[AMRadio] RE: Using a Ranger as a driver


John Coleman jc at pctechref.com
Wed Feb 13 13:18:40 EST 2008


Yes more than voltage is required; current from the modulator is also
required.  The amount of current depends on the EP/IP ratio of the final.
So as the final is loaded heavier for more IP then the modulator will be
required to produce more current as well as the previously required voltage
for 100% modulation.  

If the RF Final is tuned for less current the Modulator will also reduce its
audio current output while the audio voltage rises.  But the current will
reduce faster than the voltage increases so that the modulation power is
actually less even though the rig may now be over modulated.  This can be
proven by setting the modulation to about 50% with a tone, then make note of
the modulator current of the class B modulators.  Now while the rig is
modulating, putting power into a dummy load, reduce Ip of the final by
re-tuning a little till the current is about 50%.  Note the modulation
percentage level on the scope has increased and the class B modulators are
drawing less current.

John, WA5BXO


-----Original Message-----
From: amradio-bounces at mailman.qth.net
[mailto:amradio-bounces at mailman.qth.net] On Behalf Of Gary Schafer
Sent: Wednesday, February 13, 2008 11:20 AM
To: 'Discussion of AM Radio in the Amateur Service'
Subject: RE: [AMRadio] RE: Using a Ranger as a driver

While it is true that for 100% modulation the modulation voltage must be
equal to the final plate voltage. But it is also misleading to say that it
is strictly a function of modulation voltage.
If it were only voltage needed then we could use a very high step up ratio
modulation transformer to accomplish modulation.

The modulation transformer secondary works into a certain load impedance
(resistance) supplied by the final plate circuit of the RF stage. While it
is true that the modulation voltage must match the final plate voltage (for
100% modulation), it must also supply current to that load impedance in
order to be able to produce that voltage. Voltage times that current equals
power required of the modulator. After all, all of the side band power of
the modulated signal comes from the modulator.

If we have a transmitter that is not 100% modulated by a certain amount of
audio power and we decrease the power of the RF stage the modulation power
will remain the same but the modulation percentage will increase.

By increasing the load impedance of the RF stage (by reducing its plate
current) the modulation transformer can produce more modulation voltage
because it is working into a higher load impedance. But if we don't turn
down the modulation level the modulator will still produce the same amount
of power that it did before. The difference is the voltage out will be
higher and the current lower. This is assuming that the final is still not
being over modulated.

If we go too far at letting the modulation voltage rise on the RF stage by
further reduction of the RF stage plate current we will run out of plate
swing room on the modulator tubes. The impedance rise on the RF stage is
reflected back to modulator tubes. When that happens clipping will occur in
the modulator plates. At this point we would need to change the modulation
transformer ratio so that the modulator tubes see a lower impedance to keep
them operating within their plate current range.

73
Gary  K4FMX

> 
> 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.

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