Fundamentals of Power Electronics 1 Chapter 19: Resonant Conversion 19.4.4 Design Example Select resonant tank elements to design a resonant inverter that.
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Fundamentals of Power Electronics 1 Chapter 19: Resonant Conversion 19.4.4 Design Example Select resonant tank elements to design a resonant inverter that meets the following requirements: Switching frequency f s = 100 kHz Input voltage V g = 160 V Inverter is capable of producing a peak open circuit output voltage of 400 V Inverter can produce a nominal output of 150 Vrms at 25 W
Fundamentals of Power Electronics 2 Chapter 19: Resonant Conversion Preliminary calculations The requirements imply that the inverter tank circuit have an open-circuit transfer function of: The required short-circuit current can be found by solving the elliptical output characteristic for I sc : hence Use the requirements to evaluate the above:
Fundamentals of Power Electronics 3 Chapter 19: Resonant Conversion Matched load Matched load therefore occurs at the operating point Hence the tank should be designed such that its output impedance is
Fundamentals of Power Electronics 4 Chapter 19: Resonant Conversion Solving for the tank elements Let’s design an LCC tank network for this example The impedances of the series and shunt branches can be represented by the reactances
Fundamentals of Power Electronics 5 Chapter 19: Resonant Conversion Analysis in terms of X s and X p The transfer function is given by the voltage divider equation: The output impedance is given by the parallel combination: Solve for X s and X p :
Fundamentals of Power Electronics 6 Chapter 19: Resonant Conversion Evaluate tank element values The capacitance C p should therefore be chosen as follows: The reactance of the series branch should be
Fundamentals of Power Electronics 7 Chapter 19: Resonant Conversion Discussion Choice of series branch elements The series branch is comprised of two elements L and C s, but there is only one design parameter: X s = 733 Ω. Hence, there is an additional degree of freedom, and one of the elements can be arbitrarily chosen. This occurs because the requirements are specified at only one operating frequency. Any choice of L and C s, that satisfies X s = 733 Ω will meet the requirements, but the behavior at switching frequencies other than 100 kHz will differ. Given a choice for C s, L must be chosen according to: For example, C s = 3C p = 3.2 nF leads to L = 1.96 µH
Fundamentals of Power Electronics 8 Chapter 19: Resonant Conversion R crit For the LCC tank network chosen, R crit is determined by the parameters of the output ellipse, i.e., by the specification of V g, V oc, and I sc. Note that Z o is equal to jX p. One can find the following expression for R crit : Since Z o0 and H are determined uniquely by the operating point requirements, then R crit is also. Other, more complex tank circuits may have more degrees of freedom that allow R crit to be independently chosen. Evaluation of the above equation leads to R crit = 1466 Ω. Hence ZVS for R < 1466 Ω, and the nominal operating point with R = 900 Ω has ZVS.