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Ive got this problem with AC through capacitors (phase lag)

Discussion in 'Electronic Basics' started by Steve Evans, Nov 9, 2004.

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  1. Steve Evans

    Steve Evans Guest

    Helo,

    If Ive got a resistor, a battery and an ammeter, and I stick the
    battery across the resitor, I get a certain number of milliamps
    current going through the resistor.

    If I increase the voltage thruogh the resitor, I get more current.

    If I decreas the voltage, the resistor passes less current.

    So far, so good.

    Now, if I try to do the same thing with a capactor instead of a
    resistor, I get an initial curent flow that tapers off over time as
    the cap charges up and winds up with the same voltage acsross it as
    the battery.

    Again, so far so good.

    Now, if I try to stick a AC voltage accross the capacitor; everything
    goes tits-up. For part of the cycle, yeah, the current increases with
    the applied voltage and for other parts of the cycle, it decreases
    along with deminishing voltage.

    Fine again. (well not really -see below)

    The problem is, at *other* parts of the cycle, the current has started
    to reverse direction whilst the voltage is still increasing! And the
    bit further on again, the same thing happens vice-versa. Whats going
    on here?? I thoght current was proportional to applied volts? Thiese
    parts of the cylce are defying Ohms law! Can anyone explain?

    And how can current *lead* voltage? YOu have to have a voltage to get
    a current to flow in the first place!!


    tnx,

    steve
     
  2. The relationship between voltage and current for capacitors is:
    I=C*(dv/dt) which reads:
    The capacitor current (in amperes) is proportional to the time rate of
    change of the voltage across it (in volts per second), and that
    proportionality constant is the capacitance (in farads).

    So the peaks of current when a sine wave of voltage is applied across
    a capacitor occurs at zero volts, because that is when the voltage is
    changing the most rapidly.

    The current leads (by 90 degrees) the voltage only after many cycles
    have passed, so that only sine wave rules apply. The current is then
    driven by the energy stored during the previous peak. This is a sine
    wave only rule and is a special case of the above formula.

    But the more general formula applies all the time for any situation.
     
  3. john jardine

    john jardine Guest

    Yes. The current is due the voltage. It's traditionally called 'leading' cos
    it just looks that way on an oscilloscope screen. It's actually 'lagging'
    but anyone saying this risks being taken out and shot.
    Ohms law doesn't apply to caps and inductors. It's best to just look at 'em
    as energy storage devices. They'll take current from the supply and charge
    up during one part of a cycle and then stuff the current *back into* the
    supply the next part of the cycle. Net energy consumption is therefore
    zilch. Whereas a resistor constantly turns current and pro-rata volts drop,
    into a heat loss.
    Sinewaves are *extremely* non linear waveforms and for visualisation it's
    very, very, difficult to usefully sketch, draw, or graph out what's
    happening to voltages and current over 2 or 3 cycles. Many people give up
    at this point and just take the maths at face value.
    A very, very, useful learning aid though, is a Spice prog'. The R or C or L
    (or mixtures) currents and voltages can be watched exactly, point by point,
    from time=0. Also, perhaps more valuable from a learning POV is the use of a
    square wave voltage source and the knowledge that a fourier transform can
    neatly tie the markedly different sine/square output waveshapes together.
    regrds
    john
     
  4. Bob Myers

    Bob Myers Guest

    Minor nit here: Ohm's Law most certainly DOES apply
    to caps and inductors, and in fact to AC circuits in general;
    the only difference is that it has to be used in the proper form,
    which in the case of AC means that the voltage, current, and
    impedance (which takes the place of pure resistance when dealing
    with AC) have to be given as vectors, or complex quantities.
    Do that, and E=IR becomes E=IZ, and everything works
    just as it should.

    In the simplest terms I can think of, to answer the original
    poster's question: what you have to realize is that reactances
    (capacitors and inductors, in AC circuits; reactance is the
    "imaginary" portion of impedance, while resistance is the
    "real" portion) never dissipate any power (which is all
    that resistances CAN do). Instead, they sometimes are
    storing energy, and sometimes releasing it back into the
    rest of the circuit. The "current leads voltage by 90 degrees"
    (or vice-versa, for an inductor) happens basically because
    PART of the time (part of every cycle of AC), you've got
    an element of the circuit which is now acting as a source
    rather than a sink.

    Bob M.
     
  5. john jardine

    john jardine Guest

    Indeed yes. Ohm can be pressed into service nicely, -if- we pretend the
    capacitor is a resistor having a fixed ohms value.
    This though applies only when dealing with particular AC circuitry and using
    en-masse pure sinewaves that have had no awkward beginnings and will have no
    awkward endings.
    I.e. steady state answers that result from .AC type analysis used to
    characterise linear amps, filters, phasing networks etc. This is the case
    that always applies in textbooks and classrooms.
    But ... the capacitor as a component has ceased to exist, simplified to a
    linear impedance vector that can never accumulate/lose charge, or generate a
    current transient sufficient to cause fuses to blow. For calculating
    convenience, ohms law is being applied to a subset of special cases from the
    real world.

    In Steve's post I read he was coming in from problems figuring through the
    initial startup transient. (ie ".TRAN").
    This messy aspect is usually glossed over but needed to account for the
    effects of reactive components in real circuitry.
    I'd think it would be a perverse person (masochist!) who would wish to
    characterise or transform the non linear curve of the initial capacitor
    charging current over maybe the first 1.5 cycles, into its near infinite
    number of component sinewaves and associated phase angles and then
    manipulate and sum each of these vectors to allow forcable time varying
    application of ohms law to all the reactive impedances.
    Basically I'm saying that Ohm shouldn't even be considered ( my "ohms law
    doesn't apply"). Maybe considered only if some kind of steady (sinewave)
    state can exist. In many cases this will not occur or even be possible.
    I'd guess the most useful number till then is the slew=i/c item.
    regards
    john
     
  6. Bill Bowden

    Bill Bowden Guest

    Yes, but at other parts of the cycle, the source voltage
    is still higher than the capacitor voltage, so the cap will
    continue to charge even though the source is falling.

    The capacitor will stop charging and begin discharging
    *exactly* when the source voltage falls to the same
    potential as the capacitor.

    All you have to do is think about the difference in voltage
    at various points along the waveform to determine the direction
    of current.
    You don't really need a voltage to have a current flowing.
    Think about a tank circuit with a capacitor and inductor
    exchanging energy at some frequency. The current in the
    tank circuit will be maximum when the voltage across the circuit
    is zero, and visa versa.

    -Bill
     
  7. Steve Evans

    Steve Evans Guest

    tnx, bill, but that can't be quite right (with al due respect).
    The source voltage and the cap voltage are tied together with jumper
    wire with approx. 0 ohms resistance! Therefore, whatever the source
    votage is, the capactor voltage *must* be the same!

    Apart from that; *is* there *always* 90 degrees of phase lag/lead with
    reactive components, *or* can it vary. (i;ve simulated it in spice and
    have found varying amounts of phase diffrence dpending on hte values
    of hte cap and coil concerned. ) I guess the amount of reactance
    present determines jthe amount of phase shift (more reactance= more
    phase shift?)

    tnx,

    steve
     
  8. Steve Evans

    Steve Evans Guest

    Thnx, John.
    that's not the whole problem though!
    Just after that voltage zero-crossing point; the voltage is still
    increasing at *nearly* maximum rate (on the pos. half-cycle), but the
    current has already started to fall back! The real problem is at
    *this* point when the voltage is still rising steeply, and yet the
    current is now *falling* steeply! It doesn't make sense! Look at the
    point on the graph where current and voltage overlap and youll see
    what i mean.

    tnx,

    stve
     
  9. Joel Kolstad

    Joel Kolstad Guest

    Not true. But straightforward extension, you obtain the Laplace transform
    of your circuit (just replace all those 'j-omegas' with 's') and via
    convolution (multiplication in the s domain) you can find the output for
    relatively arbitrary input signals. The Laplace transforms for, e.g., step
    functions are quite 'reasonable' (not at all 'awkard').

    This is done quite commonly.
    Well, there's no such thing as an ideal capacitor anyway (and this fact
    become very significant at high frequencies), Ohm's law assume a 'lumped'
    network (no distributed effects -- this also becomes significant for many
    designers), Ohm's law comes from Maxwell's equations that are only an
    approximation of quantum electrodynamics (very significant to semiconductor
    guys), blah, blah, blah --> very quickly this all becomes philosophical.
    Circuit simulators use components that are meant to _model_ 'reality,' but
    obviosuly the results are no good if your model isn't any good.

    When doing a pure AC simulation, the net charge on the capacitor remains
    unchanged so it'd be silly for an AC simulator to consider this. If you
    want second by second charge, you'd use a transient analysis.
    It's not that messy. A sine wave that starts at time 0 when applied to a
    capacitor gets you a voltage that has an exponential term and the (same)
    sine wave term; this comes directly from the inverse Laplace transform.
    With a small source impedance, however, the exponential term decays very
    quickly and can be neglected.
    Again, this is undergraduate electrical circuit analysis. It isn't
    difficult.
    The question is, "What do you want to know?" AC analysis is quite useful
    for many circuit problems. So is transient analysis. Both can be performed
    easily by hand so long as you stick with resistors, capacitors, and
    inductors in your circuit. On the other hand, once you start sticking
    non-linear active devices into the mix, you either decide you're operating
    in the small signal domain and everything is still straightforward, or else
    (for networks consisting or more than a few components) you end up trying to
    solve differential equations which rapidly becomes intractable except via
    numerical methods.

    ---Joel Kolstad
     
  10. Joel Kolstad

    Joel Kolstad Guest

    Assuming an _ideal_ capacitor or inductor, _in the steady state_ there's
    always 90 degrees of phase lead or lag between the voltage across _that_
    device and the current through _that_ device. Now, if you go and put an L
    and C in series, across the entire thing there's certainly not (generally) a
    90 degree phase diference, but across each component there is.
    It could be:

    -- Not small enough time steps. Your output should look like a nice
    'smooth' waveform. During transient analysis, simulators choose time steps
    such that certain errors are minimized. While the voltage and each time
    _step_ will be correct (within that error bound), if your graphing a
    straight line between voltages at those time steps, it'll often not be at
    all 'nice' looking (sinusoidal). All simulators have a way of forcing them
    to limit the maximum time step they take so that you can get a 'nice
    looking' plot. (And unfortunately most simulators implement this feature in
    a rather inefficient way, but that's a topic for another day.)
    -- You're not looking at the steady state solution (you're looking too close
    to time=0 -- run the simulation for a few hundred cycles and then look)
    -- Numerical rounding in the simulator. But it should still be, e.g.,
    89.xxx degrees.
    With a single element, you have no control over the phase shift. Start
    stringing more elements together, and the reactances combine to determine
    the amount of phase shift (but again, the phase shift of _each component_ is
    still +/- 90 degrees).

    Have you taken any circuits classes? Do you know what a phasor is (no, not
    what they used in Star Trek)? A Laplace (or at least Fourier) transform?
    It's difficult to go from the qualitative to the quantitatitve without their
    use. (Not that it can't be done -- in many community college classes it
    is! -- but it tends to force you to accept a lot more on faith rather than
    being able to dervice the results yourself.)

    ---Joel Kolstad
     
  11. You are not as good at perceiving rate of change as a capacitor is.
    The slope of a sine wave is a cosine wave, and that cosine is exactly
    what the current through a capacitor connected across an AC voltage
    follows.
     
  12. Steve Evans

    Steve Evans Guest

    thnx, Joel.
    There\s obviously something still amiss with my understanding, then. I
    have a friend whos a radio ham and he says that to eliminate
    capacitive reactance, you "add a bit of inductance until it's
    cancelled out." Is this the same as cancelling out the phase-shift? If
    so, it suggests it is possible to have varying degrees of shift due to
    varying degrees of inductance. If that weren't the case, you could
    simply use *any* coil; its value wouldn't matter because whatever it
    was, it would only change the phase by 90'.
    Yes, i know what you mean, but the timestep is fine, thnks.
    So are you saying if I have 3 caps and 3 coils in series the overall
    phase shift will be zero because they all cancel eachother out? What
    if they're all different values? From what you say if you get 90
    degrees of shift (fixed) per element, that must be true, regardless of
    the different values of those elements:

    ac
    source------0.nF------1uH------22nF------4uH-------33pF-------180nH---->

    Regardless of the frequency of the source and the values of these
    components, the overall phase shift is zero. Is that what you're
    saying?
     
  13. The current through the capacitor is 90 degrees shifted with respect
    to the voltage across it. The current through the inductor is 90
    degrees shifted the other way with respect ot the voltage across it.
    So the two currents are 180 degrees shifted with respect to each
    other. This means the smaller current cancels some of the larger
    current. When you select the right sized components those two
    currents are also the same magnitude, and cancel each other,
    completely. With real components (that have loss) The currents are
    shifted slightly less than 90 degrees, so the cancellation is not
    perfect.
    (snip)
    The phase shift between voltage across each reactive component will be
    90 degrees with respect ot the voltage across that component. If a
    component has internal series resistance, there is no way to measure
    the voltage across only the reactive part of its impedance, so any
    measurement includes the voltage drop across the resistive component,
    also. But the current measurement is that of each of the series
    parts. Likewise, if the component contains internal parallel
    resistance, then the voltage measurement accurately represents the
    voltage across the reactive part, but the current measurement includes
    the resistive parallel current, so you will not see a perfect 90
    degree phase shift. Most real components exhibit both series and
    parallel resistive effects, so neither the voltage or current
    measurements are looking at pure reactance.
     
  14. Bill Bowden

    Bill Bowden Guest

    Maybe this will make more sense;

    You can't change the voltage on the capacitor, only the current
    can be changed. So, if the current is rising toward a peak,
    the voltage is also rising according to CE=IT. Now, as
    the current reaches a peak and the starts decreasing,
    the capacitor voltage is still going up because the current
    is still flowing the same way, even though it reached a
    peak and is decreasing. So, maybe you can see the phase
    shift is 90 degrees because the capacitor voltage will
    reach a peak exactly when the current stops at zero and
    starts going the other way. Therefore V is max when
    I is minimum, and visa versa.

    Does that make any sense?

    -Bill
     
  15. Steve Evans

    Steve Evans Guest

    thanks! That made some sense at least.

    [snip real-world stuff]

    tnx for that, but I'm only concerned here about perfect lumped
    elements. Lemme get a proper handle on that before worrying about the
    paracitics!
     
  16. Steve Evans

    Steve Evans Guest

    Yes, it does, thnx, Bill. I guess the source resistance has some
    bearing on it too.
    However, - and its a *big* however - I'm still finding I'm getting
    varying degrees of phase shift. I've posted a simple spice schematic
    to alt.binaries.schematics.electronic htat shows a 180' phase shift
    between voltage and current in one capacitor! I want to see hwere I'm
    going wrong so itll be interesting to read what everyonte has to say
    about this seemingly imposible feat. ;-)
    THere are two pictures: one for the circuit and one for the waveforms
    I get showing the current and voltage in antiphase.

    tnx,
    steve
     
  17. John Fields

    John Fields Guest

     
  18. Steve Evans

    Steve Evans Guest

    OK. Thats cool. I'm subscribed now.

    tnx,

    steve
     
  19. Joel Kolstad

    Joel Kolstad Guest

    Hi Steve,

    Yes, that's what he's saying.

    Here's a little bit of 'Phasors 101' without any proof whatsoever (you'll
    find it in any circuits text):

    -- The impedance (Z) of a resistor is R
    -- The impedance (Z) of an inductor is j*2*pi*F*L, where j is sqrt(-1), F is
    the frequency you're worrying about, and L is the inductance.
    -- The impedance of a capacitor (Z) is -j/(2*pi*F*C), same symbols as
    before.
    -- Ohms law is now V=I*Z, and all of V, I, and Z are typically complex.
    -- Complex numbers are often expressed as a magnitude and phase. E.g.,
    1.414+j1.414 is the same as '2 angle 45 degrees'

    Now, notice that for an inductor or capacitor, V/I is some purely imaginary
    number, that is, something with an angle of +/-90 degrees. This confirms
    what you know that the voltage and current in a capacitor or inductor leads
    or lags either other by 90 degrees.

    If I do something like put a resistor and capacitor in series and power it
    up from a 1V source, the current through _both_ will be
    I=V/Z=1/(R-j/(2*pi*F*C)) -- notice that it'll now be some arbitrary complex
    value. But the voltage across the resistor is just V=I*Z=I*R, so it'll just
    be that same complex value multiplied by a scalar, implying that voltage and
    current are still in phase. For the capacitor, V=I*Z=I*-j/(2*pi*F*C). Now,
    whenever you multiple a complex number by +/-j[something], the result is a
    complex number with a magnitude [something]*[magnitude of what your started
    with] and angle [+/-90 degrees]+[angle of what you started with]. I.e.,
    current and voltage across the inductor are still 90 degrees out of phase.
    What you're missing here is that while the phase across the an inductor and
    capacitor is always +/- 90 degrees, you can only 'null out' the two when the
    MAGNITUDE of the signal across them is equal as well. That is, if we're
    dealing with voltage, (5 angle +90) + (5 angle -90) is, indeed, zero, but (3
    angle +90) + (5 angle -90) is (2 angle -90), i.e., you still have something
    that looks like a capacitor or inductor. (This means that if you start
    with, say, a capacitor -- Z=-j*something Ohms -- and add an inductor --
    Z=j*something_else Ohms -- as the inductor gets bigger and bigger, you'll
    reach a point where the two are a short circuit. This is resonance. If you
    keep adding inductance, the network then starts looking inductive.)

    You can _not_ just 'add phases' across a bunch of components in series and
    get the overall phase change through the network. You can (and _must_) add
    _impedances_ .
    If the _impedance_ of all the caps and coils in series adds to zero, yes,
    they will all cancel each other out. This will only happen when the sum of
    the voltage magnitudes across the capacitors is equal to the sum of the
    voltage magnitudes across the inductors.
    Then the voltage magnitudes across the inductors won't sum to the same thing
    as that across the capacitors, and you'll end up with something like that
    likes a single inductor or capacitor (at one frequency).
    Again, only if you arrange things such that the magnitudes of the voltage
    across each one is the same as well.
    In your network here, let's assume the frequency is 1/6.28 to keep things
    simple. Hence, Z(res)=R, Z(ind)=j*L, and Z(cap)=-j/C. So, in your network,
    the series impedance is -j/0.1e-9 + j*1e-6 -j/22e-9 + j*4e-6 - j/33e-12 +
    j*180e-9 -- Obviously this won't sum to zero! If you pick the right
    frequency, though, it will. (Bonus question: What is this frequency?)
    At a certain frequency the phase shift is zero. Otherwise, no.

    You know, if you really want to get into this, just sign up for your local
    college's first engineering first 'circuits' (or 'networks') class. I took
    mine the summer of 1995, and it was really a blast. :)

    ---Joel
     
  20. Steve Evans

    Steve Evans Guest

    On Mon, 15 Nov 2004 17:41:16 -0800, "Joel Kolstad"

    [snip]
    [rest snipped]

    Okay Joel, I think I've finally gottit now. Some people seem to
    explain things more clearly than others and I guess your one of those
    good communicators! There is still a minor point or two I gotta work
    out, though, but the worst is over now, I reckon. :)

    Many thnx for your time.

    steve
     
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