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?
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?
You could use a resistive load and buy a inexpensive o-scope with a RMS watts function and play a test tone through the amp hooked to the load and bring the volume up until it clips and just back it off until the sine wave is un-clipped and see what the scope measures. Just make sure you set the scope to the proper ohms configuration.
I bought a velleman HPS40 and it works well for testing power output.
I bought a velleman HPS40 and it works well for testing power output.
Thanks for the suggestions but is there a way to do it with a reactive load?
Would this work?
1. First test output power using dummy load. Note the amount of current that the amplifier is drawing from the B+ line. Repeat test for a few other lower wattages.
2. Then connect a speaker and adjust the output until the amplifier draws those known current amounts.
With that information, I should be able to give a very accurate estimation of the actual power sent to the voice coil right?
For example, if I know that my amplifier draws 10 amps of current from the power supply when my dummy load of 4 ohms is dissipating 50 watts of energy, I can infer that when my amplifier is drawing 10 amps while driving a reactive load, it is delivering 50 watts to that load right?
Would this work?
1. First test output power using dummy load. Note the amount of current that the amplifier is drawing from the B+ line. Repeat test for a few other lower wattages.
2. Then connect a speaker and adjust the output until the amplifier draws those known current amounts.
With that information, I should be able to give a very accurate estimation of the actual power sent to the voice coil right?
For example, if I know that my amplifier draws 10 amps of current from the power supply when my dummy load of 4 ohms is dissipating 50 watts of energy, I can infer that when my amplifier is drawing 10 amps while driving a reactive load, it is delivering 50 watts to that load right?
Hi,
Its very difficult to get anything like an accurate number for this case,
unless the meters are capable of "true RMS" readings, most aren't,
they can only measure steady AC signals like the mains accurately.
Furthermore you have the problem of reactance, which draws current
from the amplifier that doesn't dissipate power in the load, and that
power must be dissipate by the amplifier itself.
TBH some simple estimations are probably better than measurements.
Music has a crest factor - which depends on its dynamic range.
Take a 100W amplifier driving 8 ohm speakers mildly clipping to
3% distortion on the peak levels of the music programme.
The continuous RMS power into an 8R resistor for this case will be something
like 5W to 25W - depending on how compressed the music material is.
A real loudspeaker probably wont dissipate any more power than an 8R,
but it will cause the amplifier to dissipate more power due to the load.
For hifi its best to assume an 8 ohm speaker is as onerous as a 4R load.
rgds, sreten.
Its very difficult to get anything like an accurate number for this case,
unless the meters are capable of "true RMS" readings, most aren't,
they can only measure steady AC signals like the mains accurately.
Furthermore you have the problem of reactance, which draws current
from the amplifier that doesn't dissipate power in the load, and that
power must be dissipate by the amplifier itself.
TBH some simple estimations are probably better than measurements.
Music has a crest factor - which depends on its dynamic range.
Take a 100W amplifier driving 8 ohm speakers mildly clipping to
3% distortion on the peak levels of the music programme.
The continuous RMS power into an 8R resistor for this case will be something
like 5W to 25W - depending on how compressed the music material is.
A real loudspeaker probably wont dissipate any more power than an 8R,
but it will cause the amplifier to dissipate more power due to the load.
For hifi its best to assume an 8 ohm speaker is as onerous as a 4R load.
rgds, sreten.
Let me see if I understand the implications.
Are you saying that an amplifier's efficiency for a given impedance can change depending on whether the load is reactive or not?
So even if I know that my amplifier draws 10 amps (at a known voltage) while delivering 50 watts to a resistive load, it is NOT delivering 50 watts to a reactive load even if the amplifier is still drawing 10 amps (at the same known voltage) from the power supply? That the amplifier's efficiency drops because of the reactive load, dissipating more energy as heat?
Are you saying that an amplifier's efficiency for a given impedance can change depending on whether the load is reactive or not?
So even if I know that my amplifier draws 10 amps (at a known voltage) while delivering 50 watts to a resistive load, it is NOT delivering 50 watts to a reactive load even if the amplifier is still drawing 10 amps (at the same known voltage) from the power supply? That the amplifier's efficiency drops because of the reactive load, dissipating more energy as heat?
Yes but if you know that your amplifier is consuming a certain number of watts (by measuring its current input and voltage), and you know what level of power output that corresponds to (by testing your amp with a dummy load), it would seem reasonable to infer that if that same amplifier is consuming that same number of watts, it would deliver the same number of watts (that was measured in the dummy load) to the reactive load, assuming the amplifier's efficiency doesn't change.
Do you agree with that statement?
So basically, let's say you have an amp that is drawing 10 amps from a 12 volt power supply. 10 x 12 = 120, so the amp is consuming 120 watts.
Now let's say you hook up a resistive load to the amp and you determine that the amp is delivering 60 watts to it.
So now we know that the amp draws 120 watts to deliver 60 watts (50% efficiency).
Now let's say you hook up a speaker to the amp. Then you turn up the volume until the amp draws 10 amps from the 12 volt power supply.
Would it not be reasonable to assume that you're delivering 60 watts to the reactive load assuming that the amplifier's efficiency hasn't changed?
It seems that sreten is suggesting that the amplifier's efficiency changes, so I'm waiting to see if he can elaborate on that or if someone else has an explanation.
Do you agree with that statement?
So basically, let's say you have an amp that is drawing 10 amps from a 12 volt power supply. 10 x 12 = 120, so the amp is consuming 120 watts.
Now let's say you hook up a resistive load to the amp and you determine that the amp is delivering 60 watts to it.
So now we know that the amp draws 120 watts to deliver 60 watts (50% efficiency).
Now let's say you hook up a speaker to the amp. Then you turn up the volume until the amp draws 10 amps from the 12 volt power supply.
Would it not be reasonable to assume that you're delivering 60 watts to the reactive load assuming that the amplifier's efficiency hasn't changed?
It seems that sreten is suggesting that the amplifier's efficiency changes, so I'm waiting to see if he can elaborate on that or if someone else has an explanation.
A useful fundamental to know is that a pure reactance, be it capacitive or inductive, can't dissipate any power at all. Power can only be dissipated in the resistive term.
With sine waves things are pretty easy, even with fixed reactive loads. I think that for a music signal into something complex like a speaker you'd have to measure the instantaneous voltage and current at a high sampling rate, and then calculate from there. It can also be done with dedicated analog circuitry. Another way is to put a thermocouple on the voice coil and get an idea from temperature rise. It's not an easy number to get with any accuracy, nor is it a very useful number.
With sine waves things are pretty easy, even with fixed reactive loads. I think that for a music signal into something complex like a speaker you'd have to measure the instantaneous voltage and current at a high sampling rate, and then calculate from there. It can also be done with dedicated analog circuitry. Another way is to put a thermocouple on the voice coil and get an idea from temperature rise. It's not an easy number to get with any accuracy, nor is it a very useful number.
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?
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?
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.
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.
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.
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.
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.
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.
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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.
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
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.
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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