Flyback turns ratio — Voltage Regulation

Along the way

1
Ideal transformer (forward). Both windings conduct together. \(V_p/V_s=N_p/N_s\), \(I_p/I_s=N_s/N_p\). A flyback does not do that in time.
2
Store, then dump. ON: only \(L_p\). OFF: only \(L_s\). Energy \( \tfrac12 L_p i_p^2 \) becomes \( \tfrac12 L_s i_s^2 \).
3
LCR trap — the square. Same core, \(L=N^2 A_L\), so \(L_p/L_s=(N_p/N_s)^2=N^2\). Measure 20 µH and 5 µH and you do not have \(N=4\). You have \(N=\sqrt{4}=2\). Voltage and current follow the side of that square, not the area.
4
Why energy still matches. \(\tfrac12 L_p i_p^2=\tfrac12 L_s i_s^2\) because \(L_s=L_p/N^2\) and \(i_s=N\,i_p\). The square in \(L\) and the square in current cancel. That is the same \(N\).
5
Current handoff. Amp-turns: \(N_p i_p \approx N_s i_s\) at the FET edge. If \(N_p:N_s=2:1\), the green pulse is twice the orange. CCM: \(V_{\mathrm{out}}=V_{\mathrm{in}}(N_s/N_p)D/(1-D)\).

What we ignored: leakage (it is not in \(A_L\)), clamp, BCM ICs. \(A_L=80\,\mathrm{nH/t}^2\) here so \(L_p\) and \(L_s\) are honest; page 2 is the parabola.

Store, then dump · N vs N²

Turns are the side of the square. Inductance ratio is the area.

Np turns Ns turns VIN V D —

Q1 on · core charging

Turns N vs inductance N²

0 %

N = N_p/N_s—

L_p—

L_s—

L_p/L_s = N²—

√(L_p/L_s)—

VOUT—

Gold packet: store then dump. Square: the side is turns ratio \(N\); the area is inductance ratio \(N^2\). An LCR meter gives you the area — take the square root.

Try it: Default 16:8 → \(N=2\), \(L_p/L_s=4\). Halve \(N_s\): \(N\) doubles to 4, but \(L_p/L_s\) goes to 16 (×4, the square). \(\sqrt{L_p/L_s}\) tracks \(N\), not the L ratio. Set 1:1 and the trap disappears.

Same-core solver

Type what you know — measured \(L\), turns, or \(A_L\). Same core: \(L=N^2 A_L\). Green cells are solved.

N_p turns N_s turns N = N_p/N_s L_p µH L_s µH A_L nH/t²

Enter any two of N_p, N_s, N — or L_p and L_s from an LCR meter.