Assessing an unknown transformer

studiot

Joined Nov 9, 2007
4,998
This link contains much of relevance to the thread.

http://www.repairfaq.org/sam/flytest.htm

You can construct a simple tester for shorted turns from a Colpits oscillator. Adjust the oscillator for stable operation and then connect the unknown transformer coil across it. A shorted turn stops or significantly reduces oscillation.
 

Thread Starter

someonesdad

Joined Jul 7, 2009
1,583
Electrician's idea of collating current manufacturing specs is an excellent one. This weekend I grabbed some data from the Triad and Stancor websites. I chose to first "specialize" in power transformers that have one primary winding and one center-tapped secondary. I entered the current and voltage rating of the secondary and the linear dimensions and mass of the transformers into a spreadsheet. These data were put into a tab-separated file and I plotted the VA rating as a function of mass, nominal surface area and nominal volume. I've attached the VA rating vs. mass graph; the other two are identical graphs (except for the axes) and don't give any new insight. The two lines are eyeballed and give an idea of the spread in data -- a regression curve through the data will get you within about a factor of 2 if you try to predict VA rating from mass alone. This will no doubt be handy in conjunction with other measured data on an unknown transformer.

It is a tiring task to enter all this data and check it, especially for my older eyes. I'd like to ask for some help -- if we can divvy up the work a bit, it eases the load. If you'd like to volunteer to enter some data, please post a message here and I will contact you via a personal message. I see a need for some representative data from other types of transformers with multiple windings. My intuition tells me their VA-mass relationship will probably be similar, but we need the data to be sure.

If you'd like to look over the manufacturer's sites, here are the links:

Stancor: http://www.stancor.com/jsp/products.jsp I've taken the graph's data from the Single Secondary document under Power Transformers.

Triad: http://www.triadmagnetics.com/home.htm and click on Products. I've taken the graph's data from the link labeled "Power Xfmr - Chassis Mount: 2.0VA to 400VA".

Also, if someone wants to gather similar data from other manufacturers, that would be great too.
 

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retched

Joined Dec 5, 2009
5,208
Can you get the output in a comma delimited file?

we could write a quick code to do the input automatically.

I guess it could also be done with a tab-delimited file...
 

KL7AJ

Joined Nov 4, 2008
2,225
One rule of thumb for POWER handling capacity, considering nothing else, is 10 watts per cubic inch of core metal.

Eric
 
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.

I've added the Triad data to the Stancor data, and produced some more plots, and I've changed the equation fitted to the data for a better fit at low power levels.

I've attached 3 images. The first shows all the data on a LogLog plot.

The second image shows all the data on a linear plot, with the 90% prediction intervals shown in blue.

The third image shows all the data plus a straight line in red which represents the 20 watt/pound approximation. It can be seen how that approximation is only good over a limited range of power levels.
 

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Thread Starter

someonesdad

Joined Jul 7, 2009
1,583
Electrician: Good stuff -- I like the simple equation:

\(P = 16 w^{1.22} - 1.25 \)

Personally, I wouldn't put too much stock in prediction limits, as the usual IID assumptions for such data are probably not satisfied here. Rather, the user just uses your formula and looks at the graph to get an idea of the spread in the data.

I think we should combine our data -- I've attached the info I typed into a spreadsheet as a CSV file. Would you mind attaching yours too?
 
Electrician: Good stuff -- I like the simple equation:

\(P = 16 w^{1.22} - 1.25 \)
My curve fit program prints all those digits; I just didn't bother changing them in the press of trying to make it all look pretty. :)

Personally, I wouldn't put too much stock in prediction limits, as the usual IID assumptions for such data are probably not satisfied here. Rather, the user just uses your formula and looks at the graph to get an idea of the spread in the data.
There are a number of problems with this data. For example in the Stancor data:

http://www.stancor.com/wrdstc/pdfs/Catalog_2006/Pg_002_4.pdf

Have a look at the two transformers, P-6134 and P-6190. They have identical power ratings, but one is larger and heavier than the other. This is due to the 5000 volt vs. 3000 volt insulation rating.

Also, there is a penalty in having a core that is substantially oversquare. I notice a few transformers where the stack height is nearly 2 times the center leg width. And, of course, there's the different weights of the mounting hardware; the larger transformers have beefier brackets (style NV).

I was thinking I might go back to the WaAc rating and see if I can get some data with less scatter.

Ultimately, though, I think the hobbyist who wants a better number for the power capability of a particular unknown transformer should start with a number from the graph, and then perform the temperature rise test I described earlier.

Another criterion Bychon mentioned is to take the load which reduces the unloaded output voltage by 10% as the rated load. I have a set of data I took a while ago on a number of transformers which includes this parameter. I'll get it together and post it.

I think we should combine our data -- I've attached the info I typed into a spreadsheet as a CSV file. Would you mind attaching yours too?
I don't see an attachment.

I have taken the data from Stancor and Triad and combined it. That's what is shown in my graphs. I've attached my data file.

I've also attached another graph showing the data points plus a two times larger and smaller limit curve. I've prettied it up and adopted your suggestion of w as the independent variable.

It's well to keep in mind that doubling the power transferred through a given transformer means doubling the output current, and this means 4 times the copper losses. The dissipation in the transformer goes up substantially faster than the transferred power ratio. A person would be well advised to choose a power rating for a given weight of transformer a little on the low side, and the perform a temperature rise test.
 

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Thread Starter

someonesdad

Joined Jul 7, 2009
1,583
Electrician:

One other question... I like the idea of determining an operating point from the temperature rise, as it tests things under real-world conditions and the user should be able to translate the information to their design.

However, I've read that there are at least two transformer insulation temperature ratings: 105 °C and 130 °C. Is there anyway for a user to distinguish the insulation type (assuming there are no markings to guide him)? If not, then the conservative approach would probably be to choose the lower number.

I've been bedeviled with yard duty tasks (busted sprinkler lines, blown tire on the lawn mower, etc.). I would like to get around to doing some experiments soon and generating some data that you and others could critique.
 
However, I've read that there are at least two transformer insulation temperature ratings: 105 °C and 130 °C. Is there anyway for a user to distinguish the insulation type (assuming there are no markings to guide him)? If not, then the conservative approach would probably be to choose the lower number.
See the thermal classes here:

http://www.stancor.com/pdfs/Catalog_2006/Pg_g.pdf

There's no good way to distinguish between class A and class B just by looking. Class H, which is what I've always designed to, can be distinguished by the polyimide bobbins (nylon bobbins, which are ok for class A, melt when used at 180 °C) and nomex insulation,

I suspect most of the transformers a hobbyist is likely to encounter are going to be class A (105 °C). I've measured the temperature rise of several of these typical transformers under rated load, and they've all had about a 40 °C rise above ambient, which is about what class A is usually designed for. Maximum ambient for these consumer grade transformers is usually taken to be 40 °C (the inside of a vacuum tube receiver case in summer), and the maximum allowable temperature rise as measured by the resistance change method is 45 °C. Comparing this allowable rise for class A with my measurements leads me to believe that standard commercial transformers we see from Mouser, Radio Shack, etc., are class A. You can get better ones, but you pay for them.

I will measure several of the ones I have around and include that info in my spread sheet to be posted later.
 
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Thread Starter

someonesdad

Joined Jul 7, 2009
1,583
Thanks; one more question. For a center-tapped secondary, I made the assumption for my graph that the user would take the secondary's rated current from both secondary circuits. On reflection, this was probably a wrong choice, as I would imagine the secondary current is defined by both the core flux levels and allowed temperature rise -- besides, if the transformer was capable of it, the marketing department would make sure people were aware of it.

Thus, please tell me if the following statement is true or false in your opinion:

If a center-tapped transformer is stated as having an output of X volts with a current of Y amps, then both of the circuits using the center tap can be used to get X/2 volts at a current of Y/2.
 
Thanks; one more question. For a center-tapped secondary, I made the assumption for my graph that the user would take the secondary's rated current from both secondary circuits. On reflection, this was probably a wrong choice, as I would imagine the secondary current is defined by both the core flux levels and allowed temperature rise -- besides, if the transformer was capable of it, the marketing department would make sure people were aware of it.

Thus, please tell me if the following statement is true or false in your opinion:

If a center-tapped transformer is stated as having an output of X volts with a current of Y amps, then both of the circuits using the center tap can be used to get X/2 volts at a current of Y/2.
Think of a center tapped transformer as having two secondary windings, but with one end of each connected together internally. Sometimes transformers (such as the P-8504) have two separate windings (primaries in this case) without having any internal connection. That way the user can connect one end of each (with due consideration of polarity) together and get a center tapped winding, or he can connect the two windings in parallel or in series (either parallel or series is what is intended for the P-8504).

Consider the P-8652 Stancor transformer. It's rated 10.0 C.T. @ 1.0. This means that you can draw 1 amp at 10 volts. What would be allowable would be to connect a 10Ω resistor across the full winding, ignoring the center tap; this would draw 1 amp. Or, you could could connect a 5Ω resistor from one end of the full winding to the center tap, and another 5Ω resistor from the other end of the full winding to the center tap. Each 5Ω resistor would be carrying 1 amp from half the full winding; this would be equivalent to the case of a 10Ω resistor across the full winding.

Is this distinction useful? Imagine you had a 24 volt C.T. transformer. You could connect a 12 volt light bulb from each end of the winding to the center tap, or you could connect the bulbs in series, and place the series combination across the full winding. In the first case, if one bulb burned out, the other would remain lit. In the second case, if one bulb burned out, they would both be unlit.

The reader should be aware that all these transformer ratings are for linear loads (resistive loads, typically). If the transformer is used in a rectifier circuit with a capacitor input filter, the ratings will be lower.

So, to be accurate, your statement should be:

"If a center-tapped transformer is stated as having an output of X volts with a current of Y amps, then both of the circuits using the center tap can be used to get X/2 volts at a current of Y."

but you can only do this into two separate circuits (connected at one point, the center tap), each drawing Y amps.

A person might think that if a transformer can put out X volts at Y amps that you should also be able to get a single output of X/2 volts at 2*Y amps, but the fact that the two halves of the center tapped winding are connected together internally (with the usual phasing) prevents you from connecting the outer ends of the full winding together (this would be a short across the full winding) to get a single winding of X/2 volts. If the two halves of the center tapped winding were not connected internally, then the user would have the flexibility to either connect the windings in parallel, or to make a center tapped winding.

I've gone into this long explanation for the benefit of beginners who might be reading this post.
 
I've made some temperature rise measurements on 3 small transformers, and here are the results.

The hobbyist will have to rely on the resistance change method of measurement. Professional transformer designers will verify a new design by embedding thermocouples in the windings when the prototypes are being made. It's very difficult to get a thermocouple into position after the transformer is built, as you can imagine! But, what I have had some success with is this: I take a bamboo chopstick and with a wood rasp, sharpen one end to a thin wedge shape. Then if the transformer wind isn't too tight, the wedge can be forced between layers in the build. (never use metal tools to press on magnet wire; the insulation will be damaged. Use only wooden tools, and exercise great care; don't push too hard). If a space can be opened up with the wedge, a thermocouple can be inserted; sometimes this can be done without damaging the transformer, sometimes not.

If a thermocouple can't be placed between layers of the windings, sometimes it can be inserted between the core center leg and the inside surface of the winding build. This position isn't measuring the hot spot temperature, which is usually in the middle of the windings, but it will typically be within 10 to 20 °C of the actual hot spot temperature.

Another property of a transformer which is helpful in determining the power capability is the regulation. This is the percentage drop in the output voltage from unloaded to loaded state. The change in output voltage is divided by the unloaded voltage (this is the british definition), with the primary voltage held constant. Ten percent regulation is typical for a class A transformer. If the regulation is greater than 10%, this may indicate that the transformer is built to a higher thermal class, or is overloaded.

I measured 3 transformers, and in two cases I was able to get a thermocouple into the actual windings; In the third I was able to place it between the windings and the center core leg. The thermocouple measurements provide a check on the resistance change method.

The room temperature for these measurements is 20 °C., and the formula from post #14 is used to calculate the temperature of the windings from the change in DC resistance.

The transformers were just sitting on the bench, cooled by convection. A muffin fan placed close to a transformer can easily increase the allowable dissipation by a factor of 2, which is an increase in power handling capability of 1.414.

The time constant for transformers of less than a few hundred watts capability is such that allowing 3 hours of heating time will get within 98% of the final temperature. After the transformer has been allowed to heat under load for 3 hours, disconnect the power and the load and measure the winding resistances quickly, within less than a minute.

Measuring the resistance of low voltage, high current windings accurately can be difficult, since the resistance will typically be under 1 ohm. The best way to do it is to pass a known DC current (I use a power supply in constant current mode, set to deliver 1.00 amps) of about 1 amp through the winding and measure the voltage drop across the winding terminals. If you have a high end DMM, such as a Fluke 187, with a "relative" measurement capability, you can zero out the resistance of the meter's test leads, and get a fairly good result without resorting to the "known" current method.

------------------------------------------------------------------------
The first transformer is a Calectro, rated for 25.2 VAC @ 2A; this is a rated power of 50 watts. This transformer weighs 2.074 pounds; the graph I posted in post #29 gives a power rating of 37.7 watts. This is the transformer where I was only able to get the thermocouple next to the center core leg. Notice that the thermocouple temperature is 10 to 15 degrees cooler than the temperature measured by the resistance change method.

The results of the load test with 115 VAC in, 2 amps out:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

  9.58  .742      11.8  .90      66.6°          80°  76°        7.4%
------------------------------------------------------------------------
The second transformer is a Stancor, rated for 12.6 VAC @ 3A; this is a rated power of 37.7 watts. This transformer weighs 1.66 pounds; the power capability from the graph is 28 watts. For this transformer I was able to slide the thermocouple into the winding. Notice the good agreement between the thermocouple measurement and the resistance change value.

The results of the load test with 115 VAC in, 3 amps out:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

 12.25  .310     14.55  .37      66.0°          67°  69°        8.5%
----------------------------------------------------------------------
The third transformer is a Calrad, rated for 24 VAC @ 1 A; this is a rated power of 24 watts. This transformer weighs .77 pounds; the power capability from the graph is 10.4 watts. I was able to slide the thermocouple into the winding for this transformer.

The results of the load test with 115 VAC in, 1 amp out:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

 67.12  3.18      88.1  4.11      103°          99°  94°       20.8%
The temperature rise for the first transformer is about 60 °C, which is somewhat higher than the usual 45 °C rule of thumb for class A transformers, but still not likely to overheat the transformer if used at typical room temperature. The second transformer, the Stancor, has about a 48 °C rise, right on the money for a class A transformer.

The third transformer has a temperature rise of about 80 °C, with an internal temperature of about 100 °C. This is right at the limit for a class A transformer, and I'm sure it's not a Class B insulation system! It's a Calrad transformer, and they don't spend extra money they don't have to.

This transformer could just barely be used at rated load if the ambient temperature is no higher than 20 °C.

The high value of regulation (20.8%) is a warning that the transformer is going to get very hot.

I reduced the load current to .707 amps (1 amp/SQRT(2)), a power output of 17 watts. This should reduce the copper loss by 1/2. The core loss (measured as 1.88 watts) remains unchanged. This reduced the internal temperature to 65 °C, as measured by the thermocouple, a temperature rise of 45 °C. This is a more reasonable rating for the transformer, and is still larger than the 10.4 watts predicted by the graph.

The graph predicts a rating of about 75% of what the manufacturer suggests for the first two transformers, and about 61% of the more reasonable rating of 17 watts for the third transformer.

A simple rule of thumb for the resistance change method is that the ratio of a winding's resistance when hot (measured after 3 hours on what is believed to be rated load) to its cold resistance should be in the range of 1.18 (45 °C rise) to 1.25 (65 °C rise). If it isn't in that range, then change the load and measure again. A couple of measurements should suffice to interpolate with reasonable accuracy the load which will give a desired temperature rise.

I conclude that for transformers whose weight is in this range, the graph will underrate a transformer by about 75%.

I will measure a couple of larger transformers and report the results later.
 
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I've measured 3 more transformer's thermal characteristics, with the results shown:

The first of the 3 is a Stancor P6378; this transformer is no longer in the Stancor catalog. It's a dual primary, dual secondary with a secondary rating of 24V@4A or 48V@2A. It weighs 4.2 pounds, and has an EI112 core with a 1 5/8 stack height.

I wired it for 24V@4A and connected a 4 amp load, allowing the transformer to heat for 3 hours and got these measurements:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

  4.05  .221      4.91  .268       N/A         74°   74°        5.7%
The temperature rise is 54 °C, 9 °C greater than the standard 45 °C for class A, but still well within a safe range, especially if used in a room temperature environment.

The second transformer is a unit with only a manufacturer's part number on it. It has an EI50 core with a .5 inch stack height. This transformer has flying leads, conforming to the standard color code for these units. The primary is the black leads and the secondary is the red leads, with a yellow colored center tap. Connecting the primary to 115 VAC, the secondary voltage is 36.2 VAC with no load. The transformer weighs .355 lb.

The weight of the transformer suggests that it should handle about 5 watts. I loaded it at .2 amps and got a primary temperature of 49.5 °C. I then increased the load to .25 amps which was accompanied by an output voltage of 25.5 VAC, and got the results shown:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

 215.5  16.42    251.0  19.07       N/A       61.9°  61.0°      29.5%
So, this transformer probably should be rated at 24 VAC @ .25 amps.

The third transformer is also an unknown with only a part number on it. It is also wound on an EI50 core with .5 inch stack height, and weighs .35 lb. This transformer also has flying leads, conforming to the standard color code for these units. The primary is the black leads and the secondary is the red leads, with a yellow colored center tap. Connecting the primary to 115 VAC, the secondary voltage is 31.6 VAC with no load.

This transformer is on the same core as the previous one, and so we would expect that it also should handle about 5 watts. I did a load test starting with a current of .2 amps. The calculated primary temperature was 58 °C. A quick calculation shows that a current of .22 amps would probably give a primary temperature of about 65 °C, which is what we want.

After 3 hours with a load of .22 amps accompanied by a loaded secondary voltage of 20.4 VAC, I got the following results:

Rich (BB code):
     Cold             Hot      Thermocouple    Calculated    Calculated
   DC resis        DC resis    temperature    temperature    regulation
  pri   sec       pri   sec                    pri   sec

 172.6  30.9    205.3  35.8       N/A       68.0°  60.0°      35%
This transformer should be rated 20 VAC @ .22 amps.

The second transformer can handle 6 watts, but the third can only handle 4.4 watts. Why the difference, since both are the same size, wound on the same core? There are a couple of reasons.

The second transformer has .54 watts of core loss, whereas the third has .98 watts core loss; that's essentially twice the core loss for the third. The manufacturer used lower quality steel laminations.

A properly designed transformer should apportion the copper losses so that about half occur in each of the primary and secondary. This means that the DC resistance of the two windings should be about the same, when the resistance of the secondary is reflected to the primary.

The way to calculate this is to multiply the resistance of the secondary by the square of the turns ratio; this gives the secondary resistance reflected to the primary side.

The second transformer has a turns ratio of 115/36.2; squaring we get 10.1. If we multiply the 16.42 ohm resistance of the secondary of the second transformer by 10.1 we get 165.7 ohms for the reflected resistance of the secondary. Compare that to the 215.5 ohm resistance of the primary and we see that they are about the same. This will cause the copper losses to be about half in the primary and half if the secondary, the condition for minimum copper losses.

For the third transformer, the reflected resistance of the secondary is (115/31.6)^2 * 30.9 = 409.2 ohms. Comparing to the 172.6 ohm resistance of the primary, we see that the difference is much greater than was the case for the second transformer.

The lower resistance of the third transformer primary, 172.6 ohms, compared to the 215.5 ohm resistance of the second transformer primsry, might have led us to think that the third transformer could handle more current. But, the much higher resistance of the third transformer secondary more than destroys the advantage of the lower primary resistance.

The third transformer uses inferior core material, and the primary and secondary copper losses are not properly apportioned; it can only handle 4.4 watts for a temperature rise of about 45 °C.

Notice that the second and third transformers both have large regulation. This is typical for small transformers, and this is why wall wart power supplies have such poor regulation (if they're not switchers).

Also, notice that the two Stancor transformers have nearly equal primary and secondary temperatures; this is because the Stancor engineers have properly apportioned the primary and secondary copper losses. I also notice that the core loss per pound of the Stancor transformers is less than the others; Stancor undoubtedly uses M6 laminations which are very good quality.

I've noticed that the curve obtained when fitted to the commercial transformer data from Stancor and Triad gives a result which would suggest that the transformers I measured should have been rated at about 75% of what I actually measured. This discrepancy is because the commercial transformers have been designed for a maximum ambient temperature of 40 °C, rather than the 20 °C ambient in which I made my measurements.

I took all the data points and multiplied the power ratings by 4/3 and replotted. This gives a curve for the ratings of the transformers as if they are only used in an ambient temperature of 20 °C.

If a hobbyist wants to know the power rating for a transformer sitting on the bench, convection cooled only (no fan), then use the curve attached to this post.

If the transformer will be enclosed in a box with no ventilation where the ambient might get a high as 40 °C, then derate the power capability to 75% of what the graph says.

I've attached 3 images.

The first shows the Stancor and Triad data, with the power ratings multiplied by 4/3, as green data points. I've shown my 6 measured transformer data points in red.

The second shows the same data as the first, but with a 20 watt/pound line also, so it can be seen where that approximation is useful.

The third image is with a white background to save ink if it's printed.

I've chosen a curve that doesn't bend down at the low power level as some of my previous curves did. I was looking at the Stancor ratings for lower power transformers and noted that a number of them are rated at a higher insulation test voltage than the standard 1500 volt rating of the larger transformers. I think that may account for the lower power rating of a number of the small transformers.
 

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someonesdad

Joined Jul 7, 2009
1,583
Excellent work! I just read through it once, but I resonate with the method -- measuring the steady state temperature increase seems to make the most sense to me (after getting in the ball park with the power vs. weight graph). I've got some small transformers I'd like to try this technique on, but I can't get to the windings, so inserting a thermocouple isn't an option (one of them is in a full metal can. But I can get to the laminations and a surface thermocouple or an IR measurement would probably be better than nothing.
 
The thermal conductivity of steel laminations is really poor. Measuring the outside temperature of the laminations isn't very helpful in my experience.

The measurement of winding resistance change is the method prescribed in Underwiter's Laboratories standards for determination of transformer winding temperature (if a buried thermocouple isn't present).
 
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