Class EF Inverter Design Chart - Examples

Example 1

Here, we want to analyze the operation of the class EF inverter shown below:

• What duty cycle \(D\) should be used to drive the switch?
• Does the switch operate in ZVS and/or ZCS?
• What will be the power delivered by the inverter to the load?

Step 1: Calculate the normalized coordinates

We simply calculate the normalized coordinates (\(r,x\)) as follows:

\[ \left\{ \begin{aligned} r &= 8.5\cdot 2\pi\cdot 30\cdot 10^6 \cdot 150\cdot 10^{-12}=0.24 \\ x &= (37\cdot 10^{-9}\cdot 2\pi\cdot 30\cdot 10^6)\cdot 2\pi\cdot 30\cdot 10^6\cdot 150\cdot 10^{-12}=0.20 \end{aligned} \right. \]

Step 2: Identify the operating point on the chart

We simply locate the point (0.24, 0.20) on the chart

Step 3: Read the normalized parameters

First observation: the operating point lies on the EF locus. The switch will therefore operate simultaneously in ZVS and ZCS (so \(v=0\) and \(q=0\)). To deduce the duty cycle and the power delivered by the inverter, just read the values of \(D\) and \(p\) from their respective curves on the chart. To do this, you have two options:

• You can directly read these values graphically, with your eyes 👀 (vector versions of the chart are downloadable in the 'Resources' tab)
• Or you can use my interactive tool available in the left tab 🤓 — just click on the operating point and read the normalized parameters at the bottom of the page

No matter which method you choose, you should get something like this:

\[ \left\{ \begin{aligned} D &\approx 33.3\ \% \\ p &\approx 1.9 \end{aligned} \right. \]

Step 4: Denormalize the normalized parameters

We simply denormalize the parameter \(p\) to find the value of the power delivered by the inverter; the duty cycle \(D\) used to drive the switch does not need to be denormalized (\(D=33.3\ \%\)):

\[ P=1.9\cdot 2\pi\cdot 30\cdot 10^6\cdot 150\cdot 10^{-12}\cdot 100^2=537\ \text{W} \]