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Old May 24th 10, 02:24 AM posted to rec.radio.amateur.boatanchors
Phil Allison[_2_] Phil Allison[_2_] is offline
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First recorded activity by RadioBanter: Mar 2010
Posts: 9
Default AM signals peak envelope power


"Antonio Vernucci"
I wonder whether someone can help me to clear an issue regarding AM signals
produced by a (class-C) tube transmitter.

It is well known that the peak-envelope power of a carrier 100%-modulated
by a pure sine wave is four times the unmodulated carrier power. This is
simply because, according to the Ohm's law, doubling voltage power gets
four times higher.

I would have no problem if a tube would behave like a normal resistor. But
a tube instead behaves like a differential resistor, and this makes a lot
of difference.

If you consider the plate characteristic of a typical pentode, you see
that plate current does not change much increasing plate voltage. So,
doubling the plate voltage, power would not get four times higher, but
just a little more that two times.

The triode plate current is instead much more dependent on plate voltage
so that, doubling plate voltage, power would grow probably at least three
times. But definitely not four times.

Can someone pinpoint what I am missing?



** I'll give it a go.

In a class C stage, the grid is overdriven with RF voltage so the output
tube is biased either fully conducting or off - as a result, the plate
voltage stays fairly low whenever the tube is conducting, which is less than
50 % of each RF cycle.

However, the actual current flow through the tube depends on the ratio of
the external circuit impedance ( ie a tuned load) to the B+ voltage. Since
the tuned load does not vary in impedance, the B+ voltage controls how much
RF current flow there is.

IOW, the output voltage from a well designed class C stage follows the B+
voltage linearly over a wide range cos to a large degree, the plate voltage
/ plate current characteristics of the tube have been rendered irrelevant.



...... Phil