ECEN4797/5797 Lecture #11 Announcements On-campus students: pick up graded HW2, turn in HW3 Homework 4 is due in class on Friday, Sept. 23. The grace-period for offcampus students expires 5pm (Mountain) on Friday, Sept. 30. I will be away on a trip next week, Sept 19 to Sept 22 No classes or office hours during this time. E-mails will be answered Make-up office hours: Friday Sept 23, 9-11am Lecture schedule and make-up classes: Today Lecture 11: 2-3pm Lecture 12: 3-4pm Friday, Sept 23 Lecture 13: 2-3pm Lecture 14: 3-4pm 1
Inclusion of Switching Loss in the Averaged Equivalent Circuit Model The methods of Chapter 3 can be extended to include switching loss in the converter equivalent circuit model Include switching transitions in the converter waveforms Model effects of diode reverse recovery, etc. To obtain tractable results, the waveforms during the switching transitions must usually be approximated Things that can substantially change the results: Ringing caused by parasitic tank circuits Snubber circuits These are modeled in ECEN 5817, Resonant and Soft- Switching Phenomena in Power Electronics
Sketch the converter waveforms The Modeling Approach Extension of Chapter 3 Methods Including the switching transitions (idealizing assumptions are made to lead to tractable results) In particular, sketch inductor voltage, capacitor current, and input current waveforms The usual steady-state relationships: v L = 0, i C = 0, i g = I g Use the resulting equations to construct an equivalent circuit model, as usual
Buck Converter Example Ideal MOSFET, p n diode with reverse recovery Neglect semiconductor device capacitances, MOSFET switching times, etc. Neglect conduction losses Neglect ripple in inductor current and capacitor voltage
Waveforms Diode recovered charge Q r, reverse recovery time t r These waveforms assume that the diode voltage changes at the end of the reverse recovery transient a snappy diode Voltage of soft-recovery diodes changes sooner Leads to a pessimistic estimate of induced switching loss
Inductor volt-second balance and capacitor charge balance As usual: v L = 0 = DV g V Also as usual: i C = 0 = I L V/R
Average input current i g = I g = (area under curve)/t s = (DT s I L + t r I L + Q r )/T s = DI L + t r I L /T s + Q r /T s
Construction of Equivalent Circuit Model From inductor volt-second balance: v L = 0 = DV g V From capacitor charge balance: i C = 0 = I L V/R
Input port of model i g = I g = DI L + t r I L /T s + Q r /T s
Combine for complete model The two independent current sources consume power V g (t r I L /T s + Q r /T s ) equal to the switching loss induced by diode reverse recovery
Solution of model Output: V = DV g Efficiency: = P out / P in P out = VI L P in = V g (DI L + t r I L /T s + Q r /T s ) Combine and simplify: = 1 / [1 + f s (t r /D + Q r R /D 2 V g )]
Predicted Efficiency vs Duty Cycle Switching frequency 100 khz Input voltage 24 V Load resistance 15 Ω Recovered charge 0.75 µcoul Reverse recovery time 75 nsec (no attempt is made here to model how the reverse recovery process varies with inductor current) Efficiency 100.00% 90.00% 80.00% 70.00% 60.00% 50.00% 40.00% Buck converter with diode reverse recovery Substantial degradation of efficiency Poor efficiency at low duty cycle 30.00% 20.00% 10.00% 0.00% 0 0.2 0.4 0.6 0.8 1 Duty cycle
Boost Converter Example Model same effects as in previous buck converter example: Ideal MOSFET, p n diode with reverse recovery Neglect semiconductor device capacitances, MOSFET switching times, etc. Neglect conduction losses Neglect ripple in inductor current and capacitor voltage
Boost converter Transistor and diode waveforms have same shapes as in buck example, but depend on different quantities
Inductor volt-second balance and average input current As usual: v L = 0 = V g D V Also as usual: i g = I L
Capacitor charge balance i C = i d V/R = 0 = V/R + I L (D T s t r )/T s Q r /T s Collect terms: V/R = I L (D T s t r )/T s Q r /T s
Construct model The result is: The two independent current sources consume power V (t r I L /T s + Q r /T s ) equal to the switching loss induced by diode reverse recovery
Predicted Efficiency vs Duty Cycle Switching frequency 100 khz Input voltage 24 V Load resistance 30 Ω Recovered charge 0.75 µcoul Reverse recovery time 75 nsec 100.00% 90.00% 80.00% 70.00% Boost converter with diode reverse recovery Substantial degradation of efficiency Very poor efficiency at high duty cycle, even with no conduction losses included Conduction losses (R L, R on, V D, etc.) can be included in the model Effic ciency 60.00% 50.00% 40.00% 30.00% 20.00% 00% 10.00% 0.00% 0 0.2 0.4 0.6 0.8 1 Duty cycle
Example: model with R L conduction loss and diode reverse recovery Power loss due to the diode d reverse recovery: V (t r I L /T s + Q r /T s ) Power loss due to the inductor R L : R L I L 2
Predicted V/V g vs duty cycle Switching frequency 100 khz Input voltage 24 V Load resistance 60 Ω Recovered charge 5 µcoul Reverse recovery time 100 nsec Inductor resistance R L = 0.3 Ω 8 7 6 5 Boost converter with diode reverse recovery With R L only V/Vg 4 3 2 With R d 1 L and diode reverse recovery 0 0 0.2 0.4 0.6 0.8 1 Duty cycle
Summary The averaged modeling approach can be extended to include effects of switching loss Transistor and diode waveforms are constructed, including the switching transitions. The effects of the switching transitions on the inductor, capacitor, and input current waveforms can then be determined Inductor volt-second balance and capacitor charge balance are applied Converter input current is averaged Equivalent circuit corresponding to the the averaged equations is constructed