Class EF Inverter Design Chart - Examples

Example 4

In this example, we revisit the case studied previously in Example 3, with the main results recalled in the figures below:

Since the operating point is located in the ZVS zone, only ZVS is reached, which can, a priori, negatively affect the overall inverter operation (because ZCS is not achieved in this case). To address this issue, we add a \(L_0\) inductance in series with the parallel load \(R_pL_p\), which will shift the operating point upwards, as shown below.

By choosing the right value for \(L_0\), it is possible to move the operating point so that it lies on the EF locus, allowing the switch to operate simultaneously in ZVS and ZCS. The following questions naturally arise:

• What value of \(L_0\) should be chosen?
• What duty cycle \(D\) should be used to ensure the inverter effectively operates in both ZVS and ZCS?
• What will be the power delivered by the modified inverter?

Step 1: Calculating \(L_0\)

Graphically, the desired operating point is at (0.19, 0.88), which means a shift of +0.40 is needed to bring our operating point onto the EF locus. Therefore, the value of \(L_0\) must satisfy the following relation:

\[ (L_0\cdot 2\pi\cdot 1\cdot 10^6)\cdot 2\pi\cdot 1\cdot 10^6\cdot 2\cdot10^{-9}=0.40\quad\Rightarrow\quad L_0=5.06\ \mu\text{H} \]

Step 2: Reading the new reduced parameters

The reduced parameters of the new operating point obtained by adding a series inductance \(L_0\) can be obtained by direct graphical reading or using the interactive tool:

\[ \left\{ \begin{aligned} D &\approx 13.9\ \% \\ p &\approx 0.14 \end{aligned} \right. \]

As a reminder, since the inverter operates in both ZVS and ZCS, the factors \(v\) and \(q=0\) are zero

Step 3: Calculating the new power delivered

The power delivered at the new operating point, obtained by adding the series inductance \(L_0\), can be calculated as:

\[ P=0.14\cdot 2\pi\cdot 1\cdot 10^6\cdot 2\cdot 10^{-9}\cdot 50^2=4.4\ \text{W} \]