DC Wave Questions

It's DC with a ripple riding on top of it. With no filtering the ripple runs down to zero.

Steve, I can't tell if you are agreeing with me or disagreeing with me. Maybe half and half? Anyway, I am not having any trouble except in convincing Bob Penoyer that his notion of the meaning of DC is flawed.

I don't see where I even mentioned sine waves above. You seem to have pulled them into the discussion out of nowhere. But of course you know, (from Fourier analysis) that any periodic signal can be represented as a (possibly infinite) sum of sine waves whose periods are multiples of the signal period. This applies to full-wave rectified AC, too. In the case of full-wave rectified AC, I don't think the sum is infinite.

Anyway, I believe I am on solid ground when I make the following points:

2) Just because the Voltage waveform at the output of a rectifier is non-negative does not mean that it is DC, though it does mean that it has a DC component.

I will try to make this my last post in this thread, or at least sub-thread.

--Mac

I like Jack's terminology. The wave itself isn't DC, but I think "fully DC" is an acceptable way of describing its location.

AC generators and transformers are usually designed to produce sine waves with no DC, but sine waves were known long before those inventions.

A wave is a succession of curves. A sine wave is a wave whose displacement follows the form of a sine. A pure acoustic tone is a sine wave regardless of ambient pressure. A ripple on a pond is a sine wave regardless of the water level.

As not all voltage variations are curves, our generic term was "waveforms". If the plate voltage of an amplifier tube varied from 998 to 1000 volts in the form of a sawtooth, we'd call that two-volt variation a sawtooth waveform. If it was sinusoidal we'd call it a sine wave.

To call a waveform an AC sine wave implies that there is no DC, but this thread is the first time I've read the claim that all sine waves are AC sine waves.

You have a 10V Dc ssignal with a superimposed 5V p-p signal. The load's impedance to DC is determined by the load's DC resistance (steady state ) The load's impedance to AC will be determined by its R (ac not Dc) ,L,C combination (steady state) The two are different and independent values.

You truly are an idiot. Bluegrass is growing all over kentucky, and some gets planted every year. Notr only that, but you can listen to it on the radio or TV.

Another waveform that's very similar ( albeit more of a sawtooth waveform rather than sine wave ) is power line ripple on a DC supply rail.

Graham

Actually, DC from a rectifier *is* "zero frequency", to the degree that it is DC. Of course until the AC is filtered out, it has both AC and DC components.

That is *precisely* correct. (It just doesn't tell enough of the story to explain the confusion of this "flows in one direction" definition of DC.)

The output of a rectifier until filtered *does* have both AC and DC, which actually is another way of saying that yes it *does* change directions.

What? you say!

The problem is that "direction" only has meaning when measured in comparison some specific point of reference. If you have three different reference points, one at the DC level, one at the peak positive swing and one at the peak negative swing, you have three very different views of "direction" for current flow:

Reference Direction Point of flow ========= =====================================

Peak Pos All Negative

DC level Equal cycles of Positive and Negative

Peak Neg All Positive

Obviously the only point of reference you are thinking about is the Peak Negative swing, and just as clearly that is *not* appropriate. The correct reference point is necessarily the DC level, and measured from that point of reference the current is going to go alternately in each direction.

Bingo, there is your AC.

This is very similar to confusing the negative side of a rectifer circuit with ground. Rectifiers are most often grounded on the negative side, but that is absolutely arbitrary. They can be grounded on the postive side, and with some effort can have ground at other levels either between those two, or even beyond either of them.

But rather than look at DC as current going all going in one direction, and AC as anything else, it is *far* easier to view it as AC is any current that is changing, and DC is anything else (i.e., the current is steady).

Technically those definitions are exactly the same, but one leads to a lot of confusion.

I think "zero" is a good reference for current flow, and that the actual (absolute) direction can be measured. Voltages have the reference issues.

This type of problem is analyzed using the superposition theorem.. One source is +10 VDC The second source is a 10 V pk-pk sine wave.

Just add the two results for the answer, into a common load. For the DC, allow time for the transient to settle....

If you can use PSPICE, or the equivalent, you can do this easily , with additional experiments which may improve your understanding......

Andy

PS It's a shame you have to weed thru all the crap from some of the posters here who have a lot of time on their hands and have no tolerance for those who are just learning their craft....

Sure... now, can you define "zero"?

E =IR

Since our resistance is fixed, it's the exact same issue, though perhaps easier to understand, with voltage. (I gave some consideration as to whether to post that with voltage or current references, and since "AC" and "DC" use the term "current", decided to go with current to avoid the easier path to the same statement you are making.)

Yes, the load impedance is a function R, L, C and frequency. L and C do matter anytime the voltage varies even on non-periodic wave and even if the voltage is always positive. On a pure DC voltage (constant, non-varying), L will represent a short-circuit while C will represent an open-circuit.

If the R+L+C are all in series, the +5V to +15V sine wave will, during the transient phase, first charge the capacitor to +10V, then (during the steady-state phase) the current will start to behave as a normal sine wave (reversing directions) even if the driving voltage never goes negative.

The circuit is pretty much unconventional (but sure smart!). No, it will not output +5V. The circuit is meant to detect only negative voltages and can only output zero or negative voltage. But I suggest that your try building it and see what happens. You might just get lucky.

You're welcome. I hope you now understand that your +5V to +15V voltage is better called "varying DC". I've never seen an EE book that will say "DC Sine Wave" for the voltage you described, and rightly so because a sine wave implies reversing directions.

Put an ammeter there and it says zero. That's zero. Electrons bouncing around in the conductor have an average net displacement, over time, of 0.

issues.

Current is a different issue from voltage because voltage is a relative quantity. It is a type of measurement of a change in field between two locations. Current is a rate of flow of charge at a single location (well, typically, through a single Gaussian surface), and is measurable at that location, and does not have the ambiguity that voltage has. It does not need a reference. If I say that my toaster is running at 120V and 8A, you may ask "120V relative to what" and I'll answer "neutral". You would not ask "8A relative to what".

"Put and ammeter there" and if it says +300mA to +800mA back and forth, then it's Alternating Current, innit?

It is equally a shame that there are those that are sometimes incapable of offering correction gracefully, eh, John? If it pains you so much to engage in your ungracious edifying, perhaps you would do well to bugger off, and leave the stress of educating imbeciles to those with more patience.

Thanks, Floyd, for that excellent and understandable post.

If we may plunge for a moment into basic boolean logic, the "OR" part is no longer necessary once one part of the proposition is shown to be true. Thus, not only was his omission convenient, it was proper.

If nothing else, your stubborn adherence to a flawed terminology and lack of openness to furthering your understanding will make you look like an idiot in a job interview, should you ever decide to pursue career advancement in the electronics industry. Please note that I am not saying you are an idiot, just that you will look like one in an interview. The interviewers will assume you know very little about the basics of the craft if you carry on like this, or at the very least will see you as a detriment to teamwork. HTH.

Is this an AC ammeter, or a DC ammeter? (And isn't that just a voltmeter anyway, in most actual cases????) Hmmm...

You can't escape the fact that voltage and current are joined at the hip, they are for all practical purposes different expressions of the same thing. Whatever affects one *has* to have affected the other.

No more or less than current. They are joined at a hip called Ohm's Law.

Relative to where?

Since we can discuss current using only voltage as the variable (resistance being a constant in this example), *anything* you can say about voltage is directly related to current.

One of the overall things that you *have* to keep in mind is that periodic reality checks are necessary. One of them is the fact, repeated by many in this thread, that "DC sine wave" is a contradiction of terms. If your definition makes it possible, your definition *can't* be right.

My point still stands, that if the current is changing, it is by definition AC, and current not changing is DC. Trying to look at it as DC is all in one direction and anything else is AC, doesn't work.