Blown primary, hopw to determine unknown secondary voltages of a mains transformer

[N Cook]

Folowing on from thread lower down. I just tried this to see what sort of
results you get simulating such a multi-secondary transformer using a known
good one but not using the primary.
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 was 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 3.4 drops to 2.99V


The size of the transformer gives an idea of the overal sum of Volt x amps
as well as repeating the above with load resistors on each secondary should
give an idea of voltages and currents per secondary, without powering the
primary.
Anyone care to make an empirical formula from the above loading data and
give an idea of its applicability to the general case ?

 

[N Cook]

Folowing on from thread lower down. I just tried this to see what sort of
results you get simulating such a multi-secondary transformer using a known
good one but not using the primary.
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 was 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 3.4 drops to 2.99V


The size of the transformer gives an idea of the overal sum of Volt x amps
as well as repeating the above with load resistors on each secondary should
give an idea of voltages and currents per secondary, without powering the
primary.
Anyone care to make an empirical formula from the above loading data and
give an idea of its applicability to the general case ?

corrections
Blown primary, how to determine unknown secondary voltages of a mains
transformer
for
8.2 ohm on 3.4 , drops to 2.55 and 3.4 drops to 2.99V
read
8.2 ohm on 3.4 , drops to 2.55 and 3.43 drops to 2.99V

 

[N Cook]

corrections
Blown primary, how to determine unknown secondary voltages of a mains
transformer
for
8.2 ohm on 3.4 , drops to 2.55 and 3.4 drops to 2.99V
read
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.
Usually you would get some idea of one rectified V from max or min, by
capactitor ratings or a regulator voltage etc.