C-Reactance of Capacitive Input Filter - Page 1
Forum section: electronics / beginners

author: FenderBender
author_rank: Super Contributor
timestamp: April 16, 2013, 01:02:36 am
content: So I think most of us know that a capacitor looks like a short circuit to AC, and an open circuit to DC. Fair enough.We know the equation 1/2pi f c, and when we increase capacitance we decrease reactance and thus its even more appealing to AC.But I really don't understand what that means. How is it that this AC component of the wave is whisked away by the capacitor? It seems that something would be...missing? Magically smooth(er) DC is created?Any ideas on how you could help me understand a capacitor filter in terms of reactance?

author: c4757p
author_rank: Super Contributor
timestamp: April 16, 2013, 01:15:33 am
content: I'm assuming you're talking about decoupling capacitors? You're just shorting the AC to ground. Another way to look at it is that you form a voltage divider between the power supply impedance and the capacitor impedance, which ends up being small:large for DC and large:small for AC. If you need really smooth DC, you can increase the impedance of the power supply line by adding a small inductor or ferrite bead, which greatly increases the effectiveness of the voltage divider.

author: MikeK
author_rank: Super Contributor
timestamp: April 16, 2013, 01:23:43 am
content: It's like the shocks (struts) of a car. The capacitor charges and discharges those fluctuations, just as the shocks dampen road bumps.

author: FenderBender
author_rank: Super Contributor
timestamp: April 16, 2013, 01:30:22 am
content: I do understand the concept capacitor providing a low impedance path to ground.It confuses me because while that portion is redirected away from the load, how is it that at the instant when the waveform dips to a low voltage, that the voltage on the load is high?I feel like I have two models. One that is saying that the capacitor provides a short circuit for AC, and another one that says it gets charged up and then constantly discharged to maintain the voltage.How can I make a connection between these two thoughts?(Just a quick side note, is this technically a RC LP filter? I'm not sure exactly what the R part is. Perhaps the Rd of the diodes?)

author: peter.mitchell
author_rank: Super Contributor
timestamp: April 16, 2013, 01:53:50 am
content: Think of it like a battery UPS on float charge; as the voltage peaks on the charger, the potential across the charger leads is higher than the charge is the battery, so it gets charged, and when the charger voltage dips, the batterys potential keeps the rails at the potential, so the dips and spikes (ac) get mitigated, the constant (dc) between the two terminals stays the same.The important thing to note is that, upon connecting a capacitor to a potential, it will look like a short, as the potential between the capacitors legs is close to 0v but the potential is high, so the capacitor charges up to that potential.A capacitor attempts to smooth out changes in voltage.Think of inductors like a small piece of rubber pipe inline with a big steel pipe carrying high pressure, when you turn on the water, it flows slowly at first because the rubber is small, letting through a little bit of water, but once the system is full, the rubber stretches, letting as much water through as the rest of the pipes. When you turn the valve at the beginning off though, the rubber stretches, pushing its stored water down the pipe.An important thing to note is that, if you remove a sink for the current, the inductor will collapse its magnetic field, spiking the voltage; eg, if you pull the pipe off the "out" end of the rubber, the rubber will quickly retract bursting out water at high pressure.An inductor attempts to smooth out current.

author: grego
author_rank: Frequent Contributor
timestamp: April 16, 2013, 01:56:39 am
content: Quote from: FenderBender on April 16, 2013, 01:30:22 amRC is for Resistor-Capacitor -- L is for inductors - hence you see a lot of talk about "LCR" meters for instance (to measure all three). In a RC circuit you have a resistor and capacitor.http://en.wikipedia.org/wiki/RC_circuitRC circuits are used a lot as filters because they can cut off frequencies (low-pass/high-pass) based on how you set up the values of your resistor(s) and cap(s).Hope I got that all right.

author: c4757p
author_rank: Super Contributor
timestamp: April 16, 2013, 02:12:33 am
content: Quote from: FenderBender on April 16, 2013, 01:30:22 amIt's technically an LP filter, but not necessarily RC. Both the resistance and inductance of the power supply lines come into play - often, the inductance more than the resistance.Quote from: FenderBender on April 16, 2013, 01:30:22 amThose models are, in fact, harmonious. For rising voltage, the capacitor acts as a short circuit to the ACbecauseit is being charged up by it, therebytakingcharge andconductingcurrent. Remember that no decoupling capacitor can ever provideperfectsmooth DC; the voltage will always rise a bit as the capacitor charges. Falling voltage can be treated like negative rising voltage - the negative current is conducted to ground as the capacitor "charges negative" (discharges).

author: Rufus
author_rank: Super Contributor
timestamp: April 16, 2013, 08:13:23 am
content: Quote from: peter.mitchell on April 16, 2013, 01:53:50 amA better hydraulic analogy for an inductor is just a pipe. A pressure differential at the ends causes the water to accelerate and gain kinetic energy. Pressure maps to voltage, current to water flow and inductance to the size of the pipe.If you want to use rubber then a rubber diaphragm inside a pipe is a good analogy for a capacitor.

author: nuhamind2
author_rank: Regular Contributor
timestamp: April 16, 2013, 08:41:16 am
content: Think capacitor like a water tank and suppose your water supply delivered from another city with truck. The size of the tank is analog to the capacitance and the frequency of the truck delivering water is analog to (of course) voltage frequency.Think how smooth you can take a bath (like taking a bath once a week is not smooth but once a day is smoother and once a month is really not smooth) with different size of tank and different frequency of truck delivering water to your home.

author: JackOfVA
author_rank: Unknown
timestamp: April 16, 2013, 10:36:19 pm
content: It's perhaps confusing because you are mixing a time domain and frequency domain analysis. While this is certainly a useful thing to do in some circumstances, it's important to keep the differences between the two in mind. Since this is the beginner's forum, a simple way to distinguish the two is that the time domain is what one normally sees on an oscilloscope, with a voltage that is, in some fashion, a function of time. The frequency domain is what one might see on a spectrum analyzer, a number of discrete spectral lines of certain frequencies with amplitude and phase relations. With full knowledge of the signal described by one domain, it is possible to transform it to the other domain without error.First, I think you are looking at a filter capacitor following a diode rectifier as might be found in a power supply. In this case, the driving voltage waveform (time domain) is a rectified sine, either full wave or half wave as the particular circuit may be.If we look at the waveform in the frequency domain, via a Fourier transform, or as you might see it with a suitable spectrum analyzer, you will find a DC component (frequency 0 Hz) plus sine waves of varying phase and amplitude at multiples of the input frequency. The capacitor does not attenuate the DC component, since it has infinite reactance at 0 Hz (1/wC = infinity when w = 0. w is omega, 2*pi*f). However, it will have a finite reactance at the other frequencies in the waveform, thereby attenuating them. Thus the RC filter alters the amplitude and phase of the non-zero frequency spectral components and the result is a different waveform, one with less AC ripple and more dominated by the DC component of the spectrum.If you prefer to look at the circuit in the time domain, then I would think of the capacitor, as has been suggested, in terms of stored charge. At any particular time, the capacitor will have a certain voltage across it and that voltage corresponds to a particular charge stored in the capacitor. Q=CV, where Q is charge in Coulombs, C is the capacitance in Farads and V is the voltage in volts. More importantly when we are looking at a time changing voltage is the change in voltage versus change in charge and current. I don't want to make this overly complex, but it requires a bit of elementary calculus to understand the relationship. I'll skip the math and concentrate on a simple physical description. As the driving voltage increases, more charge is "pushed" into the capacitor. Pushing charge into the capacitor tends to hold the voltage across the capacitor at a slowly changing value compared to how it would vary with time in the absence of the capacitor. Likewise, if the driving voltage decreases below the capacitor's stored voltage, the capacitor releases some of its stored charge and "pushes" the charge back into the circuit, thereby holding the voltage more constant with time to a greater extent than in the absence of the capacitor. A tiny bit of the calculus involved is i = dq/dt where i is current and dq/dt is the rate of change of charge. In other words, current is the change in charge with time.

author: smashedProton
author_rank: Frequent Contributor
timestamp: April 16, 2013, 11:01:17 pm
content: Some of these analogies...Who do we think we are? Theoretical physicists?

author: FenderBender
author_rank: Super Contributor
timestamp: April 17, 2013, 12:14:25 am
content: JackofVA,I'm fine with basic calculus. I know a formula I = C * dV/dt. As voltage increases, the current increases, but as voltage decreases, does the current then becomenegative? as in it flows in the opposite direction?Does this make sense.( Please scrutinize me if I am making an error in this.) So does this mean that the source charges the capacitor when dV/dt is positive, and then when dV/dt becomes negative, I (current) must become negative. So let's say that the source charged the capacitor from the left side of the capacitor to the right side, does this mean that when I becomes negative the current goes from right to left (reverse) and then discharges through the load? If so, how does it know to discharge through the load and say not through the source (maybe I need my head checked?).Thanks all.

author: peter.mitchell
author_rank: Super Contributor
timestamp: April 17, 2013, 01:22:56 pm
content: Quote from: smashedProton on April 16, 2013, 11:01:17 pmI'll just enlarge my avatar for you;

author: KJDS
author_rank: Super Contributor
timestamp: April 17, 2013, 01:28:07 pm
content: A decoupling capacitor is far better viewed when the source impedance is included in the calculations.

author: jpb
author_rank: Super Contributor
timestamp: April 17, 2013, 02:07:42 pm
content: Perhaps a simple way of viewing it is to think about charge flow onto the capacitor and the voltage across it resisting further charge flow.If you have an R in series with the C then when the C is uncharged there is no voltage across it and all the V drops across R which determines the current - it is like the C has no resistance and can take as much charge as you can give it.As soon as the C has some charge on it there is a potential across it so the potential available to drive current through the resistor drops and so does the current.Under dc the capacitor ends up grabbing all the voltage, there is no voltage drop across the R and current drops to zero.Now for AC half the cycle you are charging the capacitor and then you discharge it the other half cycle so it never accumulates charge and hence voltage.From Q = CV we have V = Q/C so the bigger C the less rapidly it develops voltage for a given charge flow (current).The higher the frequency the less the peak charge in a cycle (there is less time for charge to accumulate before it is discharged again). This is normally expressed in terms of the rate of change of the voltageQ = CV differentiates to dQ/dt = C dV/dt but dQ/dt is just the current and for a sin varying voltage V0.sin(wt) we have dV/dt = V0.w.cos(wt) which is just a phase shifted sin curve scaled by w.The ac component is not whisked away by the capacitor, it is just that the ac component of current flowing "through" (actually on and off separate plates) the capacitor doesn't give rise to much ac component of voltage across the capacitor.I hope that is not too confusing!

author: JackOfVA
author_rank: Unknown
timestamp: April 17, 2013, 02:08:24 pm
content: Quote from: FenderBender on April 17, 2013, 12:14:25 amCharge is stored in the capacitor, not the connecting wires -- well, leaving aside the few pF stray wiring capacitance, that is. Hence you have to look at the current flow into and out of the capacitor.If you look at the various wires connecting the capacitor to the rest of the circuit, the principle is that the sum of the instantaneous currents at the node is 0. Some current may flow into the node from one wire, some may flow out from another wire and some may represent charge put into or removed from the capacitor. But their algebraic sum = 0. (This is the principle behind node current analysis, after all. Look up Kirchhoff's laws for more detail.)And yes, at some time charge is forced into the capacitor and at other times charge is removed, so the sense of the current reverses.All this is stuff you learn in Introduction to Electrical Circuits 101 early in your Electrical Engineering classes - you should be able to find more than you ever wanted to know about it with a bit of internet searching. It should also be covered in the electricity and magnetism part of first year college physics.

author: FenderBender
author_rank: Super Contributor
timestamp: April 17, 2013, 02:41:37 pm
content: Thank you thank you.I'm not the best at mathematics, but sometimes I think it is very helpful to use math instead of analogies, though both have their place.I'll read over everyone's response again. But I think I have a better understanding now.