phase

A resistor will not change the phase. An inductor or capacitor will.

They do that by either the current or voltage will lead or lag the other. That is if you can put a very fast voltmeter and ampmeter in the circuit, the voltage will rise faster than the current or the current will rise faster than the current flows. It depends on if it is an inductor or capacitor as to which one reaches the maximum value first.

If I remember correctly, say you have a capacitor and put a voltage to it, the current will start to flow at a very high rate as the capacitor is like a short circuit to start with. It will take the voltage a short time to charge the capacitor. The inductor is the opposite of this.

Imagine a sinewave applied to a series resistor and capacitor. As the sinewave reaches a peak, the capacitor is not fully charged and has a voltage somewhat less than the applied sinewave. Now as the sinewave reaches a peak and starts falling, the capacitor continues to charge because the sinewave voltage is still greater than the capacitor voltage even though the sinewave is falling to a lower value. At some point, the sinewave will fall to the same voltage as the capacitor and the capacitor will stop charging and begin discharging. So, you can see the peak voltage of the applied sinewave occurs before the peak voltage on the capacitor. The amount of time shift between peaks depends on the resistor and capacitor values, and the frequency of the sinewave. Also note the amplitude of the sinewave across the capacitor decreases as the phase shift increases. By the time you get to

90 degree shift, there is no amplitude at all. The capacitor voltage at that point is just DC.

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If you don't have them yet, you should get a cheap dual trace scope, like a Tenma for $25 or so, and an audio sine wave generator.

Put the scope in dual trace chop mode, put the generator to one channel and adjust it to sync in, and then the generator output also goes to a resisto r and capacitor. Put the second scope probe to the juncion of the resistor and capacitor. Of course the loose end of the R C network" goes to the comm on gorund.

Which comes first ? The resistor or capacitor ? It doesn't matter but once you play with it one way, reverse them and see the opposite effect.

If you can measure the phase shift you can calculate the amplitude as a fra ction or percentage of the total applied signal. If you can measure the dro p in the aplitude of the signal you can calculate the phase shift. It's eit her the sin of the phase lead or lag, or the arcsine of the measued amplitu des' ratio. Of course you have to know which way to flip the formula, and I think the reality of seeing it on the scope screen helps to understand alo t of this type of stuff.

Audio generators are not expensive, so find a cap that has about the same X c as the resistor's value to start. Figure 1 kHz, that's a good middle of t he road frequency. Figure 1,000 Ohms. Now to find a capacitor with 1,000 oh ms Xc (capacitive reactance) at 1 kHz. the formula is :

Xc = 1/(2pi F C)

where:

Xc is the capacitive reactance (kinda resistance) of thre capacitor in ques tion pi is of course 3.14159265 give or take F is the frequency in Hertz C is the cpacitance in Farads

Now you are going to have a value in microfarads so you have to adjust the math accordingly. For nanofarads or picofarads, it is a matter of the decim al place in a different position.

Actually, there should be no harm in telling you that the value (I think) f or the cap will be about 0.018 uF. you'll want to figure that out for yours elf though I think. Also, I think 0.015 is a standard value and alot easier to come by. you just adjust. Refigure the Xc and make your resistor so, or vary the frequency to where Xc = R. It's all up to you.

When the Xc equals the resistance, the phase shift is 45 degrees. the volta ge at the junction of the R and C will be 0.70711 times the total - which i s the sine of a 45 greee angle. If you put in a frequency at which the rati o of Xc to R is 0.86603 and 0,5, the phase is going to lead or lag by 30 or 60 degrees, depending on which way you look at it.

Those numbers are straight out of the trig tables. And that is what you use when you do the vector algebra to figute amplitude to phase.

Lately, I have this math tutor on once in a while. Informal, when we get in the mood. He got me to read a book that started explaining just ow the tri g tables are generated. I find it fascinating, and do NOT understand it yet . But that is just someting I thought to mention because if I read your pos t right, you are not afraid of math. The more i earn the more I find I don' t know. (well that pretty much applies to everything eh ?)

But this is getting good, sines and cosines (which you will use to calculat phase shift) are all generated based on parts of a circle. Bottom line, th ere is geometry in electronics.

In his book, George F. Simmons says the waste time in the schools telling s utents to go out and figure out the height of a flagpole base on its shadow . He says that it a waste of time right in the preface. He says nobody has to do that. Nobody dios that, even architects. He said the most impotant us es of math like geometry have very little do to wiht actual shapes of thing s. electronics is a perfect example.

Oh, and after you get used to sine waves, try square waves. Theoretically t hey ahve all odd order harminics to infinity in a certain phase relationshi p. Actually, they are pretty square if you cna reproduce like the 21st harm onic.

Take and sens that thropugh your R C network. In fact, if you really want t o have a blast, send a 1 kHz square wave through a ten band graphic equalii zer. If you happen to smoke the funny stuff, hook it up first, and then pla y wiht it and see what happens to the waveform. It is alot harder to view t hat and get more understanding, but you will notice certtain things. you'll notice with the R C network as well.

It's all part of the cool stuf you can do on a dime. On eBay I am sure you can get an adequtae scope and generator for under a hundred bucks, and you know if you like electronics it woul not be like the famous oxycetylene tor ches. They were a syn=mbol of suburbia and grandeur almost. Had to have t hem. YOU CAN WELD ! So you get them, maybe fix one thing, loan them out to your brother in law once, get them back two years later and never missed th em, he retirned them because he is moving. And then they sit there for five years. Like, I have a lathe and milling machine, though I don't want to ge t rid of them they have not been turned on in months.

A scope won't be like this. for the want of a few other components like may be some coils etc., you can do all inds of cool experiments, and you can ac tualy measure and see if the math you did is correct. yup, solve it first a nd THEN observe. Good way to learn because when you make a mistake you have to find out why and how.

Readily apparent on an oscilloscope too.

The whole problem is so ill defined it's hard to say anything -- but if the circuit node in question is being fed a sinusoid through an inductor or a capacitor, putting a resistor from that node to ground will, indeed, change the phase of the voltage on that node.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

Your question is very ill-posed.

ASSUMING that the circuit in question has some sinusoidal voltages or currents running around.

And ASSUMING that the node in question is fed through some reactive component.

THEN adding a resistor to ground could change the phase of the voltage at that node.

A better question will get you better answers. One way to make your question better is to give a concrete example of the circuit you're asking the question about. If you're looking at some discussion on a web page, a URL is always a quick way to give us the necessary background.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

First, a little background. Any circuit with connected resistors, DC voltage sources, current sources etc. can be characterized by a set of equations (derived by Kirchoff's rules). You may designate any node as ground, and any node (other node than ground, if you want anything interesting) as output. Then the output is (after solving the set of equations) always a voltage source with a series resistance (Thevenin's theorem). It is also a current source with a shunt resistance (Norton's theorem); these two circuits BOTH are complete descriptions of the current/voltage curve of the output.

Now imagine that instead of DC sources, you allow exp(j * (phase +2*pi*F *T)) factor times an amplitude, and agree that only the REAL component of this complex function of time is measurable (voltage, or current). Then, the exact same solution of equations gives rise to a different solution for each value of frequency, F. For F = 0, this repeats the resistors-only case. For nonzero F values, however, there are impedances that are frequency-dependent, like capacitance (impedance of a capacitor is constant/(j*2*pi*F)), and inductance (impedance of an inductor is constant * j*2*pi*F).

Now, the Thevenin equivalent has voltage amplitude, phase (if the frequency is nonzero) and a complex output impedance, and these three things are all frequency dependent. But, you can sum multiple single-frequency solutions and get a solution for composite input signals (containing multiple single-frequency components, or even a continuum of frequencies). That linearity property is lost if you have any spark gaps, zener diodes, rectifiers... anything other than R, L, C.

'connecting a resistor or capacitor to ground' means changing an infinite impedance connection from a node to ground, to a resistor or capacitor finite impedance, and that changes the equations that have to be simultaneously solved. The circuit doesn't have a phase associated with it, though, only a single frequency component of an (output or input) signal can be assigned a phase. The 'input' phase comes from the signal source, not the network of components. The 'output' phase comes from the source(s) of some particular frequency of interest, and the network of components, not excluding any test instrument you connect to do a measurement.

I seem to have understood him just fine. It wasn't all that badly put for a newbie. This is sci.electronics.basics. BASICS. If this post was in sci.electronics.design, tear into it.

Do what I said, you'll br glasd you did. Cheap scope and audio generator. you'll learn the concept first and the math later.

And make sure it has XY operation, that you cna switch the B channel to the horizontal amp to view those Lisajouis patterns, those are neat as well.