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EE467-567'10_lect5-Fnl

EE467-567'10_lect5-Fnl - EE 467/567 INTEGRATED POWER...

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EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] EE 467/567 -- INTEGRATED POWER EE 467/567 -- INTEGRATED POWER ELECTRONICS ELECTRONICS Dr. Douglas C. Hopkins Dept. of Electrical Engineering University at Buffalo [email protected] EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Power Filtering Power Filtering Section 3.6

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EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Two approaches to filtering Two approaches to filtering The DC source and Switch produce an offset rectangular wave of voltage as shown, i.e. a 5Vdc offset with +/-5V bipolar wave as some duty cycle. Fourier Sources (Freq domain): The v D shown can be divided into a Fourier series and each source applied to the L, R filter circuit. The i out or v out is the sum of the contributions from each source Ideal Action (time domain): The effect of v D on L, R can be divided into DC and a bipolar switching wave, with v R having a DC voltage, limiting time domain analysis only to “L.” 10 V 0 0 100 μ s v D 50 % Duty L + ! 10 V f = 10 kHz 10 " v D Specified: choose L to make ! i 0.1% ripple, or ! i = 0.005A EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Filter Examples Filter Examples V IN large L I R If L is large, then I is just DC. Substitute v IN
EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Fourier Sources Fourier Sources 5 V 10 ! 0.5 A Har- monics i Linear circuit 10 V 0 0 100 μ s v D Choose L to make the maximum I out,rms of the “non-zero frequency” components less than some limit. Remember that only zero-frequency (i.e. DC) is desired. However, TOO MUCH WORK EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Ideal Action: Ideal Action: I a dc-to-dc converter, L is large and the inductor current is almost constant; so is the voltage across the load resistor. Hence, the v R voltage can be replaced with a constant voltage source. Switch on: V L = 5 V Switch off: V L = -5 V v 5 V L + " L 10 V 0 L +5 -5

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EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Ideal Action - Analysis Ideal Action - Analysis L +5 -5 V L = L di/dt If V L = 5 V = L di/dt L V 5 = di/dt = t i # # If V L = -5 V = L di/dt L V -5 = di/dt = t i # "# EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Analysis Analysis (con con ’d) # i 100 μ s 50 μ s 150 μ s 0 < i L > = 0.5 A Specified: choose L to make ! i 0.1% ripple, or ! i = 0.005A L V 5 = t # i # = 50 μ s i # i # L V 5 = 50 μ s x 50 % Duty L + ! 10 V f = 10 kHz 10 " v D
EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Analysis ( Analysis ( con con ’d) L V 5 50 x 0.005 A = x 10 ! 6 L = 5 x 10 -3 250 x 10 -6 L = 0.005 H L > 5 mH makes ! i < 0.005 A EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] Must Study: Examples Must Study: Examples Look at other examples based on the Fourier Source method, such as Example 3.6.1, page 102. Look at examples based on ideal action, such as 3.7.1 and 3.7.3, pages 103-108.

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EE467/567 Integrated Power © 2010, D. C. Hopkins [email protected] V IN I IN L i OUT #2 + !
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EE467-567'10_lect5-Fnl - EE 467/567 INTEGRATED POWER...

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