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Old 29th September 2008, 06:55 PM   #1
Rodango is offline Rodango  United States
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Default 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.

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Old 29th September 2008, 08:45 PM   #2
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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.
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Old 29th September 2008, 09:54 PM   #3
Rodango is offline Rodango  United States
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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!
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Old 29th September 2008, 11:38 PM   #4
Sheldon is offline Sheldon  United States
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Quote:
Originally posted by Rodango
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!
Yes, either will do. Pick the one closest to your expected use conditions. Ty's method is fine. If you have a true RMS multimeter, it's easier to read the output voltage directly from that, instead of the scope screen. You'll still want the scope to look for clipping onset.

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.
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Old 30th September 2008, 01:05 AM   #5
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"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
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File Type: jpg ohmite300w.jpg (60.3 KB, 179 views)
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Old 30th September 2008, 01:32 AM   #6
Rodango is offline Rodango  United States
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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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