Tranny hookup...

john65b

Member
Paid Member
2005-01-09 2:32 am
Chicago
Hello all, I have a Transformer hookup question that was most likely asked here a few times, but I cannot find the answer....

I have a Transformer (not toroidal) that has dual 120V primary and dual secondaries of 41V @ 6.1A (500VA)

The tranny can be hooked up for 240V primary too.

I am looking for a dual secondary 20V and 400VA for an aleph or a Zen V5...

Anyway, here she goes: If I connect the 120V mains to the 240V primary taps, will I get dual secondaries 20.5V at 12.2A (500VA)?

The tranny should be good for 50hz/60hz, and from what I see, the wires in transformer look uniform (240V windings look just as thick as 120V windings)

Is this possible or am I an idiot for considering it? Please be nice - I am thenthitive..

I ask since I have already done something kinda similar...at the advice of a smart guy on this forum, I took a Triad N-68X tranny which is a 230V primary to 115V secondary and hooked it up backwards to get a nice 230V supply for my tube preamp...worked perfectly -
 
oh , now i get it :rolleyes: u are using 120v for the primary...
what will the trafo drive? is it an amp?
if its an amp(not class a) then 500va will be good for more then that in music power .
so ,even in your configuration ,it will be suited for more then 300w in music power without overheating.
 
Hi,
check how warm it gets as you turn up the bias.
and measure how much the supply voltage drops as the bias is turned up.
These will give you some clue to how hard the transformer is working.

Hint,
a 6Aac transformer is at maximum rating when delivering 3Adc continuous when the load is a capacitor input filter.
It is normal to only take half the rating as maximum continuous current when powering a ClassA amplifier.
That sets the upper limit for bias at 1.5A. What had you intended?
 
Perhaps I am suffering a senior moment, but in a standard class A design, peak load current equals bias current minus some head room.

If Andrew is correct about 1.5 Amps then ((peak Amps times 1 over the sq root 2) squared times load resistance) or (1.5 * .707)^2 * 8 = 9 watts RMS output. Probably less than you had in mind. The +/- 20.5 Volts implies an amplifier of about 25 Watts, 9 is quite a bit less. Since output power is I squared, monitoring trafo temperature rise vs load would give you a reasonable means to determine max safe output.

The trafo should be comfortable at about a 40 degree(C) rise, more is possible but allowing for aging and long amplifier life means the less rise the better. The only reliable way to measure internal winding heat is to measure the winding resistance before and after a load test. Copper temperature coefficient is .0039/degree C

That same 25 watts RMS equals a peak load current of about 2.5 Amps for sine waves (also a bias current of 2.5 Amps). This also explains the massive heat sinks needed for class A amplifiers since each half of a push pull output dissipates about the same as the max output power.

Probably someone should validate my math, sometimes when I dash off a reply I miss an obvious error.
 
hermanv said:
= 9 watts RMS output. Probably less than you had in mind.

The trafo should be comfortable at about a 40 degree(C) rise,
Hi,
you're awake and correct.
P=Ipk^2 * Rload / 2

40Cdegrees measured where? internal or surface?
If the ambient alongside the transformer is 50degC and then you allow a surface temp rise of 40Cdegrees the resulting surface temperature is 90degC. What is the internal temperature of the core and windings and more importantly the insulation adjacent to the core?
 
AndrewT said:
H
40Cdegrees measured where? internal or surface?
If the ambient alongside the transformer is 50degC and then you allow a surface temp rise of 40Cdegrees the resulting surface temperature is 90degC. What is the internal temperature of the core and windings and more importantly the insulation adjacent to the core?

I recommended measuring actual secondary winding temperature rise and limiting the internal rise to 40 deg C. Internal copper resistance rise it is the only reliable method I know of.

Let's pretend the resistance started at 10 Ohms so a 40 degree rise would then be 11.56 Ohms (40 times .0039 times 10 plus the initial 10)or a change of 1.56 Ohms.