Thanks for your tips everyone! I'll perform the transformer autopsy

asap.

I experimented with a few more transformers, simulating a blown primary.

Added to my tips file off URL below

Blown primary, how to determine unknown secondary voltages of a mains

transformer

Simulating a multi-secondary transformer using a known

good one but not using the primary, to get some data.

I used a variac supply near the bottom of its range at 18volts and a 25 ohm,

20W dropper to feed 50Hz (UK)ac into a secondary. Assuming you have a

reasonable idea of the voltage of one 'unknown' secondary.

The transformer I used first was a high grade enclosed Gardners, 1.3Kg but

only 15W combined outputs,

240V (UK) with marked 2 separate secondaries of

6.3V, 0.6A and a 150-0-150 at 25mA.

With 3.43V ac on one '6.3V' secondary there was open circuit 3.40 on the

other isolated '6.3' and 161.4V end-to-end on the '150-0-150' and incidently

116.4 on the primary.

Then loading with different resistors

100K, 161.4 drops to 159.1

5.8K on 161.4 drops to 55.8, 3.43 input drops to 1.64

swapping to 5.8K on 3.4 , no change

1K on 161.4 to 12.1 and 3.43 to 0.771

swap to 1K on 3.4 , drops to 3.39

270 ohm , 161.4 to 3.34V

270 on 3.40, drops to 3.37

56 ohm on 161.4 to .704 and 3.43 to .54V

56 on 3.4 , drops to 3.28 and 3.43 to 3.42

8.2 ohm on 3.4 , drops to 2.55 and 3.43 drops to 2.99V

A bit more generalised.

Noting that for one secondary for this test transformer was rating 300V,

25mA then V/I of 12K and the 6.3V, 0.6 secondary of 10.5 ohm.

Doing as before powering a 6.3V secondary to 3.43V and '300V' was 161.4V

then loading it until the voltage ratio was 80 per cent that is 161.4V down

to 101.5V and 3.43 falling to 2.69V so 101.5/2.69 = .8 then that R is 12K.

So for similar transformer construction and high V, low I then find that

value of R for 80% then if V is known then current rating is V/R.

Doing the same for the low V,high I one then for R=10.5 ohm then

corresponding ratio drops from 1:1 ie ==3.43:3.4 down to 3.03/3.43 is 88%

for high current , low voltage.

So for similar transformer construction and high I, low V then find that

value of R for 88% then if V is known then current rating is V/R.

Other clues would be the gauge of the wires if they can be seen and the

overall size and weight giving an idea of the overall power rating.

Resistance checks would show which are more likely high V or high I.

Second test with a more basic Albion make, .8Kg, 20W open

construction 245V primary, 2 secondaries 17V,1A and 6.3V,.6A.

Again putting current into the lowest secondary giving 5.59V on

'6.3' and 14.52 on '17' (185.8V on 'primary')

6.3/.6 wire was 24thou diameter and 17V,1A wire was 27 thou diameter.

17/1 = 17 ohm. This time loading the 17V secondary with

17 ohm meant the ratio had dropped 69 per cent (15.52/5.59 to 2.778/1.544 )

Usually you would get some idea of one rectified V from max or min, by

capacitor ratings or a regulator voltage etc.

Valve radios would have one secondary connected to the

heaters so usually 6.3V. A vacuum fluorescent

display is likely to have a feed in the range only 2 to 5V

Toroidal transformer 2x 120V to 2x 15V,2A, .75Kg and 2A wires

33 thou diameter. Characteristic R = 15/2 = 7.5 ohm.

Critical ratio in this case was 82 per cent with 7.5 ohm.

15.27 input on '15' giving 15.26 on the othe rand 108V on

one of the primaries.

With 7.5 ohm 15.27 i/p drops to 3.91 and 15.26 drops to 3.19.

For a large toroidal 500W 2x 35V, 7.1A , weight 4.8 kg

Secondary wires consist of 2 paralled 56 thou diameter wires

per secondary.

Characteristic R= 35/7.1 = 5 ohm.

With 15.16 on one 'secondary' 15.1 on the other and 49.8V on

a 'primary'

15.16 dropped to 2.59V and 15.1 dropped to 2.46

so characteristic ratio is 95 percent for this transformer.