Yes, I see what you mean. The plate current peaks will not double in class
A as they would in class B. That 2x increase is not there in class A.
We were saying 50% efficiency change for class B, but efficiency should
actually double for class B, shouldn't it? 100% change? 25 watt base x2
(by plate current) x2 (by efficiency) = 100 PEP?
So in class A, that missing 100% increase from plate current would have to
be supplied by additional efficiency during modulation. This would be in
addition to the 100% efficiency increase we would normally expect during
class B.
So efficiency increase during modulation for a 100% modulated "Class A" AM
linear should be 400%. 25 watt carrier x 400% = 100 watts PEP.
Does that sound correct?
Jeff/KA5MIR
On Wednesday 15 November 2006 10:45, Gary Schafer wrote:
> It has to do with the plate load impedance that the tube sees. In order to
> handle the modulation peaks the load on the tube must show the proper
> impedance to allow the proper plate voltage swing for that power.
>
> When the load is tuned for the peak power, as it must be to avoid flat
> topping while allowing the full peaks to develope, then the load is not
> optimum for the lower power of the carrier so efficiency suffers at
> carrier power.
>
> If the amplifier is tuned for greater efficiency at carrier power then the
> tube will run out of plate voltage swing range when it tries to reach
> necessary peak power that the audio peaks demand. Flat topping will result
> which produces distortion and splatter.
>
> The 50% efficiency change in a class B amplifier happens to work out to
> 50% because of the parameters that the tube must operate at in class B. In
> the class A tube operation the efficiency change is greater but again is
> necessary to allow proper development of peak envelope power.
>
> Because of the 100% conduction angle of plate current in class A, input
> power remains constant regardless of signal input or output so the
> efficiency change range will be greater than with class B.
>
> 73
> Gary ?K4FMX
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