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Class G amplifier efficiency

Started by amal banerjee December 15, 2019
Could some electronics guru here clarify this a bit ?
The efffiiciency definition for an amplifier is
Power delivered to load/DC power supplied by power source
The Dc power supplied from power supply is 
Vdc x avg. DC current flowing into amplifier
However, Vdc changes, as per design requirements of
class G amplifier. So what then would be the efficiency
defintion for a class G amplifier.
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
On Sunday, 15 December 2019 10:07:25 UTC, amal banerjee  wrote:
> Could some electronics guru here clarify this a bit ? > The efffiiciency definition for an amplifier is > Power delivered to load/DC power supplied by power source > The Dc power supplied from power supply is > Vdc x avg. DC current flowing into amplifier > However, Vdc changes, as per design requirements of > class G amplifier. So what then would be the efficiency > defintion for a class G amplifier.
The definition is the same, the maths more complex. Do you just want eta for max amplitude sine output or more detail? NT
On 12/15/19 5:54 AM, tabbypurr@gmail.com wrote:
> On Sunday, 15 December 2019 10:07:25 UTC, amal banerjee wrote: >> Could some electronics guru here clarify this a bit ? >> The efffiiciency definition for an amplifier is >> Power delivered to load/DC power supplied by power source >> The Dc power supplied from power supply is >> Vdc x avg. DC current flowing into amplifier >> However, Vdc changes, as per design requirements of >> class G amplifier. So what then would be the efficiency >> defintion for a class G amplifier. > > 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: <https://maurmun.files.wordpress.com/2013/01/image_thumb10.png?w=743&h=472> Above half the rail voltage the supply tracks the envelope almost exactly so the rough approximation is that in that regime efficiency is 100%. 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.
On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee
<dakupoto@gmail.com> wrote:

>Could some electronics guru here clarify this a bit ? >The efffiiciency definition for an amplifier is >Power delivered to load/DC power supplied by power source >The Dc power supplied from power supply is >Vdc x avg. DC current flowing into amplifier
Not average current, rather RMS current (same as the output). It matters.
>However, Vdc changes, as per design requirements of >class G amplifier. So what then would be the efficiency >defintion for a class G amplifier.
Still power_out/power_in.
On 16.12.19 07:13, krw@notreal.com wrote:
> On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee > <dakupoto@gmail.com> wrote: > >> Could some electronics guru here clarify this a bit ? >> The efffiiciency definition for an amplifier is >> Power delivered to load/DC power supplied by power source >> The Dc power supplied from power supply is >> Vdc x avg. DC current flowing into amplifier > > Not average current, rather RMS current (same as the output). It > matters. > >> However, Vdc changes, as per design requirements of >> class G amplifier. So what then would be the efficiency >> defintion for a class G amplifier. > > 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. -- -TV
On Tue, 17 Dec 2019 10:53:15 +0200, Tauno Voipio
<tauno.voipio@notused.fi.invalid> wrote:

>On 16.12.19 07:13, krw@notreal.com wrote: >> On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee >> <dakupoto@gmail.com> wrote: >> >>> Could some electronics guru here clarify this a bit ? >>> The efffiiciency definition for an amplifier is >>> Power delivered to load/DC power supplied by power source >>> The Dc power supplied from power supply is >>> Vdc x avg. DC current flowing into amplifier >> >> Not average current, rather RMS current (same as the output). It >> matters. >> >>> However, Vdc changes, as per design requirements of >>> class G amplifier. So what then would be the efficiency >>> defintion for a class G amplifier. >> >> Still power_out/power_in. >> > >RMS is not valid here.
Bullshit.
>The 'S' in RMS is 'square', which is >valid only if the voltage and current are in fixed ratio to >each other.
No, it's *always* true. Average voltage, or current, means nothing.
>Here the voltage stays constant, and the input >power is the constant voltage times average feed current.
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)
On 12/17/2019 12:06 PM, krw@notreal.com wrote:
> On Tue, 17 Dec 2019 10:53:15 +0200, Tauno Voipio > <tauno.voipio@notused.fi.invalid> wrote: > >> On 16.12.19 07:13, krw@notreal.com wrote: >>> On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee >>> <dakupoto@gmail.com> wrote: >>> >>>> Could some electronics guru here clarify this a bit ? >>>> The efffiiciency definition for an amplifier is >>>> Power delivered to load/DC power supplied by power source >>>> The Dc power supplied from power supply is >>>> Vdc x avg. DC current flowing into amplifier >>> >>> Not average current, rather RMS current (same as the output). It >>> matters. >>> >>>> However, Vdc changes, as per design requirements of >>>> class G amplifier. So what then would be the efficiency >>>> defintion for a class G amplifier. >>> >>> Still power_out/power_in. >>> >> >> RMS is not valid here. > > Bullshit. > >> The 'S' in RMS is 'square', which is >> valid only if the voltage and current are in fixed ratio to >> each other. > > No, it's *always* true. Average voltage, or current, means nothing. > >> Here the voltage stays constant, and the input >> power is the constant voltage times average feed current. > > 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.
On 2019-12-17 13:06, krw@notreal.com wrote:
> On Tue, 17 Dec 2019 10:53:15 +0200, Tauno Voipio > <tauno.voipio@notused.fi.invalid> wrote: > >> On 16.12.19 07:13, krw@notreal.com wrote: >>> On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee >>> <dakupoto@gmail.com> wrote: >>> >>>> Could some electronics guru here clarify this a bit ? >>>> The efffiiciency definition for an amplifier is >>>> Power delivered to load/DC power supplied by power source >>>> The Dc power supplied from power supply is >>>> Vdc x avg. DC current flowing into amplifier >>> >>> Not average current, rather RMS current (same as the output). It >>> matters. >>> >>>> However, Vdc changes, as per design requirements of >>>> class G amplifier. So what then would be the efficiency >>>> defintion for a class G amplifier. >>> >>> Still power_out/power_in. >>> >> >> RMS is not valid here. > > Bullshit. > >> The 'S' in RMS is 'square', which is >> valid only if the voltage and current are in fixed ratio to >> each other. > > No, it's *always* true. Average voltage, or current, means nothing. > >> Here the voltage stays constant, and the input >> power is the constant voltage times average feed current. > > 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) >
An AC current times a DC voltage gives zero net power dissipation. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
On 12/17/2019 2:29 PM, John S wrote:
> On 12/17/2019 12:06 PM, krw@notreal.com wrote: >> On Tue, 17 Dec 2019 10:53:15 +0200, Tauno Voipio >> <tauno.voipio@notused.fi.invalid> wrote: >> >>> On 16.12.19 07:13, krw@notreal.com wrote: >>>> On Sun, 15 Dec 2019 02:07:20 -0800 (PST), amal banerjee >>>> <dakupoto@gmail.com> wrote: >>>> >>>>> Could some electronics guru here clarify this a bit ? >>>>> The efffiiciency definition for an amplifier is >>>>> Power delivered to load/DC power supplied by power source >>>>> The Dc power supplied from power supply is >>>>> Vdc x avg. DC current flowing into amplifier >>>> >>>> Not average current, rather RMS current (same as the output).&nbsp; It >>>> matters. >>>> >>>>> However, Vdc changes, as per design requirements of >>>>> class G amplifier. So what then would be the efficiency >>>>> defintion for a class G amplifier. >>>> >>>> Still power_out/power_in. >>>> >>> >>> RMS is not valid here. >> >> Bullshit. >> >>> The 'S' in RMS is 'square', which is >>> valid only if the voltage and current are in fixed ratio to >>> each other. >> >> No, it's *always* true.&nbsp; Average voltage, or current, means nothing. >> >>> Here the voltage stays constant, and the input >>> power is the constant voltage times average feed current. >> >> Wrong, it's *always* RMS voltage times RMS current.&nbsp; You don't think a >> constant voltage has an RMS value?&nbsp; There is no difference when either >> is constant.&nbsp; Power is still V(rms) * I(rms) >> > > No. Power is the integral of V(t)*I(t)*dt averaged over the period of > interest.
Correction: Power is the square root of the integral of V(t)*I(t)*dt averaged over the period of interest.