[AMRadio] RE: Using a Ranger as a driver


Jim Wilhite w5jo at brightok.net
Tue Feb 12 22:00:18 EST 2008


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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