[AMRadio] RE: Using a Ranger as a driver

John Coleman jc at pctechref.com
Wed Feb 13 10:38:59 EST 2008

```You are confusing the Z that the Final sees with the Z that the modulator
sees.  The RF final is looking into a load reflected by the tank circuit

Typically: the Z that a class C final sees for its RF load is (Ep/Ip)*2
This "Z" is used in determining the proper reactance for the tuned circuit
so they will have the proper "Q"

OTOH, the modulator is looking at a load that is determined directly by The
RF stage ratio of Ep/IP.

For some reason most people look at modulator power as a determination of
modulation percentage and this is some what misleading.  It is really easier
to just say that the peak audio voltage delivered by the modulator must
equal or exceed the DC voltage on the final RF amp for 100% modulation.

If the DC voltage is 600V on the RF class C final then the modulator must be
capable of at least 600V peak or 1200 Volts peak to peak, because for 100
percent modulation the DC on the final must vary down to 0 Volts and up to
twice the DC value.

If a XMTR is unloaded to draw less plate current (Ip) the plate voltage will
remain at the same voltage.  If there is less current being pulled by the
final but the voltage is the same then the Z seen by the modulator is higher
than before.  The higher resistance that the audio is looking into makes it
easier to produce the needed voltage.  It is easier for the one modulator
(the same mod XFMR ratio) to modulate a 60 watt rig with 600V than a 100
watt rig with 600 Volts.  Matching the impedance is not necessary. It is
often confused with the statement that says "when the load Z equals the
source Z there will be maximum power transfer".  This is a true statement.
But power transfer is not the issue here.  Getting the voltage on the final
to swing from 0 to 2*DC with out distortion, that is the issue.  Of course
we won't to do that at a power level determined by our needs.

The concern was how to reduce the power from a particular rig.  Is unloading
the final tank all that is necessary?  Perhaps it is.  But there are
over modulation can occur if not monitored.  This occurs because the
modulator is not working into as heavy a load as before and the audio
voltage at the output will rise as the load resistance goes up. Here is why.

The modulator can be looked upon as a source of voltage with a certain
amount of internal resistance.

Voltage Source (modulator fixed input with no load) = 900Volts peak

Internal resistance is a virtual value not equal to any one thing but a
result of the circuit design and tubes.

Internal resistance = 2000 OHMS

Load resistance (EP/IP of the RF final) = 4000 ohms

Total resitance is 6000 ohms

The internal resistance will drop the voltage at the load to
(4000/6000 * 900)= 600V peak

With out changing the audio input level let's raise the load resistance to
8000 ohms (equivalent to reducing plate current to 50 percent by unloading
the rig).
Total resistance = 10000 ohms

(8000/10000 * 900) = 720 volts peak

This increase in modulator audio output voltage is a result of unloading the
rig.  The audio level at the input will need to be reduced to keep from over
modulating.

BTW if the load is reduce to 2000 ohms (equal to the internal resistance)
Total resistance = 4000 ohms
Load voltage = 450 V peak 1/2 the capability of the max output voltage

Maximum power is transferred at a lower voltage output but a greater current
would be drawn by the modulators and probably a lot of distortion.

SEE
http://www.qsl.net/wa5bxo/power.html

Hope this clears some things up.

John

-----Original Message-----
[mailto:amradio-bounces at mailman.qth.net] On Behalf Of Jim Wilhite
Sent: Tuesday, February 12, 2008 9:00 PM
To: Discussion of AM Radio in the Amateur Service
Subject: Re: [AMRadio] RE: Using a Ranger as a driver

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

```