• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Transformer's primary winding power consumption

Status
Not open for further replies.

ericgibbs

Well-Known Member
Most Helpful Member
It's an internesting Topic,
If you have a question ' Why it happens? ' ; then this would be the answer...
http://www.electro-tech-online.com/threads/transformer.107849/
hi V,
I think most of know how a transformer functions, the problem starts when you want to design a transformer.

Unless the transformer you need to design is unusual its a simple matter to buy a ready made transformer, its cheaper and quicker.

If you want to modify an existing transformer it can be difficult to precisely calculate its performance, as you dont know most of the transformers materials and characteristics.

If you are designing a specialised transformer for a particular function you would purchase the materials you required after looking thru the materials datasheets and applying the standard formula's.
When you have completed and tested the transformer, the chance of it functioning to 100% of the design criteria is doubtful.

I would tend to agree with Bychon's approach when answering the OP's question.
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hello again,


I agree that for many people buying a transformer is the easiest solution. I buy transformers myself too. Somebody, somewhere however has to design them, and this is part of that process...



Lets start with a standard EI 1 1/2 core. The data for this core is attached as two drawings.
We'll also start with a square stack, which is going to be 1.5 inches high because the center leg is 1.5 inches wide.
The stacking factor we will keep at a perfect 1 for now to keep it simple.

Starting with the Faraday equation:
B=E*10^8/(28.65*F*A*N)
where
B is flux in Gauss
E is applied RMS sine voltage in volts,
F is frequency in Hertz,
A is area in square inches,
N is number of turns.

In this example:
E is 123, F is 60 Hertz, A is 1.5 inches squared (2.25 sq inches), and N is 212 so we get:
B=123*10^8/(28.65*60*2.25*212)

and after doing that calculation we end up with:
B=15000.67

which we'll round to 15000 Gauss.

Now looking at the data for the core, we see that for M6 material that curve intersects 15kG near the top of the graph (red circle) and following that vertical line down to the x axis we find it corresponds to 7 VA/Lb. Now looking at the other data for the core we find that a square stack weighs in at 5.35 pounds, so now we multiply that times 7 and we get:
VA=5.35*7=37.45
which is about 38 volt amperes.


So you see it's just a matter of doing one calculation and then looking up the final answer. The same could be done for the core loss using the other curve from one of my previous posts.
 

Attachments

Last edited:

ericgibbs

Well-Known Member
Most Helpful Member
hi Al,
As always your replies are detailed and informative, so do not think I am being negative regarding transformer design.

I have designed a number of commercial transformers in my time, so I would like to add a few words to your quote.
I agree that for many people buying a transformer is the easiest solution. I buy transformers myself too. Somebody, somewhere however has to design them, and this is part of that iterative process before the final design specification is met....
Regards
 
hi V,
I think most of know how a transformer functions, the problem starts when you want to design a transformer.

Unless the transformer you need to design is unusual its a simple matter to buy a ready made transformer, its cheaper and quicker.

If you want to modify an existing transformer it can be difficult to precisely calculate its performance, as you dont know most of the transformers materials and characteristics.

If you are designing a specialised transformer for a particular function you would purchase the materials you required after looking thru the materials datasheets and applying the standard formula's.
When you have completed and tested the transformer, the chance of it functioning to 100% of the design criteria is doubtful.

I would tend to agree with Bychon's approach when answering the OP's question.
Hi!, A good piece of Information
Near to accurate, but not accurate!
Reminds me Winston Churchill's Quote
"To improve is to change; to be perfect is to change often."
 

MrAl

Well-Known Member
Most Helpful Member
hi Al,
As always your replies are detailed and informative, so do not think I am being negative regarding transformer design.

I have designed a number of commercial transformers in my time, so I would like to add a few words to your quote.


Regards
Hi Eric,


Sure, you can add a few words if you like, or make your very own statements from scratch :)

Another use for this kind of design process is so that we can compare results from one type of material to another. For the data given in my previous post we can quickly see that M19 would make a construction that would take almost twice the volt amperes of M6 by simply comparing the intersections of both of those materials and 15kG. The dashed line (M19) at that point is about twice as high in VA than the solid line (M6) so we might consider using M6 instead of M19. It's been a while since i priced these kinds of laminations, but M6 is probably more expensive than M19 so we'd have to weigh the options. At least we would have a method to use to think about this that results in some real life usable comparative data. Without it we would have to purchase two cores and do some measurements on both cores with windings and all...quite a bit of work if you ask me compared to doing a quick calculation and looking at a graph.
If you dont agree, go ask professed magneticians Thomas and Skinner why they bother to publish this data.
 
Last edited:

The Electrician

Active Member
Now looking at the data for the core, we see that for M29 material that curve intersects 15kG near the top of the graph (red circle) and following that vertical line down to the x axis we find it corresponds to 7 VA/Lb.
You have confused the loss designation with the thickness (gauge) of the laminations.

Your red circle is on the curve for the M6 material with a thickness of 29 gauge (.014 inches), not M29 material.

M6 means .6 watts/pound loss at 15 kG, with the laminations cut so that the flux is half in the rolling direction, half at right angles to the rolling direction. This is what the M6 designation originally meant, but some suppliers are doing better than this today.

See this page:

Electronic Transformers - Core Materials

with a similar curve in figure 23 for M15 material. You can see that the loss at 15 kG is 1.5 watts per pound.

M19 material has 1.9 watts/pound; M26 material would be 2.6 watts/pound; M29 material would be 2.9 watts/pound, etc.

There are better grades than M6. Grades M2, M3, and M4 are available for a substantial cost increment. Typical good quality commercial transformers are usually built with M6 material.

See this document for a lot of good info:

http://www.electro-tech-online.com/custompdfs/2010/06/Oriented_Bulletin.pdf
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hello there,


Yes a few typos, which are now fixed.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading

 
Top