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Series-Parallel DC RC circuit

Discussion in 'Electronic Basics' started by kayvee, Mar 7, 2009.

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  1. kayvee

    kayvee Guest

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    R2 C1
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    Hi, I'm am interested in finding the voltage across the capacitor in
    this DC circuit in time. Initially, when the capacitor is uncharged,
    I know that the voltage across it is 0, and as time goes on the charge
    will build and the voltage across it will as well. But, how do I
    calculate the voltage across it as a function of time?

  2. kayvee

    kayvee Guest

    No. But I will take a look. I believe I am fairly competent at
    understanding series / parallel circuits. Mostly, I am concerned with
    finding what I can guess would be called the "effective resistance" of
    C1 in any point in time.

  3. Jon Kirwan

    Jon Kirwan Guest

    You don't mention, but it seems implied, that your driving voltage is
    DC and fixed (it could be AC and have a DC bias, as well, for example)
    and that you have a switch which you engage at some point in time, say
    at t=0.

    The initial voltage across C1 is needed. I assume we can consider it
    zero, as an initial condition. So, obviously, at t=0, V=0. That's a
    start, if trivial. At that point, R1 experiences the entire source
    voltage -- call it Vs. So the initial current is Vs/R1. No current
    flows through R2 at that first moment, since R2 has 0V across it, so
    the current initially arrives into C1. Which raises it's voltage and
    begins some current into R2.

    So, speaking in absolute magnitudes only and not necessarily getting
    the signs correct:

    I(R1) = (Vs - V) / R1
    I(R2) = V / R2
    I(C1) = C * dV / dt

    You also know that the current into node V, arriving from R1, must be
    equal to the other two currents going out. Knowing this, let's
    rewrite the above with this additional information and from the point
    of view of node V (positive incoming, negative leaving):

    I(R1) = (Vs - V) / R1 incoming
    I(R2) = -V / R2 leaving
    I(C1) = -C1 * dV / dt leaving
    I(R1) + I(R2) + I(C1) = 0

    From this, you can substitute into the last equation to arrive at:

    (Vs - V) / R1 - V / R2 - C1 * dV / dt = 0

    Vs/R1 - V/R1 - V/R2 - C1 * dV/dt = 0

    This is a simple, linear differential equation. There are lots of
    approaches to solving them.

    Before we go there, let's just sit back and take a qualitative look at
    your circuit and make some guesses based on other experience. Let's
    assume away R2 for a moment. In that case, it is a simpler RC
    charging circuit (assuming V starts at 0, of course.) The basic
    equation for that is:

    V(t) = Vs*[1-e^(t/tau)], where tau = R2*C1

    As you can see, when t=0 then V(t=0)=0. Which is expected. As t
    approaches infinity, then V(t=infinity)=Vs. Also expected.

    What might your new circuit, with R1 added back into it? Well, if you
    are familiar with basic resistor dividers, you might guess that V will
    approach the divider voltage as t goes to infinity, instead of Vs
    itself. And you'd be right. Eventually, you'd expect the node
    voltage V to approach that value. At first, of course, it will start
    at 0V. So let's substitute that divider voltage in:

    V(t) = [Vs*R2/(R1+R2)]*[1-e^(t/tau)]

    Now, we don't know tau. It was R2*C1, but now we aren't sure. If you
    are familiar with Thevenin equivalents and have some working
    experience with them, you might guess that R2 should be replaced with
    a new equivalent that is the parallel resistance of R1 combined with
    R2 -- and you'd guess right if you did imagine that.

    The full equation should be:

    V(t) = [Vs*R2/(R1+R2)]*[1-e^(t/tau)], tau = [R1*R2/(R1+R2)]*C1

    Okay, that was a bunch of intuition working. Is it right? Or just an
    argument that doesn't really mean anything and could just as well be
    wrong as right?

    Using integrating factors, try and organize the earlier equation into
    this form: dV/dt + P*V = Q:

    dV/dt + [(1/C)*(1/R1 + 1/R2)]*V = [Vs/(C*R1)]

    P = [(1/C)*(1/R1 + 1/R2)]
    Q = [Vs/(C*R1)]

    Assuming the initial condition of V(t=0)=0, the solution (if you
    remember using integrating factors) is:

    V = (Q/P)*[1 - e^(-P*t)]

    The Q/P part works out like:

    [Vs/(C*R1)] / [(1/C)*(1/R1 + 1/R2)]
    [Vs/R1] / [(1/R1 + 1/R2)]
    [Vs/R1] / [(R1 + R2)/(R1*R2)]
    Vs / [(R1 + R2)/R2]
    Vs*R2/(R1 + R2)

    Now, that looks like what we earlier expected to see. So far, so

    Now, what about tau? Well, tau=1/P. So that's not so hard:

    tau = 1 / P
    = 1 / [(1/C)*(1/R1 + 1/R2)]
    = 1 / [(1/C)*[(R1 + R2)/(R1*R2)]]
    = 1 / [(R1 + R2)/(C*R1*R2)]
    = (C*R1*R2) / (R1 + R2)

    Wow! Looks great and matches the earlier hand-waving expectations.

    There are other ways, some far more powerful to learn. Laplace
    transforms can readily handle these kinds of differential equation
    solutions. But they can also handle more complex ones, like when your
    driving voltage isn't constant but is instead a sinusoidal driver or
    other complex waveform. The Laplace form of the earlier equation
    looks like:

    s*L(V) + P*L(V) = Q/s
    L(V)*(s + P) = Q/s
    L(V) = Q/[s*(s + P)] = (Q/P)/s + (-Q/P)/(s+P)
    L(V) = Q/[s*(s + P)] = (Q/P)*(1/s) + (-Q/P)*(1/(s+P))

    The rules aren't too complex. The constant value 1 in the time domain
    is just 1/s in Laplace. And the Laplace of a constant times a
    function is just the constant times the Laplace of the function (like
    derivatives, etc), so the Laplace of k is k/s. The reverse also
    works, so the inverse Laplace of k/s is k.

    Substituting back (inverse), we get:

    V = (Q/P) + (-Q/P)*e^(-P*t)

    which, if you remember from earlier when we got:

    V = (Q/P)*[1 - e^(-P*t)]

    is the same thing.

    (The inverse Laplace of 1/(s+P) is e^(-P*t), which I didn't mention
    earlier but where you can just look that up in a table.)

    Laplace is actually something that is a lot of fun to learn about, if
    you get a chance. But so are integrating factors and simple, ordinary
    differential equations.

  4. Jon Kirwan

    Jon Kirwan Guest

    Sorry, when assuming away R2, this should actually read:

    V(t) = Vs*[1-e^(t/tau)], where tau = R1*C1

    I had mentally changed the designations.

  5. Jon Kirwan

    Jon Kirwan Guest

    And this should be R2 above, not R1. Oh, well.

  6. Jon Kirwan

    Jon Kirwan Guest

    Well, he didn't specify what R was, as I read it (I may have missed
    something, of course.) If it was R_th, and if you read his use of
    "applied voltage" as being V_th, then it could be read as equivalent.
    I didn't like the "Vc = 1 - e^ (-t/RC)" part, as the right hand
    expression is usually meant to be taken to be unitless when written
    like that. Otherwise, it's only true if the asymptote voltage at
    t=infinity is 1V. Further, he writes "RC is actually equal to time
    __in this instance__" when, in fact, the dimensions of RC is _time_ no
    matter the instance.
    (Joule-second/Coulomb^2) * (Coulomb^2/Joule) = seconds
    This is a good thing, because e^(t/RC) then means e is raised to a
    unitless power, as should be.

    For anyone wondering where the 63%, 86%, and 95% values come from,
    just imagine (1-1/e^1), (1-1/e^2), and (1-1/e^3).

  7. Jon Kirwan

    Jon Kirwan Guest

    The full equation is:

    V(t) = V_source * [R2/(R1+R2)] * ( 1 - e^( t/[C1*R1*R2/(R1+R2)] ) )

    Note that both resistor values are needed to compute the rate while
    charging. Are you suggesting that R2 does not factor into the tau
    value during charging? (It does, so I hope not.) Maybe I'm
    misunderstanding what you write, though.

  8. No no no. The shunt resistor also divides the charge /current/, so the
    capacitor charges more slowly. The Thevenin equivalent is the
    appropriate approach.

  9. Jon Kirwan

    Jon Kirwan Guest

    But that is incorrect to say.

  10. Jon Kirwan

    Jon Kirwan Guest

    Yes. I wonder why it is so hard to get right. Particularly after
    John Larkin clearly pointed out why Thevenin works (looking at things
    from the point of view of the capacitor makes it pretty clear) and I
    also took the calculations and laid them out several ways to Sunday.

    If someone wants to disagree, it would be appropriate to show the
    errors in the calculations or approach. Not just hand-wave about it.

    In any case, I guess I'm a little bit bothered seeing Dan Coby and bg
    both appearing to get this wrong. It's just an RC problem, after all.

  11. Jon Kirwan

    Jon Kirwan Guest

    But that is wrong. The time constant is NOT based only on Rseries (R1
    in the diagram by the OP.) If you think so, you've got it wrong.

  12. Jon Kirwan

    Jon Kirwan Guest

    Ah! I couldn't tell from your last sentence. Thanks.

  13. Jon Kirwan

    Jon Kirwan Guest

    Since the cap's voltage presents a voltage across the shunt resistor,
    except at t=0, there is current present in it (except at t=0.)
    That's not a very good way to look at it. But maybe I just can't read
    your English and fit it with my mental models.

    Since I already performed the complete calculations using various
    techniques for solving differential equations, each with the same
    result -- not to mention that I tested what I wrote with LTSpice
    simulation to be absolutely certain that a spice program also agreed
    with me -- Would you care to expose the calculations in detail,
    starting from basic statements?

  14. Jon Kirwan

    Jon Kirwan Guest

    Okay. We are all good, now! We have achieved a shared reality!

  15. Right ... right ... right. I was focusing on the "All the shunt
    resistor does is divide the source voltage ..." part & got sloppy. Bob
  16. Jon Kirwan

    Jon Kirwan Guest

    I mean,
    V(t) = V_source * [R2/(R1+R2)] * ( 1 - e^( -t/[C1*R1*R2/(R1+R2)] ) )

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