When a capacitor passes AC or Varying DC, it changes the impedance inside the capacitor? so the capacitor changes impedance of the circuit when there is an AC signal or varying DC apply to the capacitor?

What's the difference between an AC waveform VS a varying DC waveform?

 

The reactance of a capacitor changes with frequency:

Xc = 1/2∏fC

 

When a capacitor passes AC or Varying DC, it changes the impedance inside the capacitor? so the capacitor changes impedance of the circuit when there is an AC signal or varying DC apply to the capacitor?

What's the difference between an AC waveform VS a varying DC waveform?

The impedance of a capacitor will appear differently for signals of differing frequencies. At the extremes, signals of very low frequency and DC signals see the capacitor as virtually an open circuit (high impedance), conversely, signals of very high frequency see the capacitor as almost a short circuit (low impedance).

The impedance of an ideal capacitor is given by the following, where f is the frequency of the signal passing across (voltage) or through (current):

\(

|Z_{cap}| = \frac{1}{2\pi fC}

\)

An AC signal is a signal composed of one or more frequencies. From what I am assuming you mean by a varying DC signal, you are referring to a constant voltage that is immediately changed to a new constant value?

I am not sure of your Fourier analysis background but basically the varying DC signal will be (ideally) only DC except during the transition to the new constant DC value upon which the signal is DC again. Inherently by definition that is an AC signal especially if the voltage is consistently changed at a certain rate. That step change requires a large amount of high frequency components to occur and thus will affect the operation of the capacitor.

Long story short:

- different frequencies change the behaviour of the capacitor's impedance
- at DC the capacitor is open circuit
- capacitor appears to be an open circuit except during the transitions for varying DC signals

 

The impedance of a capacitor will appear differently for signals of differing frequencies. At the extremes, signals of very low frequency and DC signals see the capacitor as virtually an open circuit (high impedance), conversely, signals of very high frequency see the capacitor as almost a short circuit (low impedance).

The impedance of an ideal capacitor is given by the following, where f is the frequency of the signal passing across (voltage) or through (current):

\(

Z_{cap} = \frac{1}{2\pi f}

\)

Uh... the impedance of a capacitor is (1) a complex quantity and, (2) inversely proportional to the capacitance as well as the frequency.

 

What's the difference between an AC waveform VS a varying DC waveform?

Very simply an alternating waveform has positive and negative half cycles.
That is it crosses zero periodically.

A direct or unipolar waveform is all on the positive or all negative, although it may touch zero.

 

What's the difference between an AC waveform VS a varying DC waveform?

Actually there is very little difference as far as the capacitor is concerned, they both follow the same equation:

I = C dV/dT

which is a differential equation, meaning if you don't know calculus you're SOL (sadly outa luck). Also there are side effects this doesn't account for (but is the predominant relationship).

To simplify the mathematics (and who doesn't like simple?) the world has been divided into "AC" and "DC" analysis. Keep in mind there is but one world, but two ways of looking at it.

Nothing inside the cap changes. A capacitor reacts to the rate of change in the current thru (times a constant) with the voltage across it.

We call that constant "capacitance."

 

Uh... the impedance of a capacitor is (1) a complex quantity and, (2) inversely proportional to the capacitance as well as the frequency.

Late night typos. Absolute impedance and forgot 'C'. No harm done.

 

You mean "reactance", not "impedance".

 

The capacitive reactance (Xc) is 1/(2*pi*f*C) ohms, the impedance is 0 - jXc ohms. The reactance changes with frequency, thus the impedance does as well.

EDIT: Sorry guys, I have no idea how to get the latex formatting

 

Check out this intro to TEX.

X_c=\frac{1}{2 \pi f C}

\(X_c=\frac{1}{2 \pi f C}\)

 

When the DC voltage rises up or down, the capacitor will PASS the DC voltage because when the DC voltage is rising up or down it's a rate of change. This changes the impedance of the capacitor when the DC voltage is rising up or down

The Capacitors Reactance is an internal resistance

The capacitors reactance internal resistance will change will you vary the DC voltage up or down which changes the in-circuits impedance?

Is there any difference between an AC waveform VS a Varying DC voltage?

Varying DC voltage is Unipolar?

 

The Capacitors Reactance is an internal resistance

Yes and no. It's more like a resistance just for AC. Internal resistance (equivalent series resistance, or ESR) is completely separate from capacitive reactance. The ESR actually gives the capacitor an impedance of ESR - jXc, so the real part of the impedance (resistance) is the ESR, and the imaginary part of the impedance (reactance) is -jXc

 

When the DC voltage rises up or down, the capacitor will PASS the DC voltage because when the DC voltage is rising up or down it's a rate of change. This changes the impedance of the capacitor when the DC voltage is rising up or down

The Capacitors Reactance is an internal resistance

The capacitors reactance internal resistance will change will you vary the DC voltage up or down which changes the in-circuits impedance?

Is there any difference between an AC waveform VS a Varying DC voltage?

Varying DC voltage is Unipolar?

Dexter,

You are mixing up several different ideas.

It would be a good idea to let us know where you are coming from with this.
That is what subject are you studying this in and at what level?
Do you understand the discussion about j/ωC, impedance, reactance and resistance?

You have repeated your question about AC v DC. Did you not understand my post#12? Would you like more explanation?

Tell us where to start.

 

A capacitor will pass DC voltage when it is varying DC , so the reactance , ESR ,, impedance are only for an AC signal, so the capacitors reactance, eESR, impedance for varying DC or a rising DC voltage or discharging voltage will be pass through a capacitor but there is no reactance, ESR, impedance because its DC Passing through the cap?

 

A varying unipolar DC signal has an AC component (the varying part) riding on an average DC level. A capacitor will pass the varying (AC) part and block the average DC part. The AC current will be impeded by the reactance and ESR of the capacitor as determined by its capacitance and the frequency components of the AC part (as given by a Fourier analysis of the waveform).

 

as given by a Fourier analysis of the waveform

It has to be a waveform ie a wave before you can do a Fourier Analysis.

Slowly and randomly varying DC is not susceptible to such analysis.

 

You mean "reactance", not "impedance".

Either. If there is no resistance (real part) they are equivalent, and I did specify 'absolute' impedance. Bring non-ideal capacitors into the mix sure you have ESR creeping in, but let's start with the basics to get the OP on the right path.

 

the ESR of the cap should be a negligable value. if it is large enough to affect the reatance into impedance, throw the broken theing away, its bad;.

 

Note that the equation that keeps appearing with 2pi on the bottom of a fraction is only true for sine waves (like 60hz mains or other similar carrier waves).

The formula provided by Erne, above, is much better for square (digital) DC circuits because you get very fast change of voltage per unit time.

I = C dV/dT

Cheers

 

Just so as everyone knows Billy Mayo and DexterMccoy are using the same computer.

Draw what conclusions from that you will.