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Old 13th December 2012, 04:09 AM   #11
d a o is offline d a o  Denmark
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Old 13th December 2012, 06:29 AM   #12
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What's the purpose of using anything other than a proper dummy load? They are not expensive and will provide repeatable measurements.
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Old 13th December 2012, 07:49 AM   #13
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Quote:
Originally Posted by heydanchan View Post
Ok, let's say we play a 60Hz sine wave into a driver.

If the amp draws 120 watts and at that instant a dummy load is known to dissipate 60 watts, would the driver also dissipate 60 watts if the amp is still drawing 120 watts?

I understand that music is complex but what goes in must come out, and it can only come out through the driver as sound and heat or through the heatsink as heat.

So isn't it reasonable to assume that if the amp is averaging a power consumption of 120 watts, 60 watts average must be delivered to the driver?
With that same 60Hz, yes.
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Old 13th December 2012, 07:55 AM   #14
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Well, I guess a dummy load would be good for confirming an amp's rating. I'm also interested in how "hard" I'm driving my subwoofers. I want to get an idea of how much power they're dissipating. When I put a clamp meter on my setup, I could see exactly how much current was flowing at any instant, so that seems like the best way to get a "ballpark" figure.

I became curious when I suspected my 1700wattRMS@2ohm amp wasn't delivering anywhere near its rated power. When I "clamped" my system, my calculations showed that I was barely "pushing" 375 watts continuously. When I dropped the load from 1.5 ohms down to 3 ohms, I measured 243 watts.

I just got a new amp that has CEA 2006 compliant ratings, so I'm going to "clamp" that to see how it compares to my old amp, and since it is CEA 2006 compliant, I won't need to test it with a dummy load (I trust Rockford Fosgate), and it came with a birth certificate that stated how many watts it actually measured.

But all things practical aside, I'm still curious about the phase angle relationship between the current and voltage of sine waves at various frequencies. If I knew or could measure them somehow, I could calculate the power factor and get the actual wattage.

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Originally Posted by Perry Babin View Post
What's the purpose of using anything other than a proper dummy load? They are not expensive and will provide repeatable measurements.
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Old 13th December 2012, 07:24 PM   #15
sreten is online now sreten  United Kingdom
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Hi,

The basics are for a given output current (and voltage) the more reactive
the load for the given apparent impedance (Z) the less power the speaker
dissipates and the more power the amplifier dissipates compared to Z=R.

In the extreme with Z=R and the load being purely reactive all the power
is dissipated in the amplifier. For speaker impedance curves high reactance
and high phase angles are fine at high Z, but for the low Z dips of the
speakers impedance curve you are looking for moderate phase angles.

A reasonable assumption is a real speaker is as onerous a load on
an amplifier as a resistor of half the speakers nominal impedance.
In reality speakers with "nice" impedance curves somewhat more.

rgds, sreten.
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Old 14th December 2012, 07:12 AM   #16
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I'm sorry. I don't mean to argue but I don't think that is correct. If the load is purely reactive, all the power is returned to the source. It is not dissipated by the amplifier.

Based on that, a reactive speaker load should be easier to drive than a resistive load because a speaker returns some of the power it took back to the amp.

Take for example an ideal inductor (zero resistance). It stores energy and releases all of it. It consumes zero power itself. That should be a very easy load to drive.

I think this page does a good job explaining reactive power.

Power in resistive and reactive AC circuits : Power Factor

Note that in the case of a generator powering a purely reactive load, all the power is returned to the generator, making it effortless to turn the generator. In the case of car audio, it would mean that the battery would not have to work as hard (and/or the alternator), so the current meter on the B+ cable would reflect that by reading a lower current.

So based on all this, I think it's safe to say that you can get a reasonably accurate estimate of the power being dissipated by a driver if you can measure how much power the amplifier is using if you know how much power that same amplifier consumes while driving a resistive load.

And to verify that the amplifier isn't dissipating any extra energy, you can measure the temperature of the amplifier.

Quote:
Originally Posted by sreten View Post
Hi,

The basics are for a given output current (and voltage) the more reactive
the load for the given apparent impedance (Z) the less power the speaker
dissipates and the more power the amplifier dissipates compared to Z=R.

In the extreme with Z=R and the load being purely reactive all the power
is dissipated in the amplifier. For speaker impedance curves high reactance
and high phase angles are fine at high Z, but for the low Z dips of the
speakers impedance curve you are looking for moderate phase angles.

A reasonable assumption is a real speaker is as onerous a load on
an amplifier as a resistor of half the speakers nominal impedance.
In reality speakers with "nice" impedance curves somewhat more.

rgds, sreten.

Last edited by heydanchan; 14th December 2012 at 07:25 AM.
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Old 14th December 2012, 07:40 PM   #17
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Quote:
Originally Posted by heydanchan View Post
I'm sorry. I don't mean to argue but I don't think that is correct.
You have slightly misunderstood Steren position. I believe he is talking about an actual dynamic loaded speaker with music and as such there is no easy or meaningful way to say what the wattage is consistently. We have standard classroom text methods which mean little in real world environments and vice a verse.
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Old 21st December 2012, 10:39 AM   #18
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Quote:
Originally Posted by heydanchan View Post
I want to measure how much power is going into my speakers. I was going to use an AC clamp ammeter to measure the current and multiply that value by the voltage at the speakers but my friend told me that if the power factor isn't 1, I'd get an inaccurate reading.

Does anyone know what the power factor might be, if it's anywhere near 1?

I don't want to use a resistive test load because I want to know the actual power going to my speakers. How would I go about measuring that accurately?

Hi man, he's right, absolutely.

You'd have a pf clamp , that can does everything, even a coffe cup

Jokes aside, you need to measure ac volt, ac amp, and pf as well for having Z"real" e active power too.

bye
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Old 20th January 2013, 04:51 PM   #19
JMFahey is offline JMFahey  Argentina
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You didn't get the Sreten point.
The inductive load "sends back" the energy to the generator.
You understood that so far.
BUT, the "generator" in this case is not an "electromechanical generator" which by construction is inductive itself, but something very different.
The typical amplifier is made out of an energy reservoir (the PSU capacitors) whose energy is delivered in a controlled fashion to the load, through a ***resistive*** control element, be it a transistor, tube, Mosfet or anything you invent in the future.
That resistive element will dissipate power, following the I^2*R equation, no matter what the voltage and current phase difference exists in the load.
So I couldn't care less about what actual power is dissipated at the load, what will kill my amp will be instantaneous V*I at the transistors.
So in practice a Reactive load is murder for the amps.
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