Determining Ratings of used surplus transformers

I always have my eye out for surplus transformers. However usually they are unmarked andits the odd find that actaullay has the voltages labeled.
Here is a description of how I go about working out their ratings. This all assumes that the transformers have been designed properly in the first place (eg there is enough magnetising inductance in the primary for the voltage it was designed for. This may not be the case for some transformers)

Determining Wattage from core size.

This is the easy first approximation. From "Reference data for radio engineers 2nd Ed (1946)" the cross section of the core should be :

core area(sq in) = sqrt(60/f)*sqrt(wattage)/5.58

or to rearrange and metrify:
wattage = A^2 / 16042

Where A^2 = The area in square mm, squared.

Here's a chart plotting this for 50hZ and 60Hz . The chart in the ARRL handbooks in years gone by was for 60Hz. 60hz Obviously saves some iron!
Transformer wattage vs core area

Determining the core size.

The core is the actual laminations that pass through the windings. You can measure this by the height of the lamination stack by the width of the 'tounge' passing through the core. The height is easy to measure, but the width of the tounge may be difficult in some situations. However it occured to me one day that there is nearly always a constant reltionship between the width of the tounge and the width of the laminations. If you examine the plan view one set of the 'E' and 'I' laminations, you will note that the tounge is twice as wide as the side pieces. This becuase the magnetic circuit through the core via the tounge divides into two and passes around each side of the core through the ends of the 'E'. The aim is to keep the cross sectional area of the core, as far as the meagentic circuit is conecrened, as constant size. So the width of a lamination will be 4 times the width of the side pieces of the 'E' plus 2 times the gap in the 'E'. No comes the interesting part.
The lamintations appear to be stamped out of sheet steel in a pattern like this.:

This results in not wasteage of steel. However this also forces the gaps in the 'E' to be the same width as the 'I' . But, the I is part od the magnetic path and so must be the same width as the sides of the 'E'. This means the width of the laminations is now 4 times the width of the side pieces, plus 2 times the width of the I. The I and the side of the E are the same width so the total width is 6 times the sides of the 'E'. Now the tounge is 2 times the width of the sides of the 'E', so the tounge is 2/6 = 1/3 of the width of the laminations. Thus the size of the core is simply the hight of the core times 1/3 of the width of the laminations.

Voltage

The voltage of the windings can be determined by applying a low AC voltage to one of the windings and measuring the voltage appearing on the other windings. Be careful with this, as applying 6.3VAC to a 12V winding of a trnsformer will result in around 120V floating around the primary.
The voltage ratios between the windings is constant, so onnce you have measured that, you can calculate the actual voltages with 240V applied to the primary.
For example, if you apply 6V to what looks like the secondary, and you measure 60V on what you think is the primary, the ratio primary:secondary is 10:1 . With 240V on the primary, the secondary will voltage will be 1/10th which is 24V.

current

If the transformer has only one secondary winding, then once the Wattage of the transformer has been determined from the core size, the maximum secondary current is simply equal to the Wattage divided by the voltage.
If there are multimple windings it may be more complex. If the windings are all the same gauge, then the current rating for each winding will be the wattage divided by sum of the secondary voltages. If the gauges are different, its a bit more difficult.
"Reference data for radio engineers 2nd Ed (1946)" quotes a current rating for windings in transformers of 1000Amps / Sq Inch. However The Radiotron Designers handbook 4th Ed 1953 quotes from 2830 A/Sqinch to 1270 Amps/Sq inch , with the higher value for small transformers and the lower for larger.

Copper wire current carrying capacity at 1000A per sq inch

B&S Dia(mm) Amps(1000A /sq inch)Amps(1500A /sq inch)Amps(2000A /sq inch)
141.673.4 6.8
151.492.7 5.4
161.332.1 4.2
171.191.61 3.22
181.061.28 2.56
190.941.01 2.02
200.840.80 1.68
210.760.64 1.52
220.670.50 1.34

Measurement examples of 'known' transformers

These first ones are 'hobbyist' transformers and experience is that they are a bit wimpy as far as the stated ratings go.

Stadium TF2155A 15V 1A Nominally 15VA
Core Area=361 mm^2 => 8.123 VA
Wire Size = 06mm ? => 2281 A Sq Inch

DSE M6680
25V -0 25V 1.2 A ( 50V @ 1.2A) 60VA
810 mm^2 => 41 VA
Wire Size = 0.6mm ? => 2737 Amps /sq inch

M-0144 56V @ 2A = 112VA
Core area = 1024 mm^2 => 65 VA
Wire size 0.75mm =>2920 Amps/Sq inch
This is the standard transformer for 100Watt audio Amps (eg ETI 480), and I suspect that the small size of the core works OK for audio amps where the full rated current isn't drawn too often

This next one is a BIG commercial transformer designed to run 24/7 for DECADES without powerdown.

Big boy
Core printed with 650 Watts, Total sec = 64VAC , presume 10.1A
Core Area = 4560mm^2 => 1296 VA
Wire Size 2.0 mm => 2053 Amps/Sq inch (for 10A)

Ian - VK3KRI
Revision Information $Id$