Amplifier current draw/wattage


Hello All,
Trying to figure my system's total power requirement.
I'll have 12 devices plugged into the wall, or some kind
of power supply. By reading fuse sizes and/or power
consumption, I come up with about 24.75 amps (120-VAC).
This is only steady-state operation and I don't think
includes startup surges. The power amps take most of this
draw.
I feel fairly comfortable with the accuracy of this figure.
Here's my point of uncertainty: I do not know if the
amplifiers draw more current as I increase the volume.
I have 4 amps (powering 4-way speakers):
Bryston 3B-ST..............5-A fuse
Linn 5105 - 2 of these....."6.3-A MAX"
NAD 2600a..................5-A fuse
So at peak current draw the 4 amplifiers would supposedly
draw about 22.6 amps.
Can anyone tell me what class each of these amplifiers
belongs to, I know only that none are class A.
********
I can certainly supply more info if anyone asks.
Does my reasoning seem sound ?
The whole result is that it appears I need to add
a new 30-A circuit from my breaker panel, probably
with 8-ga wire. This will be a fair amount of work.
noslepums

Showing 1 response by almarg

01-17-12: Charles1dad
I don`t want to measure speakers, but want to know the power output of the amplifier at different volumes. IOW how much power is used at say SPL of 70db vs 85db(how hard is the amp working) measured in watts.
Charles, that can be calculated (to a reasonable approximation). Caveats are that the calculation assumes that speaker efficiency (or, alternatively, voltage sensitivity and impedance) is known accurately, and that the calculation neglects the contribution of room reflections to SPL.

Your Total Eclipse II's are spec'd at 94 db/1 watt/1 meter. For a box type speaker (as opposed to a planar) subtract 6 db for each doubling of distance. Add 3 db to approximately reflect the presence of two speakers.

So at a listening distance of 4 meters (about 13 feet), 1 watt per channel would result in an SPL of 94+3-6-6 = 85 db.

That figure can be adjusted to reflect different power levels based on the db relationship of the ratio of two powers, which is 10logarithm(P1/P2). From that it can be calculated that a 15 db reduction in SPL corresponds to reducing the power level by a factor of 31.6. So a 70 db SPL at 4 meters would correspond to (1 watt/31.6) = 0.032 watts.

The 8 watt maximum power rating of your Franks would correspond to an SPL at 4 meters of 85 + 10log(8/1) = 94 db.

For other listening distances, adjust up or down based on 20 times the logarithm of the distance ratio. For instance, the SPL at 3 meters would be higher than the SPL at 4 meters by 20log(4/3) = 2.5 db.

Best regards,
-- Al