"Tarmo Tammaru" wrote
"Reg Edwards" wrote

What's the problem?

You have just designed the PA.

You are familiar with its very simple internal
circuitry.

You have its component values, including the active
ones.

Hasn't it yet occurred to you engineering Ph.D's
all
you have to do is calculate it? [Source
resistance].

Reg,

I am *not* a PhD, but, I can tell you what the
problem is. You don't know
what the transistor's Norton equivalent collector
(drain) resistance is at
DC, let alone RF. It is generally not on data sheets.
They will often
specify an "output impedance". This is a convenience
for people who *think*
they are doing conjugate matching. The real component
of this has absolutely
nothing to do with a particular device, outside of
second order effects
caused by series lead inductances.

Check out Motorola application notes AN282A, 721, and
1033. The following is
a quote from AN1033:

"The output impedance of a microwave power transistor
is usually defined as
the conjugate of the load impedance required to
achieve the desired
performance. A typical output equivalent circuit is
shown in Fig1. { Shows
current source in parallel with Cout and resistor
labeled transformed load
impedance. From these two nodes there are inductor
Lc and Lcom going to the
output terminals}. The capacitor Cout is nearly
equal to the collector -
base capacitance Cob specified for the selected
transistor. Lc is the
inductance of the bond wire used to bridge from the
collector metallization
area to the package output lead, and Lcom represents
the inductive effects
of the common element bond wire"

"For correct operation of the transistor, the
ultimate load impedance must
be transformed to a real impedance across the current
generator. This real
impedance is determined by

RL = (Vcc - VCEsat)^2/(2Pout)

The load impedance presented to the package terminals
will contain the real
impedance at the current generator, transformed to a
lower value by the low
pass section formed by Cout and the parasitic
inductances Lc and Lcom.
Usually the reactive part of the load impedance is
made inductive to tune
out the residual capacitance of the device."

One of the other ap notes mentioned collector
resistance to the extent that
it is "high". This kind of analysis is not limited
to transistors. For
kicks, I calculated the load for an 813 tube, and got
the same value as a
book I was reading, which used a roundabout method.
The other ap note also
made a point of the fact that if you were actually
conjugate matched, the
efficiency could never be more than 50%.
==================================

Tam,

The only problem is that caused by entirely unnecessary
complications introduced by people being too clever.

First design a single-ended, class-A, audio amplifier
like a 6V6 beam-tetrode tube plus a transformer to
drive an 8-ohm speaker.

Then calculate the source resistance seen from the
speaker looking back into the amplifier.

To save time searching for manufacturers data, assume:

DC plate supply is 250 volts.
Peak signal at plate 200 volts.
Internal plate resistance, from mnfr's curves, is 100
K-ohms.
Audio power output = 5watts into 8-ohms load.
Peak volts across load = 8.944 volts peak.
Transformer turns ratio = 22.36

Then ask yourself "Is the source resistance of the
amplifier in the same ball-park as the 8-ohm speaker ?"

Is there anything like a conjugate match ?

Which is what it's all about.

For Class-B and Class-C conditions, just multiply
internal plate resistance by a constant which depends
on plate current operating angle. Curvature of tube
characteristics can be taken into account for a higher
order of accuracy. And all this was sorted out in the
early 1920's.

Modern Western radio engineers' education seems to have
been sadly neglected. The next generation's primary
education is taking place in the rat-ridden sewers of
Rio and in the medicine-less, water-less and
electricity-less ruins of Baghdad.

Cecil, in the meantime I'm looking forward to the day
when Texas vintages appear on the shelves of my local
supermarket. ;o)
---
Reg.