|[AMRadio] Mod xmfr?|
John Coleman, ARS WA5BXO
wa5bxo at pctechref.com
Sun Oct 10 11:04:15 EDT 2004
Actually Steve, the idea is pretty simple. When using a class C final
amplifier the output RF voltage is directly proportional to the plate to
cathode voltage. Most of the time the cathode is grounded so we just
say "plate voltage" or "plate modulation". This means that if the plate
voltage is doubled then the RF output voltage will also double. It's
sort of like saying that if we raise the voltage to a lamp it will put
out more light, etc. Power or "Wattage" is generally measured or
expressed in average values unless otherwise stated. The is the
difference from a horse pulling continuously on a wagon there by
exerting a max of one horsepower, verses jerking periodically on it. A
lamp will put out the same amount of light will 110 volts DC as it would
with 110 volts AC from the wall power. The wall power is measured in
RMS (root mean square) value. The peak voltage of the AC wave is ((sqrt
of 2) * 110 volts) or (about 1.414 * 110). The RMS voltage value of the
wave is (sqrt of .5 * peak) or ( about .707 * peak). To modulate a
class C rig to 100% (pos and neg) requires the peak voltage to be equal
that of the DC supply so the average audio power required is half the DC
power to the plate. For more on this see
I'm glad to see questions of this nature.
From: amradio-bounces at mailman.qth.net
[mailto:amradio-bounces at mailman.qth.net] On Behalf Of
StephenTetorka at cs.com
Sent: Sunday, October 10, 2004 7:12 AM
To: amradio at mailman.qth.net
Subject: [AMRadio] Mod xmfr?
I need to better understand the workings of a plate modulator for AM.
Considering the RF amp with 2 x 813's and the modulator with 2 x 813's -
are the equations and math to work out the required wattage, impedance,
And, just how much can one depart from the ideal arrangement and with
effects? ( should I not be able to find the right xmfr for this amp I'm
to build ).
PS: I am seeking that modulator xmfr at this time.
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