Calculating voltage, current, and impedance values on an old transformer

Question

I'm working on an old valve amplifier, with the intention of reusing the transformers in a new project. I've used what I know of transformers to calculate the ratio of each winding the corresponding voltages however what I calculate does not agree with what is specified in the schematic so I must be making an error somewhere. The output transformer has one primary and three secondary windings; the measured resistances are as follows:

P = 7.9
S#1 = 28.4
S#2 = 3.60
S#3 = 0.1

Given that the ratio of a transformer can be calculated as:

Vp/Vs = Ip/Is = a

The impedance of either winding can be deducted as:

Zp = (a*Vs)/(Is/A) = (a^2)Zs
Zs = (aVp)/(Ip/A) = (a^2)*Zp

So far, I know the impedance of each winding, but not the turns ratio or voltages however by rearranging the above formulas I can use the impedances to deduct the ratio:

a = Square Root (Zp/Zs)

Therefore

a(S#1) = 0.53
a(S#2) = 1.48
a(S#3) = 3.36

Using these values of a, we can predict the voltages on the secondary:

V(S#1) = 250/0.53 = 474.01
V(S#2) = 250/1.48 = 168.76
V(S#3) = 250/3.36 = 74.42

However these values are not what is specified by the schematic, which is available here. The schematic specifies 470V, -37V, and 6.3V for the first, second, and third windings respectively and even allowing for a considerable voltage drop during rectification the calculated figures do not match the specified figures for the second and third windings.

Where am I going wrong in this process?

Thanks.

Answer 1

I'm not sure you can do this using your method. You are measuring DC resistance which is not fully controlled by the number of turns - the gauge of the wire matters as well.

Think about two secondary windings - the first with 'n' turns of wire of area 'x' , and another with 2n turns of wire with 2x area. They would have the same resistance but different turns ratios.

In your case, the low voltage secondaries are likely to have heavier wire (for more current).