Years ago, I had an amp with meters. I seem to recall that when the needles hit 0db, the channels were operating at the rated max power.

Although I've never heard any distortion, I really have no idea how much of the 7500's headroom I'm using.

Is there any significance to the 0db volume setting on the 990 in relation to what is being output from the 7500?

Does anyone know of a way to "meter" the power output of a 7500 while listening to music?

If you have calibrated your 990, the 0dB setting corresponds to a peak sound pressure level (SPL) of 105 dBa at your listening position from any channel with the program material at maximum volume.
The effects channel is offset by 10 dB upward, giving a max SPL of 115 dB.

The power used to attain this sound level can vary wildly depending on room size and speaker efficiency. On the upper end, Klipshorns give about 104 dB with 1 watt input, so you'd be at just over 1 watt power. On the other end of the scale, some of the old AR (Acoustic Research) speakers were at about 82 dB at one watt. This means 23 db increase in power for 105 dB output; that's about 200x, so there's your 200 watts.

Since music is so dynamic, it's very difficult to get a good reading of the output power level. Just to make it a little more interesting, speakers are reactive, so you can't just measure voltage and current and multiply to get real power.

Someone probably sells some kind of display that will show voltage levels either tapping the preamp outputs or power amp outputs, but I've not searched for one. The amplifier power can be calculated approximately using the voltage and speaker impedance with the old V^2/R formula,
If you are using the preamp output, you can use the gain of the amp to calculate the voltage applied to the speakers and work the math from there.

Steve

As mahansm points out, the 990's volume information is unrelated to the power output being generated by the amp.
I don't know of any practical or useful such meters. It's not a direct measurement of the 7500, but some power conditioners include a readout of the amps being drawn at any given time. As an example, when playing a movie in my system at relatively high volume, I've seen the draw exceed six amps. Six amps for the entire system, which would suggest that at least with my speakers and in my room the 7500 still has a lot of headroom to spare.

Thanks!

mahansm, is there a safe way for a non-tech type to measure the gain of the amp? Would that be the V in your formula?

With 6 ohm speakers, would the R in your formula be 6?

gonk, If I'm following you, a 7700 can deliver 1400 watts at 15 amps, so approximately 93.3 total watts per amp measured? I'm assuming a 7500 is the same as a 7700 minus 2 channels.

If I was only using 2 channels, then I'd be near 190 watts per channel near 4 amps?

The 7700 and 7500, like many multichannel amps, are somewhat limited by the capacity of the electrical circuit they are fed from. A 15A circuit on a 120V system can in theory provide 1800W of power to a connected device, but in reality a 15A breaker is designed to trip before it reaches that level - probably before 1500W (12.5A). That's why the 7900 (300Wx7) has two separate power cords: you are supposed to connect each cord to a separate dedicated circuit to provide enough current to be able to achieve the amp's full potential.

You also have to take into consideration efficiency, as no machine (mechanical or electrical) can be 100% efficient. (Hmmm... "perpetual motion amp"?) That means that even with 1500W of input to the amp, you are only going to get something less out. The rest will turn into heat. How much less is difficult to say. The reason that this works out OK in practice is the nature of the source material that we are listening to. The only way to drive all seven channels to significant levels at the same time is to feed each channel a very loud test tone at the same time. Normal situations are quite different than that, with different channels having different needs. You could have the center channel (as an example) driven to "reference level" - and if you had very inefficient speakers in a very large room, the speaker might demand 190W from that channel. At the same time, though, other channels would not be seeing the same sort of demand, and you might only be delivering eight or ten watts to the surrounds.

The gain is in the spec sheet. For the 7500/7700, it's given as XLR 28dB, RCA 34dB.
Now to the strangeness: for power, 10 dB is 10x the power. For voltage, 20 dB is 10x the voltage.

Again from the spec sheet, input level for full power output is 1.34 Volts.


V is voltage. I is current. R is resistance.
P is power.

Power output of the amp is not linear with respect to input power. First, there's some idle power (bias currents, transformer losses, etc.) that is consumed even with no input signal.

There are other factors that depend on the particular class of operation (A, B, AB, C, D, G, H, etc) of the amplifier and are beyond the current scope of this discussion.

In actual practice, 91 dB is pretty loud (typical speaker efficiency at 1 watt) and most listening is below this level. The extra power is to accurately reproduce dynamic peaks in the sound without audible distortion. Back in the days of vinyl there were a few limited edition pressings that had as much as 70 dB dynamic range. Some of these had peaks 35 dB or more above the average music level. Assuming 80 dB average, you'd still need around 300 watts output to handle that peak level.

Also, on a speaker rated to handle 100 watts, the tweeter may be capable of handling 20. Overdriving an amplifier (clipping) produces a lot of distortion, most of it at high frequencies.
A 20 watt amplifier driven into hard clipping may produce 40 watts of power momentarily, most of which will go to the tweeter. This can let the magic smoke out.

(all electronics run on magic smoke. PROOF:
when you let the magic smoke out, the device quits working.)

Steve

And as my electronics professor was fond of saying, "Once you let the smoke out, there's no way to put it back in..."

Thanks for the guidance!

I wonder if anyone makes a device you could connect to the amplifier's speaker terminals that measures the RMS power while presenting a dummy load at various resistance, say 4, 6, or 8 ohms.

I saw an ad for something that looked similar for radio applications, but it seemed limited to 50 watts.

Speaker impedance alone does not define a speaker, though - just because one 8-ohm speaker (or 8-ohm measuring device) yielded a certain RMS power with a specified test tone at a specified volume level, doesn't mean that any other 8-ohm speaker will be similar. Efficiency will be a larger determining factor, as will the test tone signal being used.

I have a Kill-A-Watt meter (you can find them off Amazon), and it measures the AC power draw off the wall socket. Not the same thing, but you know the amp is providing no more than the AC draw. In general class A/B amps are around 50% efficient, so if you draw 100W you are probably getting around 50W amp power.

These devices are cheap and useful to play with. I'm using it with my computer stuff. My computer in "shut down" position is 10W while in "standby" is 12W, so I know that the computer still draws power when "shut down." However, in standby, using only 2W more, I can restart in about 2 seconds. Useful thing to have learned.

Sorry Gonk but I just can't agree with that statement. If anything the breaker will trip at a slightly higher level. Since breakers are designed and rated at a temp of 40C (104F) and usually are not in an environment of such high temps, the tripping amperage would be slightly higher. According to a typical manufacturers Trip Chart a breaker will carry the full current rating indefinitely and will carry additional current for proportionally less time. The chart shows how a typical 15 amp breaker will carry 2 times the rated current (30 amps or 3600 watts) for 10 seconds and even carry 5 times the rated current (75 amps or 9000 watts) for 1 second.

While not directly related, my compressor is rated for 15 amps at 120v and 7.5 at 220v (switchable), I can run the compressor continously on a 15 amp circuit without tripping the breaker. So maybe breakers are not created equally? Another unrelated example is one of my welders is rated at 20amps. I have plugged it in a 15 amp circuit and tacked 1/4" plate, so that tells me that a 15 amp breaker will exceed the current load for very short times.

NoMoney I think your examples are directly related to the operation of all residential thermomagnetic circuit breakers. A circuit breaker is not a current limiting device, it's just a bi-metallic strip conducting electricity..... to much current for to long and the strip heats and bends,tripping the breaker.

"Rated" is the key - does it actually consume 15 amps continuously, or is it rated to be used on a 15A circuit? Nameplate ratings don't necessarily equate to actual consumption.

To accurately measure the power output of an amplifier, all you need is a true RMS voltmeter with sufficient bandwidth and some 8 ohm power resistors of appropriate wattage rating. For a sine wave input/output, it's much easier as any RMS meter will do. Voltage squared divided by the resistance is power.

To be more accurate you could use an oscilloscope to view the waveform. This would allow you to adjust the drive until the amplifier was just below clipping. The oscilloscope will give you the peak value of the output waveform. For a sine wave, the RMS value is the peak value divided by the square root of two (about .7071). Apply the formula above and you have an answer.

McIntosh used to run clinics at various dealers where you could bring in a preamp or power amp and they would measure it and give you harmonic and IM distortion figures as well as the power output of the amplifier (at no charge). Most good repair shops have instrumentation that can measure amplifier output and harmonic distortion. For a fee, they'd probably be happy to measure and give you the numbers.

Steve

Correction to the above, before anyone else calls me on it. Square root two is 1.414; root two divided by two is the .7071 by which you multiply peak voltage to get RMS.

Sorry about that; fingers faster than brain today.

Steve