is there any way to actually tell how many watts your amp is running

[rookanga 10+ year member]

I'm not talking about a $1,000 thing you need to buy or go somewhere like an audio place. Like something you can buy and hook up.

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[PaulThompson229 10+ year member]

I remember hearing a way to do it with a dmm, not sure about how to go about it or how accurate it is though.

Or buy some watt meters like the ones that come on some McIntosh amps

 

[Donks 10+ year member]

i always wanted to know to....i heard somethin about gettin ur Amp clamped.

 

[Supergumby5000 5,000+ posts]

Over half your posts are startings of threads.

Learn to search.

This topic was brought up less than 48 hours ago.

 

[brynm 10+ year member]

use a dmm, power = voltage squared divided by impedance. Should get you close, if you disconnect your speaker leads so impedance rise isn't a factor.

 

[PaulThompson229 10+ year member]

i always wanted to know to....i heard somethin about gettin ur Amp clamped.

Yes that is the best way, but it requires special equipment and software, usually done only by pro's or some shops.

 

[alvitae 10+ year member]

Clamp meter + DMM, measure amps with clamp, volts with dmm. 60 hz tone. multiply the two readings.

I think... //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[Linkz 10+ year member]

use a dmm, power = voltage squared divided by impedance. Should get you close, if you disconnect your speaker leads so impedance rise isn't a factor.

This is what you should do.

 

[macq32 10+ year member]

wouldnt you take.. amp efficiency*volts you're running*amps(fuses)

ie. .86*14.25*120=watts

 

[oni 10+ year member]

wouldnt you take.. amp efficiency*volts you're running*amps(fuses)
ie. .86*14.25*120=watts

No. The "efficiency" takes place inside the amp. Input = Output so power in is equal to power out minus heat generated (in simple terms). The ratio of output power to input power is the efficiency. So if you're measuring the power leaving the amp (speaker leads) it will be accurate using the mentioned formula.

 

[n2audio 5,000+ posts]

use a dmm, power = voltage squared divided by impedance. Should get you close, if you disconnect your speaker leads so impedance rise isn't a factor.

This is what you should do.

I think the objective is to do a max power test on the amp, not set the gain for X power into Y impedance which is what it sounds like you're describing -- unless I'm just not following.

Clamp meter + DMM, measure amps with clamp, volts with dmm. 60 hz tone. multiply the two readings.
I think... //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

The problem with a clamp meter and a DMM measuring voltage and current through the speaker lead is you still don't know (at least not with any degree of accuracy) what ACTUAL impedance the load is presenting the amp OR at what point the amp is at full power before clipping.

If you have no way to monitor that I think it would be about as accurate to just add up the fuse value and multiply by ~10 for a/b, ~12 for D.

That's just me though.