[HBR] Filament Voltage Dropping Scheme

[Hopperdhh at aol.com]

I recently (if 2 years ago is recent) built a line operated transmitter using 
a 12SK7 oscillator and a 50L6 final.  I know this is a Home Brew RECEIVER 
site, but I wanted to tell about how I reduced the line voltage to 62.6 volts for 
the filaments.

I used capacitive reactance to do it.  There is no heat dissipated in the 
capacitor like there would be if the 120 volts were dropped with a resistor.  If 
a resistor were used here are the resistance and power dissipated in the 
resistor:

R=(120-62.6)/.15=382.67 ohms
P=V*I=(120-62.6)*.15=8.61 watts

To calculate the capacitor required for this job, here is the equation:

C=0.15/(377*sqr(120**2-62.6**2))=3.89uF

where 0.15 is the filament current and 120**2 means 120 squared, etc., and 
377 is 2 pi times 60 Hz.

You would put your values in for the current and voltage for your 
application, of course.

Obviously we can't use polarized electrolytic capacitors for this job.  I 
used mylar capacitors rated at 250 volts to come up with the right value.  
Computer circuit boards as well as solid state TV boards found at ham fests often 
have these types of capacitors.  Its easy to experimentally arrive at the right 
value after finding the target value from the equation.

The nice thing is that no power is dissipated so the capacitor runs stone 
cold as it does the job.  I don't think the power company will mind the very 
slight power factor increase.

This should work just fine in line operated receiver circuits as well.  Of 
course all the usual precautions apply for all line operated equipment.  Today, 
with houses wired for 3 wire outlets, line operated equipment is much safer 
than it was back in the 50s and 60s.

OK guys, what do you think?

Dan K9WEK