Amplifier power versus maximum speaker input power


What are the ramifications of using an amplifier that exceeds the maximum input power handling rating of a speaker? I am thinking of getting a pair of Reference 3A Dulcets -- rated at a max of 100Watts RMS into 6ohms -- yet our Threshold S/500 II puts out 250Watts at 8ohms. I normally do not listen at more than 90db. Thank you for your input.
somut

Showing 2 responses by almarg

I agree with all of the previous comments. I would add, though, that the high power capability of the amp relative to the rating of the speakers increases the importance of observing the proper turn-on and turn-off sequence. The amp should be turned on last, after all of the other components that are being used are turned on, and the amp should be turned off first, before any of the other components are turned off.

Otherwise if the design of an upstream component is such that it is not well behaved at turn-on or turn-off any transient it might generate at those times could potentially be amplified to the full power capability of the amp. If the transient is very brief, meaning that its spectral components are at high frequencies, 250 watts or more could briefly be sent into the tweeters.

I once had an S300, and I recall that it could continue to play music for about 40 seconds or so after being turned off. So after powering it down, it would also be prudent to wait for that amount of time, or longer, before turning off the other components.

Regarding the question Mark and Roscoe were discussing, while it is true that the volume control settings that will be used are a function of gain, amplifier gain and power capability will tend to have a loose correlation, so Mark's observation was not entirely invalid. Amplifiers tend to be designed to be driven to full power by input voltages that are in the same general area (1 or 2 volts or so, for unbalanced inputs), regardless of how much power they can generate. Therefore a high powered amp will more often than not require a significantly lower volume control setting than a low powered amp, everything else being equal.

Regards,
-- Al
Am I understanding input sensitivity correctly, as follows?: The higher the input sensitivity (say 2V to reach full power, as opposed to 1V to reach full power), the more travel I will have in my preamp volume control. So someone without sufficient fine tuning in volume control or who wants to move out of the 8-10 range and more into the 10-1 o'clock range should try to find an amp with higher input sensitivity.

Do I have this right?
Yes, provided that "full power" is the same in both cases.

But just to confuse the issue further, your statement would be more precise if it referred to a higher input sensitivity NUMBER, rather than to a higher input sensitivity. A higher input sensitivity number, properly speaking, corresponds to lower sensitivity. In other words, a more sensitive amp is one that requires less input voltage to reach full power.
I am not clear on if input sensitivity can give an idea of the amount of gain in dB (and what other specs might be needed to calculate dB gain from input sensitivity).
If gain is not explicitly specified, and it often is not, it can be calculated to a reasonable approximation from the specified sensitivity and the specified maximum power capability.

As you realize, gain is the ratio of output voltage to input voltage, expressed in db.

The ratio of two voltages, V1 and V2, are converted to db based on the formula (20 x logarithm(V1/V2)).

The rated maximum output power into 8 ohms can be converted into voltage based on the equation P = (Vsquared)/R, where P is power in watts, V is volts, and R is resistance in ohms.

So for example in the case of an amplifier rated at 200 watts into 8 ohms and having 2 volt sensitivity, if we represent the output voltage corresponding to the 200 watts into 8 ohms as "Vout" and the 2 volt sensitivity as "Vin," we have:

200 = ((Vout)squared)/8

From which it can be calculated that Vout = 40 volts

Therefore the gain in db is

20 x log(Vout/Vin) = 20 x log(40/2) = 26 db

That is an approximation, as I indicated, in part because it does not reflect margin that may be built into the maximum power specification. But it will generally be a reasonable approximation for most purposes.

Regards,
-- Al