DC Wave Questions

One answer. Sine waves aren't DC.

N

How come? Do you object to the term "DC" - is monophasic acceptable to you?

See also:

http://64.233.161.104/search?q=cache:SMA_gSlzQ18J:global.daikin.com/global/our_product/sp_Inverter/3_techno.html+%2B%22dc+sine+wave%22&hl=en&lr=lang_en

Impedance varies with frequency if there are reactive components, L's and C's. Since you haven't told us whether this is a series or parallel circuit of L's, R's and C's, We don't know what the impedance is at DC, zero frequency or any other frequency for that matter. If it's a parallel circuit the DC impedance is zero unless there is resistance in series with the L as is the usual case. In that case, the impedance is R at DC. If it is a series circuit, the DC impedance is infinite. SO, you have three choices, Zero ohms, Infinite ohms or R ohms depending on the connection.

A peak detector will have to work on the range of voltages expected on it's input. I can't get to the URL, sorry. Bob

That is not always true.

Take

1) A resistor of resistance R in series with a capacitor of capacitance C. 2) Another identical resistor of resistance R, but in series with an inductor L.

Make R=sqrt(L/C)

and put 1 and 2 in parallel and measure the impedance across that combination. The impedance is always R, and is independent of frequency.

A useless fact I would admit!!

Varying DC? i.e. DC varying in amplitude a manner similar to an AC sine wave. If it goes into plus and minus regions I guess we are getting pretty close to an AC waveform?

There is no such thing as a "DC sine wave." I suspect you mean what would more correctly be described as a 10 volt peak-to-peak sine wave with a +10 volt DC offset.

The principle of superposition applies: the currents and voltages in the circuit will be the sum of those that would result if the DC voltage and the AC sine wave were applied to it seperately.

That circuit (I've fixed the link) exploits the fact that the LM139 comparator has an open-collector output. It runs off a negative rail, and cannot produce a positive output voltage.

O.K. here's the combinatrics:

Combo 1: DC Sine Wave + (R+L in series with C parallel)

Combo 2: DC Sine Wave + (R+C in series with L parallel)

Combo 3: DC Sine Wave + (L+C in series with R parallel)

Combo 4: DC Sine Wave + (R, L, and C all in parallel with each other)

Combo 5: DC Sine Wave + (R, L and C all in series)

O.K., so can I correctly infer from your response that a negative peak detector will yield a value of +5V for a sine wave which varies from

+5V to +15V?

Not that it's that important, but I don't see why a "DC sine wave" is an impossible concept, considering the definition of DC as a current which flows in one direction:

A "DC Sine wave" doesn't say that current reverses direction, only that the current flow wanes and waxes.....like a river is still a river even though its flow varies with rainfall...

O.K. - now we're getting somewhere......you're saying the current and voltage (and the implied impedance Z = V/I) of the "DC sine wave" is the sum of the respective current and voltage of a +10V DC signal and a

-5V/+5V AC signal going into the same load.

Example: DC +10V into load produces 1 Amp, therefore implied resistance = 10 ohm. and AC -5V/+5V (and given frequency) into load produces 0.5 amps, therefore implied impedance = 20 ohms,

then what would the superposition prinicple predict as the resulting combined current and impednace?

read the original post - talking about a sine wave bouncing between +5V and +15V - no where near negative

That's an AC wave with a DC offset.

N

The current is simply the sum of the AC and DC components e.g. 0.5 amps peak-to-peak AC with a 1 amp DC offset Max. instantaneous current = 1.25 amps Min. instantaneous current = 0.75 amps

Impedance can be represented as a complex number: real part = reisitance = R = 10 ohms imaginary part = reactance = X

Total impedance Z = R + jX

To work out the imaginary part, you have to do a vector addition because current and voltage in a reactance are 90 degrees out of phase:

Ipk = Vpk / sqrt(X*X + R*R)

0.25 = 5 / sqrt(X*X + 10*10)

X = sqrt(300) = 17.3

i.e. Z = 10 + j*17.3

Your posts have all the characteristics that indicate you are a troll. If you aren't I suggest you quit being combative and learn from what the posters are saying.

And re the link; that refers to an inverter that uses a DC input and outputs a sinewave. You must be troll.

So the 10V p-p sinewave is riding on 10VDC. There is no requirement that a sinewave must have an absolute negative component.

Question 1: A capacitor "capacitates" whether it sees AC or DC. An inductor "inducts" whether it sees AC or DC. A resistor resists whether it sees AC or DC. You might find it beneficial to think of what happens to each component on a component level rather than thinking of total impedance. Understand what each component does, and circuit impedance will make more sense.

Question 2: 404 file not found error That said, you can peak detect on a varying DC sine. As someone else said, its AC with a DC offset.

Ed

The impedance of a set of passive devices is independent of the voltage across them. It only depends on R, L, C, and f. The fact that there is a DC component makes no difference.

An inductor will pass DC current as if it were a wire. Only differences in current cause a voltage across it. A capacitor will not pass DC, so the DC does not matter. Obviously, a resistor is a resistor, and cares nothing for ac vs dc.

This is only true for ideal components. In the real world, inductors, caps and resistors have voltage limitatations. They are usually well beyond 15V, though.

Your link has crap on the end. Here it is without the crap:

With this circuit, the input at V+ will always be outside the power rails. Thus, it will not work.

NOTE: I changed the followup-to field to sci.electronics.basics, because that is where this thread belongs. I hope you don't mind.

If you think that the term "fully DC Sine Wave" even means anything, then you have not understood the coursework. Either your teacher is incompetent, or you have been spending too much time partying and not enough time studying.

Good Luck! Rich

Bullshit. This kid is not a troll, by any means. He's just a student desperate to weasel answers to his final without having to learn the material he was supposed to have learned while partying and chasing tail.

A troll is a much more serious matter. This is just a child who needs to fail the course, have Mom and Dad scold him, and next semester, pay attention in class.

Cheers! Rich

According to Fourier analysis, any repeating waveform can be decomposed into harmonically related and appropriately phase shifted sine waves and also a DC component. If all the components involved are linear, then they react to each of these components, independently, and the result is the linear sum of all those reactions. So the capacitors react to the DC component as open circuits, and the inductors as short circuits. At all frequencies, the resistances follow ohms law, and at each AC harmonic, the inductances and capacitances react in their normal frequency dependent ways.

Throw in one nonlinear component, like a diode, and you have to do a completely different kind of analysis.

That is not a "regular" peak detector, it is a comparator used as an overcompensated opamp follower and exploits the open collector output characteristic of fast discharge and slow ( 10 second) charge of the capacitor. In concept it will work for a varying "DC sine wave" by replacing "-Vcc" with "GND" and all "GND"'s with "+15V" in that circuit diagram only. Then Vout= "Vpk,neg" =+5V.