In the following, a novel analytical model of the LLC resonant converter in OBR mode is derived. Having a closer look to the converter structure, it appears that the transformer has to be considered accordingly. This is achieved by taking the transformer turns ratio into account, subsequently transforming the load represented through \(R_{\mathrm{o,s}}\) as well as the output capacitor \(C_{\mathrm{o,s}}\) to the primary side. In the following, \(R_{\mathrm{o}}\) and \(C_{\mathrm{o}}\) are considered to be the primary sided resistance and capacitance. Furthermore, four energy storage elements appear. Setting up a differential equation in time domain would therefore lead to a system with 4th order, where also initial conditions need to be considered. Thus, it is way more convenient to switch to the Laplace domain.
By subdividing the cycle into intervals strategically in order to get rid of the non linearity caused by the full bridge rectifier (a potential nonlinear behavior of other circuit elements, e.g. due to magnetic saturation is neglected), each interval shows an individual characteristic which can be described through a linear electrical network. Applying the Laplace Transform therefore is straight forward.
3.1 Derivation of the model for interval I and III
For the model given in Fig.
4, Kirchhoff’s law provides:
$$\begin{aligned}I_{\mathrm{r}}(s)=I_{\mathrm{m}}(s)+I_{\mathrm{c}}(s)+I_{\mathrm{o}}(s)\end{aligned}$$
(4)
$$\begin{aligned}\frac{U}{s}=U_{\mathrm{C_{r}}}(s)+U_{\mathrm{C_{r}0}}(s)+U_{\mathrm{L_{r}}}(s)-U_{\mathrm{L_{r}0}}(s)+U_{\mathrm{L_{m}}}(s)-U_{\mathrm{L_{m}0}}(s)\end{aligned}$$
(5)
$$\begin{aligned}U_{\mathrm{L_{m}}}(s)=U_{\mathrm{C_{o}}}(s)+U_{\mathrm{C_{o}0}}(s)+U_{\mathrm{L_{m}0}}(s)\end{aligned}$$
(6)
$$\begin{aligned}U_{\mathrm{R_{o}}}(s)=U_{\mathrm{C_{o}}}(s)+U_{\mathrm{C_{o}0}}(s)\end{aligned}$$
(7)
Please remind that for interval
III,
\(U\) has to be replaced by
\(-U\), and
\(U_{\mathrm{C_{o}0}}(s)\) by
\(-U_{\mathrm{C_{o}0}}(s)\) due to the inverted polarity (see also Figs.
4 and
6). Eqs. (
4) to (
7) can be rewritten to
$$\begin{bmatrix}0\\ \frac{U}{s}-\frac{u_{\mathrm{{C_{r}}0}}}{s}+L_{\mathrm{r}}\cdot i_{\mathrm{r0}}+L_{\mathrm{m}}\cdot i_{\mathrm{m0}}\\ L_{\mathrm{m}}\cdot i_{\mathrm{m0}}+\frac{u_{\mathrm{{C_{o}}0}}}{s}\\ \frac{u_{\mathrm{{C_{o}}0}}}{s}\end{bmatrix}=\mathbf{A}\cdot\mathbf{x}$$
(8)
with
$$\mathbf{A}=\begin{bmatrix}-1&1&1&1\\ \frac{1}{s\cdot C_{\mathrm{r}}}+s\cdot L_{\mathrm{r}}&s\cdot L_{\mathrm{m}}&0&0\\ 0&s\cdot L_{\mathrm{m}}&-\frac{1}{s\cdot C_{\mathrm{o}}}&0\\ 0&0&-\frac{1}{s\cdot C_{\mathrm{o}}}&R_{\mathrm{o}}\end{bmatrix},$$
(9)
the state vector
$$\mathbf{x}=\begin{bmatrix}I_{\mathrm{r}}(s)\\ I_{\mathrm{m}}(s)\\ I_{\mathrm{c}}(s)\\ I_{\mathrm{o}}(s)\end{bmatrix}$$
(10)
and the initial conditions
\(i_{\mathrm{r0}}\),
\(i_{\mathrm{m0}}\),
\(u_{\mathrm{C_{r}0}}\) and
\(u_{\mathrm{C_{o}0}}\).Solving (
8) results in
$$\begin{aligned}I_{\mathrm{r}}(s)=\frac{(C_{\mathrm{r}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{m}}s+CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})U}{D(s)}+\frac{(C_{\mathrm{r}}L_{\mathrm{r}}R_{\mathrm{o}}s+C_{\mathrm{r}}L_{\mathrm{m}}L_{\mathrm{r}}s^{2})i_{\mathrm{r0}}}{D(s)}+\frac{(CL_{\mathrm{m}}L_{\mathrm{r}}R_{\mathrm{o}}s^{3})i_{\mathrm{r0}}}{D(s)}+\frac{(C_{\mathrm{r}}L_{\mathrm{m}}R_{\mathrm{o}}s)i_{\mathrm{m0}}}{D(s)}-\frac{(C_{\mathrm{r}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{m}}s+CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})u_{\mathrm{{C_{r}}0}}}{D(s)}-\frac{(CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})u_{\mathrm{{C_{o}}0}}}{D(s)}\end{aligned}$$
(11)
$$\begin{aligned}I_{\mathrm{m}}(s)=\frac{(C_{\mathrm{r}}R_{\mathrm{o}})U}{D(s)}+\frac{(C_{\mathrm{r}}L_{\mathrm{r}}R_{\mathrm{o}}s)i_{\mathrm{r0}}}{D(s)}+\frac{(L_{\mathrm{m}}+C_{\mathrm{o}}L_{\mathrm{m}}R_{\mathrm{o}}s+C_{\mathrm{r}}L_{\mathrm{m}}R_{\mathrm{o}}s)i_{\mathrm{m0}}}{D(s)}+\frac{(C_{\mathrm{r}}L_{\mathrm{m}}L_{\mathrm{r}}s^{2}+CL_{\mathrm{m}}L_{\mathrm{r}}R_{\mathrm{o}}s^{3})i_{\mathrm{m0}}}{D(s)}-\frac{(C_{\mathrm{r}}R_{\mathrm{o}})u_{{C_{\mathrm{r}}}0}}{D(s)}-\frac{(C_{\mathrm{o}}R_{\mathrm{o}}+CL_{\mathrm{r}}R_{\mathrm{o}}s^{2})u_{\mathrm{{C_{o}}0}}}{D(s)}\end{aligned}$$
(12)
$$\begin{aligned}I_{\mathrm{c}}(s)=\frac{(CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})U}{D(s)}+\frac{(CL_{\mathrm{m}}L_{\mathrm{r}}R_{\mathrm{o}}s^{3})i_{\mathrm{r0}}}{D(s)}-\frac{(C_{\mathrm{o}}L_{\mathrm{m}}R_{\mathrm{o}}s+CL_{\mathrm{m}}L_{\mathrm{r}}R_{\mathrm{o}}s^{3})i_{\mathrm{m0}}}{D(s)}-\frac{(CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})u_{{C_{\mathrm{r}}}0}}{D(s)}-\frac{(C_{\mathrm{o}}R_{\mathrm{o}}+C_{\mathrm{o}}L_{\mathrm{m}}s+CL_{\mathrm{m}}R_{\mathrm{o}}s^{2})u_{\mathrm{{C_{o}}0}}}{D(s)}-\frac{(CL_{\mathrm{r}}R_{\mathrm{o}}s^{2}+CL_{\mathrm{m}}L_{\mathrm{r}}s^{3})u_{\mathrm{{C_{o}}0}}}{D(s)}\end{aligned}$$
(13)
$$\begin{aligned}I_{\mathrm{o}}(s)=\frac{(C_{\mathrm{r}}L_{\mathrm{m}}s)U}{D(s)}+\frac{(C_{\mathrm{r}}L_{\mathrm{m}}L_{\mathrm{r}}s^{2})i_{\mathrm{r0}}}{D(s)}-\frac{(L_{\mathrm{m}}+L_{\mathrm{m}}C_{\mathrm{r}}L_{\mathrm{r}}s^{2})i_{\mathrm{m0}}}{D(s)}-\frac{(C_{\mathrm{r}}L_{\mathrm{m}}s)u_{\mathrm{{C_{r}}0}}}{D(s)}+\frac{(L_{\mathrm{m}}C_{\mathrm{o}}s+CL_{\mathrm{m}}L_{\mathrm{r}}s^{3})u_{\mathrm{{C_{o}}0}}}{D(s)}\end{aligned}$$
(14)
with
\(C=C_{\mathrm{r}}C_{\mathrm{o}}\) and with the denominator
\(D(s)\) given as
$$\begin{aligned}D(s)=(C_{\mathrm{r}}L_{\mathrm{r}}L_{\mathrm{m}}C_{\mathrm{o}}R_{\mathrm{o}})s^{4}+(C_{\mathrm{r}}L_{\mathrm{r}}L_{\mathrm{m}})s^{3}+(L_{\mathrm{m}}C_{\mathrm{o}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{m}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{r}}R_{\mathrm{o}})s^{2}+(L_{\mathrm{m}})s+(R_{\mathrm{o}}).\end{aligned}$$
(15)
Please consider that these results take already into account that the equivalent systems are excited with a voltage step generated by the half bridge.
(
15) can be rewritten to
$$\begin{aligned}D(s)=K\cdot(s-s_{1})(s-s_{2})(s-s_{3})(s-s_{4})\end{aligned}$$
(16)
with
\(K=C_{\mathrm{r}}L_{\mathrm{r}}L_{\mathrm{m}}C_{\mathrm{o}}R_{\mathrm{o}}\) and where
\(s_{i}\) with
\(i\in\{1,2,3,4\}\) are the zeros of the polynomial
$$\begin{aligned}P(s)=s^{4}+\frac{C_{\mathrm{r}}L_{\mathrm{r}}L_{\mathrm{m}}}{K}s^{3}+\frac{L_{\mathrm{m}}C_{\mathrm{o}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{m}}R_{\mathrm{o}}+C_{\mathrm{r}}L_{\mathrm{r}}R_{\mathrm{o}}}{K}s^{2}+\frac{L_{\mathrm{m}}}{K}s+\frac{R_{\mathrm{o}}}{K},\end{aligned}$$
(17)
whereas
\(D(s)=K\cdot P(s)\) is valid.
Calculating the zeros of the 4th order polynomial can be done by using methods of Ferrari and Cardano. Knowing the zeros allows a partial fractional decomposition of the state quantities in Laplace domain. The derivation of the corresponding time quantities is subsequently achieved with the Inverse Laplace Transformation. Other quantities describing the converter can simply be derived using algebraic manipulations.
Determining the time domain functions is straightforward using the partial fraction decomposition:
$$\begin{aligned}\frac{N(s)}{K\cdot P(s)}=\frac{A}{(s-s_{1})}+\frac{B}{(s-s_{2})}+\frac{C}{(s-s_{3})}+\frac{D}{(s-s_{4})},\end{aligned}$$
(18)
which is equivalent to
$$\begin{aligned}\frac{N(s)}{K}=A(s-s_{2})(s-s_{3})(s-s_{4})+B(s-s_{1})(s-s_{3})(s-s_{4})+C(s-s_{1})(s-s_{2})(s-s_{4})+D(s-s_{1})(s-s_{2})(s-s_{3}).\end{aligned}$$
(19)
The constants
\(A\) -
\(D\) are then given by
$$\begin{array}[]{l}A=\frac{N(s_{1})}{K(s_{1}-s_{2})(s_{1}-s_{3})(s_{1}-s_{4})}\\ B=\frac{N(s_{2})}{K(s_{2}-s_{1})(s_{2}-s_{3})(s_{2}-s_{4})}\\ C=\frac{N(s_{3})}{K(s_{3}-s_{1})(s_{3}-s_{2})(s_{3}-s_{4})}\\ D=\frac{N(s_{4})}{K(s_{4}-s_{1})(s_{4}-s_{2})(s_{4}-s_{3})}\end{array},$$
(20)
where
\(N(s)\) represents the numerator of the corresponding quantity. Thus, the constants
\(A,B,C\) and
\(D\) for all four state quantities can be calculated directly.
For the latter one can subsequently write
$$\begin{aligned}i_{\mathrm{r}}=A_{\mathrm{r}}\cdot e^{s_{1}t}+B_{\mathrm{r}}\cdot e^{s_{2}t}+C_{\mathrm{r}}\cdot e^{s_{3}t}+D_{\mathrm{r}}\cdot e^{s_{4}t}\end{aligned}$$
(21)
$$\begin{aligned}i_{\mathrm{m}}=A_{\mathrm{m}}\cdot e^{s_{1}t}+B_{\mathrm{m}}\cdot e^{s_{2}t}+C_{\mathrm{m}}\cdot e^{s_{3}t}+D_{\mathrm{m}}\cdot e^{s_{4}t}\end{aligned}$$
(22)
$$\begin{aligned}i_{\mathrm{c}}=A_{\mathrm{c}}\cdot e^{s_{1}t}+B_{\mathrm{c}}\cdot e^{s_{2}t}+C_{\mathrm{c}}\cdot e^{s_{3}t}+D_{\mathrm{c}}\cdot e^{s_{4}t}\end{aligned}$$
(23)
$$\begin{aligned}i_{\mathrm{o}}=A_{\mathrm{o}}\cdot e^{s_{1}t}+B_{\mathrm{o}}\cdot e^{s_{2}t}+C_{\mathrm{o}}\cdot e^{s_{3}t}+D_{\mathrm{o}}\cdot e^{s_{4}t}.\end{aligned}$$
(24)
Applying the well known correspondence
\(u_{\mathrm{L}}=L\cdot\frac{di_{\mathrm{L}}}{dt}\) while using (
21) and (
22) allows to calculate the inductor voltages
$$\begin{aligned}u_{\mathrm{L_{r}}}=L_{\mathrm{r}}(A_{\mathrm{r}}\cdot s_{1}e^{s_{1}\cdot t}+B_{\mathrm{r}}\cdot s_{2}e^{s_{2}\cdot t})+L_{\mathrm{r}}(C_{\mathrm{r}}\cdot s_{3}e^{s_{3}\cdot t}+D_{\mathrm{r}}\cdot s_{4}e^{s_{4}\cdot t})\end{aligned}$$
(25)
$$\begin{aligned}u_{\mathrm{L_{m}}}=L_{\mathrm{m}}(A_{\mathrm{m}}\cdot s_{1}e^{s_{1}\cdot t}+B_{\mathrm{m}}\cdot s_{2}e^{s_{2}\cdot t})+L_{\mathrm{m}}(C_{\mathrm{m}}\cdot s_{3}e^{s_{3}\cdot t}+D_{\mathrm{m}}\cdot s_{4}e^{s_{4}\cdot t}).\end{aligned}$$
(26)
The voltage across the output capacitor of the simplified model is equal to the voltage across the magnetizing inductor for the considered case. Thus,
$$u_{\mathrm{C_{o}}}=u_{\mathrm{L_{m}}}$$
(27)
is valid.
Finally, the voltage across the resonant capacitor
\(C_{\mathrm{r}}\) can be determined by applying Kirchhoff’s voltage law
$$u_{\mathrm{C_{r}}}=U_{\mathrm{B}}-u_{\mathrm{L_{r}}}-u_{\mathrm{L_{m}}}.$$
(28)
3.2 Derivation of the model for interval II and IV
For the separated model given in Fig.
5,
$$\begin{aligned}\frac{U_{\mathrm{B}}}{s}=\frac{I_{\mathrm{r}}(s)}{sC_{\mathrm{r}}}+\frac{u_{\mathrm{C_{r}0}}}{s}+I_{\mathrm{r}}(s)sL_{\mathrm{r}}-L_{\mathrm{r}}i_{\mathrm{r0}}+I_{\mathrm{r}}(s)sL_{\mathrm{m}}-L_{\mathrm{m}}i_{\mathrm{r0}}\end{aligned}$$
(29)
is valid.
Please remind that for interval
IV,
\(U\) has to be replaced by
\(-U\) due to the inverted polarity (see also Figs.
5 and
7).
Solving for
\(I_{\mathrm{r}}(s)\) leads to
$$I_{\mathrm{r}}(s)=\frac{U_{\mathrm{B}}C_{\mathrm{r}}-C_{\mathrm{r}}u_{\mathrm{C_{r}0}}+si_{\mathrm{r0}}(C_{\mathrm{r}}L_{\mathrm{r}}+C_{\mathrm{r}}L_{\mathrm{m}})}{1+s^{2}(C_{\mathrm{r}}L_{\mathrm{r}}+C_{\mathrm{r}}L_{\mathrm{m}})}.$$
(30)
Analog to the model valid for interval
I and
III, the denominator is given by
$$Q(s)=1+s^{2}(C_{\mathrm{r}}L_{\mathrm{r}}+C_{\mathrm{r}}L_{\mathrm{m}}),$$
(31)
which can also be written as
$$Q(s)=M\cdot(s-s_{5})(s-s_{6})$$
(32)
with
\(M=C_{\mathrm{r}}L_{\mathrm{r}}+C_{\mathrm{r}}L_{\mathrm{m}}\), where
\(s_{i}\) (
\(i\in\{5,6\}\)) are the zeros of the polynomial
$$R(s)=s^{2}+\frac{1}{M}$$
(33)
with
\(Q(s)=M\cdot R(s)\).
The two conjugate complex zeros are then given by
$$s_{5},s_{6}=\pm\frac{j}{\sqrt{C_{\mathrm{r}}L_{\mathrm{r}}+C_{\mathrm{r}}L_{\mathrm{m}}}}.$$
(34)
Following the same approach again,
$$\frac{T(s)}{M\cdot R(s)}=\frac{E}{(s-s_{5})}+\frac{F}{(s-s_{6})}$$
(35)
which is equivalent to
$$\frac{T(s)}{M}=E(s-s_{6})+F(s-s_{5})$$
(36)
can be determined and allows to find the constants with
$$\begin{array}[]{l}E=\frac{T(s_{5})}{M(s_{5}-s_{6})}\\ F=\frac{T(s_{6})}{M(s_{6}-s_{5})}\end{array},$$
(37)
where
\(T(s)\) is the numerator of the resonant current.
For the latter one finally finds
$$i_{\mathrm{r}}=E\cdot e^{s_{5}\cdot t}+F\cdot e^{s_{6}\cdot t}.$$
(38)
In these two cases, the magnetizing current is equal to the resonant current
$$i_{\mathrm{m}}=i_{r}=E\cdot e^{s_{5}\cdot t}+F\cdot e^{s_{6}\cdot t}.$$
(39)
The voltages across the inductors can again be calculated by derivation of (
39) which results in
$$\begin{aligned}u_{\mathrm{L_{r}}}=L_{\mathrm{r}}(E\cdot s_{5}e^{s_{5}\cdot t}+F\cdot s_{6}e^{s_{6}\cdot t})\end{aligned}$$
(40)
$$\begin{aligned}u_{\mathrm{L_{m}}}=L_{\mathrm{m}}(E\cdot s_{5}e^{s_{5}\cdot t}+F\cdot s_{6}e^{s_{6}\cdot t}),\end{aligned}$$
(41)
finally yielding to
$$u_{\mathrm{C_{r}}}=U_{\mathrm{B}}-u_{\mathrm{L_{r}}}-u_{\mathrm{L_{m}}}.$$
(42)
The second part of the decoupled model is given by
$$\frac{I_{\mathrm{o}}(s)}{sC_{\mathrm{o}}}+I_{\mathrm{o}}(s)R_{\mathrm{o}}-\frac{u_{\mathrm{C_{o}0}}}{s}=0$$
(43)
and the current
\(I_{\mathrm{o}}(s)\) can be calculated to
$$I_{\mathrm{o}}(s)=\frac{C_{\mathrm{o}}\cdot u_{\mathrm{C_{o}0}}}{1+sC_{\mathrm{o}}R_{\mathrm{o}}}.$$
(44)
Applying the Inverse Laplace Transformation on (
44) directly leads to
$$i_{\mathrm{o}}(t)=\frac{u_{\mathrm{C_{o}0}}}{R_{\mathrm{o}}}\cdot e^{-\frac{t}{R_{\mathrm{o}}C_{\mathrm{o}}}}.$$
(45)
The capacitor current is equal to the output current with opposite sign in the considered time intervals, i.e.
$$i_{\mathrm{c}}(t)=-i_{\mathrm{o}}(t)=-\frac{u_{\mathrm{C_{o}0}}}{R_{\mathrm{o}}}\cdot e^{-\frac{t}{R_{\mathrm{o}}C_{\mathrm{o}}}},$$
(46)
finally leading to
$$u_{\mathrm{C_{o}}}=u_{\mathrm{C_{o}0}}\cdot e^{-\frac{t}{R_{\mathrm{o}}C_{\mathrm{o}}}}.$$
(47)
3.3 Derivation of the Complete Solution
So, all quantities of the LLC resonant converter are known as a function of time, the component values, the bridge voltage \(U_{\mathrm{B}}\), as well as the initial conditions for each interval of the cycle.
For
\(t\in \textbf{I}\), the model given in Fig.
4 is valid. The bridge voltage is
\(U_{B}=U\) and the currents through the inductors, as well as the voltages across the capacitors at the end of
I can be calculated by inserting
\(t=t_{x}\). Those expressions, which are a function of the component values and the initial conditions at time instant
\(t=0\), can subsequently be inserted for the initial conditions of the separated model valid during time interval
II. Using the separated model given in Fig.
5 allows to calculate the initial conditions for the time interval
III by inserting
\(t=\frac{T_{\mathrm{S}}}{2}-t_{\mathrm{x}}\).
For interval
III, again the full model given in Fig.
6 is used. However, the bridge voltage
\(U_{B}=-U\) changes its sign. Accordingly, the polarity is inverted in Fig.
6 compared to Fig.
4. As the change of the voltage direction would discharge the output capacitor
\(C_{\mathrm{o}}\) in the used models, the initial condition for the voltage across the output capacitor
\(C_{\mathrm{o}}\) must be reversed as well. In order to compensate that, the sign of the effected quantities (output voltage
\(u_{\mathrm{C_{0}}}\) and output current
\(i_{\mathrm{o}}\) for the time intervals
III and
IV) has to be changed in the end. With this model, the initial conditions of interval
IV can finally be determined.
At this point, the converter is fully described by the component values and the initial conditions at time instant \(t=0\), assuming that \(U_{B}\) is given. Unfortunately this is not sufficient yet, as the initial conditions of the steady state operating point are not known in advance. Instead of doing several exhaustive iterations to derive the initial conditions, a novel approach is presented here.
As already mentioned, in steady state operation the voltage and current waveforms show half wave symmetry, i.e. \(f(t)=-f(t-\frac{T_{\mathrm{S}}}{2})\). The integral of each quantity over the full switching cycle must hence be zero! This relationship can be used advantageously to determine the initial conditions which provide the steady state immediately.
To find a unique solution for the four initial conditions, four equations are needed. Integrating the four currents and setting them to zero will not provide the solution, as the currents are linearly dependent on each other. To avoid that limitation, the output voltage can be included as a fifth part.
Due to analogy, the integral is only stated for the resonant current
\(i_{\mathrm{r}}\). Hence
$$\begin{aligned}i_{\mathrm{r,int}}=\int_{0}^{t_{\mathrm{x}}}i_{\mathrm{r, \textbf{I}}}\;d\tau+\int_{0}^{\frac{T_{\mathrm{S}}}{2}-t_{\mathrm{x}}}i_{\mathrm{r, \textbf{II}}}\;d\tau+\int_{0}^{t_{\mathrm{x}}}i_{\mathrm{r, \textbf{III}}}\;d\tau+\int_{0}^{\frac{T_{\mathrm{S}}}{2}-t_{\mathrm{x}}}i_{\mathrm{r, \textbf{IV}}}\;d\tau\end{aligned}$$
(48)
is calculated.
The integral is applied in the same manner for \(i_{\mathrm{m}}\), \(i_{\mathrm{c}}\), \(i_{\mathrm{o}}\) and \(u_{\mathrm{C_{o}}}\).
The lower integration limit is zero for each interval, as the related functions have been derived in a way so that the time variable is zero at the beginning of the corresponding interval.
For steady state condition,
$$\begin{array}[]{l}i_{\mathrm{r,int}}\overset{!}{=}0\\ i_{\mathrm{m,int}}\overset{!}{=}0\\ i_{\mathrm{c,int}}\overset{!}{=}0\\ u_{\mathrm{C_{o},int}}\overset{!}{=}R_{\mathrm{o}}\cdot i_{\mathrm{o,int}}\end{array}$$
(49)
must be valid accordingly.
Solving (
49) finally leads to the initial conditions
\(i_{\mathrm{r0}}\),
\(i_{\mathrm{m0}}\),
\(u_{\mathrm{C_{r}0}}\) and
\(u_{\mathrm{C_{o}0}}\), which provide steady state immediately. The expressions are not stated here as they are rather long. As a last step, the time
\(t_{\mathrm{x}}\) must be determined. The most obvious guess is
\(t_{\mathrm{x}}=\frac{T_{\mathrm{r}}}{2}\), where
\(T_{\mathrm{r}}=\frac{1}{f_{\mathrm{r}}}\). In a first step this is only a rough estimation, but it turned out to be a good starting point for an iterative process to find
\(t_{\mathrm{x}}\). The process converges within a few iterations, providing a full description of the converter in steady state condition for a dedicated operating point. As one of the initial conditions is the voltage across the output capacitor, one directly receives the output voltage of the converter if the voltage-ripple is neglected.
To derive the correct value for
\(t_{x}\), the relationship
$$i_{\mathrm{r}}(t_{\mathrm{x}})\overset{!}{=}i_{\mathrm{m}}(t_{\mathrm{x}})$$
(50)
can be used, which is equivalent to
$$f(t_{\mathrm{x}})=i_{\mathrm{r}}(t_{\mathrm{x}})-i_{\mathrm{m}}(t_{\mathrm{x}})\overset{!}{=}0,$$
(51)
as this must be valid at
\(t=t_{x}\).
The analytical expressions of \(i_{\mathrm{r}}\) and \(i_{\mathrm{m}}\) are known and Newton’s Approximation Method can be applied to calculate \(t_{\mathrm{x}}\).
Hence
$$t_{\mathrm{x},n+1}=t_{\mathrm{x},n}-\frac{f(t_{\mathrm{x},n})}{f^{\prime}(t_{\mathrm{x},n})}$$
(52)
with the iteration variable
\(n\in\mathbb{N}_{0}\), using
\(t_{\mathrm{x},0}=\frac{T_{\mathrm{r}}}{2}\) as a start value just as aforementioned has to be implemented, finally converging to the optimal value for
\(t_{\mathrm{x}}\) according to
$$t_{\mathrm{x},opt}=t_{\mathrm{x},n_{\mathrm{max}}}$$
(53)
with
$$|t_{\mathrm{x},n_{\mathrm{max}}+1}-t_{\mathrm{x},n_{\mathrm{max}}}|\leq\epsilon.$$
(54)
Simulation results in the following section verify, that this approach is describing the current and voltage conditions with high accuracy for different operating points. The models are used in a predefined order, assuming that the resonant current differs from the magnetizing current for
\(0<t<t_{\mathrm{x}}\) and
\(\frac{T_{\mathrm{S}}}{2}<t<\frac{T_{\mathrm{S}}}{2}+t\mathrm{x}\) and that the latter are the same for
\(t_{\mathrm{x}}\leq t\leq\frac{T_{\mathrm{S}}}{2}\) and
\(\frac{T_{\mathrm{S}}}{2}+t_{\mathrm{x}}\leq t\leq T_{\mathrm{S}}\) as it is depicted in Fig.
3.