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----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Jim candela Sent: Saturday, May 13, 2006 11:25 PM To: [email protected] 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 LM2941CT low dropout adjustable regulator has about 0.2 volt minimum in/out differential at 300 ma: http://www.national.com/ds/LM/LM2941.pdf LM2941CT Visual: http://pages.prodigy.net/jcandela/VFO/LM2941CT.JPG -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.1.392 / Virus Database: 268.5.6/338 - Release Date: 5/12/2006
