Ok John,
I have but a few observations and questions that might steer everyone
through this. I am having trouble understanding how the Z the modulator
see is greater if the plate load impedance goes down. If the Plate Z is
= to the Ep divide by 2 times the Ip and you change one value how does
the Z remain the same. Even if you maintain the same ratio doesn't the
plate Z change as a result? In your example you change both at the same
ration which maintains the same impedance.
Ep 600
Ip 200 ma.
Zp= 600 divided by 2 times .2 = 600/.4 = 1500
Ep 300
Ip 100 ma.
Zp = 300 divided by 2 times .1 = 300/.2 = 1500
The ratios stay the same, but what I question is if you reduce the plate
voltage of a tube but load it to the same value of Ip or near it, the
impedance changes on the secondary of the mod transformer, which I
guess, is where the divide is. I have seen many people make this
mistake. Because if you do, to maximize power or whatever reason, the
impedance seen by the primary of the mod transformer is not optimum.
In the discussion of the way to drive the amplifier no one pointed out
that you must maintain the ratio of Ep to Ip. This is where I kept
sticking. Now the question becomes, why did Johnson say to load the
Ranger to rated plate voltage and current but insert the pad between the
transmitter and amplifier? Just guessing here, but I would bet that
amateurs of the 50s were much like those of today. All knobs and stated
reading must be to the right. Or is there something that is seen on the
primary of the mod transformer by the Class A or B modulator tubes we
haven't discussed yet?
Jim/W5JO
Hi Jim,
What I said was, or tried to say, when a Class C rig is unloaded so
as to draw less current, that is, to tune the loading and plate
circuit so
that the plate dip is lower current than it was when it is tune up for
max.
The plate voltage will stay about the same but the plate current is
less and
you have less RF output as well of course. In this scenario the ratio
of Ep
/ Ip is greater. The Z that the modulator sees is greater.
Now if the plate voltage is lowered with out retuning anything, the
plate current will fall as the plate voltage falls and the ratio of
the two
remains the same.
Basically when a class C rig is set and not retuned, the Ep:Ip ratio
is set and the plate current should follow the plate voltage up and
down
linearly. The RF should follow the plate voltage up and down as well.
Some
tubes and circuits need a little help with this. Such as using grid
leak
resistance instead of a fixed supply if the stage is to be modulated.
The
grid leak resistance will allow the grid voltage to fluctuate a little
with
the audio as the plate current goes up and down. This actually helps
to
keep the ration of IP to EP constant. The grid leak resistor is
something
of a self regulator for the ratio. The screen grid tubes have a whole
other
set of things that can be done to help the plate current to plate
voltage
ratio remain constant.
There was a discussion awhile back about the plate voltage to plate
current ratios. Some one was saying that a circuit will lose
efficiency if
the plate voltage is reduced. This is only true if the person changes
the
loading or tuning. What they probably meant was that if you reduce
the
voltage and try to retune to get the same power out that you would
have less
efficiency. The only thing that should happen when plate voltage is
reduced
is that the power input goes down and the RF power also goes down. If
a rig
with a plate supply of 600V is putting out 100 watts into 50 ohms the
RF
voltage would be 70.7 volts RMA. When the plate voltage is reduced to
300V
the rig should put out 25 watts with RF voltage of 35.35 RMS on the
load.
Here is a chart.
600V EP
200ma IP
DC input power = 120 watts
RF output 100 watts
Plate dissipation = 20 watts
EFF = 83 percent
EP/IP = 3000 ohms modulator load Z
Now let's go down to 300 volts on the plate
300V EP
100ma IP
DC input power = 30 watts
RF output 25 watts
Plate dissipation = 5 watts
EFF = 83 percent
EP/IP = 3000 ohms modulator load Z
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