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15th September 2005, 09:57 PM  #11  
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Join Date: Feb 2002
Location: Richmond, VA.

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That is why a full wave (bridge or standard) gives closer to the full wave (rather than just RMS), because the cap has double the chance to charge up, since now the half waves are happening 120 times a second and are one right after the other, versus only 60 half waves with 1/120th of a second in between. A big enough cap and you can get to about 90% or more of peak. So, I quess you are correct, since at half wave there needs to be more current flow to charge the rectifier than the full wave. But I am looking at it from a voltage times current aspect, if that makes any sense. On the other hand, once the cap is full it takes less effort for the full wave to keep it charged. As for nasty spikes, a couple of small value caps (0.1 µF or so) will take care of those. Gabe
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Gabe CGV Electronics Home of the CGV300B amplifier on a budget 

15th September 2005, 11:09 PM  #12 
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Join Date: Feb 2005
Location: Zagreb

You could go at this 2 basic ways:
1) You could use the 14VAC with a resistor in series to get the required 12.6V/800mA. This gives you AC heating, but is very simple. Actually, if your 7V winding gave you 6.3V at the heaters with 1.6A, using the 7+7=14V would give you 12.6V at the heaters at 800mA. 2) DC heating  here you have a choice of 1 common rectifier or one rectifier per tube. You do need one 7812 per tube as 1A is the rated current (1.5A max). For a common rectifier you really have no other choice but to use a bridge. This will give you DC with ripple peaks at (14 * SQRT(2))1.2V, ~=18.6V. The 7812 needs a minimum of 15V on the input to function properly, so your 'sawtooth on top of DC' at the filter cap must never drop below 15V at full load. This is what gives you the size of the filter cap  if you have a scope handy, it will be a simple matter of trial and error to figure out the right value. In the mean time, here is a good approximation: Given that the voltage on the heaters is kept constant, so is the heater current. In effect, your power supply is loaded by a constant current sink of 4x0.8=3.2A. Every 10ms, the caps are charged to 18.6V approximately, and after that, discharged with a constant current of 3.2A. In order for the voltage never to fall below 15V, they must not discharge by more than 3.6V. Since C=delta(U) * I * t, putting that into the formula yields 3.6 * 3.2 * 0.01 = 115200uF! The actual capacitance will be lower because for a short time every 10ms, the rectifier feeds the cap and the load. Still, this is a lot of capacitance, and you also need some quite humongous diodes to pull it off, with very high peak currents through the winding, meaning voltage loss on the diodes, and increased transformer losses. An alternative solution would be to use 4 smaller rectifiers and caps, one for each tube. Using a halfwave rectifier does have a SLIGHT advantage that you only lose one diode drop on each peak, but you have to account with twice as long a time between replenishments of the filter cap. So, you end up with some 56000uF PER TUBE. Still not good. IMHO, I'd go with the AC heating... given that delta(V)=t*I/C, and your I=4x800mA, and delta(V)=18.615=3.6V, and t=20ms, you get C=dV*I*t=3.6 *3.2 * 0.02 ~=230000uF! 
19th September 2005, 04:14 PM  #13  
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Join Date: Oct 2002
Location: Croatia

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19th September 2005, 04:24 PM  #14 
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Join Date: Feb 2002
Location: Richmond, VA.

230,000 uF!!!!!!??????????
Yikes, man! Too much! I have 10,000 uF after a full wave (yeah, I opted for a 25 amp) bridge feeding (6) 6.3 volt 300mA tubes... very clean and very quiet. They draw 1.8 amps total. I use the same bridge/cap for another set of power output tubes that draw 3 amps total, again very clean. Also, I get the exact voltage rated. According to your calculations, he might as well get one of those 1 or 2 farad caps for automotive audio... they're at 12 volts... "Why calculate what you can measure?" Ah, thoery is a wondeful thing, isn't it?
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Gabe CGV Electronics Home of the CGV300B amplifier on a budget 
19th September 2005, 04:32 PM  #15 
diyAudio Member

so that's 3,2*.02/3.6*1000000 = 18.000uF. Per tube:4.500uF so 4.700uF should suffice. U have 2200uF now. I'll have some measurements in a sec...

19th September 2005, 04:46 PM  #16 
diyAudio Member

2200uF hardly gives a spike in the voltage. 2x2200uF gives a perfect straight line.
still with the half bridge, so will even get better! So a 4700uF or bigger cap will work fine... 
19th September 2005, 04:51 PM  #17  
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Join Date: Feb 2005
Location: Zagreb

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