OHM'S LAW ON
There are not many high-level plate modulated AM transmitters
around anymore. The Gates BC-1 series is an example of this 1950 to
1970s technology. The design typically has 2600 volts running the RF
Power Amplifier tubes. Power supplies like that need a ďbleederĒ
resistor between the high voltage and ground to bring down/bleed the
high voltage to zero when the transmitter is turned off. This should
happen in only a second or so of time. The power supply could stay hot
with high voltage for minutes or hours if the bleeder resistor fails
open. That is a serious safety issue for the engineer working on it, if
he or she fails to short the high voltage filter capacitor before
touching any part of the transmitter.
The bleeder in a Gates BC-1G transmitter is R41, a 100,000 ohm/100 watt
wire-wound resistor. You see one hand-held on the left side of the photo
in Fig. 2. Ohmís Law tells us that 2600 volts across the resistor
squared (times itself) then divided by 100,000 ohms resistance equals
67.6 watts of power dissipation required on a continuous basis on a 100
watt resistor. You would think that the 32.4-percent safety margin would
be enough. This resistor typically failed after 10 years of use. The
answer is in the ventilation the resistor gets for cooling. The 67.6
watts in heat has to go somewhere. This transmitter model has some, but
not a lot, of air flow on the bottom where the resistor is located.
My answer was to replace the 100 watt resistor with a resistor rated at
225 watts, as seen in the center of the photo. It gave more surface area
so it ran cooler, thus longer. A 100 watt resistor is $15.14 vs $18.64
for a 225 watt unit. It is only a $3.50 difference for a huge increase
in reliability and safety. The screw that holds it in place will need to
be longer if you do this modification. No big deal.
Yes, there is a meter multiplier resistor string next to the resistor
and high-voltage capacitor. It samples the high voltage for the PA
voltmeter. Dirt has accumulated on the high-voltage end of the string.
It is high voltage that attracts dirt, requiring frequent cleaning to
maintain transmitter reliability. Itís maintenance. The RF dummy
load in this transmitter has six 312 ohm/200 watt non-inductive
resistors. The transmitter sees the 52 ohms because the resistors are in
parallel. Simple math, 312 ohms divided by 6 resistors = 52 ohms. Yes,
52 ohms, 51.5 ohms, 70 ohms and other impedances were common in the past
before solid-state transmitters more or less forced the standard to be
50 ohms. Tube-based transmitters will tune into almost any load while
solid-state transmitters are designed to perform into 50 ohm loadsÖ.and
donít give me no VSWR!
OHMíS LAW ON
Letís say we know that 2 amperes of current is going into a 100
ohm resistor. What is the voltage across the resistor? The formula
is 2 amperes x 100 ohms resistance = 200 volts. From that, we can solve
for power in the resistor. It is 200 volts x 2 amperes current = 400
OHM'S LAW ON
A Continental 816R-2 FM 20 KW FM transmitter might have 7000
volts on the plate of the PA tube with 3.3 amperes of current drawn.
Ohmís Law tell us that 7000 volts x 3.3 amperes = 23,100 watts of power.
That is transmitter power input, not output. The power output is subject
to the power amplifier efficiency, which is typically 75%. Then, the
transmitter power output is 17,325 watts. That also means that 25% of
the input power is lost in heat. That is 23,100 watts of input power x
.25 = 5775 watts of heat. Be sure to check the manufacturerís data
sheets for exact numbers for each transmitter model.
Half power doesnít mean the transmitterís PA voltage is half.
If it was half, then the PA current would be half and RF output would be
one-quarter. Youíll remember when local Class 4 (now Class C) AM
stations ran 1000 watts day and 250 watts at night. A Gates BC-1
transmitter might have 2600 PA volts and 0.51 amperes of PA current
during the day. We can determine the resistance of the power amplifier
by taking the PA voltage of 2600 and dividing it by PA current of 0.51
amperes. The answer is 5098 ohms. That same PA resistance applies
regardless of the power level of this transmitter. At quarter power, the
PA voltage is 1300 volts. Ohmís law, using the same 5098 ohms, tells us
that the PA current should be 0.255 amperes. Yes, it worked out that way
in practice. The simple trick was to connect 120 VAC to the primary of
the transmitterís high voltage transformer for night operation in place
of 240 VAC in the day. With quarter power, the antenna ammeter
read half and the signal field intensity was half, not one-quarter.
Letís examine this. If you have a 50 ohm antenna and 1000 watts of
power, what is the antenna current? Using Ohmís Law, take 1000 watts
divided by 50 ohms = 20. The square root of that is 4.47 amperes. Divide
250 watts by the same 50 ohm antenna resistance and you get 5. The
square root of that is 2.236 amperes, half of the day antenna current.
Itís Ohmís Law.
Think Ohmís Law when you are on the job. It answers your questions and
makes perfect sense.
Mark Persons, ham W0MH, is an SBE
Certified Professional Broadcast Engineer; he was named the Robert W.
Flanders SBE Engineer of the Year for 2018. Mark is now retired after
more than 40 years in business. His website is www.mwpersons.com.