Forgive me if I'm still unclear. Either full wave bridge or half wave (if that's the terms we're using) will have the same unloaded voltages, or close enough, as each other, and also very similar loaded voltages. So I don't understand the perceived problem. ?
Note that a mains transformer with two 250VAC secondary windings that can be connected in parallel can be safely used to draw twice as much DC current as a single winding (if the windings are the limitation) but a single wnding of 500VAC with a center tap may (very likely) not operate safely with a full wave bridge across half of its winding into the same current load.
Transformers can be, and are, made in many ways, so details matter.
All good fortune,
Chris
Thanks Chris, so I have here a trafo with high voltage windings 2 X 250 V with a center tap and 6.3 V for heating the tubes. I was thinkin tha in case of half way rectification will use just half of the trafo - just 250 volts, but if we decide to go for a full wave /which I understand is preferable/ we have to use the center tap and two windings of 250 V each. As far as I understood in case of half way we will not multiply 250 by 1.41 after rectification / I may be wrong/
So the schematics of the Chinese engineers is based on 230 V out of the trafo and in my case it will be a bit higher, which is not an issue /probably/
So the schematics of the Chinese engineers is based on 230 V out of the trafo and in my case it will be a bit higher, which is not an issue /probably/
For new users PSUD2 will be an easier lift. Large value inductors generally come with higher DC resistance and greater B+ shifts under dynamic current loads. A second and cheaper option may be reducing the inductor value and increasing the value of C5.
Before going off and just trying various configurations, I suggest you read the following article. Pay particular attention to the second half on smoothing filter sections. Two 1H chokes with an appropriate capacitor section between them will likely give much better filtering than the single 5H/100µf filter section.
http://diyaudioprojects.com/Technical/Tube-Power-Supplies/
Sounds nice to me,the problem is I do not understand how to corelate the 2H 1 chokes with the appropriate cap ?? please help
That's why I said to read the article. The concepts and mathematics are all there.
1. Calculate primary ripple frequency
2. Calculate filter factor for the existing single LC stage.
3. Divide this number by 2 to get the required filter factor for each new stage.
3. Assume two new LC filter sections with the new factor and 1H chokes. Solve the LC filter factor equation for C and plug in the numbers.
4. Calculate reactances at primary ripple frequency and ensure the 20:1 rule is satisfied.
5. If 20:1 rule is violated, increase capacitor sizes until the rule is satisfied.
6. Calculate final filter factors with the final L and C values.
To an extent, you could compensate for the lower inductance by increasing capacitor values, especially the following capacitors after the choke. Although increasing the first capacitor value will work, it will also increase pulsating current value. Try downloading Duncan's PSUD2, it is free to use. There you can enter all your power supply components value and experiment, until you get the required ripple value.
Pozdravi!
Pozdravi!
Seems too complicated for my electroning knowlege. I am a simple old lawyer.
Благодаря и ще се опитам да навляза в дебрите на познанието
Thanks, you have in mind that my trafo has 2x250 V with central tap and I like to use just one half 250 V in order to rectfy by four diodes. And what about yhe resistors between the chokes ?
If you have a computer capable of running Windows programs PSUD2 is will clarify many of your concerns without complication. Start by modelling the original circuit and follow with 1H comparisons and different capacitor values. You'll gain a good feel for the effect of changes without math.