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Calamaro 2nd July 2009 08:23 PM

Transformer wiring
 
2 Attachment(s)
I feel that this will be a foolish question but I never saw a transformer like this... so please forgive me.

I scavanged this transformer from an old halogen lamp. I want to recycle it for my tube buffer. It's labelled "12 volts". It has 2 wires on secondary side but 3 wires on primary side. They were connected to a switch (0=no light; I=light; II=more light...).
I tried these connections:

blue and red wires to mains give 9 volts on secondary
blue and yellow wires to mains give 12 volts on secondary

If I need 12 volts can I just connect 2 wires and cut the third wire?
How this works?

Thank you in advance for any explication.

328bps 3rd July 2009 10:47 AM

YES you could very comfortably do so. what you have on your hands is a tapped primary and untapped secondary. Just take care that you insulate the unused wire cos this is part of the high voltage AC winding

AndrewT 3rd July 2009 11:06 AM

Hi,
don't cut off the spare primary wire. Insulate it or fit it into a three terminal insulated terminal strip.
Yes, it's a tapped primary.

The 12Vac output runs the transformer with fewer turns on the primary. It will draw more idle current and thus run hotter when wired for 12Vac output. It is rated for this but expect it to be pretty warm if you try to get the full 45VA rating from it.

The 9Vac output runs more turns on the primary and thus will run cooler, maybe much cooler. The maximum rated output current will still be 3.75Aac (= 34VA).
The maximum continuous DC current after rectification will be ~1.9Adc

Calamaro 3rd July 2009 02:50 PM

Thank you for your explanation and advice.
I need just 600mA for two 6N1P valves so I think that the trafo will be very good.

******************************
6N1P
From Wikipedia, the free encyclopedia

The 6N1P is a Russian-made miniature 9-pin medium gain double triode vacuum tube intended for use as a line audio amplifier and cathode driver.

Basic data:

* Uf = 6.3 V, If = 600 mA
* uM = 35
* Ia = 7.5 mA
* S = 4.35 mA/V
* Pa = 2.2 W


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