Why we use constant value of magnetic flux in transformers

Question

I have a confusion regarding the formulas of the transformer. I have seen in many places people using a constant value of the flux B in the transformer equation:

turns per volt=1/(1.44 x Bmax x A x f)

But as far as I know the value of B is equal to:

B=uI/l where u=permeability,I =current,and l= length of the coil.

This means the value of B depends on the current. As current increases the flux will increase accordingly, but then why do people consider this flux to be constant when constructing a transformer?? Is it some other value or what? Kindly help me clear up my confusion.

Secondly, suppose the B depends upon the current. Which current will we have to assume in the equation? Will it be the current on the primary side or the secondary side? The primary side is the main source of EMF and magnetization and I feel its current should be used but when current flows in the secondary coil as well will its current affect the B or flux?? Kindly help me with these two confusions.

Answer 1

Secondary load current \$\times\$ secondary turns is exactly cancelled by the load current flowing in the primary \$\times\$ primary turns hence, fluxes associated with the secondary load are cancelled.

What's left is the primary magnetization current. This current is due to the applied voltage and frequency and, the natural magnetization inductance of the primary. It has nothing to do with any load currents.

This is why we say that the flux in the transformer is constant irrespective of secondary load. Of course, we mean that the RMS value of the flux is constant. There are three currents in summary: -

• Primary magnetization current (not due to loads)
• Secondary load current
• Primary load current (N times smaller than secondary load current and opposite polarity)

Which current will we have to assume in the equation?

The primary magnetization current (which can only be measured under no load conditions).

And finally, recall that an induced EMF in a secondary winding is due to \$N\cdot\frac{d\Phi}{dt}\$ therefore, we'd have a pretty miserable transformer performance if the \$\frac{d\Phi}{dt}\$ was due to anything other than the primary magnetization current.

Answer 2

Think of the transformer theoretical model (without parasitic elements):

^{simulate this circuit – Schematic created using CircuitLab}

The ideal transformer (IDEAL TX above) has infinite magnetising inductance, and perfect coupling i.e. the energy comes in comes out completely.

From the model above, you can see that the primary current has two components:

• Magnetising current (\$i_M\$): This is the current that B-H loop is drawn for. And is determined by the well-known inductor equation: $$ V_p = L_p \ \frac{di_M}{dt} \Rightarrow i_M=\frac{1}{L_p}\int V_p \ dt $$

• Energy- (or load-) related current (\$i_{L}\$) : This is the current that is transferred to the load. And it's determined by the well-known transformer action equation:

$$ V_p \ i_L = \frac{V_p}{N} \ i_O \Rightarrow i_L=\frac{i_O}{N} $$

This means the value of B depends on the current. As current increases the flux will increase accordingly ...

An excitation by H-field generates B-field, yes. But the H-field generated by the load current will cancel out the H-field on the primary side. So all is left is magnetisation. And as can be seen from the diagram and the first equation above, magnetising current is totally independent from the load current i.e. as the load (secondary side) tries to draw more current it'll come from the energy-related component and the magnetisation stays the same because it depends on the applied voltage and the self inductance.

If you apply a voltage across the primary (self inductance or magnetising inductance) it'll generate a flux swing, \$\Delta \Phi\$ or \$d\Phi\$. This is explained by the Faraday's equation:

$$ \varepsilon = V_t = N \ \frac{\Delta\Phi}{\Delta t} = N \ A_e \frac{\Delta B}{\Delta t} $$

^{I intentionally omitted the negation as we are interested in values, not the vectoral directions.}

Since the energy-related component of the primary current has no effect on magnetisation we take the maximum flux density swing for applied voltage as a constant and use in the transformer design.

Answer 3

In power transformers (note that I'm not covering current transformer here) the flux in the core will depend on the voltage applied to the primary winding of the transformer. To be complete accurate, it will depend on the voltage at the innermost (closer to the core) winding. Supposing a sinusoidal excitation, the traditional formula to compute it is: $$\phi=\frac{U_{ph}}{\pi\sqrt{2}\ f\ N}$$ where \$\phi\$ is peak value of the flux in core [Wb], \$U_{ph}\$ is the RMS value of the voltage at the terminals of the innermost winding [V], \$f\$ is the frequency of the excitation signal [Hz] and \$N\$ is the number of turns of the innermost winding. (innermost winding can be exchanged by primary winding here, for easy of use)

The peak flux density \$B\$ in the core, given in [T], will depend on the core cross section, as per: $$B=\frac{\phi}{S}$$ where \$S\$ is the core cross section [m²].

The excitation (or magnetization) current will then be a resulting current which flows through the primary winding only and it is not compensated in any other winding. This current will be nonlinear since the relation between \$B\$ and \$H\$ is nonlinear given by the BxH curve or whatever model you use for it.

Your formula \$B=\mu\ i\ /\ l\$ seems to be right (the units are consistent), but typically you can't impose the excitation current to transformers, of course it's not a rule. However, note that in order to impose a sinusoidal excitation current to a transformer will require a non-sinusoidal voltage at the terminals of the primary.