How to find total flux inside a transformer - Page 1
Forum section: electronics / beginners

author: Neukyhm
author_rank: Contributor
timestamp: October 12, 2017, 03:02:21 pm
content: Hi users, I hope you can help me with this. I'm designing a transformer and I need to know the total flux inside the core in order to know the losses it will have.Suppose I have a circuit as attached, suppose I know all parameters of my core like Ae, Al, Bmax, effective magnetic path, volume, etc.Consider a resistive load.Now taking into account that primary and secondary currents may (not sure about this) not be in phase, I can't just add the fluxes created by primary and secondary peak currents.So please users tell me how to find the total flux inside a transformer due to primary and secondary currents. I would be really happy if you can recommend me a good book about this stuff, I have a lot but they don't explain this.

author: Benta
author_rank: Super Contributor
timestamp: October 12, 2017, 03:26:01 pm
content: The flux is proportional to the magnetizing current, which again is defined by primary voltage and inductance.Output loading causes current to flow in the secondary, theoretically reducing flux. However, due to the coupling (K=1), primary current will be increased by the same amount, canceling any influence on flux.This is the short and dirty explanation, in non-ideal transformers with K<1 and other losses, it's a bit more complicated.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 12, 2017, 05:39:28 pm
content: Since EMF ~= V(applied) (a reasonable assumption because: saturation is not specified, the primary magnetizing inductive reactance is large, and the primary / source ESR is small), we can find the peak flux by taking the time integral of the voltage (EMF).Since v(t) = A*sin(w*t), phi(t) = -A*w*cos(w*t).For common mains voltages, A should be around 340V, but you're showing 220 in the simulation. You may want to check that.The peak of cos(t) is +/- 1, so phi_peak = A*w. We need to add some conversion for frequency and RMS, and include the relations of area and turns, to get:N(turns) = V(rms) / (4.44 * F * A_e * B_max)4.44 is actually pi*sqrt(2) to reasonable accuracy.If the applied waveform were a square wave, the constant would be 4.0 instead (exercise for the student: prove it!).Tim

author: orolo
author_rank: Frequent Contributor
timestamp: October 12, 2017, 06:20:00 pm
content: What works for me is going back to basic physics. Maxwell equations, in this case. Consider a primary winding of N1 turns and a secondary winding of N2 turns, sharing all the flux (no leakage).From Faraday's law in integral form, \$\oint E \ = \ -\frac{\partial}{\partial t}\iint B \ = \ -\frac{d\Phi}{dt}\$, you know that each turn of winding contributes to the voltage the opposite of the derivative of the flux.So the voltage at the primary winding is \$E_1 \ =\ -N_1\frac{d\Phi}{dt}\$ and the flux at the secondary is \$E_2 \ =\ -N_2\frac{d\Phi}{dt}\$. From there you very easily arrive at \$\frac{E_1}{E_2} = \frac{N_1}{N_2}\$ and all that.Now take only the first equation: \$E_1 \ =\ -N_1\frac{d\Phi}{dt}\$ If you force a voltage \$E_1 = V_p\sin\omega t\$ across the winding, you get:\$ \frac{d\Phi}{dt} \ = \ -\frac{1}{N_1}V_p\sin\omega t\$Integrating,\$\Phi \ = \ \Phi_0 + \frac{V_p}{N_1\omega}\cos\omega t\$There you got the flux in terms of the forcing voltage at the winding. Since \$V_p = \sqrt{2}V_{rms}\$, you may write that as:\$\Phi \ = \ \Phi_0 + \frac{V_{rms}\sqrt{2}}{N_1\omega}\cos\omega t\$If you disregard any residual flux (Phi_0 = 0) and want maximal flux, you arrive at:\$\Phi_{max} \ = \ \frac{V_{rms}\sqrt{2}}{N_1\omega}\$To arrive to Tim's formula, change to frequency:\$\Phi_{max} \ = \ \frac{V_{rms}\sqrt{2}}{2 \pi f N_1}\$And since sqrt(2)/2 = 1/sqrt(2), you have:\$\Phi_{max} \ = \ \frac{V_{rms}}{\sqrt{2} \pi f N_1}\$Depending on the geometry of your core, you can change the flux for an adequate B times area expression.Edit: by the way, the above equations imply that the primary and secondary voltages are in phase, since both are 90 degrees out of phase with the flux. Of course, this is an ideal situation.

author: Benta
author_rank: Super Contributor
timestamp: October 12, 2017, 07:13:07 pm
content: QuoteExactly, and the currents 180 degrees out of phase, thus neutralizing effect on core flux.