Electronic – Boost converter confusion

Energy stored in a capacitor is \$ \frac{1}{2}CV^2 \$.

Energy stored in a inductor is \$ \frac{1}{2}LI^2 \$.

So the conservation of energy equation would be: $$ \frac{1}{2}LI_i^2 - \frac{1}{2}LI_f^2 = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 $$ (subscript i means initial, f means final)

Using an assumption that \$V_{out}\$ is near constant over one cycle, that is \$ V_{out} \gg V_f-V_i \$, then \$ 2 V_{out} \approx V_{f} + V_{i} \$. $$ \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 = \frac{1}{2}C(V_f^2-V_i^2) = \frac{1}{2}C(V_f+V_i)(V_f-V_i) = C V_{out} \Delta V$$ This is the answer given by jp314.

The continuous mode boost and buck-boost converters exhibit control-to-output transfer functions \$G_{vd}(s)=\hat{v}(s)/\hat{d}(s)\$ containing two poles and one RHS (right half-plane) zero, called zero of nonminimum phase.

Your original transfer funtion is:

$$G_{vd}(s)=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C\times\left ( 1-s\left ( \underline{L}/R \right ) \right )}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

Starting from a simpler function (without the RHP zero), named \$ G_{vd}'\$:

$$G_{vd}'=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

Or, placed as a standard second order t.f:

$$G_{vd}'=K_{DC}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

where the DC gain is \$K_{DC} =\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\$.

The equation can be rewritten as:

$$G_{vd}'=K_{DC}\times \frac{1}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$

With \$\omega_0=1/\sqrt{\underline{L} C}\$ and \$\omega_0Q=R/\underline{L}\$

Similarly, the original transfer function can be expressed as (RHP zero included):

$$G_{vd}=K_{DC}\times \frac{\left (1-\frac{s}{\omega_{RHP}} \right )}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$

The response will be:

$$\hat{v}(s)= \frac{K_{DC}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1}\times\hat{d}(s) - \frac{K_{DC}/\omega_{RHP}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }\times\hat{d}(s)\times s$$

The response of the original system is the sum of two components: The first is equivalent to the modified system response (without the zero) and the second is the derivative (scaled) of that one. For the case of a stable system with a step input in \$t = 0\$, this last component will have substantial influence at the beginning and then will vanishes when \$t\rightarrow\infty \$. Note the negative sign leads to a momentary opposite effect in output (nonminimum phase).

UPDATE:

The presence of zero RHP in the model is explained as follows: For the output voltage to increase, the duty cycle must be increased in such a way that the inductor will be disconnected from the load for a long time, causing the output voltage to drop (i.e. in the opposite direction to the one desired). The controller must be designed to meet the project requirements and avoid oscillations while maintaining duty cycle below an undesirable 100% - being limited by the PWM integrated circuit itself.