How to calculate the power rating of a transformer

This short video details how to calculate the Power rating of a mains frequency power transformer by using a ruler and some basic math.

[@perm.jensen7722]

How fortunate I stumpled on your channel. Thanks lot!

[@fgbhrl4907]

Interesting! For the mm version, it seems pretty straightforward to measure L*W, divide by 116, then square that answer. Seems quicker and just as easy to remember

[@Joftec]

Sure, it’s ok for estimating.

[@jumboegg5845]

Yep, (mm²/116)² =VA or (inches² x 5.58)² =VA. Six of one, or half a dozen of the other, as my father used to say.

[@jonathanperry1583]

I agree it is close,but the actual derivative is 116.139323 in mm or 5.58 in inches. For estimates 116 is fine, for designs it is much better to be 100% accurate. The difference is quite small, but can be important when physically fitting the secondary ( usually, as the wire is thicker) winding onto the bobbin. The actual area available for the copper as apposed to how much copper can be wound in that area, can be and is different. During the design process, this is one of many factors which have already been taken into account when the finished product is estimated. Like everything else copper, laminations and time are expensive and transformers are for sure designed to be as cheap as possible, this means that they will perform at 100% of their design capacity with an agreed temperature rise in an agreed cabinet size/airflow. The copper and lamination quantity increase with size, therefore smaller is cheaper, but smaller is also hotter electrically and cooling efficiency so there is an exponent there too. Different manufacturers in different regions may also have different Bobbin/lamination sizes available,in the UK and both metric and inch sizes were freely available when I was doing this , this can make subtle differences in the area available for copper. Different regions have different standards for the insulation thickness on the copper, and even differing Amps per square mm for the copper itself, which is nuts, but true. So in summary, for estimates, which is the purpose of the video, I agree 116 for mm is acceptable. For designs 116.129323 is better, and for simplicity 5.58 is the best, for me anyway.

[@jumboegg5845]

@@jonathanperry1583 That doesn't make sense to me, using 6 decimal places in mm, and only 2 decimal places in inches. 0.1mm equals approx 0.004 inches. if you wanted to be that precise, inches would need to have more decimal places. But the calculation was being used as a ballpark estimation of VA anyway, 0.1 mm makes little difference in the calculation. Then you would down rate the estimate by 10-20% or so, don't want to push the transformer to 100% of its estimated maximum VA.

[@chongli297]

A nice shortcut would be to multiply the two dimensions in mm together directly, then divide this quantity by 115.62, then square the result. This will give you the same answer but with fewer steps! 115.62 is just (25.4)^2 / 5.58

[@Joftec]

The actual metric value for the equation is 116.129323 as shown in the video . In very basic terms because the core you measure is not actually the core

[@Ozzy3333333]

Good stuff, thanks for sharing. How does the third dimension of the cord affect power? For example, if you transformer had the same x and y but the transformer was only half as tall (low profile), how do you factor that in?

[@Joftec]

Hi, interesting question. As I touched on in the video , normally when designing a transformer the turns per volt and current required are calculated. The cross sectional area of copper is then calculated, and a suitable bobbin is selected based on that requirement.The actual math behind the calculation is relatively complicated, and the method shown is a simplified shortcut. Bobbin and lamination manufacturers follow the rules determined by the math allowing the shortcut shown to become true. Therefore in effect the only measurements needed are the ones shown and the overall width is irrelevant for the shortcut. As far as I am aware the overall width of the lamination is determined by the width of the central section of the E. I believe that roughly speaking the area of the outer “arms “ of the E are close to the inner section of the E. Therefore the overall width is roughly the centre plus 2x the arms plus 2x the gap for the winding. The gap for the winding is based on the area of the gap and its relationship to the current handling capability of copper wire for a given area. The really interesting part is that that value can and is adjusted depending on many factors including desired operating temperature, enclosure type, and cooling, for example . Hope that helps.

[@Ozzy3333333]

@@Joftec Wow, thanks, yes that helps!

[@peterdkay]

Would it not be easier to weigh transformer and multiply that by a constant to get power?

[@Joftec]

I think it is far easier and entirely more accurate to measure . I have seen toroidal transformers gestimated by weight, as it is practically impossible to measure the core, but not EI transformers where core measurements are relatively easy.

[@timc3600]

Is there an equivalent calculation for toroidal transformers ?

[@Joftec]

Hi, yes there is . However this is far more problematic for several reasons, a couple being , 1. It is almost impossible to measure the cross sectional area of the core of a wound toroid as it is completely covered by the winding(s),insulation etc. 2. There tends to be a wider range of core materials, and thickness of coatings on the cores, which can mean large differences in core efficiency. I therefore do not want to endorse reverse calculation on a toroidal transformer. Hope that helps

[@timc3600]

@@Joftec Thanks and I understand the logic. The reason I asked is that most of the transformers in my junk/spares/gold box are toroidal

[@dl5244]

where does the 5.58x "constant factor" come from?

[@Joftec]

It is basically derived from the formula shown near the beginning of the video.

[@dl5244]

@@Joftec ok. at 3:18 I see you assuming 50Hz and 2.60 turns/volt. But then at 15:43 there is a logical jump that eludes me

[@Joftec]

The actual design process is not covered in this video, the purpose of the video is to estimate the design power dissipation of an already designed transformer. The example at the start is an example of how the process works . For example when you measure the core size as shown in the video, you are not actually measuring the core because for the laminations to work they are individually insulated, this insulation therefore occupies space in the core and reduces the magnetic material in the core. These details are taken into account within the math . Hope that helps