Actually, I -am- confused about Phil's post.

Phil Leigh;292086 Wrote: 
> 
> As an example, if your power amp can swing 40V from its PSU rails, has
> an impedance of 2 ohms and is running into an 8 ohm speaker, the max
> voltage the amp could deliver to the speaker would be:
> 40*(8/8+2) = 32 volts
> The power the amp could deliver into the speaker would be (32*32)/8=128
> watts
> and
> the power lost inside the amp by heating up the heatsinks would be
> (32*32)/2=
> 512 watts of heat
> 
> That's roughly 25% efficiency. This is for a Class A amp with full
> power delivery at all times (not class B!!!)
> 

In your examples you treated the amp as a voltage source and the output
impedance as if it were in -parallel- with the load.  But (regardless of
the class) I thought output impedance is defined as the part of the
internal resistance that's in -series- with the load (again, treating
the amp is a voltage source, as you did).

If that's right, an amp with an output impedance of 2 ohms driving an 8
ohm speaker will only lose (40-32)*(40-32)/2 = 32 watts of power, not
512, and the efficiency goes to 1 as the output impedance goes to
zero.

The reason class A amps have low efficiency is because they have an
output transistor which dumps current to ground even when the signal is
zero.

Right?


-- 
opaqueice
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