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

• posted

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.

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John Popelish```
• posted

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 t*ts-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

```--

Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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

• posted

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.

• posted

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

• posted

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

• posted

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.

• posted

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.)

• posted

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.

```--
John Popelish```
• posted

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

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Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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

```--

Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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.
```--
John Popelish```
• posted

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

• posted

thnx, Joel. Theres 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?

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Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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and forth.```
• posted

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!

```--

Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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

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Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

OK. Thats cool. I'm subscribed now.

tnx,

steve

```--

Fat, sugar, salt, beer: the four essentials for a healthy diet.```
• posted

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

• posted
[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. :-)

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