How to measure actual power output of amplifier? - diyAudio
Go Back   Home > Forums > General Interest > Car Audio

Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 29th November 2012, 02:50 AM   #1
diyAudio Member
 
Join Date: Nov 2004
Location: US
Default 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?
  Reply With Quote
Old 29th November 2012, 03:13 AM   #2
diyAudio Member
 
seankane's Avatar
 
Join Date: May 2010
Location: Washington State
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.
__________________
Pyramid 52 Amp Power Supply----Fluke 787 Processmeter----Tektronix 465 Scope----Velleman HPS40 Scope----Soldering Iron----Desoldering Iron----Hejet HJ 500-S Heatgun----DM4070 Digital LCR Meter----Propane Torch
  Reply With Quote
Old 29th November 2012, 03:19 AM   #3
diyAudio Member
 
Join Date: Nov 2003
Location: Louisiana
Page 102 of the car audio site (link in sig line below).
  Reply With Quote
Old 12th December 2012, 07:15 AM   #4
diyAudio Member
 
Join Date: Nov 2004
Location: US
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?

Quote:
Originally Posted by seankane View Post
You could use a resistive load...
Quote:
Originally Posted by Perry Babin View Post
Page 102 of the car audio site (link in sig line below).
  Reply With Quote
Old 12th December 2012, 10:28 AM   #5
sreten is offline sreten  United Kingdom
diyAudio Member
 
Join Date: Nov 2003
Location: Brighton UK
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.
__________________
There is nothing so practical as a really good theory - Ludwig Boltzmann
When your only tool is a hammer, every problem looks like a nail - Abraham Maslow
  Reply With Quote
Old 13th December 2012, 01:01 AM   #6
diyAudio Member
 
Join Date: Nov 2004
Location: US
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?

Quote:
Originally Posted by sreten View Post
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.
  Reply With Quote
Old 13th December 2012, 01:39 AM   #7
diyAudio Member
 
nigelwright7557's Avatar
 
Join Date: Apr 2008
Location: Carlisle, England
It will be hard to get a true reading for a speaker as load.
A speakers impedance varies wildly across the frequency spectrum.
  Reply With Quote
Old 13th December 2012, 01:50 AM   #8
diyAudio Member
 
Join Date: Nov 2004
Location: US
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.

Quote:
Originally Posted by nigelwright7557 View Post
It will be hard to get a true reading for a speaker as load.
A speakers impedance varies wildly across the frequency spectrum.
  Reply With Quote
Old 13th December 2012, 02:05 AM   #9
diyAudio Member
 
Join Date: Mar 2007
Location: Canandaigua, NY USA
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.
__________________
May the root sum of the squares of the Forces be with you.
  Reply With Quote
Old 13th December 2012, 02:11 AM   #10
diyAudio Member
 
Join Date: Nov 2004
Location: US
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?

Quote:
Originally Posted by Conrad Hoffman View Post
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.
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
So how do you measure power amplifier PSRR then? GK Solid State 23 30th March 2013 06:27 PM
How to measure output power EddieRich Tubes / Valves 18 7th June 2012 02:10 PM
Max/Actual Output Level vs. Distortion gootee Solid State 3 13th February 2010 07:41 AM
PPI Promos 25 - Actual output? Sub Ace Car Audio 12 19th January 2010 08:21 PM
How to measure amp power output hernanstafe Parts 2 16th February 2007 01:27 AM


New To Site? Need Help?

All times are GMT. The time now is 02:29 PM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright 1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2