You are here:

Electronics/basic transformer question


QUESTION: My understanding of how a transformer works: In a transformer, flux density B equals uNI (permeability, wire turns, current)and the flux through the core is BA (A = cross-sectional area of core). The voltage on the primary and secondary is
N x d(flux)/dt. If the number of turns is the same for primary and secondary, then the two AC voltages are equal. A load on the secondary causes a current through the secondary that creates a revervse flux that the primary must cancel in order to maintain its reverse EMF, so the current in the primary will equal the secondary current to create the same amount of flux. BUT, what if the cross-sectional area of the primary is double that of the secondary? Doesn't that mean the primary only has to drive half the amount of current to create the same amount of flux, and if so, doesn't that make the efficiency of the transformer is 200% (assuming no other losses)? Thanks for your time and expertise to address this question.

ANSWER: Wow! you like to make it complicated.

Here'a the reality of transformers.  The theory of flux density, induction, etc. are nice to explain the physics of it all, etc....  But it is not an ideal model.  

In reality as the current in the primary increases, so also does the current increase on the secondary side.  Eddy current losses, etc.  will all have their share of the effect for lost power but the biggest loss is generated in the i**2 power dissipation in the turns.  At small or "0" current in the primary the losses are wasted in some heating of the iron core - whatever type of iron is used.  As the current increases the i**2 dissipation in the coil begins to rise significantly and eventually it is the dominant or primary loss - which in turn dictates the efficiency of the transformer (where EFF = Pout/Pin X 100).

This help?

---------- FOLLOW-UP ----------

QUESTION: Thank you for your quick response, but I believe you have overlooked the crux of my question. If only the cross-sectional area of the primary is doubled and the number of turns remains the same, then the primary can generate twice the amount of flux for the same input current. Yes, the I**2 R losses would increase, but these losses would be very small compared to the increase in efficiency caused by the ratio of areas between the primary and the secondary. I was hoping you could help me understand where I was going wrong with my understanding of the physics.

ANSWER: Doubling the area only increases the magnetic strength but not the number of turns. I think your logic about the flux being a factor in efficiency is wrong;  the increased area will help reduce the losses - which effect the efficiency of the transformer.  It cannot be more than 100%.  The actual losses in real world metal core transformers are eddy currents, hysteresis and i**2 losses.  The eddy currents are set up inside the iron core material and dissipate heat due to the movement of current in the iron.  Hysteresis losses are caused by reversing power to get the current to go the other direction as the sine wave passes through zero and up the other side.

See this web page for a good review of transformer action.  Look at page 3-10:

Hope this helps.

---------- FOLLOW-UP ----------

QUESTION: Yes, I know the efficiency can't be greater than 100%, which means my understanding of the physics must be wrong. I was hoping you could clear up my misconception. All textbook calculations of efficiencies assume that the primary and secondary have the same cross-sectional area, so the A in the formulae just falls out. I still don't see what I am doing wrong.
The secondary creates a flux density B=unI multiplied by its C-S area equals the flux=unIA. The primary must cancel this reverse flux by driving a current. But since A(p) = 2A(s), then I(p) = 1/2 I(s).
Why is this wrong?
Thanks for your patience.

How could the cross-sectional area become different between the two windings?'s_law

Lenz law will also help you in understanding the physics of electromagnetic induction.

Maybe the last paragraph in this web page will help:

Hope this helps.

PS:  I don't find your calculating formula anywhere in the literature or text books that I looked at.  What is the source of your using your equations in the efficiency calculations?  


All Answers

Answers by Expert:

Ask Experts




Electronics questions about AC, DC and digital theory.


Graduate electrical engineer with over 40 years in electronic design, manufacturing, project organization and patent review. Experience in fields of industrial and consumer electronics (audio, video, acoustics, etc.)

IEEE (Institute of Electrical and Electronics Engineers); Senior Life member AES (Audio Engineering Society), Fellow Life member

BSEE University of North Dakota

©2016 All rights reserved.