# 1w=????

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#### SteveG

When measuring the output power of an amp, or designing an amp to put out a certain amount of power, do you look at RMS power, peak power, or what?
Specifically, I am having a difficult time figuring out how much power supply I will need for an Aleph-60. The circuit seems to be able to swing 64V peak to peak, which equates to 22VRms. Using the formula P=V^2/R, I get a power output into my speaker (minimum impedance of 3.2 ohms) of about 150 watts (RMS). Using the formula I=sqrt(P/R), I get a current of about 6.8A. Now, I am assuming that this is an RMS amperage, as the voltage that I was using to get 6.8A was an RMS voltage. Does this mean that my power supply will have to deliver 19.23 amps peak to peak?!! If so, is that 19.23 Amps from both rails, or half from each rail?
On a similar note, I have always assumed that speaker manufacturers rate their sensitivity based on 1W RMS. Is this also correct?
Isn't there a standard for how power is rated? If anyone knows where I can read about it, please let me know.
Steve

#### AGGEMAM

SteveG said:
On a similar note, I have always assumed that speaker manufacturers rate their sensitivity based on 1W RMS. Is this also correct?

Normally, yes, they do, or more precisely at 2.83 (2 times 2 squared). And usually they measure sensitivity at 1 meter, note that some mostly pro-speaker manufacturers measure it at 3 feet which gives another and higher result, and consider that each time you double the distance the SPL drops by 3 db.

#### seangoesbonk

AGGEMAM said:
Normally, yes, they do, or more precisely at 2.83 (2 times 2 squared).
This might confuse someone who is just still learning...
He means 2 times the square root of 2 volts, but less confusingly:
P = (V^2)/R
If P = 1 watt and R = 8 ohms, then V = sqrt(1*8) = 2.828 volts
So the 2.83v develops 1w in a 8 ohm driver.

#### wuffwaff

Hello Steve,

normally you measure rms watts and with a stable power supply these should be the same as peak power.

With an Aleph you've got to take a different route.

Start with the bias current you want and then multiply this with the power supply voltage and you'll get the watts the transformer will deliver all the time. Multiply this with 2 (or 3) and you'll get the wattage for the transformer. (This factor depends on taste and transformer reserves)

So for an Aleph60 with +-32V and 2,4A bias you'll get 2x32x2,4x2(3) =
307(461) VA

Your calculation for peak to peak current has a small error in it cause there isn´t any peak to peak current!
Peak current will be 1.41x6.8=9,61A peak wich means about half of this as bias current in an Aleph60 (4,8A ) wich would mean about 800mA per fet and 64x4.8=308watts dissipation per channel. This means heat sinks of about 0,08°K/W per channel and a tranformer of between 600 and 900 VA per channel.

This will be a nice project although not really an Aleph60 anymore

william

#### SteveG

Thanks, William. Your calculations agree with what I concluded on my own. It's nice to hear someone confirm what I concluded.
Why wouldn't you consider this to be an Aleph 60 anymore? It's basically the same thing, only with additional bias current.
Steve

#### wuffwaff

Hi Steve,

that´s true but with this bias and power supply it will become something else and will deserve a new name

william

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