Transformer testing help

Discussion in 'Electronic Design' started by YoScotyBoy, Dec 15, 2003.

1. YoScotyBoyGuest

Over some time I have amassed a very large and heavy box of transformers that
I've used in projects and then pulled when the project was done. A few are well
maked (manuf./model # etc) others are more generic. These I can't pull up
specs for them (volts/uamps ect). I can't remember how to test a transformer
to find out the output current. The best I remember was I had to use a couple
of resistors, I measured volts/amps and then had a formula to do.

Does anyone know the way for me to test these. I promise I will mark them with
their specs.

I know, it's proabaly an easy do, really, I can't remember how!

2. Robert BaerGuest

Assume you do not know which winding is the primary.
A dirty trick is to use a 60 watt light bulb in series with the
winding (and the power line), and measure voltages to derive voltage
ratios (and perhaps find primariy/primaries.
Next, a very conservative way to get power rating (for E/I
laminations, 60Hz), is to determine the cross sectional area of the
inner or middle "E" leg insquare inches.
Square the area and divide by 0.084 to get the power rating; power
rating = (Design#)*(Design#).
From calculated Design#: Turns/Volt = 24/Design#; 110V primary turns =
2600/Design#.
I use wire cross-sectional area incircular mils to determine "no loss"
current rating; i never ran out of space when re-winding (power)
transformers using these formulas.
Hope this helps.

3. John WoodgateGuest

I read in sci.electronics.design that YoScotyBoy <>
I don't think there is an accurate way of doing what you want just with
electrical measurements. The current rating of a transformer basically
depends on how hot it gets (and that depend on how it is installed in
the product as well as what the designer intended). You can get a rough
idea by comparing the size and weight of an unknown transformer with
those of a known one. Equal size and weight *normally* indicates equal
VA ratings - secondary voltage x secondary current.

Another 'rule of thumb' way is to measure the open circuit output
voltage and then put various resistors across the secondary winding
until the voltage drops by a certain amount; 10% for 'small'
transformers, falling to 5% for 'large' ones. A 'small' transformer has
no dimension exceeding 3 ins/75 mm. The reason for the difference is
that large transformers have proportionally less surface area for
cooling than small ones.

Yet another way, which is a bit more scientific, is to measure the open-
circuit output voltage and the resistances of the primary and secondary
windings. Then you 'refer' the primary winding resistance to the
secondary by dividing by the square of the turns ratio, which you get
from the open-circuit voltage. For example, with 120 V in and 12 V out,
you have a turns ratio of 10. If the primary resistance is 60 ohms,
referred to the secondary it is 60/(10^2) = 0.6 ohm. This *should
normally* be similar to the resistance of the secondary winding itself,
so that the total effective resistance is 1.2 ohms. This resistance
drops 10% of 12 V with 1 A secondary current, so the transformer is
probably rated at 10.8 V/1 A at full load.

It gets complicated to do this if the transformer has several
secondaries of different voltages.

4. Robert BaerGuest

And we get your precision scientific method *after* i gave useable
equations that have worked since the 1940s?

5. John WoodgateGuest

I read in sci.electronics.design that Robert Baer
You may have seen it after you wrote your article, but I wrote it before

QUOTE

Next, a very conservative way to get power rating (for E/I
laminations, 60Hz), is to determine the cross sectional area of the
inner or middle "E" leg insquare inches.
Square the area and divide by 0.084 to get the power rating; power
rating = (Design#)*(Design#).
From calculated Design#: Turns/Volt = 24/Design#; 110V primary turns =
2600/Design#.
I use wire cross-sectional area incircular mils to determine "no loss"
current rating; i never ran out of space when re-winding (power)
transformers using these formulas.

UNQUOTE

How do you determine the area of the middle leg (if there is one - think
TU lams or toroid) of a finished transformer?

Using yoru first step:

Square the area and divide by 0.084 to get the power rating;

OK, I have a 1 square inch leg: that gives me 1^2/0.084 = 11.9 W. I
think a 1 square inch leg is good for at least 30 W (using 1940s iron
data) or 50 W these days, especially at 60 Hz. None of my transformers
burn up, in spite of the 5:1 difference in our figures.

You would need to explain what this is supposed to mean:

power rating = (Design#)*(Design#).

This is a special mathematical notation you have invented?

seems futile to try to work out where those empirical numbers came from.
I suspect 10 kGs (1 T) and 2000 A/in^2 (or even less - the British
Admiralty used 1000 A/in^2). - good 1940s values but 60 years out of
date.

You can only determine current ratings from wire size if you can
actually see and measure the wire, and that the wire you can see is the
actual wire used for the winding and not a thicker one used to give a