Determining power output

[atmars]

I have read that a way to determine the approximate output of an amp in watts is to put a volt meter across the speaker leads, play for a while and watch the meter and take the result, say 12 volts-ish, square that number, divide by the speaker ohms, and you got the watts output. Accurate?

[Enzo]

Yep, that gives a ball park figure. Of course that depends on what you are playing and your meter. My old Fluke is great on AC up to about 1000Hz, after that it starts to roll off. SO with that if I am shredding up on the 15th fret on the high E, then maybe not, but in general, yes.

[cminor9]

Yep, that gives a ball park figure. Of course that depends on what you are playing and your meter. My old Fluke is great on AC up to about 1000Hz, after that it starts to roll off. SO with that if I am shredding up on the 15th fret on the high E, then maybe not, but in general, yes.

I have to admit, I read this post and expected someone to debunk it. Obviously, I don't know everything.

So p=v^2/r. Also, p=v*i. Therefore, v*i = v^2/r. This puts i inversely proportional to r, IOW ohms law. Behold! The power of maths!

Yet another seemingly obvious thing I have learned here. But this is interesting. By this logic, halving the impedance doubles the wattage. Suppose I have a 15w amp expecting an 8 ohm load. Suppose I want to use a 4 ohm speaker. Should I then double the power rating on what I would otherwise buy?

[Enzo]

Halving the impedance would double the power - IF THE OUTPUT VOLTAGE REMAINED CONSTANT.

SOlid state amps work fairly consistently this way. Set up a 4 ohm speaker to draw 100 watts, then replace it with a 8 ohm speaker, and now the draw is 50 watts. AMps with 8 and 4 ohm ratings might not be in a ratio of 2. AMps may be derated at the lower impedance, since the output may not be rated for enough current at the tougher load. And the circuit has losses. And the amp has to cool itself, etc. SO you might see an amp with a 100 watt rating at 8 ohms and a 150 watt rating at 4 ohms. I made the example up.

Now if I have a 100 watt amp rated for 4 ohms, I can;t just slap 2 ohms on it and expect 200 watts. the load would try to draw that, sure, but the amp would not in all likelihood be able to supply it.

Tube amps are different in this regard. The load is reflected to the power tubes through the transformer. When you alter the load impedance, what you are doing is sliding the operating point down the tube curves. SO they don;t respond the same way.

[cminor9]

now see, that makes sense. Thanks, Enzo!