On Tue, 24 Jan 2006 16:07:30 -0800, Don Bowey
wrote:
On 1/24/06 2:31 PM, in article ,
"Gary Schafer" wrote:
On Tue, 24 Jan 2006 13:40:54 -0800, Don Bowey
wrote:
On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:
"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).
Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:
"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."
Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.
Pete
Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.
In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.
Don
I don't remember the 32v3 specs but a pair of 6146B's is rated for
120 watts carrier output on AM. 6146A's are rated for 100 watts output
on AM.
My Viking had a 4D32 final and it would load to well over 100W.
Assuming the 120 watts carrier output, when modulated 100% the voltage
doubles and the current also doubles on modulation peaks. Doubling the
voltage and doubling the current works out to 4 times the power. This
is of course Peak Envelope Power of the signal which would be 480
watts.
Where does the double voltage come from at 100% modulation? I can only
account for a 50% rise in voltage.
You can not just add the audio power to the carrier power to find PEP.
You must first add the voltages together.
Good idea, if one knows the voltages......
Peak envelope power is what the FCC is concerned with for maximum
allowable power of 1500 watts.
Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.
73
Gary K4FMX
My point is that listing the PEP capability of an AM transmitter isn't as
useful as stating it can output about 100 watts.
Don
I agree stating PEP output of an AM transmitter does little. But a
properly operating transmitter should be able to give pep at 4 times
the carrier power. Some transmitters do not have that capability
because of a poor modulator or too small finals, or power supply etc.
I, like Peter, was trying to dispel the somewhat misleading add of the
original poster.
As far as the voltage doubling with modulation, you only need to look
at the output on an oscilloscope at the composite signal and you will
easily see that it does. Set the scope to show the carrier level at
say 2 divisions on the screen. With modulation you will see the
positive peaks reach 4 divisions on the scope. The negative peaks will
reach zero on the scope.
Another way to look at this is when modulating the final the peak
audio voltage must equal the plate voltage for 100% modulation. In
order for the modulation to go to negative 100% the audio voltage must
cause the plate voltage to swing down to zero. By the same note in
order to reach 100% positive modulation the audio voltage must cause
the plate voltage to go to twice the dc voltage.
It may seem confusing because if you add the average output power up a
100 watt transmitter is only 150 watts. 100 watts carrier and 25 watts
in each side band. However if you add the voltage of the carrier plus
the voltage of each audio side band and then calculate the power you
will see that it is 4 times the carrier power.
100 watts into 50 ohms = 70.7 volts
25 watts into 50 ohms = 35.35 volts
25 watts into 50 ohms = 35.35 volts
Total voltage = 141.4 volts (which is 2 x carrier
voltage)
P= E squared / R
141.4 x 141.4 = 19994
19994 / 40 = 400 watts
73
Gary K4FMX