testing amp wattage
Printed From: the12volt.com
Forum Name: Car Audio
Forum Discription: Car Stereos, Amplifiers, Crossovers, Processors, Speakers, Subwoofers, etc.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=93572
Printed Date: April 27, 2024 at 3:14 PM
Topic: testing amp wattage
Posted By: idigmusic
Subject: testing amp wattage
Date Posted: May 03, 2007 at 1:41 AM
So I couldn't find this in the forums and I'm surprised, sorry if I sound dumb...
I am curious how much power my amps are putting out so should I just measure the current going into the speaker wire (for a mono) using a DMM and multiply that by my battery/alternator voltage or is there a better way? Thanks
Replies:
Posted By: Big Dog
Date Posted: May 04, 2007 at 2:07 PM
Here's what I usually do: ((Amp fuse/2) ^2 x impedance) / 2 So it goes something like this. Let's say you have a 40 amp fuse in the amp and you want to use a 4 ohm sub. ((40/2) ^2 x 4) / 2 40/2 = 20 20 ^2 = 20 x 20 = 400 400 x 4 = 1600 1600 / 2 = 800 watts Works for me! ------------- Prepare your future. It wasn't the lack of stones that killed the stone age.
Posted By: Alpine Guy
Date Posted: May 04, 2007 at 2:31 PM
This is a crap shoot, but easier. Amp fuse times voltage, so 40amp x14 volts. 40A X 14V=560 Watts at 100% efficiency, then multiply that by an estimated efficiency of your amp, say 85% efficient, so take 480 Watts x 85%, or .85=476 Watts And RMS is 70.7% of the peak wattage so 476 Watts x .707= ~336 Watts RMS.
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2003 Chevy Avalanche,Eclipse CD7000,Morel Elate 5,Adire Extremis,Alpine PDX-4.150, 15" TC-3000, 2 Alpine PDX-1.1000, 470Amp HO Alt.
Posted By: idigmusic
Date Posted: May 04, 2007 at 3:43 PM
I know how to find the theoretical wattage of my amp, but I'm looking to see exactly what it's putting out. Just like you could say your battery puts out 12V because it says 12V on it, but maybe it actually puts out 11.6V due to imperfections in the manufacturing process. Sorry if I wasn't clear earlier.
Posted By: haemphyst
Date Posted: May 04, 2007 at 5:15 PM
Without an oscilloscope to actually SEE the onset of clipping in the waveform, you can't. You must know the EXACT output rail voltages to determine actual power, as well as the EXACT impedance the amplifier is seeing at the clipped frequency. As you know, loudspeakers are dynamic impedance devices, they will have minimum and maximum impedance peaks throughout their usable bandwidth. This will affect power output at each frequency. I suppose you COULD use a shunt resistor, (this would be able to give you the current being delivered to the speaker) and a second (storage) 'scope, and plug the numbers in that way, but FAR more expensive. If you know two of the three specifications, (voltage, resistance, and/or current) you can always figure POWER.
As a VERY CLOSE approximation, you could measure the DC rails WITHIN the amplifier, but this will involve opening the amplifier, and voiding any warranty you might still have. Using Ohm's Law you can plug this number in, and determine how much power the amplifier CAN make into a RESISTOR, not a loudspeaker.
If you do it this way, your numbers will be a LITTLE bit high, because the transistors will have a small drop across them, affecting the REAL output, but it'll be close enough.
Why the need to know so exactly?
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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: idigmusic
Date Posted: May 04, 2007 at 5:34 PM
I'm just curious because it's an Alpine PDX 1.600 I got off ebay and I'm wondering if there's a reason ebay is so much cheaper for the "same" stuff (i.e. broken/defective, counterfeit, maybe didn't quite meet alpine's standards to sell in stores).
The reason I don't want to go by what fuses I have and what Alpine says the efficiency should be is that if someone slapped a 100A fuse in a $10 counterfeit amp and said it puts out 100A*12V*(50% efficiency)=600W rms, while it probably doesn't do that in real life. I was just thinking if I could measure the current coming out of the amp into the speaker I could figure out the power using the voltage the amp should be running at (battery/alternator voltage). It sounds like that won't work from your post haemphyst; why is that?
Posted By: haemphyst
Date Posted: May 04, 2007 at 7:27 PM
Because the numbers regarding input vs. output calculations are suggestions. They are in NO way concrete, as EVERY manufacturer's efficiency will be a little bit different, based on topology, output devices chosen, power supply driver transistors, quality of the traces or bussbars internally, how MUCH Class A (in percentage) will the amplifier run... ALL of those things will have an effect on it's output - the "theoretical" output.
If mfr A says this is a 500 watt amplifier, but it is only 50% efficient, it will have +/- X rail voltage, right? This rail voltage will be limited by the input fuse. If mfr B also makes a 500 watt amplifier, but it it 60% efficient, the output rails will be the same, (they HAVE to be, right) but the input fuse can be smaller because more power is tranferred to the output buss. It's just that mfr A chose a topology slightly less efficient than mfr B - maybe a different switching frequency, maybe a smaller toroid, larger filter caps... MANY things can affect efficiency, and since nobody advertises efficiency numbers, you'll have to use a number "that's close enough for government work", right?
Since REAL output is based ON THE RAIL VOLTAGE AVAILABLE, you have to know that voltage, right? Meters won't get it, because you can be into a 10 or 15% clipped situation, before the meter even registers a difference in the voltage.
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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: sedate
Date Posted: May 05, 2007 at 1:47 AM
idigmusic wrote:
It sounds like that won't work from your post haemphyst; why is that?
Perhaps a simpler way to explain this is like this: The *power* the amplifier produces is not the same as power that can be used the drive speakers. Power ratings are given in dense code, and without an input voltage, a power output, a THD at that output (and really other specs are helpful) the specification is rather useless. As haemphyst so polysyllabically illuminated for us, the actual useable power the amplifier is producing is highly dependent on all of these things that go into the design of the amplifier so there really isn't a way of calculating some reference efficiency with any real certainty from those numbers either. Anyway, you could get a net power used simply by using a DMM in the power input side of the amp... but... this really doesn't tell you anything about the speaker power at the channels... Which is what you would be concerned about if you were ordering the possibly counterfiet amp from e-bay. ------------- "I'm finished!" - Daniel Plainview
Posted By: haemphyst
Date Posted: May 05, 2007 at 10:11 AM
sedate wrote:
As haemphyst so polysyllabically illuminated for us, the actual useable power the amplifier is producing is highly dependent on all of these things that go into the design of the amplifier so there really isn't a way of calculating some reference efficiency with any real certainty from those numbers either.
Polysyllabically? {ROFL} I never knew I was a polysyllabic! ------------- It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: gk_thin
Date Posted: May 11, 2007 at 9:50 PM
Posted By: haemphyst
Date Posted: May 11, 2007 at 10:47 PM
Again, RF is in the LEAD with misleading information... to whit:
Rockford Fosgate wrote:
N O T I C E
The method described in this document is not designed to determine the actual "rated," "peak," or "dynamic" output power of an amplifier. It is specifically written for the "do it yourself" car audio fanatic as a troubleshooting method designed to verify power output of a properly working amplifier. This document recommends using test equipment and measurement procedures that are fairly easy to work with, versus professionals who use oscilloscopes, specific voltage measurements, distortion limits and signal-to-noise parameters to accurately test and measure the actual output power of amplifiers per CEA-2006 guidelines.
Right there on the TOP of the page! Had you read both of my previous posts, you'll see WHY this method WILL NOT WORK. Also, NO meter will be accurate at 1K, so, yet again, RF is spewing crap. (I hate that company - sometimes they are as bad as Audiobling) ------------- It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: bassmechanik
Date Posted: June 25, 2007 at 12:09 PM
haemphyst wrote:
Because the numbers regarding input vs. output calculations are suggestions. They are in NO way concrete, as EVERY manufacturer's efficiency will be a little bit different, based on topology, output devices chosen, power supply driver transistors, quality of the traces or bussbars internally, how MUCH Class A (in percentage) will the amplifier run... ALL of those things will have an effect on it's output - the "theoretical" output.
If mfr A says this is a 500 watt amplifier, but it is only 50% efficient, it will have +/- X rail voltage, right? This rail voltage will be limited by the input fuse. If mfr B also makes a 500 watt amplifier, but it it 60% efficient, the output rails will be the same, (they HAVE to be, right) but the input fuse can be smaller because more power is tranferred to the output buss. It's just that mfr A chose a topology slightly less efficient than mfr B - maybe a different switching frequency, maybe a smaller toroid, larger filter caps... MANY things can affect efficiency, and since nobody advertises efficiency numbers, you'll have to use a number "that's close enough for government work", right?
Since REAL output is based ON THE RAIL VOLTAGE AVAILABLE, you have to know that voltage, right? Meters won't get it, because you can be into a 10 or 15% clipped situation, before the meter even registers a difference in the voltage.
Does anyone know how to get the actual efficiency # on a class D amp or is it just around a standard percentage say 85%? I called my manufacturer and they didn't know. Is there some other calculation to get this figure?
Posted By: haemphyst
Date Posted: June 25, 2007 at 12:36 PM
Step 1: You must know the RMS output power, first. Figure it however you need to.
Step 2: Once you know that, than you can easily figure the efficiency, as calculating the input power is easily done. DC current at DC voltage, equals input power.
Step 3: Output power divided by input power equals efficiency.
That formula will give you overall amplifier efficiency, including DC-DC conversion, and output stages all figured in. Obviously, you need to figure your input power AT THE SAME OUTPUT POWER that you discovered in step 1.
-------------
It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: bassmechanik
Date Posted: June 25, 2007 at 12:51 PM
Do you know which amp brands have the highest efficiency? I calculated 62.5 on my Kendwood 9152. So CEA on this puppy says 900rms and in all actuality it's 540rms. Is that legal for CEA compliancy?
Posted By: haemphyst
Date Posted: June 25, 2007 at 3:56 PM
No, you're using the wrong numbers. If the amp is 900 watts RMS, and it has a claimed 62.5%, then the INPUT side would be 1440 watts. 1440 watts, divided by 14.4 volts SHOULD be meaning 100A worth of fusing on the power cable.
If you are not seeing 100A of fusing or MORE, then either the efficiency has been stretched, or the real output power has been, or a possible combination of both. That's all there is to it.
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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: bassmechanik
Date Posted: June 25, 2007 at 4:00 PM
Ok bare with me here as I'm new to all the tech stuff. Kenwood 9152 900rms / 60a fuse / 14.4 is the voltage rating, what do you come up with and what calc do you use?
Posted By: hustlin247
Date Posted: June 25, 2007 at 4:02 PM
gk_thin wrote:
https://www.rockfordfosgate.com/scripts/rightnow.cfg/php.exe/enduser/popup_adp.php?p_sid=VrNE2Fwi&p_lva=311&p_admin=&p_li=&p_faqid=468&p_created=1119555993 Here is a link for the do-it-yourselfer to estimate the output wattage of an amplifier.
That tests the output voltage. :/ ------------- '94 Ford Explorer / Kenwood KVT-815DVD / RF Power T1682C 6x8 (all doors) / RF Power T10001 / 12" Kicker L5 (x4) / Optima Yellow Top Battery
Posted By: bassmechanik
Date Posted: June 25, 2007 at 4:02 PM
bassmechanik wrote:
Ok bare with me here as I'm new to all the tech stuff. Kenwood 9152 900rms / 60a fuse on the amp/ 14.4 is the voltage rating, what do you come up with and what calc do you use?
Posted By: bassmechanik
Date Posted: June 25, 2007 at 4:04 PM
Yeah I need the voltage meter to work with that option.
Posted By: wormy
Date Posted: June 25, 2007 at 4:30 PM
voltmeters do tend to help...lol. Get one. Best, cheap way to do most of the short cuts...lol. ------------- ...typically, I just run whatever I randomly pick up off the floor.
1995 Ford Ranger Supercab
MECA member
Team CSS
Posted By: haemphyst
Date Posted: June 25, 2007 at 4:38 PM
bassmechanik wrote:
Ok bare with me here as I'm new to all the tech stuff. Kenwood 9152 900rms / 60a fuse / 14.4 is the voltage rating, what do you come up with and what calc do you use?
14.4v*60a=864 watts in.
900 watts out/864 watts in = 104.2 percent efficient.
12.0v*60a=720 watts in.
900 watts out/720 watts in = 125 percent efficient.
Both of those results show more power OUT than power IN. Impossible, and welcome to the world of Kenwood. Like I said, either the power rating or the efficiency "truth" was stretched... There's my math, you decide. ------------- It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."
Posted By: aznboi3644
Date Posted: June 25, 2007 at 7:41 PM
haemphyst wrote:
No, you're using the wrong numbers. If the amp is 900 watts RMS, and it has a claimed 62.5%, then the INPUT side would be 1440 watts. 1440 watts, divided by 14.4 volts SHOULD be meaning 100A worth of fusing on the power cable.
If you are not seeing 100A of fusing or MORE, then either the efficiency has been stretched, or the real output power has been, or a possible combination of both. That's all there is to it.
haem...does this factor affect the fuse ratings
https://www.bcae1.com/fuses.htm
"Fuse Opening Time:
A fuse does not blow when the current reaches its rated current. It is designed to pass its rated current without opening. A fuse will take varying times to blow under different conditions. A fuse will pass significantly more than its rated current for a very short time. It may take 10 minutes or more to blow a fuse at 25% over its rated current. The table below is an example of the specifications for a slow blow fuse. You can see that a 20 amp fuse may pass 40 amps of current for as long as 5 minutes before blowing although it probably wouldn't take a full 5 minutes to blow. The times for other fuses will be slightly different."
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