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Class EF Inverter Design Chart - Examples





Example 3



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

Example_3_circuit
Studied EF inverter





Step 1: Find the equivalent series circuit



Unlike previous cases, the load presented to the inverter is not a series connection of an inductance and a resistance, but an inductance \(L_p=7\ \mu\)H in parallel with a resistance \(R_p=110\ \Omega\), which slightly complicates matters. Therefore, the first step of our study is to express this parallel load \(R_pL_p\) as an equivalent series load \(RL\), where \(R\) and \(L\) are given by:

\[ \left\{ \begin{aligned} R &= \frac{R_p\cdot(2\pi\cdot F\cdot L_p)^2}{R_p^2+(2\pi\cdot F\cdot L_p)^2}=15.19\ \Omega \\ L &= \frac{R_p^2\cdot 2\pi\cdot F\cdot L_p}{R_p^2+(2\pi\cdot F\cdot L_p)^2}=6.03\ \mu\text{H} \end{aligned} \right. \]

These formulas can be easily derived by expressing the equivalent impedance of the parallel load \(R_pL_p\) observed at the switching frequency of the circuit. Consequently, the studied circuit can be simply reduced to previous cases. We then just need to follow the same steps as in Examples 1 and 2 to predict the inverter operation using the design chart.



Example_3_circuit
Equivalent circuit of the studied inverter





Step 2: Calculate the normalized coordinates



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

\[ \left\{ \begin{aligned} r &= 15.16\cdot 2\pi\cdot 1\cdot 10^6 \cdot 2\cdot 10^{-9}=0.19 \\ x &= (6.03\cdot 10^{-6}\cdot 2\pi\cdot 1\cdot 10^6)\cdot 2\pi\cdot 1\cdot 10^6\cdot 2\cdot 10^{-9}=0.48 \end{aligned} \right. \]



Step 3: Identify the operating point on the chart



We simply locate the point (0.19, 0.48) on the chart

Example_3_chart
Operating point position on the chart
Example_3_vs_is
Schematic shape of \(v_s(\omega t)\) and \(i_s(\omega t)\)





Step 4: Read the normalized parameters



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

\[ \left\{ \begin{aligned} D &\approx 34.8\ \% \\ p &\approx 0.54 \\ q &\approx 0.24 \end{aligned} \right. \]



Step 5: Denormalize the normalized parameters



We simply denormalize parameter \(p\) to find the power delivered by the inverter. The duty cycle \(D\) and ratio \(q\) do not need denormalization (\(D=34.8\ \%\) and \(q=0.24\)):

\[ P=0.54\cdot 2\pi\cdot 1\cdot 10^6\cdot 2\cdot 10^{-9}\cdot 50^2=16.96\ \text{W} \]