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Old 04-13-2008, 01:25 AM   #1 (permalink)
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Default power factor and amps

so if we have an amp, a driver, and a connecting cable.....
and we use a wideband sound pressure meter to determine a level at which we drive the output device [driver]. and we measure the wattage and VAR's of the amp output at this level we can calculate the amplifiers power factor in this circuit. change driver or amp or wire and repeat SPL measurement and recalculate power factor. repeat as needed till options exhausted.....

what have we determined?? have we determined how well an amp likes a particular driver? does a higher power factor mean cleaner sound?? longer component life???

or is this a futile exercise?
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Old 04-13-2008, 07:51 PM   #2 (permalink)
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Whut!? Can you try to clarify that question a little ... or even a lot?
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Old 04-13-2008, 08:20 PM   #3 (permalink)
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I think you just calculated the efficiency of the speaker (how much energy from the amp becomes sound/how much is dissipated as heat). I've never heard of something called "power factor."
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Old 04-13-2008, 11:50 PM   #4 (permalink)
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Quote:
Originally Posted by m3_arun View Post
I think you just calculated the efficiency of the speaker (how much energy from the amp becomes sound/how much is dissipated as heat). I've never heard of something called "power factor."
no i think speaker efficiency would just be sound pressure vs. watts.
power factor is like phase angle i think you could say....

if you measure the power factor its a percentage of efficiency of the energy generator in the powered circuit. so if a set of cables is better than another it should raise the power factor maybe?

power factor is drawn as a triangle. watts on one side, var's on the other.
solve for hypotenuse and divide answer into watts. answer is power factor.

counter EMF from the driver against the amp and how it is affects the phase is what we are talking about here i guess. certainly there is more than impedance shown to an amp from a driver. and it's level driven.

check wiki for an explanation of VAR's if your interested in this. hard to explain here.
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Old 04-14-2008, 03:04 PM   #5 (permalink)
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oh yea, and the SPL would be the constant here. re-driving to a predetermined SPL before re-measuring the circuit. even if there is no measurable change in sound level per watt, other characteristics of the sound could be changed by component changes, not measured as sound pressure or freq response.

phase and time alignment maybe. i dunno. maybe there is a way to see what spending 30,000.00 on a 6' pair or speaker cables actually did rather than just subjective observation.

maybe there is a way to see how electronics get along with each other, and what works best with what. see if something really broke in or it didn't matter.

this is not a question really i am making, just trying for a little brain storming. just pub talk i guess.......
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Old 04-14-2008, 04:29 PM   #6 (permalink)
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Aha! A little pub talk you want on power factor! Sheesh, I better turn in my pocket protector because that's not the kind of stuff Italk about over a beer. You gotta be a bit more level headed for this discussion.

Your interest is a well directed one, however, and apropos to the problems of amps actually driving loads with authority. I'm really no expert at all on the issue other than I kinda get the gist of it. Llet me point you to a very good article.

Stereophile: Heavy Load: How Loudspeakers Torture Amplifiers

And the thread on their forum about the article in which I blabber a bit.


Stereophile Forums: Speaker load
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Old 04-14-2008, 06:40 PM   #7 (permalink)
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I think practically speaking, the amp produces Volt-Amps, period. It's the load that determines what percentage of the VA's turn into Watts.

Most definitely, a speaker load would be considered an inductive reactance, due to the back EMF produced by the speaker cone return. So, the power factor will be less than 1 for an amplifier-speaker circuit. This means that although the amp produces a rated voltage and a rated current, the actual power consumed by the speaker is much less.

In large industrial/commercial distribution systems, power factor correction is often applied at the service entrance with banks of capacitors. This is because the utilities charge much more if the power factor is less than 0.9 (in most cases). This occurs often in the case of manufacturing plants, where the windings in many motors that exist throughout a mfg plant add up to a significant inductive load.

Essentially, poor power factors by the end users mean the utility must install a much larger base of power generation. So, it's in everbody's interest to correct for power factor. In my early days of consulting engineering, I sometimes helped audit power consumers for ways to reduce the power factor and lower their utility bills.

In the case of speakers, I believe a Zobel network performs the same function: a capacitor is added to counteract the effect of the inductive speaker coil. This ensures that the amp is delivering as much power as it can, rather than wasting VA's against a reactive load.
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Old 04-15-2008, 12:33 PM   #8 (permalink)
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excellent responses. i will dig into the stereophile url's today. good info there for sure.

and yes i was looking at the setup as basically a power generation circuit in my mind and thats how i got on this power factor thing. now i wonder if the PF has an affect on sound. is tweeking a setup to achieve a better PF going to be worth the time and effort. could one spend money on new gear only to realize a lower fidelity due to a decreased PF?

if so maybe adding some correction would be a benefit, or even a new revenue source for someone down the road.

in the early 80's i had a kenwood ka-1000 amplifier which had what they called sigma drive. it amounted to 4 terminals at the amp to connect each speaker. they were absorbing the counter EMF or something on a pair from each speaker. or maybe they were using the signal to tweek the amp for a better PF. i dunno.

i do know that a generator running solely on a distribution network will have a var measurement that can only be changed by adding capacitor banks as you describe located near the offending device. however when a generator is tied to the grid and a distribution network the generator var's can be changed by varying the exciter voltage, thus stealing var's if you will from the grid. the grid maintains the voltage. untied from the grid the exciter must be set to maintain voltage.

so maybe there was something similar going on in the kenwood amp to vary the var's. i really dont know......

and while speaker has been assumed in this discussion i was really thinking headphones all along. here i assume a bigger variance in PF, between say an IEM and like an HD-600. i'm thinking these are way different animals.

but maybe not i really dont know this stuff. just wondering. looking for the final tweek
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Old 04-15-2008, 12:50 PM   #9 (permalink)
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I think you know much more than you're telling.

Generators tied into a grid can be very complex - literally and practically. It's one reason loads are often sized for single diesel generators in the case of emergency backup. It's not just the cost, it's the complexity of the paralleling gear and whether multiple generators can sync enough to tie into an existing grid.

Since yesterday, I've learned that the impedance of the amp must be taken into account, too.
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Old 04-15-2008, 01:02 PM   #10 (permalink)
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I'll add here that power factor in terms of a power distriibution system is usefull because you're only talking about one frequency. So, the power factor calculation comes up to be one nice, steady number. If you were to distribute power as a broad band signal (pink noise, lets say) you'd have a different situation completely. (Not to mention that motors pretty musch wouldn't work at all!)

So the power factor of an amp driving a transducer (whether it be a speaker or headphone) is going to vary significantly from one instant to the next depending on what just happened and what the amp is trying to make make happen. So power factor is continually changing and sometime the amp is 'ahead' of the speaker and sometimes not.
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