-----Original Message-----
From: Jon Ogden [SMTP:jono@webspun.com]
Sent: Tuesday, March 24, 1998 4:52 AM
To: Rich Measures; amps@contesting.com
Subject: [AMPS] Re: Drive Level in amplifiers
>>One other point on drive level:
>>
>>If the point were true that input impedance varies with drive, then in
>>order to have an amplifier system that is operating properly, one must
>>always operate it at maximum drive.
>>If not operated then at maximum
>>drive, the matching network is not properly attenuating harmonics, not
>>providing the proper "flywheel" affect, etc.
>
>- The flywheel makes things mo betta, and it's always there.
Again though, my point on matching network impedances still stands. If
at a low drive level my tube impedance is say 150 Ohms and at high drive
levels it drops to say 75 Ohms then a matching network design to match a
150 Ohm impedance to a 50 Ohm impedance will NOT work properly with a
load of 75 Ohms.
In fact, the input impedance of a tuned circuit designed to match 50 Ohms
on its input to some output impedance may actually look MUCH worse with a
different impedance on it's output. And you won't get proper power
transfer either. Optimal power transfer to a a load (such as a tube)
occurs when the impedance that load sees looking back toward the source
is the complex conjugate of it's impedance. For example if the tube has
an impedance of 125-j200 Ohms, then for proper power transfer from
matching network to tube, the tube wants the matching network to have an
output impedance of 125+j200 Ohms.
[Richard W. Ehrhorn]
Oh, boy! I've been biting my lip and staying out of the discussion of Zin
variation versus drive power - and I agree with George.
BUT I can't resist commenting on the "conjugate match" issue, even though I
do know better.
Unless the basic electrical engineering principles that I was taught in
EE101 (probably out of Terman or Ryder) have been repealed, Thevinen and
similar equivalent circuit principles on which the "conjugate matching"
idea is based are specifically defined as applying only to LINEAR devices
and circuits.
In class AB, B, and C power amps such as discussed here THE TUBES (or
transistors) ARE NOT LINEAR DEVICES. Plate current flows over much less
than 360 degrees of the rf cycle and is OFF the rest. The tube is as much a
switch as a linear device.
In fact, since the source R and load R are equal in a conjugate match, half
the total available power is dissipated in Rs and only half can be
delivered to RL: maximum theoretical efficiency = 50%. A tube running class
A draws relatively uniform Ip over the entire rf cycle; it IS a linear
device. If I remember right, maximum efficiency typically realized in a
class A amp is around 30%.
IMHO the most fundamental way to define the function of the input matching
network of a class AB2 or B amp is that it (plus all the circuitry
intervening between it and the plate/collector/drain of the driver devices)
must transform the input impedance of the power tube(s) into whatever load
impedance those driver devices must see to deliver their desired output.
Alternately, since nearly all transceivers are designed to work best into a
nominal 50 ohm resistive load, the power amplifier input network should
present a 50 ohm resistive input.
I know there are many fans of the conjugate matching idea out there, but no
one has explained credibly to me how a 50 ohm SOURCE resistance would allow
more than 50% of the total drive power to be delivered to the load. The
remarkable concept of a NON-DISSIPATIVE Rs has been offered, but it's sort
of beyond my comprehension.
During preparation of the amps chapter of the 1995 ARRL Handbook I included
this argument, but the issue apparently had been so controversial that
League diplomats thought it best to forgo that discussion. Probably I'll do
the same hereafter!
73 to all! Dick W0ID (ex-W4ETO)
If the matching network therefore is
designed for an output impedance of 125+j200 Ohms, but the tube changes
to something like 80-j130 Ohms, optimal power transfer will not exist.
And if the power isn't transferred to the tube, where does it go? It
goes BACK to the exciter and hence appears as an SWR.
So just having inductors and capacitors on your circuit doesn't
necessarily make things always better.
>.
>>It is not always beneficial to run an amplifier at maximum drive.
>
>During ssb modulation, the drive varies widely during each spoken word.
>However, max. peak drive and max. anode-I is essential to correct tank
>adjustment.
What does correct tank adjustment have to do with the input impedance?
>
>
>Please consider that: Below a certain drive level, zero grid current
>flows. At max. drive, grid current flows.
I would say that this depends on how the tube is biased, but for argument
sake, let's assume you are right.
However, what does the flow of grid current have to do with tube
impedance? Maybe I am showing some ignorance here, but again, I haven't
seen this.
I guess we will see eventually who is correct here. If I am able to take
a noise bridge, measure the tube impedance and match it up, then I will
be correct. If however, it doesn't work than I am wrong. Also, then I
will conclude that 5 watts into the tube would be considered a large
signal condition.
73,
Jon
KE9NA
-------------------------------------
Jon Ogden
KE9NA
http://www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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