Determining primary transfomer current with only partial secondary loading

[iwire]

See above.

Thanks, I did not see it.

 

[gar]

111128-0844 EST

T.M.Haja Sahib:

Suppose we could have a core with no hysteresis or eddy current losses, but the core still goes into saturation, then the current is non-siusoidial. No hysteresis does not imply no saturation. But there is no such practical material.

If we take a practical transformer and do not run it into saturation, and could ignore the core losses, then the magnetizing current would be low enough to be of no concern.


Consider my sample transformer with 175 VA rating. At 120 V this is 1.46 A. At 120 V no load input current is 0.16 A RMS. Using your equation and an assumed input current from a 50% load of 0.73 A the total current would be 0.747 or 2.3 % higher. Probably of no practical concern. But really some of this was a result of losses, not just magnetization current. In other words a moderate amount of the 0.16 A current is from losses and not magnetizing current.

If I do a reverse calculation using your equation and assume a 1% effect from the core on the current, then 0.73 + 1% is 0.7373. This means the no load current needs to be about 0.104 A. This I get at about 110 V excitation. But again it includes more than magnetizing current.

If I am allowed to eliminate losses, then I can build a transformer as larger as I choose, and therefore make the magnetizing current as small as desired, or make the permeability as high as i desire (Supermalloy has a max value of about 1,050,000). I do not see any value in using your equation, and I think it is more valuable for someone to get a better understanding of how practical transformers work than plugging values into equations without any understanding of the assumptions behind the equation.

A perfectly accurate calculator does not mean that the result of a calculation on the calculator is an accurate or meaningful calculatuion.

.

 

[T.M.Haja Sahib]

111128-0844 EST

T.M.Haja Sahib:

Suppose we could have a core with no hysteresis or eddy current losses, but the core still goes into saturation, then the current is non-siusoidial. No hysteresis does not imply no saturation. But there is no such practical material.

If we take a practical transformer and do not run it into saturation, and could ignore the core losses, then the magnetizing current would be low enough to be of no concern.


Consider my sample transformer with 175 VA rating. At 120 V this is 1.46 A. At 120 V no load input current is 0.16 A RMS. Using your equation and an assumed input current from a 50% load of 0.73 A the total current would be 0.747 or 2.3 % higher. Probably of no practical concern. But really some of this was a result of losses, not just magnetization current. In other words a moderate amount of the 0.16 A current is from losses and not magnetizing current.

If I do a reverse calculation using your equation and assume a 1% effect from the core on the current, then 0.73 + 1% is 0.7373. This means the no load current needs to be about 0.104 A. This I get at about 110 V excitation. But again it includes more than magnetizing current.

If I am allowed to eliminate losses, then I can build a transformer as larger as I choose, and therefore make the magnetizing current as small as desired, or make the permeability as high as i desire (Supermalloy has a max value of about 1,050,000). I do not see any value in using your equation, and I think it is more valuable for someone to get a better understanding of how practical transformers work than plugging values into equations without any understanding of the assumptions behind the equation.

A perfectly accurate calculator does not mean that the result of a calculation on the calculator is an accurate or meaningful calculatuion.

.

No.The no-load loss i.e core loss component is less than 10 % of no-load current.The rest is magnetizing current.

 

[gar]

111128-1213 EST

If the inductance of the unloaded primary is infinite, then the magnetizing current is zero.

.

 

[T.M.Haja Sahib]

111128-1213 EST

If the inductance of the unloaded primary is infinite, then the magnetizing current is zero.

.

What is the need for assuming infinite inductance?

 

[gar]

111128-2326 EST

To show you that the magnetizing current is a function of the inductance. Not full load current.

Then when you have saturation during a portion of the cycle there is a nonlinear inductance.

.

 

[T.M.Haja Sahib]

111128-2326 EST

To show you that the magnetizing current is a function of the inductance. Not full load current.

Then when you have saturation during a portion of the cycle there is a nonlinear inductance.

.

Thanks,gar.Your replies excite me to go deeper into the subject.Well,the magnetizing current is a function of inductance.But it can also be considered a function of full load current phenomenologically.
We can operate the transformer below the knee point of B-H curve i.e below saturation without any distortion in current at steady state.

 

[gar]

111129-0808 EST

T.M.Haja Sahib:

I will suggest a book, but probably not available to you.

"Electric and Magnetic Fields", by Steven S. Attwood, 1932, 1941, 1949, John Wiley & Sons.

See the section on "The Ferromagnetic Field", p 325 ....

Normal power transformers are designed with magnetic material with a relatively soft saturation characteristic, and at nominal input voltage are driven slightly into saturation producing a non-sinusoidal waveform for the magnetizing current as shown in the waveform to which I previously referenced you. The design has evolved over many years to achieve an economic balance of material, and operational cost.

.

 

[T.M.Haja Sahib]

111129-0808 EST

T.M.Haja Sahib:

I will suggest a book, but probably not available to you.

"Electric and Magnetic Fields", by Steven S. Attwood, 1932, 1941, 1949, John Wiley & Sons.

See the section on "The Ferromagnetic Field", p 325 ....

Normal power transformers are designed with magnetic material with a relatively soft saturation characteristic, and at nominal input voltage are driven slightly into saturation producing a non-sinusoidal waveform for the magnetizing current as shown in the waveform to which I previously referenced you. The design has evolved over many years to achieve an economic balance of material, and operational cost.

It is a matter of economy to run the transformer that way.But for our purpose to have practically purely sinusoidal transformer primary current by operating it below the knee point of B-H curve i.e at its linear portion,do you have any theoretical objections?

 

[gar]

111129-2229 EST

T.M.Haja Sahib:

In the first post a real transformer was being described. This would most certainly be designed to partially saturate.

In the original post Pitt wanted to ignore transformer losses. If you eliminate those losses and further assume no saturation by using some lower level of excitation, then the magnetizing current can probably be totally ignored.

I did not see the value of bringing in an equation that probably really did not apply. In the non-saturation condition you could probably measure the primary inductance with a bridge circuit and separate out the losses. Then from this inductance calculate a current, but this would not really be the magnetizing current because the saturation portion was not included.

The original question might have been more realistic to just assume no I^2*R losses, and let the magnetizing current measurement include the core losses. But still the sq-root of the sum of the squares does not apply.

.

 

[T.M.Haja Sahib]

111129-2229 EST

T.M.Haja Sahib:

In the first post a real transformer was being described. This would most certainly be designed to partially saturate.

In the original post Pitt wanted to ignore transformer losses. If you eliminate those losses and further assume no saturation by using some lower level of excitation, then the magnetizing current can probably be totally ignored.

I did not see the value of bringing in an equation that probably really did not apply. In the non-saturation condition you could probably measure the primary inductance with a bridge circuit and separate out the losses. Then from this inductance calculate a current, but this would not really be the magnetizing current because the saturation portion was not included.

The original question might have been more realistic to just assume no I^2*R losses, and let the magnetizing current measurement include the core losses. But still the sq-root of the sum of the squares does not apply.

.

The magnetizing current per se does not practically have harmonics at steady state operation of the transformer.It is 'contaminated' by the harmonics from the core.Kindly see that.So the components of no-load current i.e core loss component and magnetizing component can be treated as sinusoidal quantities as may be seen from text books on transformers and hence the sq-root of the sum of the squares does apply.

 

[Besoeker]

111129-0808 EST

T.M.Haja Sahib:

I will suggest a book, but probably not available to you.

"Electric and Magnetic Fields", by Steven S. Attwood, 1932, 1941, 1949, John Wiley & Sons.

See the section on "The Ferromagnetic Field", p 325 ....

Normal power transformers are designed with magnetic material with a relatively soft saturation characteristic, and at nominal input voltage are driven slightly into saturation producing a non-sinusoidal waveform for the magnetizing current as shown in the waveform to which I previously referenced you. The design has evolved over many years to achieve an economic balance of material, and operational cost.

Another book that's directly related to the subject is the J&P Transformer Book. The one that sits on my shelf is the eleventh edition by A.C. Franklin and D.P. Franklin.
I'm sure there are more up to date versions.

 

[gar]

111130-0818 EST

T.M.Haja Sahib:

We have different definitions and I see no reason to continue the discussion.

.