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Speaker sensitivity, impedance, and calculating amp power

This is an elementary question, but I'm not quite understanding how to match amp power to speakers. When I searched around on this forum, I found many discussions which went deep into the weeds. I am hoping for a way of calculating the level of amplifier power I need for speakers with different sensitivities and impedances.

If you have the patience, here's the basic question. So, I've learned that one must consider a number of factors to calculate the amount of amplifier power to drive the speaker:

Sensitivity of the loudspeaker
Loss of db at the listening position
SPL desired at listening position
Amount of headroom desired

Most discussion of the demands a speaker will make on an amp focus on the speaker sensitivity. But the speakers I'm considering vary also in their impedance. How would I use both of those factors to estimate necessary amplifier power to drive them with a comfortable amount of headroom?


128x128hilde45
tomic601's avatar
tomic601
12,733 posts
 
Impedence and phase angle, there is always that nasty and...

I always liked the late great Roger Modjeski advice on the subject ( which you find w search this site ), measure SPL and use a DMM at speaker terminals.

OP there was nothing wrong with your post, as you can see there are some awesome contributors this site
hilde45's avatar
hilde45 OP
4,773 posts
 
@tomic601 There sure are!
gdnrbob
4,406 posts
 
@tomic601 ,
+1
Bob
jchiappinelli's avatar
jchiappinelli
116 posts
 
Erik (or anyone),

You stated:
So, if you have an 89 dB sensitive speaker, and apply 100 watts, the output at 1 meter will be 109 dB. 
What is the formula for calculating what the SPL would be at 3 meters (typical listening position) rather than 1 meter?

Thanks,
J.Chip
almarg's avatar
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almarg
9,670 posts
 
What is the formula for calculating what the SPL would be at 3 meters (typical listening position) rather than 1 meter?

For a single speaker, and assuming it is a conventional box-type (i.e., it is non-planar and non-line source), and neglecting room reflections, the attenuation of SPL resulting from the increase in distance is:

20 x log(distance from listener/1 meter)

where "log" is the base-10 logarithm.

So in this case, and under those assumptions:

20 x log(3 meters/1 meter) = 9.5 db of attenuation relative to the SPL produced at 1 meter (rounding off slightly).

For Erik’s example that you quoted:

109 db - 9.5 = 99.5 db SPL at 3 meters.

The presence of a second speaker and the effects of room reflections will typically add several db to that.

The 20 x log formula is built into the online calculators which have been linked to in some of the posts above.

Regards,
-- Al


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