I am trying to derive the equation of small signal vout/d for a CCM flyback converter. In order to do this, I need the small-signal average model. I am having difficulty seeing how the small signal model for the flyback (or buck-boost) is derived from the flyback converter:

I recognize that the Vout/Vin for the flyback and buck-boost include a D in the numerator like a buck converter, and a 1-D in the denominator like that boost converter, and that is why the small signal model has a buck type stage (1:D) and a boost type stage ((1-D):1). I just wish to be able to be able to derive the small signal model directly from the normal circuit schematic.

## Best Answer

The primary turns, \$Np\$ are selected to satisfy the AC voltage stress (volt-seconds) and the core AC saturation properties:

\$\frac{V\:T\:}{B\:Ae}=Np\$

Energy in the primary coil is transferred to the secondary coil during the "off" state of the flyback operation.\$E = \frac{1}2LI^{2}\$ (joules)

\$V_{L} = (V_{i}-V_{o})D-V_{o}\$

then modulating duty cycle, \$D\$ with signal \$\delta\$ to get AC voltage \$v_L\$ on \$L\$

\$v_L = (V_i+V_o)\delta - V_o(1-D)\; ≈ (V_i+V_o)\delta \;\$

\$i_L = \frac {v_L}{j\omega L} = -j \frac{V_i+V_o}{\omega L}\delta \$

during the "off" state \$I_o=I_L (1-D)\$ then differentiating ac current with resulting \$i_L\$ having in-phase and 180' out of phase with \$\delta\$

\$i_o=-j\frac{(V_i+V_o)(1-D)}{\omega L}\delta -I_L\delta =-j \frac{V_i}{\omega L}\delta -I_Ld\$

The RHP Zero becomes \$ \omega_z=\frac{V_i}{LI_L}=\frac{R_oV_i^2}{LV_o(V_i+V_o)}\$

By using primary current sensing and current control loop, stability improves greatly.

I would like to give credit to above from one of the best SMPS designers I have briefly worked with in my past life as Test Eng Mgr at Burroughs.(mid'80's)

I hope this encourages you to buy one of his many books on the subject of SMPS Design... Tony.

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