# Class G amplifier efficiency

Started by 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.
```