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.
Home
Help
Search
Login
Register
