Class EF Inverter Design Chart - Examples

Example 2

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:

\[ \left\{ \begin{aligned} r &= 13\cdot 2\pi\cdot 0.8\cdot 10^9 \cdot 8.1\cdot 10^{-12}=0.53 \\ x &= (4.3\cdot 10^{-9}\cdot 2\pi\cdot 0.8\cdot 10^9)\cdot 2\pi\cdot 0.8\cdot 10^9\cdot 8.1\cdot 10^{-12}=0.88 \end{aligned} \right. \]

Step 2: Identify the operating point on the chart

We simply locate the point (0.53, 0.88) on the chart

Step 3: Read the normalized parameters

First observation: the operating point lies in the ZCS region. Therefore, the switch will operate only in ZCS (so \(q=0\)). To deduce the duty cycle to use, the power delivered by the inverter, as well as the voltage across the switch at turn-on, simply read the values of \(D\), \(p\), and \(v\) from their respective curves on the chart:

\[ \left\{ \begin{aligned} D &\approx 14.0\ \% \\ p &\approx 0.22 \\ v &\approx 0.25 \end{aligned} \right. \]

Step 4: Denormalize the normalized parameters

We simply denormalize parameters \(p\) and \(v\) to find the power delivered by the inverter and the voltage across the switch at turn-on. The duty cycle \(D\) used to drive the switch does not need denormalization (\(D=14.0\ \%\)):

\[ \left\{ \begin{aligned} &P=0.22\cdot 2\pi\cdot 0.8\cdot 10^9\cdot 8.1\cdot 10^{-12}\cdot 28^2=7.02\ \text{W} \\ &V_0=0.25\cdot 2\cdot 28=14\ \text{V} \end{aligned} \right. \]