# How to measure the maximun of an old and big transformer without open it?

Discussion in 'General Electronics Chat' started by ldrueda, Jun 3, 2016.

1. ### ldrueda Thread Starter New Member

Jun 3, 2016
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I have a very old tarnsformer 220-110V with a primary and secondary coils. It looks of the same size of another new transformer that support 13A . I would like to measure the max current that the old one can support. Can you suggest me what can I do?. I was thinking to connect to resistantce that support high currents until see that the output current start decresing. What do you think? Could you suggest me something better?

regards

dqniel

Jul 18, 2013
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The Kva decider is usually the core size and what current the wire gauge supports on the secondary.
Max.

3. ### dl324 AAC Fanatic!

Mar 30, 2015
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Welcome to AAC!
That's the typical way of determining max current for an unknown transformer. The rule of thumb I've seen is to increase load current until open circuit voltage drops 10%.
EDIT: changed transistor to transformer; can't even blame a spell checker...

Rather than trying a lot of different power resistors, I'd opt for one resistor, or small number of different values, with a regulator to vary the voltage drop across the resistor.

Last edited: Jun 3, 2016

Jul 18, 2013
18,221
5,592
There is no definitive numbers for transformer design, per-se, The voltage will drop with load as is normal, how do you know what is acceptable?
I usually match to one of the same physical characteristics/size where in all probability the manuf. has done some definitive tests.
Max.

5. ### Lestraveled Well-Known Member

May 19, 2014
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I fully agree. You will be guessing at the current rating based on data you get from the tests and measurements you make.
Another way of looking at the transformer is temperature. Measure the change in temperature at different loads. If it too hot to touch, it is over loaded.

Good luck and be safe.

6. ### andre_teprom Member

Jan 17, 2016
31
9
For Silicon steel laminated cores you can use a quite simple formula containing just the "Power" and "Cross sectional area" of the central leg ( which implicitly carries information about its weight ). You did not mention anything about the mains frequency, but at least for 60Hz it is used the following equation:

7. ### Lestraveled Well-Known Member

May 19, 2014
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You formula is not visible.

8. ### #12 Expert

Nov 30, 2010
18,076
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Evaluating unknown transformers is very difficult. You can vaguely relate its weight to its power capability.
For instance, if you have a 3 pound transformer that is rated for 50 watts, a mystery transformer that looks like it has the same kind of core material and weighs 3 pounds will probably transfer about 50 watts.
50 watts/output voltage = current.

If your transformer has more than one output winding this evaluation job becomes almost impossible.
@andre_teprom
ps, what place does "BR" represent?
Britain? British Rhodesia? Belgian Republic?

Last edited: Jun 4, 2016
9. ### Ramussons Well-Known Member

May 3, 2013
776
140
Thumb Rule: 1 sq inch core area = 30 VA
2 sq inch = 120 VA

VA is prportional to Square of the Area

#12 likes this.

Oct 9, 2007
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11. ### #12 Expert

Nov 30, 2010
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How on Earth did you figure out I was Bychon?
It's true, but how did you figure it out?

Next question: How did I establish two login identities from the same computer and the same IP address without the Mods objecting? (I don't expect you to answer that.)

Last edited: Jun 4, 2016
12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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It may have been on another forum that you explained that Bychon was your dog.

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13. ### andre_teprom Member

Jan 17, 2016
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@Lestraveled
Trying to upload the picture again, but now in another format.
By the way, that's the rule of thumbs for calculating the standard laminated 60Hz transformer's core size in function of the power required:

@#12
Profile now filled.

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14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The power handling capability of a transformer is not determined just by the area of the core. It's determined by the area of the core multiplied by the area of the winding window. This is called the area product WaAc; see figure 5.1 here: http://coefs.uncc.edu/mnoras/files/2013/03/Transformer-and-Inductor-Design-Handbook_Chapter_5.pdf

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15. ### #12 Expert

Nov 30, 2010
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You have been here longer than I so I must concede that you have noticed things I forgot about.
And, yes, I was on electrotech for a while.

16. ### andre_teprom Member

Jan 17, 2016
31
9

I Fully agree with that, however based on the initial description "I have a very old tarnsformer 220-110V" we could assume ( with a certain shot of guess ) that OP is referring to the standard laminated core for converting 110/220 volts and vice versa, when there was not so common to have switched equipment bi-volt. We can see these transformers winded either in E-E or E-I cores, and its reel has almost the same window factor. I also don’t like simplistic approaches, but in this particular case, we can confidingly use the above formula for 60Hz. For 50Hz, should change the constant factor but I’m unaware of what is value. If you're not yet confident, I invite you to confer with a transformer that you have easily accessible and perform the measurements to check how accurate it can be.

Oct 9, 2007
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Jan 17, 2016
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19. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The entire last half of the thread is relevant, but particularly post #26 where I say:

"I gave a reference in post #14 to an article describing how the WaAc product is the determinant of power handling capability.

The WaAc product is the product of two areas, so that it has units of length raised to the 4th power.

The weight of a transformer is proportional to the average density times the volume. The volume involves length raised to the 3rd power. This means that volume or weight can't approximate the power handling capability except over a limited range; the dimensionality isn't right."

Your rule of thumb involves units of length squared, so it can only approximate the power handling capability over a limited range of transformer sizes. Also, a formula involving units of length squared will only work properly for a certain bobbin squareness ratio.

This graph from post #36 of that thread shows that, for example, the approximation of 20 watts per pound crosses the line which fits all the transformer data at only one point. The twenty watts per pound approximation only works well at 1 pound.

I have measured many transformers. I gave the measurements for some of them in that thread, and plotted the data points in red along with all the various manufacturers data on the graph. It can be seen that an approximation that involves units of length cubed (the 20 watts per pound rule), rather than units of length raised to the fourth power, only work over a limited range.

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