Capacitors and Inductors

[peter d]

As I mentioned in this thread , I failed at my attempt to become an electrical engineer.

Anyway, I'm trying to wrap my head around some principles that admittedly are a little difficult to understand to someone with an incomplete engineering background.

I understand the basic principles of operation of capacitors and inductors, but I'm still not satisfied with my level of understanding.

So I pose my questions to the fourm:

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• <font size="2" face="Verdana, Helvetica, sans-serif">Why do capacitors oppose changes in voltage?</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">Why does current come first with C?</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">Why do inductors oppose changes in current?</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">Why does voltage come first with L?</font>

<font size="2" face="Verdana, Helvetica, sans-serif">
I had an EE explain the answers to these questions for me before, but I'm the type of person that needs difficult topics like this explained several times to fully understand them.

[ January 05, 2005, 09:53 PM: Message edited by: peter d ]

 

[physis]

Re: Capacitors and Inductors

I'm not an engineer either Peter but I'll see if I can at least fill some of the holes with putty.

[1] Capacitors opposing voltage:

If you start with a discharged capacitor.
Then you apply a voltage across it.

What happens?

Initially, the current is huge, it acts like a short. The voltage slowly rises as the plates charge and the current drops.

Remove the voltage source.

What happens?

The capacitor begins discharging, initially keeping the voltage up, decreasing over time.

It's opposing the change in voltage.

I'll be back for some of the other stuff. The girl wants me to make dinner.

 

[physis]

Re: Capacitors and Inductors

Oh, current comes first because it's initially a short. Volts down, current up.

Edit: It's no different in AC except that it never gets static or normalizes.

[ January 05, 2005, 10:29 PM: Message edited by: physis ]

 

[jtester]

Re: Capacitors and Inductors

Peter I didn't fail at becoming an ee, and on my good days things still are foggy. You don't need to apologize or explain. I'll take a shot at the capacitor.
Imagine a capacitor is a water tank with a pipe in and a pipe out. When you turn the water on, the pipe has to fill the tank before you can measure pressure between the inlet pipe and the outlet pipe.
The same thing happens with a cap. When you energize it, it has to "fill with charge" before there is a voltage across it. When the 1/2 cycle changes, the cap has to drain and then refill. That is why you get a current flow before there's voltage across it.
I'll try an inductor in another post, I've got to get to work. Hope that helps.
Jim T

 

[steve66]

Re: Capacitors and Inductors

I'll take a stab at:

Why do inductors oppose a change in current?

If you start with a inductor, and apply a voltage, as the current starts to flow, it produces a magnetic field around each turn of wire. As the current increases, the magnetic field increases. As the magnetic field increases, it cuts through other turns of the inductor (i.e. when the magnetic field grows, it wraps around additional turns of the inductor). When a magnetic cuts a conductor, a voltage is induced in that conductor.

The result is that as current starts to flow, it induces a counter-voltage in other turns of the inductor that resist the applied voltage, and any increase in current.

The same thing happens when the voltage applied is decreased, except the magnetic field collapses, and the energy in the field is returned to the circuit.

Steve

 

[peter d]

Re: Capacitors and Inductors

Thanks for the responses so far. I should state that I am a lot more confused by capacitors than inductors.

Steve,

That makes sense. The back EMF produced by the coil opposes the voltage that produced it in the first place, and that initially limits the current until a "steady state" is reached, and the current begins to flow.

Apply more current, and more back EMF is produced, which again limits the current flow for 5 time constants.

Correct??

 

[g1matrix]

Re: Capacitors and Inductors

Peter:

I remember the catch phrase, ELI the ICE man, in order to remember that in an inductor voltage leads current. In other words what the current i does with respect to the voltage.
to define the voltage across an inductor v = L(di/dt) this gives an indication how the inductor responds to current and voltage.
This says that the voltage across an inductor is proportional to the time rate of change of the current through it. Meaning in order to have a voltage across an inductor there must be a time varying current hence AC. Also it shows that if there is no current change in the circuit, there is no voltage across the coil. Then this inductor will look like a short circuit to DC. of course once the inductor has charged.
Another fact is that a sudden change in current must be associated with an infinite voltage across the inductor.
Another way to look at the definition is take the (inductance x 'delta' i/ 'delta' t) any change in current for a finite change in time will cause a large voltage to be present across the coil.( a good example of this abrupt change would the result of opening a field switch to a DC motor, this shows the whole principle of the energy stored in a coil, by opening the switch this is an abrupt change in 'i' and in a very short period of time and the Arc is incredible for a brief moment until the field coil collapses and dumps all of its energy, of course the motor is now very unstable and will runaway with itself , unless the switch is reclosed, but for the purposes of demonstration this is classic)
ELI comes from the fact that the change in current must be accomplished first before there is a voltage to be considered.

Gary

 

[rattus]

Re: Capacitors and Inductors

I don't think anyone has quite said it right yet, so I will put in my own 5 cents worth:

Inductors:

A changing current in an inductor creates a changing magnetic field in the inductor core which INDUCES a counter emf in the windings and the counter emf OPPOSES the change in current. This counter emf is the reason that the current cannot change instantaneously. In all cases this emf is:

1) emf = -L(di/dt);

where di/dt is the change in current per unit time in Amps/Sec, and L is the inductance in Henrys.

In a pure inductor, this emf would be equal and opposite to the applied voltage. Therefore, the rate of change in current (di/dt) is limited by the applied voltage and the inductance.

If we apply DC voltage to a pure inductor, the current is:

2) i = (V/L)t

Theoretically, the current continues to rise forever.

If we apply AC to an inductor, the inductor presents an inductive REACTANCE which is measured in OHMS, but dissipates no energy and forces the current to lag the voltage by 90 degrees. This reactance is given by:

3) Xl = 2 x Pi x f x L

Where f is the frequency in Herz ( cyles per second)

Capacitors:

The capacitor stores charge, that is electrons, lots of electrons, and the amount of charge determines the voltage on the capacitor. The current into a cap is given by:

4) i = C(dv/dt)

which leads to:

5) v = q/C

Where q is the charge in Coulombs, and C is the capacitance in Farads.

If we apply a direct current to a capacitor, the voltage is:

6) v = (I/C)t

Theoretically the voltage rises in a straight line forever.

If we apply AC to a capacitor, a capacitive reactance, measured in OHMS, but dissipating no energy and causing the current to LEAD the voltage by 90 degrees.

7) Xc = 1/(2 x Pi x f x C)

Where C is the capacitance in Farads.

By the way, the charge on one electron is -1.6 x 10exp(-23) Coulombs

[ January 06, 2005, 03:03 PM: Message edited by: rattus ]

 

[g1matrix]

Re: Capacitors and Inductors

Rattus
oh yeah that clarifies it.


Gary

 

[rob123]

Re: Capacitors and Inductors

physis wrote:

Remove the voltage source.

What happens?

The capacitor begins discharging, initially keeping the voltage up, decreasing over time.

actually, for an ideal capacitor, it won't self-discharge. so for the sake of understanding things better, start with the ideal device. your discharge cycle is equivalent to adding a resistor in parallel with the cap. so, your description is not of one component but two. now when you look at equations for a cap, it will never quite make sense; it will never equate.

so it would be as bad as, oh say....dividing by zero!

 

[rattus]

Re: Capacitors and Inductors

Gary et al,

To simplify things a bit. It is all about energy.

The current in an inductor is an indication of the energy in the magnetic field.

The voltage on a capacitor is an indication of the energy in the electostatic field.

In both cases, it takes time to change these energy levels. Is that not crystal clear?

 

[peter d]

Re: Capacitors and Inductors

Originally posted by rattus:
In both cases, it takes time to change these energy levels. Is that not crystal clear?

Yes, very clear. I assume this is the root cause of phase shift, because it takes some time for the current to build up in an inductor, and time for voltage to build up in a capacitor.

I like the mechanical analogies (the water tank, etc). They really help.

It never ceases to amaze me how with 3 and only 3 electrical properties we can do such marvelous things.

 

[physis]

Re: Capacitors and Inductors

You're absolutely right Rob.

I neglected to add some other circuit to the capacitor to illustrate it's discharging behavior.

Good catch.

Hi Rob

 

[physis]

Re: Capacitors and Inductors

What do you think. Put a high value of resistance across it?

Edit: I see you've already done that. I guess we agree.

[ January 06, 2005, 03:36 PM: Message edited by: physis ]

 

[rob123]

Re: Capacitors and Inductors

hi physis,

yes, for the real capacitor, you add resistors & inductors to the ideal.

so for the cycle you described, when voltage source is removed after charging, ideal cap is static. real cap will discharge (as you said). But ideal equation alone will never explain real behavior.

with R in par. with C, during charge, current flows into C and R.

with source removed, current flows out of C thru R.

for AC, it's all the same, only backwards.


and by the way, as t goes to positive infinity, we have another limit condition, but it doesn't involve divide by zero.

 

[rattus]

Re: Capacitors and Inductors

In my youth I worked with some 16KV dc-dc converters. I would charge my bod up to 16KV, then walk around on my rubber soles and shock someone. One's bod is a pretty good capacitor. Other caps are so good that bleeder resistors must be placed across them to reduce the shock hazard.

 

[physis]

Re: Capacitors and Inductors

As with B+

Edit: Wait a minute Rattus. Your bod is only one plate, and your shoes and socks the dielectric.

Your bod's just a good plate. Or is it your shoes and socks are a good dielectric?

[ January 06, 2005, 06:38 PM: Message edited by: physis ]

 

[rattus]

Re: Capacitors and Inductors

Physis, The rubber soles of my Hush Puppies were the main dielectric, then there was the air, and the other plate was the conductive concrete floor.

Perhaps though, as they say in the physics books, I was just another "charged body".

 

[coulter]

Re: Capacitors and Inductors

Peter -

You really don't want an engineer, you want a physicist. These are the guys and girls that understand the relationship between the Laws of God and physics and womenkind (okay - mankind too).

Engineers only understand how to use the mathematic models describing the thing they want to use - and electricals are the worst, cause the math closely fits the real world observations. The rest of the disciplines aren't that lucky.

So, given that, I have absolutely no clues as to why God (or Chaos as fitting your preference) decided this phenomena would work this way. But we have had the extraordinary good luck that physicists discovered the phenomena, did experiments to show it was repeatable, and developed the math model to describe it.

I know my explanation will sound really simplistic, but bear with me - it works for me. Snickers are acceptable and understandable - but please, save the outright laughter until the end.

Capacitors:

What they (the physicists) found was that if you separate two conductors with an insulator, that energy can be stored. If an electric source is connected across the plates, current is pumped from one plate to the other and the voltage across the plates goes up. Remove the electric source and the voltage stays - showing there is stored energy. So they came up with a unit of Capacitance that made the equation come out easy:

Q = CV where

Q = electric charge,
C = unit of capacitance
V = volts

So, if you pump charge in, voltage goes up. If charge is pumped out, voltage goes down. The bigger the capacitor, the more charge it takes to move the voltage any certain amount. Nice, linear, clean, and works every time (that's the repeatable part)

Given that, the rest is just math - okay, a touch more than High School algebra, but not much.

Lets look at what happens if we start changing the electric charge (Q) in the capacitor. The rate of change of Q, with respect to time, is the current (I). That's what current is, a movement of electric charge. We can write that as Delta(Q)/Delta(Time) = I. It is just a convention, a definition, it means Delta (Q) divided by Delta (Time), where Delta is a change, so it is a change in Q divided by the amount of time it took the electric charge to change.

If we let the time period be long, say 1 second, the charge (Q) can take really big jumps and that is hard to handle. So let the time frame be really small, so small we almost can't measure it. So we are looking at successive snapshots of the charge (Q) taken close together. Now the rate of change of the charge (Q) changes smoothly. To show that we are using small time bites, we write the equation as d(Q)/d(t) = i. the lower case letters just mean small time bites and smooth changes.

Before anybody bites me about dividing by zero, please read it to the end.

That aside, we have a major breakthrough in the math:
d(Q)/dt = i and Q = CV
Substituting for Q, d(CV)/dt = i

C is a constant, so we can take it out and get

i = C dv/dt

That says the capacitance times the rate of change of the voltage equals the current.

If we believe that to be true, that is all we need to answer the questions:

Why do capacitors oppose changes in voltage? Well they don't, they just oppose instantaneous jumps in voltage - it can be a very steep change, it just takes a lot of current. It takes energy to pump charge into a capacitor and that raises the voltage across the cap and stores the energy. When you stop pumping energy in and start letting current flow out, the voltage starts dropping. Small current flow, then small voltage change. Large current flow, large voltage change. Steep as you want, just not instantaneous.

Don't forget, when I am talking about current flowing out of a capacitor, I really mean the current is flowing out of one plate, through an external circuit, and back to the other plate.

Why does current come first with C?. Well it doesn't, but sometimes it looks like it is.

Remember the equation, i = C dv/dt. If we put a battery across the cap, then the volts are constant (unchanging) so the rate of change = 0, so the current is zero. That is what constant means, no change. Now change to a DC current source, harder to find - but available, and pump in a constant current. That means the rate of change of the voltage is constant. That means the voltage is ramping up nice and linear and just keeps climbing forever (well until the power supply crashes or we arc over in the cap). These two skull numbing, boring cases point out that the voltage is not following the current, rather the voltage is related to the current by an equation - dull but important.

Now lets put a sine wave voltage source (2V peak to peak) across the cap - this is what we see every day, 60Hz, caps, over excited synchronous motors; and the other side, reactors, induction motors.

The sine wave starts out at zero volts (0 deg), climbs smoothly to level out at the top (+1V at 90 deg), smoothly drops off to zero volts (180 deg) and continues to drop to the bottom (-1V at 270 deg), leveling out agan at the bottom, climbing back to zero (360 deg). We could have used time instead of degrees, but it gets hard to remember what part of the sine wave we are on after a few days. And the degrees keep repeating.

Now lets look at what the rate of change of the sine wave, cause that is what determines the current in the cap. At 0 deg, the sine wave is ramping up, so the rate is positive, in fact the rate happens to be 1. At 90 deg the sine wave is flattening out and turning down. Well flat is no change and the rate is zero. At 180 deg, the sine wave is headed down so the rate is negative and happens to be -1. At 270 deg the sinewave has bottomed out and the rate is again zero. At 360 deg the sine wave is headed up so the rate is positive and is +1 again.

Sketch this out and amazement will follow.

deg 0 90 180 270 360

V(sine) 0 +1 0 -1 0

I(rate) +1 0 -1 0 +1

The rate , which is the current, is the same curve as the voltage just displaced back in time by 90 deg - so the current looks like it is doing its thing right ahead of the voltage wave.

If you feed the cap with a square wave voltage source, the current is huge positive spike falling immediately to zero on the positive leading edge and a huge negative spike on the negative going trailing edge of the pulse. If you feed the cap with a triangular wave voltage source, the current is a square wave. And if you feed the cap with a sinewave, the current is a sine wave displaced back in time 90 deg.

Bit all of you, you just did some differential calculus. If you differentiate a sine wave, you get a (drum roll please) cosine wave. And the MEWs loved it.

You will have to read the next post for the divide by zero story and the MEWS definition

carl

 

[g1matrix]

Re: Capacitors and Inductors

Carl; wrote

"So, if you pump charge in, voltage goes up. If charge is pumped out, voltage goes down. The bigger the capacitor, the more charge it takes to move the voltage any certain amount. Nice, linear, clean, and works every time (that's the repeatable part))

So if this be true, then you have positive current entering one plate representing a positive charge moving toward that plate through its terminal lead' this charge cannot pass through the interior of the capacitor(since there is a huge resistance between the plates) and it therefore accumulates on the plate. Which is defined by i = dq/dt.
If this plate is considered an overgrown node and apply Kirchhoffs current law it appears not to hold, ie current is approaching the plate from external circuit, but it is not flowing out of the plate into the 'internal circuit'.
Why is this ?

Gary