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Old 19th November 2003, 02:38 AM   #1
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Location: Sacramento, CA
Default Ohm's Law 101

Quote:
I=V/Z is a linear relationship BUT
the Z of a driver (or any reactance)
is (by definition) not. Something to
do with 'rate of change'
So what if the Z isn't linear versus frequency? That's not what the author said. What the author said was that current and voltage will only have a linear relationship if the circuit is purely resistive.

Here, read it again:

By Ohm's law, the current in a speaker voice coil is the amplifier output voltage divided by the speaker impedance. Therefore, the current can only be linearly related to the voltage if the speaker presents a purely resistive load to the amplifier.

I'm sorry, but Ohm's Law is a linear equation whether you're using resistance or reactance or impedance and therefore there will be a linear relationship between current and voltage even in a reactive circuit.

I = V/R, V = I x R, R = E/I

I = V/X, V = I x X, X = V/I

I = V/Z, V = I x Z, Z = V/I

se
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Old 19th November 2003, 03:04 AM   #2
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isn't Z a complex number ?
it contains a 'j factor'
-1 ^ 0.5
I'm out on a limb here but
maybe because of the phase?
in a resistor i and e are in phase
notso with reactance
possibly 'Ohms' (as applied to
reactance is sort of a misnomer)
NO
it's because the final result of a
amp/transducer is WATTS !
so the i and e being out of phase
( p = i * e )
is complicated .........
elaborations wellcome
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Old 19th November 2003, 03:27 AM   #3
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Quote:
Originally posted by hitsware
isn't Z a complex number ?
it contains a 'j factor'
-1 ^ 0.5
Yeah, it's a complex number, the sum of the real part (resistance) and the imaginary part (reactance). But for a given Z, voltage and current still maintain a linear relationship so it doesn't matter whether you measure distortion based on voltage or distortion based on current.

Quote:
I'm out on a limb here but
maybe because of the phase?
in a resistor i and e are in phase
notso with reactance
possibly 'Ohms' (as applied to
reactance is sort of a misnomer)
Sure, the phase changes but you still have a linear relationship between voltage and current so it doesn't matter if you measure distortion based on voltage or distortion based on current.

Quote:
NO
it's because the final result of a
amp/transducer is WATTS !
so the i and e being out of phase
( p = i * e )
is complicated .........
But by Ohm's Law, I and E still maintain a linear relationship.

For a given Z, if you measure x amount of voltage across it, you can assume that there's y amount of current flowing through it by way of I = E/Z.

Quote:
elaborations wellcome
Well, is it possible that the guy who wrote that article is well, wrong? I mean, he says that voltage and current will only have a linear relationship in a purely resistive circuit yet every instance of Ohm's Law I've seen with regard to reactive AC circuits says the same thing. And they're all linear equations.

se
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Old 19th November 2003, 03:31 AM   #4
SY is offline SY  United States
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Linear is what is the key. Linear in voltage. Linear in impedance. Nonlinear in frequency (except for special cases).
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Old 19th November 2003, 03:37 AM   #5
jam is offline jam  United States
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Steve,

From http://www.fighting.org/cgi-bin/disc...=441&post=1277

Another "Law" Bites The Dust

(Ohms Law full of crap)
This post is dedicated to otherpower.com

I'm transfering this info from the last JCP show that otherpower deleted.
Some of you have seen it before. But it's proof that "laws" are FULL OF ****.
__________________________

15 Negative Difference

“You will find below a very simple experiment that anyone can perform with few and cheap electronic components. This experiment will demonstrate you how a very simple oscillator can be built with a component which shows a Negative Differential Resistance ( NDR ) effect when it is used properly.

There is no overunity effect here but this device is worth to be known because in this case the NDR component used does not seem to agree with the basic Ohm's Law : According to the Ohm's law an increase of the voltage produces an increase of the current, in this case, in the negative resistance region of the NDR characteristic curve, an increase of the voltage produces a decrease of the current.”



Build and test your own Negative Differential Resistance Oscillator with a Negistor
http://jnaudin.free.fr/cnr/negosc.htm
_____________________

This guy "Jean Naudin" has been missing for a few weeks by the way, no one has heard from him.
Hope he's okay.
He has shown quite a few secrets others don't want you to know.
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Old 19th November 2003, 03:48 AM   #6
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Quote:
Originally posted by SY
Linear is what is the key. Linear in voltage. Linear in impedance. Nonlinear in frequency (except for special cases).
Well impedance won't be a linear function of frequency. But the voltage and current relationships across a given impedance will be.

se
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Old 19th November 2003, 03:52 AM   #7
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Quote:
Originally posted by jam
Steve,

From http://www.fighting.org/cgi-bin/disc...=441&post=1277

Another "Law" Bites The Dust

(Ohms Law full of crap)
This post is dedicated to otherpower.com

I'm transfering this info from the last JCP show that otherpower deleted.
Some of you have seen it before. But it's proof that "laws" are FULL OF ****.
__________________________

15 Negative Difference

“You will find below a very simple experiment that anyone can perform with few and cheap electronic components. This experiment will demonstrate you how a very simple oscillator can be built with a component which shows a Negative Differential Resistance ( NDR ) effect when it is used properly.

There is no overunity effect here but this device is worth to be known because in this case the NDR component used does not seem to agree with the basic Ohm's Law : According to the Ohm's law an increase of the voltage produces an increase of the current, in this case, in the negative resistance region of the NDR characteristic curve, an increase of the voltage produces a decrease of the current.”



Build and test your own Negative Differential Resistance Oscillator with a Negistor
http://jnaudin.free.fr/cnr/negosc.htm
_____________________

This guy "Jean Naudin" has been missing for a few weeks by the way, no one has heard from him.
Hope he's okay.
He has shown quite a few secrets others don't want you to know.
Hahahaha! Thanks, Jam! Gotta love these folks.

I like this one:

As you can see, this kid knows something is not right with what they are ordered to teach in schools.
It's not a mistake, it's a premeditated crime. You're being robbed.


Rather smacks a bit of the same verbiage spewed by some of the charlatans in this business.

se
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Old 19th November 2003, 04:19 AM   #8
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>No, if you feed a sinusoid into a reactive circuit you'll get a sinusoid output.

a sinewave is a 'special case' try a squarewave
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Old 19th November 2003, 05:05 AM   #9
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I think what he is saying is this:
Say we sit down to design a new amp.
What's out first step?
We hook it up to a 8 Ohm resistor to test it.
But the position of the cone (as noted by you
in another space) is proportional to the current
through the coil (not the voltage accross it)
(AND isn't the position of the cone the be all/end all)?
SO
Our analysis of the system is flawed
(our scopes and distortion analysers work on voltage)
.........SO
Our amplifier models (feedback included)
are based on voltage .... only a part of
power (which ultimately does the work)

hope this helps
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Old 19th November 2003, 05:05 AM   #10
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"So what if the Z isn't linear versus frequency?"
-then you bust out the diffy-Q skills... (differential equations, my teach would write out the abbreviation as Diffy-Q instead of diff-EQ) and then you solve some KVL and KCL. remember that at any time V = I Z, so if Z changes so will I or V or both. for goofy changes in Z is this easy? depends on how much you like laplace or how good you are at diffy-Q.
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