Circuit Protection Using Capacitor

No, the voltage waveform at nodes A and B have absolutely nothing to do with the value of the capacitor. The voltage will be 230V/50HZ. Now if your voltage source output impedance is within an order of magnitude of the application's input impedance, then and only then might the capacitor value have an effect on the voltage at AB. As drawn though the source output impedance is the theoretical ideal value of zero.

How silly of me, its directly connected at the points so yes the voltage would be the same. But would the waveform also be exactly the same sinusoidal 50 Hz?

My basic understanding of capacitors is that they give a resistance (reactance) of

1/[2(pi)fC]. In this case the greater the capacitance, less the reactance (at a particular frequency like 50). Right? So how does the capacitor work for "protection" exactly?

First, the waveform will be exactly the same waveform as the voltage source.

Second, the Vishay article you're referring to discusses placing capacitors directly across the AC power line to reduce Electro-Magnetic Interference (EMI) signals on the power line and/or application load. EMI is a complicated collection of possible added voltage and current sources that happen on a power line in addition to the actual power line energy of 220 VAC 50 HZ. Since these added sources are not dependent energy sources of the power line power, the rules of superposition apply in doing a circuit analysis. These EMI signals can come from the power line side or even your application load side of your schematic. The capacitor is preventing power line EMI from interfering with the application load while also preventing application load EMI from interfering with the power line and other loads attached to the power line.

I applaud you trying to understand this application note. However, there are many jargon simplifications and other things that can baffle a novice in this paper.

Third, ideal capacitors do not produce any resistance. They produce reactance. The terms while related are not synonyms. Resistance and reactance are both types of impedances. The impedance of a capacitor is Xc=1/(jωC). The impedance of a resistor is Xr=R. The impedance of an inductor is Xl=jωL. Depending on which electronics or physics class you were taking, not including the square root of -1 (j or i) might give you partial credit. In my class saying that the resistance of a capacitor is 1/(2[pi]fC) will give you no credit at all.

The steady state waveform is the same on either side of the cap if the wires have zero impedance but caps will suppress transients because of dV/dT = I/C if the source impedance is not zero.

So the capacitor is only there for transients?

The capacitor is there for power factor correction only.

Not in this case. Read the application note. It deals mostly with RFI and not power factor correction. Although many parts of this application note will transfer to power factor correction.

What kind of protection are you thinking about? I don't understand what you mean, "protecting an application against the mains voltage."

MOVs (metal oxide varistors) are probably the simplest protection against high voltage transients. Littlefuse has a nice overview paper, "The ABCs of MOVs".

I don't think I've seen any method using just resistors and caps. However, I've forgotten just about everything I ever learned about AC theory. There may well be some such method.

The schematic reminds me of a question I have. I am running my whole shop on one thirty amp breaker, fluorescent lights and power tools. I'm concerned that something will get damaged by the drop when I turn on a power saw. I have lots of caps that I have saved from air conditioners and lighting fixtures.

So, can I improve my power by using caps, either in series or parallel?

Sigh, no.

Capacitors operating on an AC distribution grid are useful to change the power factor the utility sees. Power factor is related to Power but it is not the same thing. If you have any location fed by only one power line that maybe undersized then you need more power lines provided. You can likely improve the local grid by doing a better balancing of loads with multiple power feeds, regardless of this being a three phase or split phase configuration.

Thanks, Redfred, I'm still trying to get a handle on just what capacitors do for a circuit. If they are supposed to help a motor get started (start cap) and both the motor and the cap get the power at the same time, how can it help?

The lights dim for a second while a motor is starting in my shop. I was hoping a cap would help that by supplying power while the motor spins up.

Also, I'm testing used florescent fixtures, and they all seem to not to want to start until I touch or rub the tube. These units don't look as old as other units that are starting just fine in my shop.

I plan to check the voltage at the cord I'm using with my DVM. Gotta get a cheap one for the shop, analog is not always accurate enough.

Capacitors are used in some AC motors during the starting cycle only to get the rotor up to speed. In one scenario there's a separate set of starting windings that go through a capacitor and centrifugal switch that opens once the rotor is up to speed. In another scenario there no speed switch but a timer relay and a capacitor.

To answer your implied question though about how a capacitor works with AC, it changes when current runs through the windings of a motor. The capacitor's value must match the inductance of the motor windings to be effective. So you cannot just capriciously add some random capacitor value and get an improvement. You also have the complication that the mechanical load on a motor changes when current runs through an AC motor.

One thing I forgot to mention. For seeing how much your voltage dips at start up you'll probably do better using an analog meter instead of a digital meter. This is because the response time of a cheap DVM will be so slow that all that you'll see is a blur of changing numbers. With a real analog, moving needle meter you'll be able to visually track how far the needle moved and get a realistic number. Now if you get a good digital or analog storage oscilloscope and really learn how to set up the trigger circuitry, then you can get a very precise analysis of what your starting circuitry is doing to your local grid. This will be above and beyond what you need to know but I thought you'd like to know what it will take to get a precise result.

Thanks, that would be a good thing to learn. I'll read the instructions for my good digital meter.