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29th September 2008, 06:55 PM  #1 
diyAudio Member

Measuring Output Power
I have my DIY bass amp head all tuned up and sounding good and very loud. It was inspired by the 300W Fender 300PS but is there any way short if a boatload of expensive instruments to determine the actual power output capability? Can it be calculated?
This is kind of like when I had my '68 GTO dynoed. I was sure that street/strip 455 was going to put me into the 400HP bracket. But instead, I had 265 HP on the floor. Thanks 
29th September 2008, 08:45 PM  #2 
diyAudio Member
Join Date: Mar 2007
Location: Newark, DE

Sure. P=I*V and V=I*R. Therefore, I=V/R and P=(V^2)/R.
Get a sine generator. Hook it up to the input of your amp. Connect a known load to the output, probably an 8 ohm dummy load made out of sufficient quantity of cemented resistors to handle the expected output power. Connect your oscilloscope across the dummy load, power everything up, and watch the waveform. Slowly increase the gain until you observe signs of clipping (peaks start to flatten) or any other unwarranted distortion. At this point, read the voltage from the zero line to the peak of the sine wave. This is your peak voltage. Divide by the square root of two (1.414) to account for the area under the curve of the sine function. Take the result, square it, and divide by the load resistance. This is your output power. If you are hoping to get 300 watts out of the thing, you'll need to see just under 70 volts zero to peak driven into an 8 ohm load. 
29th September 2008, 09:54 PM  #3 
diyAudio Member

I have a huge 6.8 ohm and a huge 10 ohm resistor. Would either of those be acceptable? The amp has an 8 ohm output.
Thanks! 
29th September 2008, 11:38 PM  #4  
diyAudio Member
Join Date: Dec 2003
Location: San Diego

Quote:
Sheldon edit: BTW, as long as you are at it, make note of the signal generator level that you use for your output measurement. I like to record this, so that I know amplifier gain. Comes in handy when you want to substitute amps. 

30th September 2008, 01:05 AM  #5 
diyAudio Member
Join Date: Jan 2008
Location: Eureka, CA

"I have a huge 6.8 ohm and a huge 10 ohm resistor. Would either of those be acceptable?"
Depends on what you mean by huge. Here is a picture of a 300 watt resistor. This is a compact one. I use this type for dummy loads because they're rugged and cheap. They also have an optional sliding contact you can use to get smaller resistance values. I use a 10 ohm with the contact that allows me to set to 8.0 ohms. This reduces the power rating of the resistor correspondingly. Also, as these heat up the resistance increases; I don't know how much because I run them cool. That amp is like an Ampeg SVT, right? Might be worth trying both 6.8 and 10R loads to see what you get. Real speakers will vary with frequency anyway, and what you have is probably about the right range for an 8 ohm instrument speaker. Cheers, Michael 
30th September 2008, 01:32 AM  #6 
diyAudio Member

Yes, they're 300W and I just ordered an 8 ohm 300W resistor too. So I should be all set.
And yes, its similar to the power of an SVT. We'll see how powerful. It REALLY puts that quad of KT88s through its paces... 
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