Hi Jeff,
No that's not really the case. The plate current peaks have to go just as
high in class A as they do in class B. The difference is that on the
negative swing of grid drive the class A amplifier is always conducting
where in the class B amp it does not conduct. That lowers the conduction
time of the class B amp and improves its efficiency.
On efficiency change % it depends on the starting point from which you look
at it from. With the class B amp if you start at full power as the reference
then the AM carrier efficiency would be down 50% from there. If you start at
the carrier as a reference point then I guess you could say that the peak
efficiency doubles or goes up by 100%. Most texts use full power out as the
reference point and talk about how many % down things are from there.
Kind of like the merchant marking prices up or marking them down. If you
mark $1.00 down by 50% you get 50 cents. If you mark 50 cents up by 50% you
get $1.00. To mark the 50 cents up you do not add to it but divide it by
50%. But if you just add 50% to 50 cents you then only get 75 cents. One is
called "mark up" and the other is called "add on".
All in how you work the numbers.
73
Gary K4FMX
> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
> Behalf Of KA5MIR
> Sent: Wednesday, November 15, 2006 12:36 PM
> To: amps@contesting.com
> Subject: Re: [Amps] Class A for AM
>
> 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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