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tube rectifier arrangement

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Joined 2004
The second arrangement is a common one. But I found a PSU with the first one, taking the HT from the CT of the transformer.
Which is better and why?
 

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The cathode in the 5U4 and similar rectifier tubes is a resistor end to end. If you pull the DC output from the tube out of one end of the cathode the most cathode emission will be concentrated on the end of the cathode you are using as the output. Over time the cathode will wear out faster because the other end of it will not be stressed as highly and the emissive layer coating will still be fresher than the side that was connected to the load. By using a center tapped filament transformer more equal cathode loading is achieved and longer tube life is the result. The center tapped filament winding costs more money to make the transformer so it is seen less often in consumer gear.
 
Disabled Account
Joined 2004
My preamp needs about 60mA and I'll use 1*10u and 3*220u caps and 47u near anodes of each triode.One triode at each channel.SE.
I'll go with 5U4G.
My phono needs 30mA and I'll use 1*10u and 5*100u.Shindo style(in both projects).And some small about 10-20u at each triode anode of two stages.
I'll go with 5V4.

Why with these?I'd like to have plenty of current.I'm a little bit extreme with PSUs but not with capacitance of caps.Just what is needed for uV of ripple.
 
Kuei Yang Wang said:
Konnichiwa,

The current will flow equally but it will flow in opposite directions through the heater winding halves, so it is cancelled

Sayonara

Directions opposite - thats right. But phases different.
Current flows by pulses one polarity in different time intervals.
Would they be different polarity you'd be right.

May be (even sure that not) it is not so important, but I have
some doubts about it. :) It was such a question.
 
Konnichiwa,

igormak said:
Directions opposite - thats right. But phases different.
Current flows by pulses one polarity in different time intervals.
Would they be different polarity you'd be right.

May be (even sure that not) it is not so important, but I have
some doubts about it. :) It was such a question.

Look at it again. No matter how you resolve it, averaged over one full cycle of the mains the net of all currents is zero, nothing remains, so no problem.

Sayonara
 
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