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Old 15th September 2005, 09:57 PM   #11
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the xformer has 7-3.15-0-3.15-7volt windings, so 7-0-7 yoelds 14 volts ac. Rectified with one diode it drives the xformer incredibly warm.

So which is kind of logical, as the xformer has to produce all it's power in a half wave. Plus it gets all these nasty spikes from the rect.
Hmmmmmm. This seems wrong to me. Half wave rectification will give you a final voltage of if I recall about 20-30 percent less than RMS, simply because of the recovery rate. When the other half wave is going the there is no current flow so the load draws current. Voltage goes down until the next positive peak reaches the same level and returns the voltage to the mean level (average, which is less than RMS?), whatever that is. But it is definitely less than the RMS value of 14, unless you have a ridiculously large capacitor that maintains much of its current.

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 F or so) will take care of those.

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Old 15th September 2005, 11:09 PM   #12
ilimzn is offline ilimzn  Croatia
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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 half-wave 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)=18.6-15=3.6V, and t=20ms, you get C=dV*I*t=3.6 *3.2 * 0.02 ~=230000uF!
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Old 19th September 2005, 04:14 PM   #13
moamps is offline moamps  Croatia
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Originally posted by ilimzn
.....given that delta(V)=t*I/C, and your I=4x800mA, and delta(V)=18.6-15=3.6V, and t=20ms, you get C=dV*I*t=3.6 *3.2 * 0.02 ~=230000uF! [/B]
C=I*t/dV
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Old 19th September 2005, 04:24 PM   #14
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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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Old 19th September 2005, 04:32 PM   #15
beamnet is offline beamnet  Netherlands
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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...
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Old 19th September 2005, 04:46 PM   #16
beamnet is offline beamnet  Netherlands
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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...
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Old 19th September 2005, 04:51 PM   #17
ilimzn is offline ilimzn  Croatia
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Originally posted by moamps
C=I*t/dV
Argh, of course... sorry for that blunder, my bad
I knew I should have had more sleep that day
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