How to measure actual power output of amplifier?
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?
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.
Page 102 of the car audio site (link in sig line below).
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?
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.
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?
It will be hard to get a true reading for a speaker as load.
A speakers impedance varies wildly across the frequency spectrum.
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.
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.
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?
|All times are GMT. The time now is 01:00 AM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
Copyright ©1999-2017 diyAudio