[AMRadio] Oscillator Filament Voltage change and frequency drift

Jim candela jcandela at prodigy.net
Sun May 14 09:26:43 EDT 2006

```After sleeping on this idea, a few things emerged from my dreams. The peak
diode current will be considerably higher than 0.3 amp because it will be
pulsed every half cycle, and on that 1/2 cycle it only conducts on a portion
of the rising edge up to the peak. So the diode peak current may be upwards
of 1 ampere when the RMS current is 300 ma. The curve for the 1N5817 shows
about 450 mv diode drop at 1 ampere. That's not too bad. If you want to get
that back down to 300 mv @ 1 ampere, then you could just use 2 or 3 diodes
in parallel.

The 6AU6 filament might work fine at say 5.5 volts instead of 6.3 volts. If
so, this increases the headroom by 6.3-5.5=0.8 volts. That is a bunch, and
likely worth a try. To regulate this circuit at 5.5 volts just substitute
the 3.9K for a lower value which in this case is 3.3K.

I might add here that this circuit is unproven, so take this as an idea
subject to further refinement, or possibly need major revision to a half
wave doubler input. The problem with the doubler besides the extra parts is
that the regulator will then have too much voltage to drop (input-output),
and then heating becomes an issue. Extra heat in the VFO compartment is not
always a good idea unless that heat is somehow thermally regulated to
maintain a constant temperature.

Jim
JKO

-----Original Message-----
[mailto:amradio-bounces at mailman.qth.net]On Behalf Of Jim candela
Sent: Saturday, May 13, 2006 11:25 PM
Subject: RE: [AMRadio] Oscillator Filament Voltage change and frequency
drift

Ok, Here is my first stab at this. The idea here is that the VFO uses 6.3
volts AC, and one side is grounded. This makes it tough to regulate unless
you use a half wave voltage doubler. Trying to use simple half wave
rectification and regulate has in the past failed because the rectifier
dropped nearly 1 volt, and the linear regulator needed at least 2 volts
in/out differential to regulate. So: 6.3 X 1.414 - .9 = 8 volts. On the
surface this will barely work with minimum 2 volt drop across the regulator,
but the input voltage will sag down considerably between the 60 hz half
cycles depending on load current, and input capacitor value. Therefore the
output will be pulsating at a 60 hz rate as the regulator keeps going in and
then out of regulation. This is a good idea that don't work.

In the circuit shown below, the diode drops only .3 volt at 300ma (6AU6
filament), and the regulator only needs about 0.2 volts in/out differential
to regulate at 300 ma. So 6.3 x 1.414 - .3 = 8.6 volts. This should work
since the regulator only needs .2 volts head so long as the input to the
regulator when dipping during the half cycle when the diode is reverse
biased can only decay to 6.5 volts.

I have a question for you mathematicians out there. What is the minimum
value of the input capacitor required in order to maintain regulation of the
LM2941CT? I picked 4700 uf @ 10 volts as a SWAG because it was as big as it
gets before size, and cost become factors. Assume the load current is a 6AU6
filament. The *AU6 tubes don't come in a 5 volt version, just 3, 4, 6, 12.

All these parts are in either Mouser or Digikey, and the major cost will be
shipping.

Jim
JKO

Schematic:
http://pages.prodigy.net/jcandela/VFO/VFO_Fil_Reg.JPG

1n5817 has forward voltage of about 0.3 volts @ 300 ma forward current:
http://www.fairchildsemi.com/ds/1N%2F1N5818.pdf

differential at 300 ma:
http://www.national.com/ds/LM/LM2941.pdf

LM2941CT Visual:
http://pages.prodigy.net/jcandela/VFO/LM2941CT.JPG

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