An estimator is used to estimate the inductor current and facilitate the design of the parameter adaptation law for the estimate of $P$. To this end, the following estimator is considered.

$\begin{array}{l}{\stackrel{\dot{}}{\hat{x}}}_1=-\frac{R_L}{L}{\hat{x}}_1-\left(1-u\right)\frac{{\hat{x}}_2}{L}+\frac{E}{L}\\ {\stackrel{\dot{}}{\hat{x}}}_2=\left(1-u\right)\frac{{\hat{x}}_1}{C}-\frac{\hat{P}}{C{x}_2}+K_2\left(x_2-{\hat{x}}_2\right)\end{array}$ (3)

where ${\hat{x}}_1,{\hat{x}}_2$ and $\hat{P}$ are the estimates of $x_1,x_2$ and $P$ respectively with $K_2>0$ the estimator gain. Let ${\tilde{x}}_1=x_1-{\hat{x}}_1$, ${\tilde{x}}_2=x_2-{\hat{x}}_2$ and $\tilde{P}=P-\hat{P}$ then using (1) and (3), we obtain the following error equations

$\begin{array}{l}{\stackrel{\dot{}}{\tilde{x}}}_1=-\frac{R_L}{L}{\tilde{x}}_1-\left(1-u\right)\frac{{\tilde{x}}_2}{L}\\ {\stackrel{\dot{}}{\tilde{x}}}_2=\left(1-u\right)\frac{{\tilde{x}}_1}{C}-\frac{\tilde{P}}{C{x}_2}-K_2{\tilde{x}}_2\end{array}$ (4)

To generate the adaptation law for the estimate of $P$, we consider the following quadratic Lyapunov function

$V=\frac{1}{2}L{\tilde{x}}_1^2+\frac{1}{2}C{\tilde{x}}_2^2+\frac{1}{2{\gamma }_p}{\tilde{P}}^2$ (5)

where ${\gamma }_p>0$ is a design parameter. Its time derivative along the solutions of (4) is

$\stackrel{\dot{}}{V}=-R_L{\tilde{x}}_1^2-K_{2}C{\tilde{x}}_2^2-\tilde{P}\left[\frac{{\tilde{x}}_2}{x_2}+\frac{\stackrel{\dot{}}{\hat{P}}}{{\gamma }_p}\right]$ (6)

The adaptation law is determined by cancelling the term in brackets and is given by

$\stackrel{\dot{}}{\hat{P}}=-{\gamma }_p\frac{{\tilde{x}}_2}{x_2}$ (7)

With the adaptation law (7), we now have

$\stackrel{\dot{}}{V}=-R_L{\tilde{x}}_1^2-K_{2}C{\tilde{x}}_2^2$ (8)

Using Lasalle invariant principle, we can conclude that ${\tilde{x}}_1\to 0,{\tilde{x}}_2\to 0$ aymptotically. Since $\tilde{P}\to 0$ we have $\hat{P}\to P$.

We consider the following switching surface

$\sigma ={\hat{x}}_1-\beta +K_d{z}_2$ (9)

where

$\begin{array}{l}\beta =\frac{E-\sqrt{E^2-4R_L\hat{P}}}{2R_L}\\ z_2={\hat{x}}_2-V_{ref}\end{array}$ (10)

The controller $u$ is derived by differentiating $\sigma$ with respect to time and setting $\stackrel{\dot{}}{\sigma }=0$. Using (3), (9) and (10), the control is given by

$\begin{array}{c}u=\frac{C{\hat{x}}_2+\left(R_{L}C-K_{d}L\right){\hat{x}}_1-EC+LC\stackrel{\dot{}}{\beta }+\frac{K_{d}L\hat{P}}{x_2}}{C{\hat{x}}_2-K_{d}L{\hat{x}}_1}\\ -\frac{K_2{K}_{d}LC{\tilde{x}}_2}{C{\hat{x}}_2-K_{d}L{\hat{x}}_1}-\frac{K_s\sigma }{C{\hat{x}}_2-K_{d}L{\hat{x}}_1}\end{array}$ (11)

where

$\stackrel{\dot{}}{\beta }=\frac{\stackrel{\dot{}}{\hat{P}}}{\sqrt{E^2-4R_L\hat{P}}}$ (12)

The last term of the controller u given by (11) is added to ensure the sliding mode condition.

The substitution of the controller (11) into the derivative of (9) leads to

$\stackrel{\dot{}}{\sigma }=-K_s\sigma$ (13)

and therefore we have $\stackrel{\dot{}}{\sigma }\sigma =-K_s{\sigma }^2<0$, which ensures the sliding mode condition. Equation (13) admits as a solution

$\sigma \left(t\right)=\sigma \left(0\right){\text{e}}^{-K_{s}t}$ (14)

For a choice of the initial condition of the adaptation law $\hat{P}\left(0\right)$, the initial conditions of the estimator ${\hat{x}}_1\left(0\right)$ and ${\hat{x}}_2\left(0\right)$ are selected to guarantee the occurrence of sliding mode at $t=0$ with the reaching phase completely eliminated with

$\begin{array}{l}{\hat{x}}_2\left(0\right)=V_{ref}\\ \beta \left(0\right)=\frac{E-\sqrt{E^2-4R_L\hat{P}\left(0\right)}}{2R_L}\\ {\hat{x}}_1\left(0\right)=\beta \left(0\right)\end{array}$ (15)

This yields $\sigma \left(0\right)=0$ and in view of (14), $\sigma \left(t\right)=0$ for all $t\ge 0$. In this case we have

${\hat{x}}_1\left(t\right)=\beta -K_d{z}_2\left(t\right),t\ge 0$ (16)

The parasitics of the converter increase the natural damping of the system [1] , [17] and reduce the destabilizing effect of the CPL. Therefore, we set $R_L=0\Omega$ for which the destabilizing effect are more pronounced for the stability analysis.

For $R_L=0\Omega$ we consider $\beta =\frac{\hat{P}}{E}$. Using (7), $\stackrel{\dot{}}{\beta }=\frac{\stackrel{\dot{}}{\hat{P}}}{E}=-{\gamma }_p\frac{{\tilde{x}}_2}{E{x}_2}$ and equation (16) becomes

${\hat{x}}_1\left(t\right)=\frac{\hat{P}}{E}-K_d{z}_2\left(t\right),t\ge 0$ (17)

Substituting the controller (11) into the second equation of (3) and using (17) with $\sigma =0$ yields the following equation

${\stackrel{\dot{}}{\hat{x}}}_2=\frac{\left[EC-\frac{\hat{P}L{K}_d}{x_2}\right]\left[\frac{\hat{P}}{EC}-\frac{K_d{z}_2}{C}\right]}{C{\hat{x}}_2-K_{d}L\left(\frac{\hat{P}}{E}-K_d{z}_2\right)}-\frac{\hat{P}}{C{x}_2}+{\Psi }_1$ (18)

where

${\Psi }_1=\frac{L{\gamma }_p\frac{{\hat{x}}_1{\tilde{x}}_2}{E{x}_2}+K_2{K}_{d}L{\hat{x}}_1{\tilde{x}}_2}{C{\hat{x}}_2-K_{d}L\left(\frac{\hat{P}}{E}-K_d{z}_2\right)}+K_2{\tilde{x}}_2$ (19)

Multiplying both sides of (18) by the denominator $C{\hat{x}}_2-K_{d}L\left(\frac{\hat{P}}{E}-K_d{z}_2\right)$ yields

${\stackrel{\dot{}}{\hat{x}}}_2\left[C{\hat{x}}_2-K_{d}L\left(\frac{\hat{P}}{E}-K_d{z}_2\right)\right]=-K_{d}E{z}_2+{\Psi }_2$ (20)

where

${\Psi }_2={\Psi }_1\left[C{\hat{x}}_2-K_{d}L\left(\frac{\hat{P}}{E}-K_d{z}_2\right)\right]+\frac{\hat{P}}{x_2}{\tilde{x}}_2$ (21)

Using ${\stackrel{\dot{}}{\hat{x}}}_2={\stackrel{\dot{}}{z}}_2$ and $z_2={\hat{x}}_2-V_{ref}$ we can write (20) as

$\begin{array}{l}C{\stackrel{\dot{}}{z}}_2\left[z_2+V_{ref}\right]-K_{d}L{\stackrel{\dot{}}{z}}_2\left[\frac{\hat{P}}{E}-K_d{z}_2\right]\\ ={\stackrel{\dot{}}{z}}_2\left[\left(C+K_d^{2}L\right)z_2+\left(C{V}_{ref}-K_{d}L\frac{\hat{P}}{E}\right)\right]\\ =-K_{d}E{z}_2+{\Psi }_2\end{array}$ (22)

and

$\begin{array}{l}{\stackrel{\dot{}}{z}}_2=\frac{-K_{d}E{z}_2+{\Psi }_2}{\left(C+K_d^{2}L\right)z_2+C{V}_{ref}-K_{d}L\frac{\hat{P}}{E}}\hfill \end{array}$ (23)

Since ${\tilde{x}}_2\to 0$, ${\Psi }_1\to 0$ and therefore ${\Psi }_2\to 0$. For $K_d>0$ we want the denominator of (23) to be positive to establish stability. Using $z_2={\hat{x}}_2-V_{ref}$ we want

$\begin{array}{l}\left(C+K_d^{2}L\right)\left({\hat{x}}_2-V_{ref}\right)+C{V}_{ref}-K_{d}L\frac{\hat{P}}{E}>0\hfill \end{array}$ (24)

The crucial case in satisfying condition (24) is when ${\hat{x}}_2<V_{ref}$ starting with ${\hat{x}}_2\left(0\right)=V_{ref}$ since the first term and the last term of (24) are both negative. This situation occurs with the existence of undershoots with the maximum possible magnitude undershoot ${\hat{x}}_2\left(\min \right)=p{V}_{ref}$, $0<p<1$. The first undershoot being the most critical in the determination of $K_d$ for stability as subsequent undershoots, if they exist, will be decreasing since ${\text{Ψ}}_2$ is decreasing with time as ${\tilde{x}}_2\to 0$. In this case, it will be shown that there exist a gain $K_d$ such that condition (24) is satisfied for all $t\ge 0$ and asymptotic stability is achieved with $z_2\to 0$, ${\hat{x}}_2\to V_{ref}$ and $x_2\to V_{ref}$. From (17) ${\hat{x}}_1\to \frac{P}{E}$ and $x_1\to \frac{P}{E}$ since ${\tilde{x}}_1\to 0.$

Assume that ${\hat{x}}_2\left(t\right)$ undershoots to ${\hat{x}}_2\left(\min \right)=p{V}_{ref}$, $0<p<1$. The worst case for condition (24) to hold is when $\hat{P}={\hat{P}}_{\max }$ and in this case condition (24) becomes

$\left(C+K_d^{2}L\right)\left({\hat{x}}_2-V_{ref}\right)+\left(C{V}_{ref}-K_{d}L\frac{{\hat{P}}_{\max }}{E}\right)>0$ (25)

Substituting ${\hat{x}}_2=p{V}_{ref}$, condition (25) is written as

$K_d^{2}L{V}_{ref}\left(p-1\right)-K_{d}L\frac{{\hat{P}}_{\max }}{E}+Cp{V}_{ref}>0$ (26)

The quadratic equation in $K_d$ given by (26) admits two roots $K_{d_1}<0$ and $K_{d_2}>0$ with

$\begin{array}{l}{K}_{d_1}=\frac{\frac{L{\hat{P}}_{\max }}{E}+\sqrt{{\left(\frac{L{\hat{P}}_{\max }}{E}\right)}^2-4L\left(p-1\right)V_{ref}^{2}pC}}{2L\left(p-1\right)V_{ref}}\\ K_{d_2}=\frac{\frac{L{\hat{P}}_{\max }}{E}-\sqrt{{\left(\frac{L{\hat{P}}_{\max }}{E}\right)}^2-4L\left(p-1\right)V_{ref}^{2}pC}}{2L\left(p-1\right)V_{ref}}\end{array}$ (27)

For the condition (26) to be satisfied the gain $K_d>0$ should be chosen as

$0<K_d<K_{d_2}$ (28)

and (26) is satisfied for ${\hat{x}}_2\left(t\right)\ge p{V}_{ref},0<p<1$.

It is straightforward to show that $K_d$ given by (28) will also satisfy condition (25) for the case of an overshoot of ${\hat{x}}_2$. Therefore, for ${\hat{x}}_2\ge p{V}_{ref}$, $p>0$ condition (25) is satisfied and asymptotic stability is achieved with $x_2\to V_{ref}$, $\hat{P}\to P$ and finally $x_1\to \frac{P}{E}$.

Remark 1: The determination of the gain $K_d$ depends on the knowledge of ${\hat{x}}_2\left(\min \right)$. The auxiliary diode $D_1$ constrains the output $x_2$ to

$x_2={\hat{x}}_2+{\tilde{x}}_2\ge E$ (29)

From (23) to ensure that the estimator has much faster dynamics than the controlled feedback system, the gain of the estimator $K_2$ must be chosen much greater than $K_{d}E$ to ensure that ${\tilde{x}}_2\to 0$ at a much faster rate than the controlled output to $V_{ref}$. Please also note that initially we have $x_2\left(0\right)=E$ and ${\hat{x}}_2\left(0\right)=V_{ref}$. In this case, we can assume that the maximum possible undershoot of ${\hat{x}}_2$ is constrained to

${\hat{x}}_2\left(\min \right)=E$ (30)

Remark 2: A possible general guideline for the choice of the controller and estimator gains is:

1) Choose $K_s=1$ to satisfy the sliding mode condition. Choose a high gain ${\gamma }_p$ for the adaptation law (7). The higher the gain, the faster the convergence of the estimated $\hat{P}$ to the actual $P$. However, if the gain is too high, there will be small oscillations in the output response during load power disturbances.

2) Choose ${\hat{x}}_2\left(0\right)=V_{ref}$ and a value for $\hat{P}\left(0\right)$ and determine ${\hat{x}}_1\left(0\right)$ given by (15) to guarantee that the sliding mode starts at $t=0\text{s}$ with the reaching phase completely eliminated.

3) Choose ${\hat{P}}_{\max }$ to be 3 to 5 greater than the expected maximum load power $P$.

4) As discussed previously, the crucial factor that determines the stability of the closed-loop system is the existence of an undershoot of maximum amplitude ${\hat{x}}_2\left(\min \right)=p{V}_{ref}=E$. In this case, we have

$p=\frac{E}{V_{ref}}$ (31)

5) Calculate $K_{d_2}$ using (27) and choose the gain $K_d$ to be

$K_d=K_d^{\text{*}}=0.5K_{d_2}$ (32)

In this case, the maximum possible magnitude of the undershoot ${\hat{x}}_2\left(\min \right)=p_1{V}_{ref}$ with $p_1<p$ for which stability is maintained is determined by $p_1$ given by

$p_1=\frac{K_d^{*2}L{V}_{ref}+\frac{K_d^{*}L{P}_{\max }}{E}}{K_d^{*2}L{V}_{ref}+C{V}_{ref}}$ (33)

6) In view of Remark 1, choose the gain of the estimator $K_2=50\times m\times K_{d}E$ with $m\ge 1$ to ensure that the estimator has faster dynamics than the controlled feedback system.

Shown in Figure 2 is the block diagram of the proposed controlled system.