    Jon
     
  14. Thanks Art. I enjoyed thinking about it and writing it. The exponential
    function is everywhere in nature, and, despite all the mathematical
    machinery required to analyze it in detail, its a pretty simple concept.

    Regards,
    Bob Monsen


    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  15. Jim Black

    Jim Black Guest

    First, I should make a few corrections to what you've said. The
    voltage doesn't increase 63% during each time-constant period. If it
    did, it would become 1.63 times its original value after the first
    period, then 1.63*1.63 = 2.66 times, then 1.63^3 = 4.33 times, etc.
    The voltage would increase forever, and the circuit would fry itself.

    I'm assuming you're talking about a battery connected to a resistor
    and capacitor (or inductor) in series, or similar circuits. In the
    case of the capacitor, the voltage across the capacitor starts from
    zero, and in the first time-constant period, the voltage across the
    capacitor rises to 63% of the voltage across the battery. In each
    time-constant period thereafter, it increases by 63% of the difference
    between the battery voltage and its previous voltage.

    Sorry if you knew this already, but it's hard to tell from just
    reading a post.

    It also isn't exactly 63%. The number predicted by theory is
    63.21205588285576784044762298385 ... %. But that number will not be
    exact either, due to the fact that the models which predict it assume
    perfect wires, perfect resistors, etc., which is never the case. At
    some point one has to round.

    Let's examine such a circuit at at some time after the switch has been
    flipped. We'll call the voltage of the capacitor at this point V and
    the voltage of the battery E, the value of the resistor R and the
    value of the capacitor C.

    The voltage across the resistor is E-V, making the current through it
    (E-V)/R. The charge on the capacitor is CV. If the capacitor were
    fully charged, it would have a charge CE. That means the capacitor
    needs CE-CV more charge to be fully charged. If the current through
    the resistor continued to flow at its current rate, the time it would
    take to supply this charge would be CR. This time is called the time
    constant.

    It's crucial to notice that the time constant we just computed does
    not depend on how far the capacitor has been charged. At any time, it
    is correct to say that if the current through the capacitor would only
    stay the same, it would be fully charged after an amount of time later
    given by the time constant.

    Of course, the current through the resistor will not continue to flow
    at its current rate. As capacitor is charged, the voltage across the
    resistor decreases, and with it, the current.

    Let's pretend the current really does stay constant for a small
    fraction 1/n of the time constant, that is, for the short time CR/n.
    In this time, the capacitor's voltage would increase by 1/n of the
    difference between the battery's voltage E and its present voltage V.
    The voltage E-V across the resistor would decrease by the same amount,
    making it (1-1/n) of what it was before. After one time-constant, the
    voltage across the resistor would have decreased by the same ratio n
    times, making the voltage across the resistor (1-1/n)^n times its
    previous value.

    This all still an approximation; the current doesn't really stay
    constant for any length of time. But if CR/n is very small, the drop
    in current is insignificant. If we choose larger and larger values of
    n, the answer we get becomes better and better. If we choose an
    extremely large n, the error will go away with rounding, and we will
    get (1-1/n)^n = 0.3678794... . And if the difference between the
    battery voltage E and the capacitor voltage V is 37% of what it was
    before, that means that the voltage of the capacitor increased by 63%
    of that difference.

    The reciprical of 0.3678794..., 1/0.3678794... = 2.7182818..., appears
    in the answers to a lot of problems, and so it has been given a
    special name: "e." It is near [n/(n-1)]^n, for large n. We can
    simplify this formula even further by replacing n by m+1, making it
    (1+1/m)^(m+1), and dividing by (1+1/m), which is very near 1, to get
    (1+1/m)^m, for large m.

    While in principle one could compute e by choosing a very large n, in
    practice you need such a large n to get a correct answer that the
    calculation that ensues becomes tedious. But, using the binomial
    theorem, we can find:

    (1+1/m)^m = 1 + m(1/m) + [m(m-1)/2!](1/m)^2
    + [m(m-1)(m-2)/3!](1/m)^3 + ...

    (1+1/m)^m = 1 + 1 + (1-1/m)/2! + (1-1/m)(1-2/m)/3! + ...

    (1+1/m)^m ~ 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ... = e

    We can do a similar calculation to find a series for 1/e from "near
    (1-1/n)^n for large n." It all works out the same, except that we're
    dealing with powers of (-1/n) instead of (1/n). Terms that had odd
    powers have their sign flipped:

    1/e = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7! + ...

    or

    1/e = 3/3! - 1/3! + 5/5! - 1/5! + 7/7! - 1/7! + 9/9! - 1/9! + ...

    1/e = 2/3! + 4/5! + 6/7! + 8/9! + 10/11! + 12/13! + ...

    1/e = 1/1!/3 + 1/3!/5 + 1/5!/7 + 1/7!/9 + 1/9!/11 + 1/11!/13 + ...

    With just the first two terms, which you can add in your head, you can
    get 1/e accurate to two significant figures.

    --
    Jim Black

    "Within the philosophy system, it is quite correct. Let's try this: if
    it was in single quotes it would 'mean' "chaos". As it is not, there
    'is some form'." -- Peter Kinane
     
  16. Steve Nosko

    Steve Nosko Guest

    Jonathan,
    you put this to some nice words.
    Steve K9DCI
     
  17. John Larkin

    John Larkin Guest


    How about this:

    Charge a 1 farad capacitor to 1 volt and slap a 1 ohm resistor across
    it. The resistor current is 1 amp, so the cap discharges, and the
    voltage is at first declining at a rate of 1 volt per second. But
    1/100 of a second later, the voltage is 0.99 volts, so the current is
    only 0.99 amps, so the rate of discharge is only 0.99 volts per
    second.

    So we write a Basic program:

    v = 1 ' charge the cap

    for t = 1 to 100 ' then, for 1 second at 0.01 sec steps,

    v = v - 0.01 * v ' discharge the cap by 1%

    next

    print v ' voltage is this, 1 second later


    which simulates what I was doing above, but for a full second. The
    value of v at the end is 0.36603 volts. That's close to 1/e, not exact
    because I took 100 discrete steps, as an approximation to
    continuous-time math. With 1000 steps, simulating 1 second of
    discharge in 1 millisecond steps, you get 0.367700, even closer.

    'e' is just nature's answer to a natural discharge curve.

    John
     
  18. Bill Bowden

    Bill Bowden Guest

    It just turns out to be 63% if you add up all
    the little voltage changes on the capacitor as it
    charges during one time constant. The short basic
    program below uses 10000 samples and keeps track
    of the resistor voltage at each step and then prints
    out the final capacitor voltage of 63%. The accuracy
    can be improved by taking more samples. Change time
    to a smaller value for a better approximation.

    Voltage =1
    Resistance =1
    time = .0001
    Limit = 1/time
    For n = 1 to Limit
    Current = Voltage/Resistance
    Voltage = Voltage - (Current*Resistance*time)
    Next n
    Print "Capacitor voltage = "; (1-Voltage)*100;"%"

    'Answer = 63.2139444%

    -Bill
     
  19. John Larkin

    John Larkin Guest

    Hell, the average programmer couldn't set himself on fire.

    Actually, I've done a lot of pll and control system simulation in
    Basic. I really began using an HP 9100 desktop calculator (for
    steamship throttle control system simulation), graduated to FOCAL,
    then RSTS/E Basic+, QuickBasic, and now PowerBasic.

    I agree, C isn't logical. Why this bizarre and ugly hack of a PDP-11
    assembler got to be the programming standard is beyond me. We're
    paying the price big time.

    John
     
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