Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: kagliostro on January 15, 2013, 02:45:04 am
-
In a traditional PT we have a CT on the HV winding
if the transformer is a 300v-0-300v @ 100mA
(so, on each winding the thickness of the wire admit a current of 100mA max)
and we don't connect the CT to ground and connect a diode to one extremity we obtain 600v @ 100mA (Single Wave Rectify)
connecting the CT to ground and a diode at any branch we have 300v @ 100mA (Full Wave Rectify)
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
But if instead of a 300v-0-300v @ 100mA 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
Instead to connect in parallel the two winding is also possible to rectify each 0-300v winding separately with a bridge than parallel the exit of the two bridges and obtain 0-300v 200mA
Is all this reasoning correct ?
Expecially, is correct to say that the currents from two different windings in a CT PT can't be summed ?
Thanks
K
p.s.: May be we discuss about this previously, but I'm not able to remember well or to find that discussion
-
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.
-
This I've perfectly understand
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.
---------- OMISIS --------------
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.
Also this was perfectly understand
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
-------------- OMISSIS --------------
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.
Also this was understand
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.
This I haven't understand :w2: :dontknow: :w2:
... 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.
When I told is also possible to rectify each 0-300v winding separately
I mean a transformer that has two completely separated windings (that can be put in AC parallel respecting the relatives phases of the winding)
So I don't understand why I can't rectify it separately and parallel de exit of the two bridges to obtain 300v @ 200mA
See the attached schematic, be patient and if you can, try to explain me why two different windings can be put in parallel for AC but not rectified and put in parallel the exit of the bridges
I can understand the reason why this can't be done with a PT that has a CT, but I'm not able to understand what happen with the two separated windings and rectify - SORRY
K
-
This I haven't understand :w2: :dontknow: :w2:
... 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.
Our problem is different assumed situations.
If you have two independent windings, yes, you can rectify separately and stack the outputs for more voltage, or parallel for same voltage at more current. This is common in many kinds of electronic devices.
I simply meant there would be no way to separately rectify the halves of a center-tapped winding to place them in parallel. With a single center-tapped winding, you can either use a full-wave rectfier which delivers half the winding voltage and all the current (e.g., 300v 100mA) or use a bridge rectifier for the whole winding to get double the voltage at half the current (e.g., 600v 50mA).
Maybe I made wrong assumptions about your question, since you had asked in an earlier thread about ways to rectify and parallel the halves of a single center-tapped secondary.
Also, I was trying to highlight a subtle fact about a center-tapped winding using a full-wave rectifier, which is that only half of the total winding is being used at any moment, but all the current capability is being used. So if you pretend that you were in a magical world where you could parallel the halves of the secondary and stay at 300v, you would still only get 100mA out.
But since that connection is not possible in a center-tapped secondary, I guess my extra point just confused things.
-
Ciao HotBluePlates
Many thanks for your answer
As I have a great consideration of what you say when you give your answers and councils
I was really astonished and confused about that matter, now I took a sigh of relief
the thing was only misunderstand
Meanwhile I've do a research and I've find some interesting info
unfortunately also some bad news, see here
(http://s9.imagestime.com/out.php/i815675_Methodofdeterminingsecondarycurrentratings.png)
See the difference in current request due to different rectify options
with a bridge rectifier and a capacitor input :sad2: the AC current necessary is 1.6 till 1.8 the DC current :cussing:
For those who can be interested here some documentation
Here look to the last pages
http://www.belfuse.com/Data/UploadedFiles/signalcatalog.pdf (http://www.belfuse.com/Data/UploadedFiles/signalcatalog.pdf)
Here look to chapter 5 and 7
http://www.introni.it/pdf/Motorola%20-%20Rectifier%20Applications%20Handbook.pdf (http://www.introni.it/pdf/Motorola%20-%20Rectifier%20Applications%20Handbook.pdf)
Must Read
Very interesting page here
http://web.archive.org/web/20080322224537/http://www.atc-frost.com/products/design/va.htm (http://web.archive.org/web/20080322224537/http://www.atc-frost.com/products/design/va.htm)
and here
http://s9.imagestime.com/out.php/i815684_Hammond.png (http://s9.imagestime.com/out.php/i815684_Hammond.png)
Many thanks again HBP
Franco
-
unfortunately also some bad news, see here
(http://s9.imagestime.com/out.php/i815675_Methodofdeterminingsecondarycurrentratings.png)
See the difference in current request due to different rectify options
Don't get hung up on that chart. It is providing an answer to a question you didn't ask.
Specifically, how to compute the RMS current of a winding in a situation where you know the expected d.c. load current. Or, looking at it another way, how to work backwards from the d.c. load current rating provided by the manufacturer to compute the RMS current so that you can calculate the VA drawn by the winding (RMS volts * RMS amperes).
They are mixing the conversion factors for a couple different things to provide you an easy reference. If you're not confident on what the numbers mean, it's best to figure them out the hard way, step by step, until you can see how the shortcut was derived.
-
Take a look at this link. This amp has some circuitry worth discussing on the forum. ...
This includes,
1. xlr input both balanced and unbalanced.
2. ss control of a triode tube.
3. ss control of the voltage to the heater circuit
4. ss control of the power tubes. (I would like to discuss this one as a possible method of feeding class B tubes without a transformer PI)
This is probably a hifi application, but look at the ideas incorporated into the layout circuits.
It is outside the scope of this thread, but the things being done are not that uncommon (just not typical in guitar amps):
1. That an XLR is being used means nothing, as it is just a connector. The balanced/unbalanced switch presents signals present on pins 2, 3 to the grids of a differential amplifier in the balanced position, while presenting a ~600Ω input impedance, which assumes a 600Ω balanced line signal. In the unbalanced position, the amp assumes a high impedance unbalanced input signal, and passes the input on pin 2 to the upper triode of the differential amp and presents the incoming signal a 100kΩ input impedance. The lower triode is effectively "grounded" through the 620Ω resistor (pin 3 is assumed connected to ground (pin 1) anyway, so the resistor is unimportant).
2. Not solid-state control. A JFET is being used as a constant current source to present a large apparent cathode load for the differential amp and enforce good balance between the triodes.
3. Not "solid-state control" exactly of the heater circuit, but simple voltage regulation.
4. Not "solid-state control" exactly, but use of a voltage regulator to attempt to enforce a constant current in the output tubes, and keep the bias from drifting with applied signal. May or may not be applicable to guitar amps, especially since many players seem to like the effect of sag and output stage operation changes with big peak signals.
-
so that you can calculate the VA drawn by the winding
But the VA drawn by the winding didn't say me the current I must have disposable at the AC winding to have the DC current I need for my circuit ?
Applying that (Full Wave Bridge - Capacitor Input)
if my circuit request is 100mA DC I need 160mA to 180mA of AC current, or not ?
Am I omitting/forgetting something ?
Thanks
K