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
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 🙂
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 🙂
hitsware said:
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.
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.
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.
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
Linear is what is the key. Linear in voltage. Linear in impedance. Nonlinear in frequency (except for special cases).
Steve,
From http://www.fighting.org/cgi-bin/discus/show.cgi?tpc=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.
From http://www.fighting.org/cgi-bin/discus/show.cgi?tpc=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.
Attachments
SY said:
Well impedance won't be a linear function of frequency. But the voltage and current relationships across a given impedance will be.
se
jam said:
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
>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
a sinewave is a 'special case' try a squarewave
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 🙂
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 🙂
"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.
-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.
hitsware said:
What is a square wave but a series of sinewaves?
Sure, depending on the Q of the reactive circuit's resonance you can get some ringing. But even then, your voltage is still going to be linearly proportional to current.
se
>Yes. That's because pretty much all loudspeakers are designed to be driven by a voltage source rather than a current source.
Even though "designed" for voltage, there is an incongruity here. (bassed on current doing the moving)
BTW
Someone once told me that with a perfect xfmr......
If you hook the secondary to a scope (infinate Z)
You will not see any voltage (primary driven by sig gen)
UNTIL
You load the secondary.....
I.E. xfmrs work by current ..
Does that make any sense ?
Even though "designed" for voltage, there is an incongruity here. (bassed on current doing the moving)
BTW
Someone once told me that with a perfect xfmr......
If you hook the secondary to a scope (infinate Z)
You will not see any voltage (primary driven by sig gen)
UNTIL
You load the secondary.....
I.E. xfmrs work by current ..
Does that make any sense ?
hitsware said:
That's easy. Fire up the bong. 😀
Sure, if you're wanting to test it driving an 8 ohm resistive load.
I thought we were testing the amplifier?
How is our analysis of the system flawed? What exactly are you wanting to analyze? If you want to analyze distortion, how does it matter whether you're measuring it by way of current or voltage? As long as there is a linear relationship between current and voltage, you'll get the same result.
se
theChris said:
Diffy Q's cool. I like how they put those little loops in the tops of their ice cream cones. 😀
se
hitsware said:
How is there an incongrity?
As long as I = E/Z and E = I x Z, then I will always be linearly proportional to E and vice versa.
Well yeah, since transformers work by way of inductive coupling and since you need a current flowing to produce the magnetic field...
Of course that rather presupposes that you're driving the transformer with a voltage source.
se
>How is our analysis of the system flawed? What exactly are you wanting to analyze? If you want to analyze distortion, how does it matter whether you're measuring it by way of current or voltage? As long as there is a linear relationship between current and voltage, you'll get the same result.
There is NOT a 'linear relationship' between
the output voltage (which our meters show) and the audio out
(position of the cone)
NOR
quite just the current ...
It's more complex ....
That's all the guy was sayin'
( Quite eloquently ) IMHO
There is NOT a 'linear relationship' between
the output voltage (which our meters show) and the audio out
(position of the cone)
NOR
quite just the current ...
It's more complex ....
That's all the guy was sayin'
( Quite eloquently ) IMHO
>I thought we were testing the amplifier?
In this mode we're testing the amplifier for
things that it doesn't really do.
In this mode we're testing the amplifier for
things that it doesn't really do.
hitsware said:
Huh? Did you miss this part?
The basic conflict of interest is that (electrodynamic) speakers are inherently current controlled transducers, the force accelerating the diaphragm(s) and thus controlling the air pressure fluctuations we call sound, is ideally related to current through the formula:
F = (Bxl)*I
That's a linear equation as well. Since I has a linear relationship to E, then F must have a linear relationship to both I and E.
I'm sorry, but I can't seem to find any evidence to support his claim that I can only have a linear relationship to E in a purely resistive circuit. And since that claim seems to form the foundation of the rest that follows, then the rest that follows seems to be flawed as well.
se
Hmmm .... We may be getting into sort of a 'time domain'
vs 'freqency domain' thing
Actually (which oftimes happens)
I think we're squabblin' over semantics .....
vs 'freqency domain' thing
Actually (which oftimes happens)
I think we're squabblin' over semantics .....
- Status
- Not open for further replies.