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Announcements. Homework #1 statistics (for on-campus students): Average = 65.2/80 = 82% Homework #3 due Friday, Feb. 8 for on-campus students Correction to Problem 19.6:

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Announcements
• Homework #1 statistics (for on-campus students):
• Average = 65.2/80 = 82%
• Homework #3 due Friday, Feb. 8 for on-campus students
• Correction to Problem 19.6:

Part (a) asks you to derive expression for Voc and Isc in terms of the variables F = fs/ finf, Vg, n = Cs / Cp, and Rinf. There is a square root missing from Rinf, i.e. it should read

Same as dc-dc series resonant converter, except output rectifiers are replaced with four-quadrant switches:

19.4.4 Design Example
• Select resonant tank elements to design a resonant inverter that meets the following requirements:
• Switching frequency fs = 100 kHz
• Input voltage Vg = 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
Calculations

The required short-circuit current can be found by solving the elliptical output characteristic for Isc:

hence

Use the requirements to evaluate the above:

Solve for the open circuit transfer function
• The requirements imply that the inverter tank circuit have an open-circuit transfer function of:
• Note that Voc need not have been given as a requirement, we can solve the elliptical relationship, and therefore find Voc given any two required operating points of ellipse. E.g. Isc could have been a requirement instead of Voc
Solve for matched load (magnitude of output impedance )
• Matched load therefore occurs at the operating point

Hence the tank should be designed such that its output impedance is

• Let’s design an LCC tank network for this example

The impedances of the series and shunt branches can be represented by the reactances

Analysis in terms of Xs and Xp
• The transfer function is given by the voltage divider equation:

The output impedance is given by the parallel combination:

Solve for Xs and Xp:

Analysis in terms of Xs and Xp
• The transfer function is given by the voltage divider equation:

The output impedance is given by the parallel combination:

Solve for Xs and Xp:

DiscussionChoice of series branch elements
• The series branch is comprised of two elements L and Cs, but there is only one design parameter: Xs = 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 Cs, that satisfies Xs = 733 Ω will meet the requirements, but the behavior at switching frequencies other than 100 kHz will differ.
• Given a choice for Cs, L must be chosen according to:

For example, Cs = 3Cp = 3.2 nF leads to L = 1.96 mH

Rcrit
• For the LCC tank network chosen, Rcrit is determined by the parameters of the output ellipse, i.e., by the specification of Vg, Voc, and Isc. Note that Zo is equal to jXp. One can find the following expression for Rcrit:

Since Zo0 and H  are determined uniquely by the operating point requirements, then Rcrit is also. Other, more complex tank circuits may have more degrees of freedom that allow Rcrit to be independently chosen.

Evaluation of the above equation leads to Rcrit = 1466 Ω. Hence ZVS for R < 1466 Ω, and the nominal operating point with R = 900 Ω has ZVS.

Converter performance
• For this design, the salient tank frequencies are
• (note that fs is nearly equal to fm, so the transistor current should be nearly independent of load)

The open-circuit tank input impedance is

So when the load is open-circuited, the transistor current is

Dynamic Modeling and Analysisof Resonant Inverters

Closed-loop control system to regulate amplitude of ac output

(Lamp ballast example shown, but other applications have similar needs) (frequency modulation control shown)

• Issues for design of closed-loop resonant converter system:
• Need open-loop control-to-output transfer function to model loop gain, closed-loop transfer functions
• How does control-to-output transfer function depend on the tank transfer function H(s)?
Spectrum of v(t)

The control input is varied at modulation frequency fm

This leads to sidebands at frequencies fs ± fm (true for both AM and narrowband FM)

The spectrum of v(t) contains no component at f = fm.

• Effect of the tank transfer function H(s) on the output:
• Changing the amplitude of the carrier affects the steady-state output amplitude
• Changing the amplitudes of the sidebands affects the ac variations of the output amplitude— i.e., the envelope
• The control-to-output-envelope transfer function Genv(s) depends on the tank transfer function H(s) at the sideband frequencies fs ± fm. It doesn’t depend on H(s) at f = fm.
How frequency modulation of tank input voltage introduces amplitude modulation of output envelope
Poles of Genv(s)

Example:

H(s) has resonant poles at f = fo

These poles affect the sidebands when fs ± fm = fo

Hence poles are observed in Genv(s) at modulation frequencies of fm = fs – fo

Outline of discussion

DIRECT MODELING APPROACH

• How small-signal variations in the switching frequency affect the spectrum of the switch network output voltage vs1(t)
• Passing the frequency-modulated voltage vs1(t) through the tank transfer function H(s) leads to amplitude modulation of the output voltage v(t)
• How to recover the envelope of the output voltage and determine the small-signal control-to-output-envelope transfer function Genv(s)

PHASOR TRANSFORMATION APPROACH

• Equivalent circuit modeling via the phasor transform
• PSPICE simulation of Genv(s) using the phasor transform