The idea is to have the transformer operate without getting too hot.
I think you're good for about 18VAC @5A = 90W supply.

With the 8V winding, you don't want the loaded voltage to fall below 6VAC.
I think this would be good enough for a 5VDC @1A supply.

 

got it.. ill post reports from the center tapped coil ( the on im usein for my dual supply), my 8.1vac coil and the 38.5vac coils soon... thanks again

 

ok heres my results..I have 3 secondaries, one with a center tap that im using for my main supply. ill post them as unloaded vac, 190w load and a 360w load.. so here we go.

unloaded sec. with center tap. center tapwas common and mesurements was taken on both sides of it.

22.5vac each side unloaded, 190w 20.5vac 4,7a each side.360w 18.15vac 8.7a
8.1 sec. winding unloaded, 190w 7.8vac 2.8a, 360w 7.39vac 5.16a
38.5vac unloaded. 190w 13.7vac 3.75a, 360w 5.56vac 4.3a
each test had a duration of 5 minutes and the transformer didn't even get warm.... I hope ive provided enough info.. if more is needed I can add more load at 190w at a time.. thanks ass always/

 

what does that ac current mean for dc current? wiil I be able to get my 4a that it want?

 

each test had a duration of 5 minutes and the transformer didn't even get warm....

You need to wait about 2 hours to see how hot the transformer will get.

You can get an idea about how hot the transformer hot spot is by measuring the resistance of a winding before and after heating it up.

Have a look at this thread:

http://forum.allaboutcircuits.com/showthread.php?t=38273

 

...
22.5vac each side unloaded, 190w 20.5vac 4,7a each side.360w 18.15vac 8.7a

You need to be careful! 22.5v each side unloaded sounds about right.

But 20.5v each side loaded at 4.7A sounds like a significant overload to me! That sized transformer did not come out of a 100+100W amp. More like a 60+60W amp.You should not be drawing more than about 3A continuous from the output at approx +/- 21v.

8.1 sec. winding unloaded, 190w 7.8vac 2.8a, 360w 7.39vac 5.16a

This is just silly. That secondary rail was for the low voltage control electronics, it will have fine winding wire and is not designed for high amperage!

And what do you mean by the 190W and 360W numbers?

38.5vac unloaded. 190w 13.7vac 3.75a, 360w 5.56vac 4.3a
...

Even more silly. It's a wonder you didn't burn this winding out! You dropped a 38v winding down to 5.5v with a severe overload!

I told you previously what currents to safely expect from the output windings. Why ignore that completely and do really silly and possibly destructive tests?

Just because 5 pounds of transformer didn't get hot in a few minutes is not proof that you are operating it safely.

You have been lucky up to this point that Marantz build a good level of safety margin into their transformers.

 

I was just doing as I was instructed,, The load I was using was with 190w headlamps. The transformer didn't show any signs of fatigue. I just want a 4a adjustable PSU. After these tests I believe I can.

 

Maybe this will help you make some decisions..

 

I still haven't found an ac to dc amperage conversion method..

 

I still haven't found an ac to dc amperage conversion method..

In my post above, it has the formulas you need. IDC means amps dc, IAC means amps ac.

 

You need to be careful! 22.5v each side unloaded sounds about right.

But 20.5v each side loaded at 4.7A sounds like a significant overload to me! That sized transformer did not come out of a 100+100W amp. More like a 60+60W amp.You should not be drawing more than about 3A continuous from the output at approx +/- 21v.



This is just silly. That secondary rail was for the low voltage control electronics, it will have fine winding wire and is not designed for high amperage!

And what do you mean by the 190W and 360W numbers?



Even more silly. It's a wonder you didn't burn this winding out! You dropped a 38v winding down to 5.5v with a severe overload!

I told you previously what currents to safely expect from the output windings. Why ignore that completely and do really silly and possibly destructive tests?

Just because 5 pounds of transformer didn't get hot in a few minutes is not proof that you are operating it safely.

You have been lucky up to this point that Marantz build a good level of safety margin into their transformers.

Im not trying to be "silly" , I was merely doing measurements that were suggested to me.. I did read your numbers. and I took them into account when measuring my own. you gave me a reasonable idea,, I wanted real world numbers from my transformer. and I think I got what I was looking for. the 8.1 winding is not intended for high current use.. I was going to make it a 5vdc 1a supply.. and the 38vac winding isn't even planned for anything yet. I know I plan on adding an o-scope to the finished product so maybe ill power it with that coil.. Im learning while in doing this so mistakes are bound to happen,, I didn't hurt my transformer from the tests I did.. I plan on further testing giving it a couple hours to measure temperature. What do you recommend I use on the windings that are safer?

 

In my post above, it has the formulas you need. IDC means amps dc, IAC means amps ac.

got it, I read the link but didn't understand a few things..

 

I still haven't found an ac to dc amperage conversion method..

Have a look here:

http://forum.allaboutcircuits.com/showthread.php?t=68839

 

got it, I read the link but didn't understand a few things..

Tell us what you don't understand

 

I didn't understand some of the terminology like idc iac......is there an actual formula for converting ac current to dc?

 

I didn't understand some of the terminology like idc iac......is there an actual formula for converting ac current to dc?

The thread I linked in post #33 is long, but if you will read it you will understand how the DC out of your rectifier circuit is related to the AC out of your transformer.

 

In the first example it shows IDC = .64* sec IAC..

what this means is the current (amps) dc is equal to .64* current ac.
So if you have a transformer with an output of 1 amp ac, you would multiply that by .64.
So you would then have IDC =.64* 1ac.
Your answer would then be .64amps (640milliamps)

 

Thank you,, that what I was looking for.....

 

so my main sec. has 8.7a at 18.1vac.. so this would be 5.5a at 25.2vdc correct?

 

so my main sec. has 8.7a at 18.1vac.. so this would be 5.5a at 25.2vdc correct?

Go back and look at the sheet. It depends on what type of rectification you're using. Find the diagram that fits what you have and use that formula