But if ... we have a 0-300v @ 100mA and a 0-300v @ 100mA we can connect the two 0-300v winding in parallel as to sum the available current and obtain a 0-300v 200mA
Yes, but you have to be careful of the polarities of the two windings.
Imagine an instant in time, where one end of 300v secondary #1 is positive relative to the other end of that winding. You'll need to find which end of 300v secondary #2 is positive at that same instant in time. So now you have two secondaries, which you can mark the + and - for each winding.
So you can connect secondary #1 + to secondary #2 +, and secondary #1 - to secondary #2 - to place the windings in parallel. You will get 300vac 200mA for the paralleled windings.
If you made a mistake and paralleled the windings by placing + to -, you'll get 0vac out because the voltage of the two windings are opposite polarities and trying to cancel each other.
Assume the 100mA quoted is the a.c. rating of the secondaries; 300v * 0.1A = 30VA, and so you have 60VA total power available from these windings.
Or you could arrange the windings in series, by stacking secondary #1 on top of secondary #2, with secondary #1's - to secondary #2's +. You will add #1's voltage to #2's voltage, and get 600v 100mA because each winding is delivering 100mA, but it happens to be the same 100mA because the windings are in series. But checking VA, 600v * 0.1A = 60VA, which is the same as we got before.
You still must watch polarity, because if you had:
+ sec#1 - -----> - sec#2 +
the voltages of each winding are in opposition at every moment, and you get 0vac out.
As far as I can know there is no a way to have from such architecture (300v-0-300v @ 100mA) 300v 200mA because of the way the two windings are connected inside the transformer
Because of the
way they are connected inside the transformer.
The two 300v sections are in series, like our 600v example above:
+ sec - -----> + sec -
With the arrow being an internal connection. If you try to bring the two outside leads together to place the secondary halves in parallel, you have + connected to - on both ends, which we showed results in 0v output when discussing parallel windings above.
... is also possible to rectify each 0-300v winding separately ... than parallel the exit of the two bridges and obtain 0-300v 200mA
No. You could parallel and get 300v 100mA.
What happened?
You started with an example transformer which has two independent windings, which the transformer designer assumes will both be used to full capacity at all times. So two 30VA windings, for a total of 60VA.
You
assumed a 300v-0-300v winding delivering 100mA a.c. is equivalent to a pair of 300v 100mA windings; indeed, it looks like that is what I proved above (60VA = 60VA).
Get a diagram of a 300-0-300v transformer using a full-wave rectifier. Mark one end of the secondary "+" and the other end "-". Now look at the diodes, and figure which one is forward-biased and therefore turned on; the other diode will be turned off. You will noticed the path for current goes from:
"+" end of the secondary -> diode -> filter cap "+" -> filter cap "-" -> secondary CT -> secondary "+"
Only one-half of the seondary voltage is being used, but it draws all the current the total winding can pass!So this entire secondary winding could be call 300v*0.1A = 30VA, or 600v*0.05A = 30VA. This PT is only able to deliver half the VA and half the current of your example.
The transformer designer knows only half the winding is working at any moment in a CT transformer using a full-wave rectifier, so they make the core smaller than a similar transformer with two 300v 100mA windings. That reduces costs, saves money, make the transformer cheaper and sells more transformers.
And, as you said, there is no way use separate rectifier circuits to connect the two halves of a 300-0-300v winding to get anything other than 600v, or +300v/-300v. Again, this is due to the incompatible internal connection of the secondaries.
Topic: PT question (Read 3143 times)