Class G amplifier efficiency

Since voltage to the amplifier dynamically changes, you just multiply voltage and current with high time resolution and average, thus you get input power

Cheers

Klaus

The definition is the same, the maths more complex. Do you just want eta for max amplitude sine output or more detail?

NT

as an approximation for the efficiency at max sine wave output if you have a single transition boost amplitude cut-in at half the total supply voltage, say:

Above half the rail voltage the supply tracks the envelope almost exactly so the rough approximation is that in that regime efficiency is

Below that the signal looks somewhat like a square wave. So my guesstimate for full output sine-wave input power at the full rail voltage is the input power of e.g. a class AB amp driving a square wave into the load at half the rail voltage times some fudge factor like 0.8.

a weighted average of an AB amp's efficiency/output power into the same load at the lower voltage, at max RMS sine wave output (fundamental) and the AB input power at the amplitude of the first two harmonics can maybe in turn approximate that.

Not average current, rather RMS current (same as the output). It matters.

Still power_out/power_in.

RMS is not valid here. The 'S' in RMS is 'square', which is valid only if the voltage and current are in fixed ratio to each other. Here the voltage stays constant, and the input power is the constant voltage times average feed current.

Bullshit.

No, it's *always* true. Average voltage, or current, means nothing.

Wrong, it's *always* RMS voltage times RMS current. You don't think a constant voltage has an RMS value? There is no difference when either is constant. Power is still V(rms) * I(rms)

No. Power is the integral of V(t)*I(t)*dt averaged over the period of interest.

An AC current times a DC voltage gives zero net power dissipation.

Cheers

Phil Hobbs

Correction: Power is the square root of the integral of V(t)*I(t)*dt averaged over the period of interest.

You were right the first time.

Cheers

Phil Hobbs

I think you mean that V(rms) * I(rms) = VA if you are measuring using something with a long-ish time constant such as a Fluke meter.

Oh, well, if it's DC we're talking about then, yes, this works for average power.

No, it doesn't. With a constant DC voltage, any AC current component contributes zero average power dissipation, but does increase the RMS current.

Cheers

Phil Hobbs

Yes correct. FLAT DC with no ripple works though.

DC without AC component

?? What means "net" ??

A DC voltage has that voltage as its RMS voltage, obviously.

Sure, there is that PF thing in there but it's *not* I(avg) * V(avg).

That is not the same as V(averaged over time) * I(averaged over time).

If you have a 10V supply, and you draw a current

I(t) = 10A * sin(2 pi * 60 Hz * t), you have

Vrms = 10V

Irms = 7.07 A,

so Vrms*Irms = 70.7 W.

Instantaneous power is

P(t) = 100W * sin (2 pi * 60 Hz * t),

which over an integer number of cycles (or as the averaging time becomes long) gives

Pavg = 0.0 W.

Cheers

Phil Hobbs

When V(t) is constant, the power *is* V * I(averaged) due to the linearity of multiplication and summing.